THE NEW METAL WORKER PATTERN BOOK ^^KjrBiKn THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA Arthur H. Memmler 1874-1956 THE NEW METAL WORKER PATTERN BOOK A TREATISE ON THE PRINCIPLES AND PRACTICE OF PATTERN CUTTING AS APPLIED TO SHEET METAL WORK. BY GEO. W. KITTREDGE. . NEW YORK: DAVID WILLIAMS COMPANY, 232-238 WILLIAM STREET. 1901. THE NEW METAL WORKER PATTERN BOOK, COPYRIGHTED, 1896, BY DAVID WILLIAMS. THE METAL WORKER PATTERN BOOK, COPYRIGHTED, l88l, BY DAVID Wll.LlAMS. GIFT CONTENTS, PAGE. Introduction 9 CHAPTER I. Terms and Definitions ALIMIABETIOAL LlST OF TERMS - - 15 CHAPTER II. Drawing Instruments and Materials 17 CHAPTER III. Linear Drawing - CHAPTER IV. Geometrical Problems CONSTRUCTION OF REGULAR POLYGONS - 43 TIIK KLLIVSE . - 59 TIIK VOLUTE - - - - - 67 CHAPTER V. Principles of Pattern Cutting 71 PARALLEL KOKMS - 72 REGULAR TAPERING KOKMS - 79 IRREGULAR KOKMS - - 86 CHAPTER VI. Pattern Problems - 96 SECTION 1. PARALLEL FORMS (MITEK CUTTING) (' SECTION 2. REGULAR TAVKRING FORMS (FLARING WORK) 240 SECTION 3. IRREGULAR FORMS (TKIAXGULATION) 306 Index of Problems - 421 766 FOR the benefit of those who may contemplate making- use of this work, wholly or in part, it is well to lay before them at the outset a general statement of the plan upon which it is written, together with some advice for the use and study of the same, which may not properly belong under any of the several headings comprising the subject matter. A glance at the table of con- tents immediately preceding will give at once a clear idea of its scope and arrangement. From this it will be seen that the first five chapters are theoretical or educational in their nature, while the last chapter is devoted to practical work; and further, that the book does not presume upon anv previous technical knowledge upon the part of the beginner, but aims to place before him in the preliminary chapters all that is necessary to a thorough understanding of the work performed in the last chapter, which constitutes the bulk of the book. A very important feature of the work is the classification of the problems. The forms for which patterns mav be required are divided, according to the methods employed in developing their surfaces, into three classes, and the problems relating to each are arranged in three corre- sponding sections of the last chapter, thus bringing near together those in which principles and methods are alike. In Chapter V (Principles of Pattern Cutting) this classification is defined and the principles governing each class are explained and illustrated under three sub-headings of the chapter. The third subdivision treats of the method of developing the surfaces of irregular forms by Triuiir/ulation, a subject not heretofore systematically treated in any work on pattern cutting. A chapter on drawing (Chapter III) has been prepared for the benefit of the pattern cutter especially interested in cornice wcrk, and though he may not intend to become a finished archi- tectural draftsman, this chapter will render him valuable assistance in reading the original drawings received from architects, from which he is required in many cases to make new drawings adapted to his own peculiar wants. The New Metal Worker Pattern Book, besides being a systematic treatise on the principles of pattern cutting, is also valuable as a reference book of pattern problems and as a fund of information on the subject treated, to be drawn from at convenience, and is so written that each problem, or chapter of descriptive matter, can be read independently of the others ; so that the student whose time is limited can turn to any portion of the work the title of which promises the information sought, without feeling that he must read all that precedes it. The relative importance of the chapters depends, of course, upon the individual reader, and will be determined by what he considers his weakest points. However, it is advisable in the study of all works of a scientific nature to begin at the beginning and take everything in its course. If, therefore, the study of this work can be continued progressively from the first, much advantage will be gained. The statement of each problem in prominent type appears at the head of the demonstration, iv Tlie Neiv Metal Worker Pattern Book. and every problem is numbered, by which arrangement the problems are well separated from each other and easily found. While each demonstration is considered complete in itself, some are necessarily carried farther into detail than others, and references arc made from one problem to another, pointing out similarity of principle, where such comparison would be advantageous to one who is looking for principles rather than for individual solutions. In preparing the diagrams used to illustrate the solutions of the problems, forms have been chosen which are as simple in outline as the case will admit, upon the supposition that the reader will be able to make the application of the method described in connection with the same to his own especial case, which may embody more complicated forms. It must also be noted that, owing to the small scale to which the drawings in this work are necessarily made, extreme accuracy in the operations there performed is impossible. In many instances the length of the spaces used in dividing the profiles is much too great in proportion to the amount of curvature to insure accuracy. Therefore if apparent errors in measurements or results are found, they must not be considered the fault of the system taught. If such errors arc discovered the student is recommended to reconstruct the drawing upon his own drawing board in accordance with the demonstration given and to a scale sufficiently large to insure accurate results, before passing judgment. In the preparation of this book, the former Metal Worker Pattern Book has been made the basis, to a certain extent, of the new work.* Such problems or portions of the former work as were found satisfactory have been assigned to their proper places in the new work without change. In the case of most of the problems, however, the demonstrations have been revised and the drawings accompanying them have been amended or corrected in accordance with the text, and in many cases entire new drawings have been made. To these have been added a large number of new problems based upon inquiries and solutions that have appeared in the columns of Tin- MHul WrL-<'i- since the former work was published. Much new explanatory matter not in the former work has also been added in the preliminary chapters, prominent among which are Chapter III, and the principles of Triangulation in Chapter V. p]special care has been taken in the composition of the book to have each engraving and the text referring to it arranged, as far as possible, on the same page or upon facing pages, so as to obviate the necessity of turning the leaf in making references. A great advantage is gained over the former work by the classification and numbering of the problems, which, in connection with the table of contents, renders any desired subject or problem easily found. In regard to the system of reference letters employed in the drawings, it should be said that the same letter has been used so far as possible to represent any given point in the several views or positions in which it may occur, the superior figure or exponent being changed in each view. To fully comprehend this the reader must carry in mind the concrete idea of the form under considera- tion, just as though he held in his hand a perfect completed model of the same, which lie turned this way or that to obtain the several views given. Any point, therefore, which might on the model be marked by a letter A, would be designated in one of the views as A. while in other views or places where it might appear it would be designated as A 1 , A 2 , etc., or as A', A", etc. In the * Publisher's Note. The author of this book, George W. Kittredge, prepared the drawings and outlined the demonstrations of all but a few of the less important problems in The Metal Worker Pattern Book, which was published in 1881, and also prepared portions of. the introductory chapters of that wcrk. Introduction* solution of problems by triangulation, dotted lines arc alternated w-itli solid linos, as lines of meas- urement, merely for the sake of distinction and to facilitate the work. Occasions arise in the experience of every pattern cutter wherein some portion of the work before him, of relatively small importance, is BO situated that the development of its pattern by a strietly accurate method would involve more labor and time than would be justified by the value of the part wanted. It is the purpose of this work to teach the principles of pattern cutting, leav- ing the decision of such questions to the individual. Nevertheless, if one is thoroughly con- versant with pattern cutting methods and familiar with pattern shapes it may be possible in such cases to obtain accurately the principal points of a required pattern and to complete the same by the eye with sullicient accuracy for all practical purposes. As intimated above, some of the demonstrations are' necessarily made more explicit than others. In the longer demonstrations and those occurring near the ends of the Sections, less important details x of the work are sometimes omitted and certain parts of the operation arc only hinted at or are described in a general way, upon the supposition that the simpler problems in which the demon- strations are carried further into detail would naturally be studied first. Although the principles of pattern cutting here set forth may at times be regarded as somewhat intricate, it is believed that any one possessed of a fair degree of intelligence and application can easily master them. Notwithstanding the great care which has been used in the preparation of this work, it is possible that errors mav have found their way into its columns. Should errors be discovered by any of its readers, information of such will be gladly received. CHAPTER I. Pattern cutting as applied 1<> sheet-metal work. by its verv nature, involves ilic application v such their ap- plication is easily made. For the benefit of those, however, who have not had such advantages, this work purposes to make specilic application of those princi- ples in a way to lie readily understood bv the mechanic. While throughout the work the use of an unnecessarv number of technical terms and words not in common use among mechanics will be carefullv avoided, it must be here noted that precise lnf .<,////;<<<*, under which name its principles are now being taught to a great extent in schools of practical instruction. Articles made from sheet metal are hollow, being only shells, and must, therefore, be considered in the process of pattern cut- ting as though they wen.' the coverings or casings stripped from solids of the same shape. 3. A Point is that which has place or position with- out magnitude, as the intersection of two lines or the center of a circle ; it is usually represented to the eye by u small dot. LINES. 4. A Line is that which has length merely, and may be straight or curved. 5. A Straight Line, or, as it is sometimes called, a right line, is the shortest line that can be drawn between two given points. Straight lines are generally desig- nated by letters or figures at their extremities, as A B, Fig. 1. (i. A Curved Line is one which changes its direc- tion at every point, or one of which no portion, how- Fig. 1 A S raight Line. ever small, is straight. It is therefore longer than a straight line connecting the same points. Curved lines Fiy. 2. Curved Lines. are designated by letters or figures at their extrem- ities and at intermediate points, as A ~R C or I) K I 1 ', l-'iu'. 2. 7. Parallel Lines are those which have no inclina- tion to cadi other, being everywhere equidistant. A B and A 1 B' in Fig. 3 are parallel straight lines, and -B -B C' D' Fig. 3. Parallel Lines. can never meet though produced to infinity. G D and C' D' are parallel curved lines, being arcs of circles which have a common center. 8. Horizontal Lines are lines parallel to the hori- zon, or level. A Horizontal Line in a drawing is indi- cated by a line drawn from left to right across the paper, as A B in Fig. 4. Horizontal Fig. 4. Names of Lines by Direction. 9. Vertical Lines are lines parallel to a plumb line suspended freely in a still atmosphere. A Verti- cal Line in a drawing is represented by a line drawn up and down the paper, or at right angles to a hori- zontal line, as E C in Fig. 4. 10. Inclined or Oblique Lines occupy an interme- diate between horizontal and vertical lines, as C D, Fig. 5. Perpendicular Lines. Fig. 4. Two lines which converge toward each other and which, if produced, would meet or intersect, are said to incline to each other. 11. Perpendicular Lines. Lines are perpendicular in each other when the angles on either side ol the point of meeting are equal. Vertical and horizontal lines are always perpendicular to each other, but per- pendicular line.s are not always vertical and horizontal, but may be at any inclination to the horizon, provided that the angles on either side of the point of intersec- tion are equal. In Fig. 5, C F, D II and E G are said to be perpendicular to A B. Also in Fig. G, B F Fig. 6. Perpendicular Lines. C D and E F are perpendicular to A B. Line? per- pendicular to the same line are parallel to each other, as C F and D H, Fig. 5, which are perpendicular to A B. 12. An Angle is the opening between two straight linos which meet one another. An angle is commonly designated by three letters, the letter designating the point in which the straight lines containing the angle B meet being between the other two letters, as the angle BCD, Fig. 4. 13. A Right Angle. 'When a straight line meets another straight line so as to make the adjacent angles equal to each other, each angle is a right angle, and the straight lines are said to be perpendicular to each other. (See C B E or C B D, Fig. 7.) 14. An Acute Angle is an angle less than a right angle, as A B D or ABC, Fig. 7. 15. An Obtuse Angle is an angle greater than a right angle, as A B E, Fig. 7. 7'rrrnx uml Definitions. 3 STRAIGHT SIDED FIGURES. 16. A Surface is that which has length and breadth without thickness. 1 7. A Plane is a surface such that if any two of its points be joined by a straight line, such line will be wholly in the surface. Every surface which is not a plane surface, or composed of plane surfaces, is a I'l-rri'd surface. 18. A Single Curved Surface is one in which only certain points may be joined by straight lines which Fig. 8. An Equilateral Triangle. Fig. 9. An Isosceles Triangle. B B Fig. 10. A Scalene Triangle. Fig. 11 Right- Angled Triangles. shall lie wholly in its surface. The rounded surface of a cylinder or cone is a single curved surface. 19. A Double Curved Surface is one in which no two points can be joined by a straight line lying wholly in its surface. The surface of a sphere, for example, is a double curved surface. 20. A Plane Figure is a portion of a plane termi- nated on all sides by lines either straight or curved. 21. A Rectilinear Figure is a surface bounded by straight lines. (See Figs. 8, 16, 21, etc.) 22. Polygon is the general name applied to all rectilinear figures, but is commonly applied to those having more than- four sides. A regular polygon is one in which the sides are equal. 23. A Triangle is a flat surface bounded by three straight lines. (Figs. 8, 9, 10, 11, 13, etc.) 24. An Equilateral Triangle is one in which the three sides are equal. (Fig. 8.) 25. An Isosceles Triangle is one in which two of the sides are equal. (Fig. 9.) 26. A Scalene Triangle is one in which the three sides are of different lengths. (Fig. 10.) 27. A Right-Angled Triangle is one in which one of the angles is a right angle. (Fig. 11.) 28. An Acute-Angled Triangle is one which has its three angles acute. (Fig. 12.) 29. An Obtuse-Angled Triangle is one which has an obtuse angle. (Fig 13.) Fig. IS. An Acute-Angled Triangle. Fig. IS. An Obtuse-Angled Triangle. Apex orVcrtX Base B Fig. 14. Names of the Sides of a Fig. IS. Names of the Parts Right- Angled Triangle. of a Triangle. 30. A Hypothenuse is the longest side in a right- angled triangle, or the side opposite the right angle. A C, Fig. 14. 31. The Apex of a triangle is its upper extremity, as B, Fig. 15. It is also called vertex. 32. The Base of a triangle is the line at the bottom. B C and A C, Figs. 14 and 15. 33. The Sides of a triangle are the including lines. A C, A B and B C, Figs. 14 and 15. 34. The Vertex is the point in any figure opposite to and furthest from the base. The vertex of an angle is the point in which the sides of the angle meet. B, Fig. 15. 35. The Altitude of a triangle is the length of a perpendicular let fall from its vertex to its base, as B D, Fig. 15, 36. A Quadrilateral figure is a surface bounded by four straight lines. There are three kinds of Quadri- Tlie \'.'- M' "/.f its sides parallel. (Fig. n;.) 38. The Trapezoid has only two of its sides parallel. (Fig. IT.) 3i. Tlu- Parallelogram has its opposite sides par- allel. There are four varieties of parallelograms: The Rhomboid, the Rhombus, the Rectangle and till- Square. 411. A Heptagon is a plane figure of seven sides. 47. An Octagon is a plane figure of eight sides. (Pig. 25.) 4S. A Decagon is a plane figure of ten sides. (Fig. 2<>.) 49. A Dodecagon is a plane figure of twelve sides. (Fig. 27.) 54. The Center of a circle is a point within the circumference equally distant from every point in its circumference, as A, Fig. 29. Terms and Definitions. 55. The Radius of a circle is a line ilra\vn from dicnlar to the radium drawn to the point of tan- the center to any point in the circumference, as A 15. gency. Tims K I) is perpendicular to !' Hand A Fig. ait, that is, half the diameter. The plural of radius to F 15. is rinlii. 56. The Diameter of a circle is any straight line drawn through the renter to opposite points of the cir- cumference, as (' I), Fig. L'!. 57. A Semicircle is the half of a cirele, and is Dismeter B Fig SO. A Semicircle. Fiy. SI Stymints. bounded liv half the circumference and a diameter (Fig. 3i.) ' .V>. A Segment of a circle is anv part of its sur- face cut oil' l.v a straight line, as A K 15 and C F D Fig- 31. .V.i. An Arc of a circle is any part of the circum- ference, as A B E and C F I), Fig. 3:2. Fig. fig Arcs and Chords. Fiij. .:.',. Sec/o .1. (!S. A Degree. The circumference of a circle is considered as divided into 3t>< equal parts, called i/cr/rei-a Neiu Metal Worker Pattern ffook. (marked ). Each degree is divided into 60 minutes (marked ') ; and each minute into 60 seconds (marked "). Thus if the circle be large or small the number of divisions is always the same, a degree being equal to Fig. 40. A Circle Divided into Degrees for Measuring Angles. part of the whole circumference ; the semicircle is equal to 180 and the quadrant to 90. The radii drawn from the center of a circle to the extremities of a quadrant are always at right angles with each other; a right angle is therefore called an angle of 90 (A E B, Fig. 40). If a right angle be bisected by a straight line, it divides the arc of the quadrant also into two equal parts, each being equal to one-eighth of the whole circumference, or 45, (A E F and FEE, Fig. 40) ; if the right angle were divided into three equal parts by straight lines, it would divide the arc into three equal parts, each containing 30 (AEG, G E H, H E B, Fig. 40). Thus the degrees of the circle are Fig. 41. Complement. Fig. 42. Supplement. used to measure angles, therefore by an angle of any number of degrees, it is understood that if a circle with any length of radius be struck with one foot of the compasses in its vertex, the sides of the angle will intercept a portion of the circle equal to the num- ber of degrees given. Thus the angle A E II, Fig. 40, is an angle of 60. In the measurement of angles by the circumference of the circle, and in the various mathematical calculations based thereon, use is made of certain lines known as circular functions, always liearing a fixed relationship to the radius of the circle and to each other, which gives rise to a number of terms, some of which, at least, it is desirable for the pattern cutter to understand. 69. The Complement of an arc or of an angle is the difference between that arc or angle and a quad- rant. In Fig. 41, A D B is the complement of B D C, and vice versa. To. The Supplement of an arc or of an angle is the difference between that arc or angle and a semicircle. Fig. 4$. Diagram Showing the Circular Functions of the Arc A H or Angle A C H. In Fig. 42, B D C is the supplement of A I) B, and vice versa. 71. The Sine of an arc is a straight line drawn from one extremity perpendicular to a radius drawn to the other extremity of the arc. (II B, Fig. 43.) 72. The Co-Sine of an arc is the sine of the com- plement of that arc. H K, Fig. 4:1. is the sine of the arc A H. 73. The Tangent Of an Arc is a line which touches the arc at one extremity, and is terminated by a line passing from the center of the circle through the other extremity of the arc. In Fig. 43, A E is the tangent of A H or of the angle A C H. 74. The Co-Tangent of an arc is the tangent of the complement. Thus F G, Fig. 43, is the co-tan gent of the arc A H. Terms ami Definitions, 75. Tlic Secant of an arc is a straight lino drawn from the center of a circle through one extremity of that arc and prolonged to meet a tangent to the other extremity of the arc. (K 0, Fig 48.) 70. The Co-Secant of an arc or angle is the secant of the complement of that arc or angle, as V C, Fig. 43. 77. The Versed Sine of an arc is that part of the radius intercepted between the sine and the circumfer- ence. (A B, Fig. 43.) 78. An Ellipse is an oval-shaped curve (Fig. 44), ix. and the straight lino (CD) the directrix. In this lignre anv point, as N or M, is equally distant from F ami the nearest point in I>, as II or K. (See defi- nition 1 1 '.',. i SO. A Hyperbola (A B. Fig. 4ti) is a curve from any point in which, if two straight lines be drawn to two fixed points, their difference shall always be the same. Thus, the difference between E G and G L is II L. and the difference between E F and F L is B L. II L and B L are equal. The two. fixed points, E and L, are called /oa. (See definition 113.) Fig. 44. .In KlHpse. Fig. 40. A Hyperbola. Fig. 45. A Parabola. Fig. 47.Evolute and Involute. Fig. 4S. A Trian- gular Prism. from any point in which, if straight lines be drawn to two fixed points within the curve, their sum will be always, the same. These two points are called foci (F and II). The line A B, passing through the foci, is called the major or transverse a.r/.s. The line E G, per- pendicular to the middle of the major axis, and extend- ing from one side of the figure to the other, is called the minor or conjugate axis. There are various other definitions of the ellipse besides the one given here, dependent upon the means employed for drawing it, which will be fully explained at the proper place among the problems. (See definition 113.) 79. A Parabola (A B, Fig. 45) is a curve in which any point is equally distant from a certain fixed point and a straight line. The fixed point (F) is called the 81. An Evolute is a circle or other curve from which another curve, called the involute or evolutent, is described by the aid of a thread gradually unwound from it. (Fig. 47.) 82. An Involute is a curve traced by the end of a string wound upon another curve or unwound from it. (Fig. 47.) (See also Prob. 84, Chapter IV.) SOLIDS. 83. A Solid has length, breadth and thickness. 84. A Prism is a solid of which the ends are equal, similar and parallel straight-sided figures, and of which the other faces are parallelograms. 85. A Triangular Prism is one whose bases or ends are triangles. (Fig. 48.) 8 n<- \>->r .\f't,il Pattern *;. A Quadrangular Prism is one whose bases or 97. A Truncated Cone is one whose apex is out off ends are quadrilaterals. (Fig. 49.) l>y a plane parallel to its base. (Fig. .~>7. ) This 1'iLnire 87. A Pentagonal Prism is one whose bases or ends is also called a /;;/.-////// of a cone. A pyramid may also are pentagons. (Fig. 50.) be ti:in<<-9 and ~o and dctinition 88. A Hexagonal Prism is one whose bases or ends 112.) are hexagons. (Fig. 51.) 9S. A Pyramid is a solid having a straight-sided s'.t. A Cube is a prism of which all the faces arc base and triangular .sides terminating in one point or squares. (Fiir. ~>~2. i apex. Pyramids arc distinguished as /rim n/u/rn-. f/nn(>. A Cylinde.",or properly sjieakingaCirCUlarCylin- i-"ii>//t/ti>; JM iitnijnniil. lir.rni/niiiil. etc., according as the der, is a round solid of uniform diameter, of which the base has three sides, four sides, live sides, six sides. ends or bases are equal and parallel circles. (Fig..V!.j etc. (Figs. .V s ;. 59 and 60.) Fig. 49. A Quad- rangular Prism. Fig. 50. A Pent- Fig. 51. A Hex- agonal Prism. agonal Prism. Fig. 5S.A Cube. Fig. 53. A Cylinder. Fig .14. A Cone. BOM Fig 55. A Right Cone. Fig. 56 An Oblique or Scalene Cone. Fig.H7.-A Tiun- cated Cone. 91. Aii Elliptical Cylinder is one whose bases are ellipses. 92. A Right Cylinder is one whose curved surface is perpendicular to its bases. 93. An Oblique Cylinder is one whose curved sur- face is inclined to its base. 94. A Cone is a round solid with a circle for its base, and tapering uniformly to a point ac the top called the apex. (Fig 54.) 95. A Right Cone is one in which the perpendicular let fall from the vertex upon the base passes through the center of the base. This perpendicular is then called the axin of the cone. (Fig. 55.) '.<;. An Obliqe Cone or Scalene Cone is one in which the axis is inclined to the plane of its base. (Fig. 5G.) Fig .o. -.1 Ti-ian- Fig. '>!>. .4 Quadran- Fig.6().An,Octag- gular Pyramid. gular Pyramid. onal Pyramid. Fig. 61 A Right Pyramid. / Fig. r,2. Altitude Fig <;.! Altitude of a Cone. of a Pyramid. \ \ fig. r,4.-AUitude of a Priam. Fig. r,5 Altitude of a Cylinder. Fig. 66. A Sphere, or Globe. 99. A Right Pyramid is one whose base is a regular polygon, and in which the perpendicular let fall from the apex upon the base passes _ through the center of the base. This perpendicular is then called the ".'/.-- of the pyramid. (Fig. 61.) 100. The Altitude of a pyramid or cone is the length of the perpendicular let fall from the apex to the plane of the base. The altitude of a prism or cvlinder is the distance between its two bases or ends. and is measured bv a line drawn from a point in one base perpendicular to the plane of the other. (Figs. :,'2. A Sphere or Globe is a solid hounded bv a uniformly curved surface, anv point of which is equally distant from a point within the sphere called tin- center. (Fig. 66.) 103. A Polyhedron is a solid hounded by plane ligurcs. There are live regular polyhedrons, vi/,. : 104. A Tetrahedron is a .solid hounded by f<> in- equilateral triangles. It is one form of triangular pyramid. (.Fiji. ( >7.) 105. A Hexahedron is a solid bounded by six squares. The common name for this solid is which see. (Fig. 52.) io each other as to appear as though one passes through the other. The intersection of their surfaces forms the basis of the greater part of the problems of Chap. VI. 1 1 >. The Frustum of a Cone or Frustum of a Pyra- mid is that portion of the original solid which remains after the apex has been cut away upon a plane parallel to the base.. (Figs. .">". r architects and cornice makers. The cornice of clas- sical architecture may contain simply a bed mold, planceer and crown mold, or it may contain, in addi- tion, a dentil course or a modillion course, or both. 117. The Entablature was used by the ancients to finish a wall or colonnade (more especially the latter), and consisted of three parts, the cornice, the frieze and the architrave. (Fig. 75.) 118. The Architrave, the lower division of the entablature, was in reality a lintel used to span the space between the columns, but its form was main- tained when used above a wall. In modern imitations of the antique styles the molded portion is frequently used without the fascias, in which case it is commonly known as the foot mold. (Fig. 75.) The term arch- itrave is also used to designate the molding and fascias running around an arch or a window opening. 119. The Frieze, the middle division of the entab- lature, is really a continuation of the wall surface to add hight and effect to the building, and was originally intended for the display of symbols, inscriptions, orna- ments, &c., appropriate to the use of the building of which it was a part. It is sometimes treated very plainly and sometimes receives considerable ornamen- tation, being subdivided into panels or enriched by ENTABLATURE < CORNICE < ..... . , J CROWN , PLANCEER I MODILLION BAND 3 __]' IUvQ/^t) V| // -k ( MODILLION ==?i^^ f COURSE DENTIL BAND T 1 1 1 1 1 1 J DENTIL I DENTIL STILE PANEL ARCHITRAVE < STILE FASCIA FASCIA FASCIA Fig. 75. The Entablature and Its Parts. of which, according to modern ideas, might be considered as a cornice. The distinction is shown in Fig. 75, which shows an entablature of a design adaptable to sheet metal construction, and in which all enrichment scrolls, etc. The terms plain frieze, designating a frieze devoid of ornamentation, md. frieze-piece or ///'<./ panel, are used to designate one of the parts of which a frieze is constructed. (Fig. 7/5.) Ti'/-iiis ii/nl Definitions. 11 120. ArCi. Tin 1 curved top of an opening in a wall. Tlio arch of masonry is constructed of separate blocks and is supported only at the extremities. The joint lines between the bloeks are disposed in the direc- plan, designed as a support for an entablature. It con- sists of three parts : a base, a shaft and a capital. 12-2. An Engaged Column is a column placed against the face of a wall or other surface, from which Fig. 76. A Semicircular Arch. Fig. 77. A Pointed Arch. Fig. 80. A Pilaster. tion of radii of the curve, thus enabling the arch to support the weight of the wall above the opening. When in classical designs its face is finished with moldings their proper profile is that of an architrave. (Fig. 75.) The level lines at which the curve of the arch begins are called the springing lines. Sometimes the lower stones of the arch rise vertically a short distance from the supports before the springing lines are reached, in which case the arch is said to be stilted. it projects one-half or more than one-half its diam- eter. 123. A Pilaster differs from a column in that it is square in plan instead of round and is usually engaged within a wall. (Fig. 80.) 124. Pedastal. A structure designed to support a column, statue, vase or other object. It is by some described as the foot of a column, but is, properly speaking, not a part of it. It consists of three Fig. 81. An Angular Pediment. Fig. 78. A Moresque Arch. Fig. 79.-A Flat Arch. Fig. 82. A Segmental Pediment. The stones composing the arch are called the voussoirs, and the middle or top stone is called the keystone. The supports below the ends of the arch are called imposts. Arches are usually semicircular (Fig. 76), semi- elliptical, segmental, pointed (Fig. 77) or Moresque (horse- shoe) (Fig. 78) in shape, according to the style of archi- tecture with which they are used. The top of an opening may be perfectly level and yet composed of wedge-shaped blocks so combined as to be self-supporting, in which case it is called a flat arch. (Fig. 79.) 121. A Column is a vertical shaft or pillar round in parts, a base, a middle portion cubical in shape called a die and a cap or cornice. It is also used as a finish at the ends of a balustrade course. Fig. SS.A Broken Pediment. 125. A Pediment is a triangular or segmental orna- mental facing over a portico, door, window, etc. (Figs. 81, 82 and 83. 12 TJie Xeir Wor/,-> Ifook. 126. A Broken Pediment is one. cither in the frm of a gable <>r a segment, which is cni away in its central portion for the purpose of ornamentation. (Fig. 83.) \-2~t. A Gable is the vertical triangular end of a house or other building, from the cornice or eaves to the top. I'JS. A Lintel Cornice is a cornice above or some- times including a lintel. This term is very generally used to designate the cornice used above the lirst story of stores. (Fig. S4.) 1 2!). A Deck Cornice or Deck Molding is the cornice Fig. S4.A Lintel Cornice. Fig. 85. A Bracket. or molding used to finish the edge of a flat roof where it joins a steeper portion. 130. A Bracket, as used in sheet metal work, is simplv an ornament of the cornice. Brackets in stone architecture were originally used as supports of the parts coming above them. Hence modern architecture has kept up that idea in their designs. (Fig. So.) 1:51. Modillions are also cornice ornaments, and differ from brackets onlv in general shape. (Fig. 86.) While a bracket has more' depth than projection, modil- lions have more projection than depth. 132. A Dentil is a cornice ornament smaller than a modillion, which i,i shape usuallv represents a solid with plain rectangular face and sides. Dentils are never used singly, but in courses, the spaces between them being less than their face width. (Fig. 76.) 1:;:;. A Cortd is a modified form of bracket. It is used to terminate the lower parts of window caps, and also forms the support for arches, etc., in gothic forms. Fig. S6.A Modillion. Fiij. X7.A Head Block. 13-t. A Head Block or Truss is a large terminal bracket in a cornice, projecting suflicientl v to receive all tlu> moldings against its side, thus forming a finish to the end of the cornice. (Fig. ST.) 135. A Stop Block is a block-shaped structure, vari- ously ornamented, which is placed above the end Fig. 88. A Stop Mock. Fig. 89. A Pinnacle. bracket in a'cornice, and which projects far enough to receive against its side the various moldings occurring above the brackets, forming an end linish. (Fig. *V i 136, A Pinnacle is a slender turret or part of a building elevated above the main building. A small spire. (Fig. s!.) Definitions. 13 137. A Finial is an ornanie-nt. variously designed, placed at tlic apex of a pediment, gable, spire or roof. 13.s. Capital. The upper member or head of acol- iiniii <>r pilaster. It may vary in character according to the style of architecture with which it is employed, from a few simple projecting moldings around the top of tin- column to an elaborately foliated ornament. The lower- 'A8ACUS IVOLUTE Fig. 00. Capitals, most mold is called the ni'i'l: muld and the uppermost member sustaining the weight of the lintel or arch above is called the ulnu-ii*. (Fig. 90.) 1:5!). Panel. A sunken compartment having molded cdu'cs used to ornament a plane surface', as a frieze ceil- ing, planceer or tvmpanum. A panel may, however, be raised instead of sunken. The margin or space between the sides of the pane! and the edges of the surface- in which it is placed is usuallv made equal all around and is called the stil< j . 140. A Volute is a spiral scroll used as the principal ornament of a capital and is placed under the corners of the abacus. For inethe>il of drawing the volute see Probs. 81 and 8i>, Chap. IV. 14-1. A Molding is an assemblage of forms projevt- ing bevoiid the wall, column or surface- to which it is affixed. (See tirst part of Chap. V.) 142. Crown Molding is the term applied to the upper or projecting membe-r of a cornice. (Fig. 75.) 143. Planceer or PlanCher is the ceiling or under side of the projecting part of a cornice. (Fig. 75.) 144. The Bed Moldings of a cornice are those mold- ings forming the lower division of the cornice proper, and which are made up of the bed course, modillion course and dentil course. (Fig. 75.) 145. The Bed Course is the upper division of the be-d moldings, the part with which the bracket heads and modillion heads ordinarily correspond, and against which they miter. (Fig. 75.) 14f a cornice embraces the modillions and all the moldings which are imme- diately back of and below them. The plain surface lying back of or between the modillions is called in she-et metal work the ni'/'l.illinii lauid. anil the molding immediatelv below them the ///l, is the molding around the upper part of a bracket, ami which generally members with the bed molding, against which it finishes. (Fig. 75.) 150. A Gable Molding is an inclined molding which is used in the finish of a gable. 151. A Ridge Molding is a mohling used to cap or tinish a rielge. It is also culled a //'/of. It is very fre- quently included in the more general term '''ut of higher price, consists of a walnut head with a Lard- wood blade, edged with some oilier kind of wood. Still another variety has a mahogany blade edged with ebony. T-squares constructed with cast-iron head open work finished by japanning with nickel-plated steel blade, are also to be had from dealers. They are also made with a hard rubber blade, of which Fig. 104 is an illustration. The liability to fracture, however, by dropping necessitates the greatest care in use: otherwise hard rubber makes a very desirable article and is the favorite material with many draftsmen. As to size, T-squares should be selected with reference to the use to be made of them. Generally, the blade should be a very little less in length than the width of the table or board upon which it is to be Fig. W4.t-Square with Fixed and Swivel Bead. used. Where a large board or a table is used it will be found economical to have two instruments of dif- ferent sixes. The Ste.l Square. One of the most useful tools in connection with the pattern cutter's outfit is an ordinary steel square. The divisions upon it concern him much less than its accuracy. He seldom requires other divisions than inches and eighths of an inch; therefore in selection the principal point to be considered is that 21 of aeeuracv. The finish, bowever, is a imittcr not to he overlooked. Since a nickel-plated square costs hut a trilling advance ii|nin the plain article, it is' cheaper in the long' run to have the plated tool. A convenient method of testing the correctness of the outside of a square, and one which can be used at the time and place of purchase, is illustrated in Fig. 105, Testing the Exterior Angle of a Steel Square. Tig. H>.">. Two squares are placed against each other and against a straight-edge, or against the arm of a third square. If the edges touch throughout, the squares mav be considered correct. Having procured a square which is accurate upon the outside, the correctness of the inside of another square may he proven, as shown in Fig. 10t>. Place mi" square within the other, as shown. If the edges Fig. Uifi.Ttas ing the Interior Angle of a Steel Square. fit together tightly and uniformly throughout, the square may he considered entirely satisfactory. An accurate square is especially desirable, as it ail'ords the readiest means of testing the "[-square and the drawing table and beard, as elsewhere described. The greatest care should be given, therefore, to the selection of a square. For all ordinary. purposes the two-foot size is most desirable. In some cases the one-foot sixe is better suited. Many pattern cutters on cornice work like to have both sixes at their com- mand, making use of them interchangeably, according to the nature of the work to he done. Triangles, or Set Squares. In the selection of tri- angles, the draftsman has the choice in material be- Fig. 107. Open Hard Rubber Triangle or Set Sfauare, 45 x 43 x 90 Degrees. tween pear wood ; mahogany, ebony lined ; hard rub- ber ; German silver, and steel, silver or nickel plated. In style he has the choice between open work, of the form shown in Fig. 107, and the solid, as in Fig. 108. In shape, the two kinds which are adapted to the pattern cutter's use are shown in Figs. 107 and 108, the latter being described as 30, 60 and 90 de- grees, or 30 by 60 degrees, and the former as 45, 45 and 90 degrees, or simply 45 degrees. The special uses of each of these two tools are shown in the chapter on Geometrical Problems (Chap. IV). In size, the pattern cutter requires large rather than small Fig. lOS.Hard Wood Triangle or Set Square, SO x GO x 00 Degrees. ones. If he can have two sizes of each, the smaller should measure from 4 to 6 inches on the side, and the larger from 10 to 12 inches; but if only a single size is to be had, one having dimensions intermediate to those named will be found the most serviceable. The value of a triangle, for whatever purpose used, depends on its accuracy. Particularly is this to be said of the right angle, which is used more than either of the others. A method of testing the accuracy of the right angle is shown in Fig. 109. Draw the line A B 22 Tin' Xc Mdal Worker Pattern Boole. with an accurate ruler or straight-edge. Place the right angle of a triangle near the center of this line, as shown by D C B, and make one of the edges coincide with the line, and then draw the line D C against the B C Fig. 109. Testing the Right Angle of a Triangle. other edge. 'Turn the triangle into the position indi- cated by D C A. If it is found that the sides agree with A C and C D, it is proof that the angle is a right angle and that the sides are straight. Besides the kinds of triangles described above, a fair article can be made by the mechanic from sheet zinc or of heavy tin. Care must, however, be taken in cutting to obtain the greatest possible accuracy. For many of the purposes for which a large size 45 degree a profile line is called spacers, as illustrated and de- scribed below. A pair of compasses consists of the parts shown in Fig. 110, being the instrument proper with detach- able points, and extras comprising a needle point, n pencil point, a pen and a lengthening bar, all as shown to the left. In selection, care should be given to the workmanship; notice whether the parts lit together neatly and without lost motion, and whether the joint works tightly and yet without too great friction. A good German silver instrument, although quite ex- pensive at the outset, will be found the cheapest in the end. A pencil point of the kind shown in our engraving is to be preferred over the old style which clamps a common pencil to the leg. The latter is not nearly so convenient and is far less accurate. Of dividers there are two general kinds, the plain dividers, as shown in Fig. Ill, and the hair-spring dividers, as shown in Fig. 112. The latter differ from the former simply in the fact of having a fine spring and a joint in one leg, the movement being controlled by the screw shown at the right. In this way, after Fig. 110. Compasses with Interchangeable Parts. Fig. 111. Plain Dividers. Fig. 112.- Hair-Spring Dividers. Fig. US. Steel Spring Spacers. triangle would be used the steel square is available, but as the line of the hypothenuse is lacking, it can- not be considered a substitute. Compasses and Dividers. The term compasses is applied to those tools, of various sizes and descriptions, which hold a pencil and pen in one leg, and are used for drawing circles, while dividers are those tools which, while of the same general form as compasses, have both legs ending in fixed points, and are used for measuring spaces. A special form of dividers used exclusively for setting off spaces, as in the divisions of the instrument has been set approximately to the dis- tance desired, the adjustable leg is moved, by means of the screw, either in or out, as may be required, thus making the greatest accuracy of spacing possible. Both instruments are found desirable in an ordinary set of tools. The plain dividers will naturally be used for larger and less particular work, while the hair- spring dividers will be used in the finer parts. It fre- quently happens that two pairs of dividers, set to dif- ferent spaces, are convenient to have at the same time. A pair of spacers, shown in Fig. 113, is almost Tool* inn! 23 indispensable in a pattern cutter's outfit. He will find advantageous use for this tool, oven though possess! IILT Imtli pairs of dividers described above. In size they are made less than that of the dividers. The points should be needle-like in their fineness, and should lie capable of adjustment to within a very small distance of each other. It is sometimes desirable to divide a given profile into spaces of an eighth of an inch. The spacers should be capable of this, as well as adapted to spaces of three-quarters of an inch, without being too loose. As will be seen from the engraving, this instrument is arranged for minute variations in ad- justment. Beam Compasses and Trammels In Fig. 114 is shown a set of beam compasses, together with a portion of the wooden rod or beam on which they are used. The latter, as will be seen by the section drawn to one side (A), is in the shape of a f. This form has con- siderable strength and rigiditv, while at the same timr it is not clumsy or heavy. Beam compasses are pro- vided with extra points for pencil and ink work, as shown. While the general adjustment is effected by means of the clamp against the wood, minute varia- tions are made by the screw shifting one of the points, as shown. This instrument is quite delicate and when in good order is very accurate. It should be used only for line work on paper and never for scribing on metal. A coarser instrument, and one especially designed for use upon metal, is shown in Fig. 115 and is called Fig. 114. Beam Compasses, n trammel. It is to be remarked in this connection that the name trammel, by common usage, is applied to this instrument and also to a device for drawing ellipses, which will be found described at another place. There are various forms of this instrument, all being the same in principle. The engraving shows a form in common use. A heavier stick is used with it than with the beam compasses, and no other adjustment is provided than that which is all'orded by clamping against the stick. In the illustration a carrier at the Fig. 115. Trammel. side is shown in which a pencil may be placed. Some trammels are arranged in such a manner that either of the points may be detached and a pencil substituted. A trammel, by careful management, can be made to describe very accurate curves, and hence can be \ised in place of the beam compasses in many instances. For all coarse work it is to be preferred to the beam compasses. It is useful for all short sweeps upon sheets of metal, but for curves of a very long radius a strip of sheet iron or a piece of wire will be found of more practical service than even this tool. The length of rods for both beam compasses and trammels, up to certain limits, is determined by the nature of the work to be done. The extreme length is determined by the strength and rigidity of the rod itself. It is usually convenient to have two rods for each instrument, one about 3$ or i feet in length and the other considerably longer as long as the strength of material will admit. In the case of the trammel, by means of a simple clamping device, or, in lieu cf better, by use of common wrapping twine, the rods may be spliced when unusual length is required; but a strip of sheet iron or a piece of fine wire forms a better radius, under such circumstances, than the rod. 24: The New Metal Worker Pattern book. The Protractor is an instrument for laying down and measuring angles upon paper. The instrument consists of a semicircle of thin metal or horn, as rep- resented in Fig. 116, the circumference of which is divided into 180 equal parts or degrees. The princi- ples upon which the protractor is constructed and used are clearly explained in the chapter on Terms and Defi- nitions (Def. 68 " Degree "). The methods of employing it in the construction of geometrical figures are shown Fig. 116. Semicircular Protractor. in Chapter IV among the problems. For purposes of accuracy, a large protractor is to be preferred to a small size, because in the former fractions of a degree are indicated. While a number of geometrical problems are con- veniently solved by the use of this instrument, it is not one that is specially adapted to the pattern cutter's use. All the problems which are solved by it can be worked out by other accurate and expeditious methods, which, in most cases, are preferable. It is one of the instruments, however, included in almost every case of instruments sold, and the student will find it advantageous to become thoroughly familiar with it, whether in practice he employs it or not. Besides the semicircular form of the .,,, protractor shown, corresponding lines and Ln iilinn divisions to those upon it are sometimes put upon some of the varieties of scales in use, as shown in Fig. 120. Scales. Many of the drawings from which the pattern cutter works that is, from which he gets di- mensions, etc., are what are called scale drawings, being some specified fraction of the full size of the object represented. Architects' elevations and floor plans are very generally made either -J- or -J inch to the foot, or, in other words, -5*3- or -fa full size. Scale details are also employed quite extensively by architects, scales in very common use for the purpose being H inrhes to the foot and 3 inches to the fool, or, in other words, ^ and -J- full size respectively. It is essential that the pattern cutter should be familiar with the various scales in common use, that lie may be able to work from any of them on demand. Sev- eral of the scales are easily read by means af the com- mon rule, as, for example, 3 inches to the foot, in which each quarter inch on the rule becomes one inch of the scale; also, 1-J- inches to the foot, in which each eighth of an inch on the rule becomes an inch of the scale; and, likewise, f inch to the foot, in which each sixteenth of an inch on the rule becomes an inch of the scale. However, other scales besides these are occasionally required, which are not easily read from the common rule, and sometimes special scales are used, which are not shown on the instruments, espe- cially calculated for the purpose. Accordingly, it is sometimes necessary for the pattern cutter to construct his own scale. The method of constructing a scale of 1 inch to the foot is illustrated in Fig. 117, in which the di- visions are made by feet, inches and half inches. In constructing such scales, it is usual to set off the di- visions representing feet in one direction (say to the right) from a point marked 0, while the divisions for inches and fractions thereof are set off the opposite way (or to the left from 0) as shown in the illustra- tion. In using the scale, measurements are made by placing one point of the dividers at the number of feet required ; the other point can then be moved to the other side of the to the required number of inches, thus embracing the entire number of feet and inches between the points of the dividers. Besides scales of the kind just described, which Inches Feet Fig. 117. Plain Scale (1 inch to the Foot ) are termed plain divided scales, there are in common use what aBe known as diagonal scales, an illustration of one of which is shown in Fig. 118. The scale rep- resented is that of 1-j- inches to the foot. The left- hand unit of division has been divided by means of the vertical lines into 12 equal parts, representing inches. In width the scale is divided into 8 equal parts by means of the parallel lines running its entire length. Next the diagonal lines are drawn, as shown. J)ra>ci>ig 7W< and Materials, Bv a moment's inspection it will bo seen that, by means of these diagonal lilies, one-eighth of an inch and multiples thereof are shown on the several hori- zontal lines. A distanee equal to the space from A to 11. as marked on the scale, is read (first at the right for feet) '2 feet (then to the left for inches by means of Aflat scale is also manufactured in both boxwood and ivory. Fewer scales or divisions can be put upon it than upon the triangular scale, yet for certain pur- poses it is to be preferred to the latter. There are less divisions to perplex the eye in hunting out just what is required, and accordingly, there is less lia- Fent Fig. US. Diagonal Scale inches to the foot). the vertical lines ligured both at top and bottom) 6 inches (and last, by means of the diagonal line, figured at the end of the scale, for fractions) and three-eighths. The top and bottom lines of the scale measure feet and inches only. The other horizontal lines measure feet, inches and fractions of an inch, each horizontal line having its own particular fraction, as shown. Such scales are frequently quite useful, as greater accuracy is obtained and, as the reader will see, may be con- structed by any one to any unit of measurement, and divided by the number of horizontal lines into any de- sired fractious. A scale in common use, and known as the tri- angular scale, is shown in Fig. 110. The shape of this scale, which is indicated by the name, and which bility to error in its use. However, the limited num- ber of scales which it contains greatly restricts its usefulness. Fig. 120 shows another form of the flat scale, in quite common use in the past, but now virtually dis- carded in favor of more convenient dimensions and shapes. This scale combines with the various divi- sions of an inch the divisions of the protractor, as shown around the margin. The fact that the divisions of an inch for purposes of a scale are located in the middle of the instrument, away from the edge, which makes it necessary to take off all measurement with the dividers, renders the article awkward for use, and Fig. 119 Triangular Boxwood Scale. is also shown in the cut, presents three sides for divi- sion. By dividing each of these through the center lengthways by a groove, as shown, six spaces for divisions are obtained, and by running the scales in pairs that is, taking two scales, one of which is twice the size of the other, and commencing with the unit at opposite ends the number of scales which may be put upon one of these instruments is increased to twelve. This article, which may be had in either boxwood, ivory or plated metal, and of 6, 12, 18 or 24 inches in length, is probably the most desirable for general use of any sold. s 2 j \ ' S \ i 1 f ' , / / / / ' - ., f ^ -^ ~i'f^ -^ \ .. i > -V - '',r ^ ^ Q// ^ J 2 f< t; 1 2 | -1 B -I 7 e ^ 10 1 1 12J 13 M n 16 *' a I-.-, 1 2 3 4 B f j \\ s 1 2 I 4 ,'_ . 1lN (j 1 2 ; = s Fig. ISO. Flat Scale with Divisions of the Protractor on the Margins. the arrangement of the divisions of the circle, on the margins, is less satisfactory for use than the circular protractor. Lead Pencils. Various qualities of pencils are sold, some at much lower prices than others, but, all things considered, in this as in other case's, the best arc the cheapest. The leading brands arc made in two grades or qualities. The ordinary grades employ numbers. 1, 2, 3, etc., to indicate hardness of lead, No. 1 being the softest, and No. 5 being the hardest in common use. A finer grade of pencils, known as poligrades, is marked by letters, commencing at the softest with B B, T/ie New Metal Worker Pattern, Book. and ending at the hardest with H H II II H H, while other makes of pencils are marked by systems peculiar to their v manufacturer. The draftsman has the choice of round or hexagon shape in all except the finest grades, the latter being made exclusively hexagon. Whatever kind of pencil the draftsman or mechanic uses, he will require different numbers for different I mi-poses. For working drawings, full-sized details, etc., on manila paper, a No. 3 (or F) is quite satis- factory. Some like a little harder lead, and therefore prefer a No. -i (or II). For lettering and writing in connection with drawings upon manila or ordinary detail paper, a No. 2 (II B) is usually chosen. For fine lines, as in developing a miter, in which the great- est possible accuracy is required, a No. 5 is very gen- e rally used, although many pattern cutters prefer the liner grade for this purpose and use a H II II H H. The quality and accuracy of drawings depend, in a considerable measure, upon the manner in which pencils arc sharpened. A pencil used for making fine Fig. 121. Two Views of Pencil Sharpened to a Chisel Point. straight lines, as, for instance, in the various opera- tions of pattern cutting, should be sharpened to a chisel point, as illustrated in Fig. 121. Pencils for general work away from the edges of the T-square, triangle, etc., should be sharpened to a round point, as shown in Fig. 122. It facilitates work and it is quite eco- nomical to have several pencils at command, sharpened in different ways for different purposes. Where for any reason only one pencil of a kind can be had, both ends may be sharpened, one to a chisel point and the other to a round point. For keeping a good point upon a pencil, a piece of fine sand paper or emery paper, glued upon a piece of wood, will be found very serviceable. A flat file, mill-saw cut, is also useful for the same purpose. Sharpen the pencil with a knife, so far as the wood part is concerned, and then shape the lead as required upon the file or sand paper. Drawing Pens. Although most of the pattern cutter's work is done with the pencil, there occasion- ally arise circumstances under which the use of ink is desirable. Tracings of parts of drawings are frequently required which can be better made with ink than with pencil. The drawing pen or ruling pen, as illustrated in Fig. 123, is used for drawing straight lines. The drawing pen, whether as a separate instrument or as an attachment to compasses or beam compasses for drawing curved lines, consists of two blades with steel points, fixed to a handle. The blades are so curved that a sufficient cavity is left between them for ink when the points meet close together or nearly so. The space between the points is regulated by means of the screw shown in the engraving, so as to draw lines of any required thickness. One of the blades is provided with a joint, so that, by taking out the screw, the blades may be completely opened and the points readily cleaned after use. The ink is put between the blades with a common pen, or sometimes by a small hair brush. In using the drawing pen it should be slightly inclined in the direction of the line to be drawn, and should be kept uniformly close to the ruler or straiglit- Fig. IS?. Pencil Sharpened to a Round Point, Fig. 123. Ruling Pen. edge during the whole operation of drawing a line, but not so close as to prevent both points from touch- ing the paper equally. Keeping the blades of the pen clean is essential to good work. If the draftsman is careless in this par- ticular, the ink will soon corrode the points to such an extent that it will be impossible to draw fine lines. Pens will gradually wear away, and in course of time they require dressing. To dress up the tips of the blades of a pen, since they are generally worn unequally by customary usage, is a matter of some nicety. A small oil stone is most convenient for use in the operation. The points should be screwed into contact in the first place, and passed along the stone, turning upon the point in a directly perpendicular plane until they acquire an identical profile. Xext they arc to be unscrewed and examined to ascertain the parts of unequal thickness around the nib. The blades are, then to be laid separately upon their backs upon the stone, and rubbed down at the points until they are brought up to an edge of uniform fineness. It is well to screw them together again and pass them over the Tun/.* mill Materials. 27 stone once or twice more to bring up anv fault ami t<> retouch them also at tin 1 outer and inner side of each blade to remove barbs or framing, and linally to draw them across the palm of the hand. India Ink. .For tracing*, and fur some kinds of drawings, which the ])att( i rn cutter is obliged to make occasionally. India ink is niueh better than the pencil, which is used for the greater part of his work, ('arc is to lie exercised in the selection of ink, as poor grades arc sold as well as -good ones. Some little skill is required in dissolving or mixing it for use. India ink is sold in cakes or sticks, of a variety of shapes. It is prepared for use by rubbing the 1 end of the stick upon the surface of aground glass, or of a porcelain slab or dish, in a very small quantity of water, until the mixture is siitlieiently thick to produce a black line as it Hows from the point of the ruling pen. Tin 1 qualitv of ink may generally be determined bv the price. The common si/e sticks are about 3 inches long. Inferior grades can be bought as low as 40 cents per stick, while a good quality is worth $(1.50 to *_' per stick, and the very best is still higher. How- ever, except iu the hands of a responsible and expe- rienced dealer, this method of judging is hardly satisfactorv. To a certain extent ink may be judged by the brands upon it, although in the ease of the higher qualities the brands frequently change, so that this test may not be infallible. The quality of India ink is quite apparent the moment it is used. The best is entire! v free from grit and sediment, is not musky, and has a soft feel when wetted and smoothed. The color of the lines may also be used as a test of quality. \Vith a poor ink it is impossible to make a black line. It will be brown or irregular in color and will present an irregular edge, as though broken or ragged, while an ink of satisfactory quality will produce a clean line, whether drawn very fine or quite coarse. Various shaped cups, slabs and dishes are in use for mixing and containing India ink. In many re- spects thev arc like those used for mixing and holding water colors. Indeed, in many cases the same articles are employed. The engraving (Fig. 1'24) shows what is termed an India ink slab, with three holes and one slant. This article is in common use among draftsmen and serves a satisfactory purpose. In order to retard evaporation, a kind of saucers, in sets, is frequently used, so constructed that one piece will form a cover to the other, and which are known in the trade as cabinet sets or cabinet saucers. They are from 2 to 3 inches in diameter and come in sets of six. In the absence of ware especially designed for the purpose, India ink can be satisfactorily mixed in ;ind used from an ordinary saucer or plate of small size. The articles made especially for it, however, are convenient, and Front with Cover On. Top with Cover Off. Fig. U4. India Ink Slab. in facilitating the care and economical use of the ink are well worth the small price they cost. Several makes of liquid drawing ink are also to be had, which possess the advantage of being always ready for use, thus doing away with the rubbing process. The ink costs about 25 cents a bottle, keeps well, and will answer almost every purpose quite as well as the stick ink. Thumb Tacks or Drawing Pins, both names being in common use, are made of a variety of sizes, ranging from those with heads one-quarter of an inch in diam- Fig. 125. Thumb Tacks, or Drawing Pins. eter up to eleven-sixteenths of an inch in diameter. They are likewise to be had of various grades and qualities. The best for general use are those of Ger- man silver, about three-eighths to five-eighths of an inch in diameter, and with steel points screwed in and riveted Those which have the points riveted only are of the second quality. The heads should be flat, to allow the "T-square to pass over them readily. In the 28 The New Metnl \\'<>rke,- Pattern llonlc. annexed cut. Fig. 125, are shown an assortment of kinds and sixes. Those which are beveled upon their upper edges are preferable to those which are beveled underneath. A Box of Instruments. Fig- 126 shows a box of in- struments of medium grade, as made up and sold by the trade general Iv. While it contains some pieces that the pattern cutter has no use for, it also contains the principal tools he requires, all put together in com- pact shape, and in a convenient manner for keeping the instruments clean and in good order. The tray of the box lifts out, there being a space underneath it in which may be placed odd tools, pencils, etc. Tools may be selected, as required, of most of the large dealers in drawing instruments. It will be found ad- vantageous to the pattern cutter to buy his instruments singly as he requires, them, as by so doing he will get only what he requires for use, and will probably secure Fig. 126. A Box of Instruments. a better quality in the tools. After he has made his selection, a box properly fitted and lined should be provided for them and can be obtained at a small cost, or made if desirable. India Rubber. A good rubber with which to erase erroneous lines is indispensable in the pattern cutter's outfit. The several pencil manufacturers have put their brands upon rubber as well as upon pencils, and satisfactory quality can be had from any of them. The shape is somewhat a matter of choice, flat cakes being the most used. A very soft rubber is not so well adapted to erasing on detail paper as the harder varie- ties, but is to be preferred for use in fine drawings on good quality paper. Paper. The principal paper that the pattern cutter has anything to do with is known as brown detail paper, or manila detail paper. It can be bought of almost any width, from 30 inches up to 54 inches, in rolls of 50 to 100 pounds each. It is ordinarily sold in (lie roll by the pound, but can be bought at retail by the yard, although at a higher figure. There are dilVerent thicknesses of the same qualitv. Sonic dealers indi- cate them by arbitrary marks, as XX, XXX. XXXX; others by numbers 1, 2, 3; and still others as thin, medium and thick. The most desirable paper for tin- pattern cutter's use is one which combines several good qualities. It should be just as thin as is consistent with strength. A thick paper, like a still' card, breaks when folded or bent short, and is, therefore, objection- able. The ] taper should be very strong and tough, as the requirements in use arc quite severe. The surface should be very even and smooth, yet not so glossv as to be unsuitcd to the use of hard pencils. It should be hard rather than soft and should be of such a texture as to withstand repeated erasures in the same spot without damage to the surface. White drawing paper, which the pattern cutter has occasionally to use in connection with his work. can be had of almost every conceivable grade and in a variety of sixes. The very best qualitv, and the kinds suited for the finest drawing*, come in sheets exclu- sively, although the cheaper kinds are also made in the shape of sheets as well as in rolls. White drawing paper in rolls can be bought of different widths, rang- ing from 36 to 54 inches, and from a verv thin grade up to a very heavy article, and of various surfaces. It is sold by the pound, in rolls ranging from 30 to 40 pounds each, and also at retail by the yard. A kind known as eggshell is generally preferred bv architec- tural draftsmen. Drawing paper in sheets is sold by the quire, and at retail by the single sheet. The sixes are generally indicated by names which have been applied to them. The following are some of the terms in common use. with the dimensions which they represent placed op- posite : Cap. 13 x 17 I Elephant 23 x 28 Demy 15 x 20 Atlas 21! x 34 17 x 22 Columbier 23 x 35 Medium Eoyal 19 x 24 | DoubleElephant. 27 x 40 Super Royal 1!> x 27 Antiquarian 31 x 53 Imperial 22 x 30iKmpcror 48 x , being the upper limit of a curve, is entitled to representation. but being so situated as to be invisible when viewed from a point in front of the mold, the line is properly made dotted, The lines of projection from the section to the elevation are also shown dotted in the engrav- ing. A vertical line terminates the elevation of the mold at the right or end nearest the section, while the absence of such a line at its left end indicates that it extends indefinitely in that direction. It would also be proper, upon that supposition, to finish the eleva- tion at the left with a broken line. Referring now to Fig. I'-'*, it is most likely that the front elevation would be next drawn after the plan. For this purpose the plan should be so placed upon the board that the part representing the front should be turned toward the bottom of the board, in which position it appears to be turned toward the observer. Place the "["-square so that the blade lies vertically upon the board that is, crossing it from front to back and bringing it to the different angles or points of the front side of the plan, draw a line ver- tically from each, through that portion of space upon Fig. 1S7. Elevation Projected from Section. the paper allotted to the elevation, all as shown by the dotted lines. Thus each point of the elevation comes directly over the point which represents it in the plan, and the horizontal distance across any part of the new elevation thus becomes exactly the same as that of the plan. The question of bights is here a matter of de- sign and is governed by specifications supplemented by the designer's judgment. With the plan and the front elevation complete the drawing of any other ele- vations or sections is entirely a matter of projection, except as new features might occur in those views which would not appear in either of the views already drawn. If an elevation of the right side is about to be con- structed, lines would be projected horizontally to the right from every point in the front elevation of the ob- ject which would be visible when seen from the right side, thus locating all the hights in the new view. As the horizontal distances in this view must agree with distances from front to back on the plan, they may best lie obtained by turning the plan (or so much of itas nec- essarv to this view) one-quarter around to the right, so 32 The Sew Metal \\~ui-ker I'atttni il<><>k. ^ ^~- y / ^ z ^--'' ," < X- ai ,'' _i UJ X K I [I oc HI ee - " 1~ ,-' J- ^J - n ^-_- -t T T~l , '' -- -='4c) ' 1 S V ' x .,'Z-z.* 1U '','*;'' / **v / '?/ -+- 1 [_ / //// ^* //// ;>T" //// /V ,s~ \ I ... --- uj k\s"M k^-^jj^g u i ^ // / / / / / // / / / / / nil ' f / \ I I t / *""| ! ' '/ ' / ' x< : ' II / ' / / /f , 1 r- 1 - 1 > 1 1 // /' ! ' LL. 1 p 1 X Z 1 1 1 X.VlVc".".',' , ,':!,'. . 1 L',V,;V;VA '; S -h i 1 L ---{ 4- 0. _1_ t H > 1 <- -h S; 1 r ^ > | 21 r 1 ! * -1 n o " " i ~ ui a 1 o u| TIT r ? z H IT i i --f z -I- L -^^- * < .? { o " " ? " ~~T~ "*~ "^S: t MJ ^y Q - h~ i u. ~ *\~/*\7 \ 3 S id ^V^ 4 '33r i i i LL 1- i n \ \ i i i i i I o 2j \ ) o \ UJ \ \ en / / 1=^ p ^ v 1- \ u. \ !2 < / \ 1 i U h A \ 1 UJ \ %. i " j_i \ to 1- u. UJ _1 L Linear Drawing. 33 tluit tlic side of \vliidi the new elevation is 1<> lie drawn will he toward the bottom 'or near side, o[ the lioanl. as shown at (i : after which lines inav lie projected with the T-square from the points of the plan into the eleva- tion, intersecting with corresponding lines, as shown. The same result ma\ he accomplished hv projecting the lines to the right from the side of the plan, as shown in the top view, until thev reach anv line paral- lel to the side, as II I. PYoin this line tliev may be carried around a quarter circle from any convenient center, as N, arriving at a horizontal line, N M, and thence dropped downward, intersecting as before. It will thus he seen that the elevation of the right hand side of anv object comes naturally at the right of the front elevation, and the left side elevation, at its left. This idea is best illustrated hv supposing that the ohject in question he placed in a glass box of the dimensions of the base II I J K of the top view, and that thi' elevation of cadi side of the object be pro- LONGITUDINAL SECTION ON C-D TRANSVERSE SECTION ON A-B Fig. 129. Vertical Sections Derived from Fig. 128. jected upon the adjacent parallel side of the box at right angles to the same, and that afterward all the sides (supposing them to be hinged at the corners) be opened out into one plane, as shown by K L, II O and O 1' (the top face of the box being opened upward), thus displaying all the views in one plane as repre- sented by Fig. 12S. This idea should not be carried so far as to open the bottom face of the box downward, because this would produce a plan as seen from below, which is never done except in the case of a design of a ceiling or sollit. when it should be spoken of as an inverted plan. In Fig. 129 the transverse section is shown at the right of the longitudinal section, because the view in it is from the right, or in the direction of the arrow in the longitudinal section, showing what would he seen if the house were cut in two on the line A B of the plan and the right hand portion removed. The longi- tudinal section is for the same reason placed at the left, of the transverse section that is, it is a view from the left of the house when placed in the' position shown hv the transverse section. From the foregoing it, is to he understood, therefore, that when a view appears to the ri^ht of another it is supposed to show what would be seen when the object is viewed from the right hand end or side of what is shown in the other, the other (or front) view being at the same time a view of the left side of what is shown by the right side elevation. In this class of drawings various kinds of lines are used, cadi of which possesses a certain significance. The general outlines of the different views should be linn and strong enough to he distinctly visible, with- out being so broad as to leave any doubt as to the ex- ad dimensions of the part shown when the rule is ap- plied for purposes of measurement. It is not always necessary that all the lines of projection should be shown. When shown they may appear as the finest possible continuous lines, or as dotted lines such as are shown in Figs. 127, 128 and 129. Lines used in carrying points from a profile to a miter line, or from one line to another for any purpose, are rea 1 1 v lines of projection, and for the pattern drafts- man's purposes it may be said that the liner they are drawn the greater the accuracy ob- tained (see Chapter II under the head of Lead Pencils). Dotted lines are also used to represent portions which are out of sight that is, back of or underneath the other parts which constitute the view under consideration, but which it is necessary to show, as, for instance, a portion of the chimnev in longitudinal section Fig. 129 and, points 1) and F in the profile of the mold in Fig. 127. Dotted lines are also used to show a change of position or an alternate position of some part, as, for example, the lines L K and .J L show that the side .1 K of the top view has been swung around on the point K until it occupies the position shown by L K, its ex- tremity J traversing the line J L. When it is neces- sary to use two kinds of dotted lines, those used for one purpose may be made in line or short dots, while the others may be made a series of short dashes. Lilies showing the part of a view through which a. section is taken are composed of a series of dots and dashes, as shown by A B, (.' 1>, etc., in Fig. 12S, and when further distinction is required may be made by two dots alternating with a short or long dash. 34 The Xcw Metal UW/yr I \Mvrn Book. When it is desirable to omit the drawing of a con- siderable portion of any view it is customary to termi- nate the incomplete side of such view by an irregular line, as shown above the plan G in Fig. 128. It is customary in all sectional views for the parts which are represented as being cut to be ruled or lined with lines running in an oblique direction, as in Fig. 129. When the section comprises several different, pieces lying adjacent to one another, each different part should be lined in a different direction. This rul- ing is understood to mean solidity. In Fig. 129 the walls and base in the different sections are represented as though made of some solid material, as wood or stone, and ruled accordingly. Where it is necessary to represent different kinds of material in the same sec- tion, different systems or kinds of lines may be used for the purpose. Thus solid and dotted lines may be used alternately, as in the base. Coarse and fine rul- ing, or stippling, may also be employed, according to the size of the part, or very small parts may be shown solid black, as window weights, piping or hinges. A heavy line is the only way that a thickness of metal can properly be shown in a section. In the case of a sectional view of a cornice or molding where nothing but the sheet iron appears, it is customary to make use of section lines close to the metal surface, but not to extend them clear across the space which should be filled if the moldings were of stone or other solid mate- rial. By this means a section may be distinguished from what might otherwise be taken for an elevation of a return. In the case of elaborate drawings prepared by an architect color is frequently resorted to as a means of showing the different materials as they appear in the sectional view, yellow or differing shades of brown being used for various kinds of wood, while blue is generally used for iron, gray for stone, red for brick, etc. In the case of drawings showing many different materials it is usual to place a legend in one corner of the drawing showing what each color or style of ruling indicates. It is always advisable to keep the different views, which it is necessary to construct, separate and dis- tinct from one another, drawing them as near to- gether as circumstances will permit, but never allow- ing one view to cover any part of the space upon the paper occupied by the other view if it can be avoided. One notable exception to this rule is to be observed. It frequently occurs in drawing an elevation of a large surface, as a pediment or side of a bracket, that it is necessary to indicate that some part of it is recessed or raised, or that a certain edg'e is molded or chamfered, when it would not be necessary to construct an entire sectional view for this purpose alo;ie. To this end it is customary to draw through such mold, chamfer or recess a small section, in which case, if tin- depression or mold runs horizontally, the section is turned i<> the. right or left, according to convenience, <>r if it runs obliquely, it is turned in the direction the mold runs. In such a section the line which represents the. plane surface also shows the direction of the cut across the mold or line upon which the section is taken. In Fig. i:;n is shown an elevation of a portion of a pediment, in which a small section, A B G, is introduced to show the profiles of the moldings. The line B C, which Fig. ISO, Elevation with Section of Parts. represents the profile of the stile around the panel, shows the line upon which, or the direction in which, the section is taken, said section being turned upon this line obliquely to the left. It is necessary to rule or line this section, the ruling being kept close to and inside the outline or profile. By placing the ruling in- side the profile no doubt can exist as to which parts are raised and which are depressed, for if at D the ruling were upon the other side of the line from that shown the section D would indicate a depressed panel instead of a raised one. In the solution of the class of problems treated in Chapter VI, Section 1 (Miter Cutting), confusion often arises in the mind of the pattern cutter as to the proper position of a profile or of a miter line, which Linear Drawing, 35 confusion could never occur if all the necessary views were lirst drawn In accordance with the principles which this chapter is written to explain. A pro- file is always a section, and a miter line is either a part of an elevation projected from the section or part of another section bearing certain relations of hight or breadth to the first. A pattern is likewise always pro- jected that is, carried off by right lines from an elevation or plan the same as an elevation is projected from a section. Jt should also be remembered in this connection that the operation of developing a pattern is not com- pleted until its entire outline is drawn. The line form- ing its termination at the end opposite the miter cut, although simply u straight line, is properly derived from the elevation or plan used, the same as all points and other lines of the pattern. Much trouble is experienced through lack of knowledge of the principles of Linear Drawing, which if thoroughly understood could never result in such mistakes as producing a face miter where a return was intended or using the piece of metal from the wrong side of the miter cut. Too much emphasis cannot be placed upon the im- portance of thoroughly understanding the subject treated in this chapter, as such a knowledge comprehends within itself an answer to the many questions continually arising in the course of the pattern draftsman's labors. CHAPTER IV. In presenting this chapter to the student no at- tempt has been made to give a complete list of ^co- metrical problems, but all those have been selected which can be of any assistance to the pattern drafts- man, and especial attention has been given in their solution to those methods most adaptable to his wants. They are arranged' as far as possible in logical order and are classified under various sub-heads in such a manner that the reader will have no difficulty in finding what he wishes by simply looking through the pages, the diagrams given with each being sufficient to indi- cate the nature of the problem and, as it were, form a sort of index. 1. To Draw a Straight Line Parallel to a Given Line and at a Given Distance from it, Using the Compasses and Straight-Edge. In Fig. 131, let C D be the given line parallel to which it is desired to draw another straight line. Take any two points, as A and B, in the given line as centers, and, with a radius equal to the given distance, describe the arcs x x and y y. Draw a line, touching these arcs, as shown at E F. Then E F will be parallel to C D. 2. To Draw a Line Parallel to Another by the Use of Triangles or Set-Squares. In Fig. 132, let A B be the line parallel to which it is desired to draw another. Place one side of a triangle or set-square, F 1 , against it, as indicated by the dotted lines. While holding F 1 firmly in this position, bring a second triangle, or any straightedge, E, against out 1 of its other sides, as shown. Then, holding the second triangle firmly in place, slide the first away from the given line, keeping the edges of the two triangles in contact, as shown in the figure. Against the same edge of the first triangle that was placed against the given line draw a second line, as shown by 1). Then C 1) will be parallel to A B. In drawing parallel lines by this method it is found advantageous to place the longest edges of the tri- angles against cadi other, and to so place the two in- struments that the movement of one triangle against the other shall be in a direction oblique to the lines to be drawn, as greater accuracy is attainable in this way. 3. To Erect a Perpendicular at a Given Point in a Straight Line by Means of the Compasses and Straight- Edge. In Fig. 133, let A B represent the given straight line, at the point C in which it is required to erect a perpendicular. From C as a center with any convenient radius strike small arcs cutting A B, as shown \)y D and B. With D and B as centers, and with any radius longer than the distance from each of these points to C, strike arcs, as shown by x.rand // //. From the point at which these ares intersect, E, draw a line to the point C, as shown. Then E C will be per- pendicular to A B. 4. To Erect a Perpendicular at or near the End of a Given Straight Line by Means of the Compasses and Straight-Edge. First Method. In Fig. 134, let A B be the given straight line, to which, at the point P, situ- ated near the end, it is required to erect a perpendic- ular. Take any point (C) outside of the line A B. With C as center, and with a radius equal to thr distance from C to P, strike the arc, as shown, cutting the given line A B in the point P, continuing it till it also cuts in another point, as at E. From E, through the center, C. draw the line E F, cutting the arc, as shown at F. Then from the point F, thus determined, draw a line to P, as shown. The line F P is perpendic nlar to A B. 5. To Erect a Perpendicular at or near the End of a Given Straight Line by Means of the Compasses and Straight-Edge. Second Method. In Fig. 135, let B A be the given straight line, to which, at the point P, it is required to erect a perpendicular. From the point 1', witli a radius equal to three parts, by any scale, de- scribe an arc, as indicated by ./ ,r. From the same point, with a radius equal to four parts, cut the line B A in the point C. From the point 0, witli a radius equal to live parts, intersect the arc first drawn by the Geometrical Problems. 87 arc y ?/. From the point of intersection T) draw the line 1) P. Then I) P will l.e perpendicular to H A. 6. To Draw a Line Perpendicular to Another Line by the Use of Triangles or Set Squares. In Fig. 1 .">;. let (' I) be the given line, perpendicular to which it is required to draw another line. Place one side of a triangle. R, against the given line, as shown. Rring another triangle, A, or any straight edge, against tin- long side or hypothenuse of the triangle R, as shown. Then move the triangle R along the straight edge or triangle A, as indicated hv the dotted lines, until the opposite side of R crosses the line (! 1) at the required sides of the given line A B. A line drawn through these points of intersection, as shown by G H, will liiseet the line A 15. or, in other words, divide it into two equal parts. 8. To Divide a Straight Line into Two Equal Parts by the Use of a Pair of Dividers. In Fig. 13S, it is re- quired to divide the line A R into two equal parts, or to lind its middle point. Open the dividers to as near half of the given line as possible by the eye. Place one point, of the dividers on one end of the line, as at A. Rring the other point of the dividers to the line, as at C, and turn on this point, carrying the first /',;/. ;?/ To Draw a Straight Line Par- allel to a Given Straight Line, and at a d'ifeti Distance from it, Using the, Com- j>ttfsr,es and a Straight-Edge. Fig. 1.12. To Draw a Line Parallel to Another by the Use of Triangles or Set Squares. Fig. 1S3 To Erect a Perpendicular at a Given Point in a Straight Line, Using the. Compasses and Straight- Edgt. \ < 9 'iy. 1;'4 To Erect a perpendicu- lar at or near the End of a Given Straight Line, Using the Compassi x anil Straight-Edge. First Method. D/ ^~ x X P 1 Parts C Fig. ISS. To Erect a Perpendicular at or near the End of a Given Straight Line. Second Method. Fig. 1X6. To Draw a Line Perpen- dicular to Another by the Use of Triangles. point. When against it, draw the line E F, as shown. Then K F is perpendicular to C D. It is evident that this rule is adapted to drawing perpendiculars at anv point in the given line, whether central or located near the end. Its use will be found (-specially convenient for erecting perpendiculars to lines which run oblique to the sides of the drawing board. 7. To Divide a Given Straight Line into Two Equal Parts, with the Compasses, by Means of Arcs. In Fig. 137, let it be required to divide the straight line A R into two equal parts. From the extremes A and R as centers, and with any radius greater than one-half of A R, describe the. arcs rf/and a e, intersecting each other on opposite around to D. Should the point D coincide with the other end of the line, the division will be correct. Rut should the point I) fall within (or without) the end of the line, divide this deficit (or excess) into two equal parts, as nearly as is possible by the eye, and extend (or contract) the opening of the dividers to this point and apply them again as at first. Thus, finding that the point D still falls within the end of the line, the first division is evidently too short. Therefore, divide the deficit I) R by the eye, as shown by E, and in- crease the space of the dividers to the amount of one of D E. Then, commencing again at A, step off as before, and finding that upon turning the dividers 38 'Hie New Worker PnW'rn Honk-. upon the point F the other point coincides with the end of the line B, F is found to be the middle point in the line. In some cases it may be necessary to repeat this operation several times before the exact center is obtained. 9. To Divide a Straight Line into Two Equal Parts by the Use of a Triangle or Set Square. In Fig. 139, let A B be a given straight line. Place a J-square / \ / \ \ 1 1 1 \ \ \ \ I t 1 1 1 I R i i \ i \ \ \ i f \ t \ / \ \ \ e'H-f Pig. 1S7.TO Divide a Straight Line into Two Equal Parts by Means of Arcs. or some straight edge parallel to A B. Then bring one of the right-angled sides of a set square against it, and slide it along until its long side, or hypothenuse, meets one end of the line, as A. Draw a line along the long side of the triangle indefinitely. Keverse the position of the set square, as shown by the dotted lines, bringing its long side against the end, B, of the given Fig. lS8.To Bisect a Straight Line by the Use of the Dividers. t straight line, and in like manner draw a line along its long side cutting the first line. Next slide the set square along until its vertical side meets the intersec- tion of the two lines, as shown at C, from which point drop a perpendicular to the line A B, cutting it at D. Then D will be equidistant from the two extremities A and B. 10. To Divide a Given Straight Line into Any Num- ber of Equal Parts. In Fig. 140, let A B be a given straight line to be divided into equal parts, in this case eight. From one extremity in this line, as at A, draw a line, as cither A C or A D. oblique 1<> A 15. Set tlie dividers to anv convenient spaee. HIM! step oil' tlie oblique line, as A C, eight divisions, as shown by a I cd, etc. From the last of the points, //, thus obtained, draw a line to the end of the given line, as shown l>v h h*. Parallel to this line draw other lines, from each of the other points to the given line. The divisions K Fig. 1S9.TO Bisect a Straight Line by the Use of a Triangle. thus obtained, indicated in the engraving by a" V c 4 , etc., will be the desired spaces in the given line. It is evident by this rule that it is immaterial, except as a matter of convenience, to what space the dividers are set. The object of the second oblique line in the en- graving is to illustrate this. Upon A C the dividers were set so as to produce spaces shorter than those re- g" b* c d* e* / Fig. 340. To Divide a Given Straight Line into Any Number of Equal Parts quired in the given line A B, while in A D the spaces were made longer than those required in the given line. By connecting the last point of either line with the point B, as shown by the lines h h' and h 1 h', and draw- ing lines from the points in each line parallel to these lines respectively, it will be seen that the same divi- sions are obtained from either oblique line. 11. To Divide a Straight Line Into Any Number of Equal Parts by Means of a Scale. It may be more con- venient to transfer the length of a given line to a slip of Geometrical I 'ml, I nun. 3'J paper, rind by laving the paper across a scale, as shown in Fig. 141, mark the required dimensions upon it, and afterward transfer them to a given line, than to divide the line itself by one of the methods explained for that purpose. It mav also occur that it is desirable to divide lines of different lengths into the same num- ber of equal parts, or the same lengths of lines into different numbers of equal parts. Such a scale as is shown in Fig. 141 is adapted to all of these purposes. The scale may be ruled upon a piece of paper or upon a sheet of metal, as is preferred. The lines may be all of one color, or two or more colors may be alternated. in order to facilitate counting the lines or following them by the eye across the sheet. In size, the scale should be adapted to the special purposes for which it is intended to be used. By the contrast of two colors Fig. HI. To Divide a Straight Lin* into Any Number of Equal Parts by Means of a Scale. in ruling the lines, one scale may be adapted to both coarse and fine work. For instance, if the lines are ruled a quarter of an inch apart, in colors alternating red and blue, in fine work all the lines in a given space may be used, while in large work, in which the di- mensions are not required to be so small, either all the red or all the blue lines may be used, to the exclusion of those of the other color. Let it be required to di- vide the line A B in Fig. 141 into thirty equal 1 parts. Transfer the length A B to one edge of a slip of paper, as shown by A 1 B 1 , and placing A 1 against the first line of the scale, carry B' to the thirtieth line. Then mark divisions upon the edge of the strip of paper opposite each of the several lines it crosses, as shown. Let it be required to divide the same length A B into fifteen equal parts by the scale. Transfer the length A B to a straight strip of paper, as before. PlacJ A' against the first line and carry B 2 against the fifteenth line, as shown. Then mark divisions upon the edtre of the paper opposite each line of the scale, as shown. 12. To Divide a Given Angle into Two Equal Parts. In Fig. 142, let A B represent any angle which it is Fig. US. To Bisect a Given Angle. required to bisect. From the vertex, or point C, as center, with any convenient raditis, strike the arc D E, cutting the two sides of the angle. From 1) and E as centers, with any radius greater than one-half the length of the arc D E, strike short arcs intersecting at G, as shown. Through the point of intersection, G, draw a line to the vertex of the angle, as shown by F C. Then F C will divide the angle into two equal parts. 13. To Trisect an Angle. No strictly geometrical method of solving this problem has ever been discov- ered. The following method, partly geometrical and partly mechanical, is, however, perfectly accurate and can be used to advantage whenever it becomes neces- sary to find an exact one-third or two-thirds of an angle : Let ABC, Fig. 143, be the angle of which it is Fig. 143. To Trisect a Given Angle. required to find one-third. Extend one of its sides beyond the vertex indefinitely, as shown by B E, and upon this line from B as center with any convenient radius describe a semicircle A C D, cutting both sides of the angle. Place a straight edge firmly against the 40 Tlie Xew Metal Worker Tiook. extended side as at F, and a pin at the point C. On another straight edge (Ci) having a perfect corner at K, set oil' from one end a distance equal to the radius of the semicircle as shown by point x\ and placing this straight edge, with the end upon which the radius was set oil', against the other straight edge (F) and its edge near the other end, against the pin at the point C, all as shown, slide it along until the mark x comes to the semicircle establishing the point 1). Draw the line I) B, then the angle D E B will be one-third of the angle ABC, and C D B will be two-thirds of it. 14. To Find the Center from which a Given Arc is Struck. In Fig. U4, let A B C represent the given Fig. 144. To Find the Center from which a Given Arc is Struck. Fig. 146 The Chord and Right of a Segment of a Circle Being Given, to find the Center from which the Arc may be Struck. are,, the center from which it was struck being un- known and to be found. From any point near the middle of the arc, as B, with any convenient radius, strike the arc F G, as shown. Then from the points A and C, with the same radius, strike the intersecting arcs I II and E D. Through the points of intersec- tion draw the lines K M and L M, which will meet in M. Then M is the center from which the given arc was struck. Instead of the points A and C being taken at the extremities of the arc, which would be quite inconvenient in the case of a long arc, these, points may be located in any part of the arc which is most, convenient. The greater the distance between A and B and 15 and C, the greater will be the accu- racv of succeeding operations. The essential feature of this rule is to strike an arc from the middle one of the points, and then strike; intersecting ares from the other two points, using tiie same radius. It is not necessary that the distance from A to B and from 15 to C shall be exact! v the same. 15. To Find the Center from which a Given Arc is Struck by the Use of the Square. In Kig. u.v, let A B G be the given arc. Kstablisli the point B at pleas nre and draw two chords, as shown l>v A 15 and B C. Misect these chords, obtaining the points E and D. Place the square against the chord 15 C, as shown in the engraving, bringing the heel against, the midd:- point, D, and scribe along the Made indefinitely. Then place the square, as shown by the dotted lines, with the heid against the middle point, E, of the second chord, and in like manner scribe along the blade, cutting the first line in the point F. ThenF' will be the center of the circle, of which tue are A 15 C Fig. 145. To Find the Center from which a Given Are. is Struck, by the Use of the Square. is a part. This rule will be found very convenient for use in all cases where the radius is less than 24 inches in length. 16. The Chord and Right of a Segment of a Circle being Given, to Find the Center from which the Arc may be Struck. In Fig. 14-C., let A B represent the chord of a segment or arc of a circle, and 1) C the rise or hight. Tt, is required to find a center from which an arc, if struck, will pass through the three points A, D and B. Draw A D and B D. Bisect A D, as shown, and prolong the line 11 L indefinitely. Bisect D B and prolong I M until it cuts II L, produced in the point E. Then E, the point of intersection, will be the center sought. It will be observed that by producing DC, and intersecting it by either H L or 41 I M prolonged, the same point is found. Therefore, if preferred, ihr bisecting of either A Dm- 1.) 15 ma\ he dispensed with. A practical application of this rule occurs quite frequently in cornice work, in the construction of window c:ips and other similar forum, to lit frames already made. In the conveying of or- ders from the master builder or carpenter to the cor- nice worker, it is quite customar\ to describe the shape of the head of the frames which the caps are to lit by stating that the width is, for example-. :;tl inches, and that the rise is 4 inches. To draw the shape thus de- scribed, proceed as follows: Set oil' A 1> equal to :'>('> inches, from the center of which erect a perpendicular, D C, which make equal to 4 inches. Continue D C in the direction of K indeiinitely. Draw A 1), which bisect, as shown, and draw 11 L, producing it until it- cuts D C prolonged, in the point E. Then with E as center and K ]) as radius, strike the arc A D B. 17. To Strike an Arc of a Circle by a Triangular Guide, the Chord and Hight Being Given. In Fig. 147, let Fig. 14?. To Strike an Arc of a Circle by a Triangular A D be the given chord and B F the given hight. The iirst step is to determine the shape and size of the tri- angular guide. Connect A and F, as shown. From F, parallel to the given chord A D, draw F (1, making it in length equal to A F, or longer. Then A F (!. as shown in the engraving, is the angle of the triangular guide to be used. Construct the guide of any suitable material, making the angle of two of its sides equal to the angle A F G. Drive pins at the points A, F and D. Place the guide -as shown. Put a pencil at the point F. Shift the guide in such a manner that the pend! will move toward A, keeping the Lruide at all times against the pins A and F. Then reversing, shift the guide so that the pencil at the point F will move toward I), keeping the guide during this operation atrainst the pins F and D. By this means the pencil will be made to describe the arc A F D. It may be interesting to know that if the angle F of the triangu- lar guide be made a right angle, the .arc described by it will be a semicircle. By these means, then, a steel square- may be used in drawing circles, as illustrated in Fig. 148, the pins being placed at A, B and C. 18. To Draw a Circle Through any Three Given Points not in a Straight Line. In Kig. im, let A. Hand K lie any three given points not in a straight Hue, through 'which it. is required to draw a circle. Con- nect the given points by drawing the lines A I) and 1) K. Bisect the line A 1) by F C, drawn perpendicular to it, as shown. Also bisect D K by the line (! C, as shown. Then the point C, at which these lines meet, is the center of the required circle. 19. To Erect a Perpendicular to an Arc of a Circle, without having Recourse to the Center. In Fig. 15o, let A D B be the arc of a circle to which it is required to erect a perpendicular. With A as center, and with any radius greater than half the length of the given arc, describe the arc x x, and with B as center, and with the same radius, describe the arc y y, intersecting Fig. IfS. To Describe a Semicircle with a Steel Square. the arc first struck, as shown. Through the points of intersection draw the line F E. Then F E will be perpendicular to the arc, and if sufficiently produced will reach the center from which the arc A B is drawn. 20. To Draw a Tangent to a Circle or Arc of a Circle at a given Point without having Recourse to the Center. In Fig. 151, let A D B be the arc of a circle, to which a tangent is to be drawn at the point D. With D as center, and, with any convenient radius, describe the arc A F B, cutting the given arc in the points A and B. Join the points A and B, as shown. Through D draw a straight line parallel to A B, as shown by I 1 ] II, then E II will be the required tangent. 21. To Ascertain the Circumference of a Given Circle. In Fig. 152, let A D B C be the circle, eqaal to the circumference of which it is required to draw a straight line. Draw any tw:> diameters at right angles, as shown by A B and 1) C. Divide one of the four ai'cs, as, for instance, D 1!. into eleven equal parts, as shown. The New Mi'tal MV/'vy l\iU<-,-n Hook. I'Yom !', tlie second of these divisions from the point P>. let fall a. perpendicular to A 1>, as shown l>y '' I''. To three times the diameter of the circle (A J5 or DC) add the length !) F, and the result will he a very c.h>se approximation to the length of the circumference. This rule, upon ;\. diameter of 1 foot, gives a length of about T 3 T ths of an inch in excess of the actual length of the circumference. 22. To Draw a Straight Line Equal in Length to the Circumference of any Circle or of any Part of a Circle - dividers arc placed upon the line, no perceptible curve shall exist between them, and, beginning at one end of the curve, step to the oilier end of the same, or so near the end tliat the remaining space shall be less than that between the points of the dividers, then beginning at the end of any straight line step oil' upon it the same number of spaces, after which add to them the remain- ing small space of the curve hv measurement with the dividers. This will be found the quickest and most accurate of any method for the pattern cutters' use. Fig. 149, To Draw a Circle Through any Three Given Points Not in a Straight Line. Fig. 150. To Erect a Perpendicular to an Arc of a Circle. Fig. 151. To Draw a Tangent to a Circle or Arc. ./ft Fig. IBS. To Ascertain the Circumfer- ence of a Given Circle. Fig. 15S. To Inscribe an Equilateral Triangle within a Given Circle. Fig. 154. To Inscribe a Square ivithin a Given Circle. Various approximate rules, similar to the one given in the problem above, for performing these operations are known and sometimes used among workmen, but can- not be recommended here because in using them con- siderable time and trouble is required to obtain a result which is not accurate when obtained, thus rendering such methods impracticable. The simplest and most accurate method for obtaining the length of any curved line is as follows : Take between the points of the dividers a space so small that when the points of the The most common rules in use for the construc- tion of polygons, whether drawn within circles or erected upon given sides, are those which einplov the straight-edge and compasses only. Other instruments may also be employed to great advantage, as will be shown further on, leaving the student to decide which method is the most suited to any case he may have in hand. Accordingly, the construction of polygons will be treated under three different heads arranged accord- ing to the tools employed. I I! THE CONSTRUCTION OF REGULAR POLYGONS. I BY THE USE OF COMPASSES AND STRAIGHT-EDGE. 23. To Inscribe an Equilateral Triangle within a Given Circle. In Fig. Ij;;, let A B D be any given circle within which an equilateral triangle is to be dr:iwn. From any point in the circumference, as K, with a radius equal to the radius of the circle, describe the arc D C B, cutting the given circle in iho points 1) and B. Draw the line D B, which will be one side of the required triangle. From D or B as center, and with 1) B as radius, cut the circumference of the given cirele, as shown at A. Draw A B and A D, which will complete the iignre. 24. To Inscribe a Square within a Given Circle. In Fig. 154, let A C B 1) bo any given circle within which it is required to draw a square. Draw any two Circle. In Fig. i:.i;, let A B D K F G be any given circle within which a hexagon is to be drawn. From any point in the circumference of the circle, as at A, with a radius equal to the radius of the circle, de- scribe the arc C B, cutting the circumference of the circle in the point B. Connect the points A and B. Then A B will be one side of the hexagon. With the dividers set to the distance A B, step off in the cir- cumference of the circle the points G, F, E and D. Draw the connecting lines A G, (I F, F E, E D and 1) B, thus completing the figure. By inspection of this ligure it will be noticed that the radius of a circle is equal to one side of the regular hexagon which may be inscribed within it. Therefore set the dividers to the radius of a circle and step around the circumfer- ence, connecting the points thus obtained. Fig. 155 To Inscribe a Regular Penta- gon within a Given Circle. Fig. 15fi.To Inscribe a Regular Hexa- gon within a Given Circle. Fig. 157. To Inscribe a Regular Hepta- gon \uithin a Given Circle. diameters at right angles with each other, as C D and A B. Join the points C B, B D, D A and A C, which will complete the required figure. 25. To Inscribe a Regular Pentagon within a Given Circle. In Fig. loo, A D B G represents a circle in which it is required to draw a regular pentagon. Draw any two diameters at right angles to each other, as A B and D C. Bisect the radius A II, as shown at E. With E as center and E D as radius strike the arc D F, and with the chord D F as radius, from D as center, strike the arc F G, cutting the circumference of the given circle at the point G. Draw D G, which will equal one side of the required figure. With the dividers set equal to D G, step off the spaces in the circumference of the circle, as shown by the points I K and L. Draw D I, I K, K L and L G, thus com- pleting the figure. 26. To Inscribe a Regular Hexagon within a Given 2f. To Inscribe a Regular Heptagon within a Given Circle. In Fig. 157, let F A G B II I K L D be the given circle. From any point, A, in the circumfer- ence, with a radius equal to the radius of the circle, describe the arc BCD, cutting the circumference of the circle in the points B and D. Draw the chord B D. Bisect the chord B D, as shown at E. With D as center, and with D E as radius, strike the arc E F, cutting the circumference in the point F. Draw D F, which will be one side of the heptagon. With the dividers set to the distance D F, set off in the circum- ference of the circle the points G II I K and L, and draw the connecting lines F G, G II, II I, I K, K L and L D, thus completing the figure. 28. To Inscribe a Regular Octagon within a Given Circle. In Fig. 158, let B I D F A G E II be the given circle within which an octagon is to be drawn. Draw any two diameters at right angles to each other, 44 The New Metal Worker Pattern Book. as 15 A ami 1) E. Draw the chords D A and A E. Bisect 13 A, as shown, and draw L II. Bisect A K and draw K 1. Then connect llie several points in the circumference thus obtained liv drawing the lines I) I, I 11, B II, II E, K G, G A, A F and F D, which will complete the figure. 29. To Inscribe a Regular Nonagon within a Given Circle. ! n Fig. l.V., let M K F E l>e the given circle. Draw nnv two radii at right angles to^ach other, as 1> (} and A C, and draw the chord B A. From A as center, and with a radius equal to one-half the chord A B, as shown l>y A D, strike the arc D E, cutting the circumference of the circle at the point E. Draw A E, which will he one side of the nonagon. Set the di- viders to the distance A E and step off the points M, II, K, G, I, F and L, and draw the connecting lines, as shown, thus completing the figure. Circle. In Fig. 101, let B D A L ho any given circle in which a regular iignre of eleven sides is to be drawn. Draw any diameler, as J> A. and draw a radius, as D G, at right angles to B A. Bisect C A, thus obtaining the point E. From E as center, and with K D as radius, describe the arc D F, cutting P> A in the point F. With D as center, and 1) F as radius, describe the arc F G, cutting the circumference in the point. G. Draw the chord G D and bisect it, as shown by II C, thus obtaining the point K. From D as center, and with D K as radius, cut the circumference in the point I. Draw I D. Then I D will be equal to one side of the required figure. Set the dividers to this space and step off the points in the circumference, as shown by N, E, S, M, P, L, 0, T and V, and draw the connecting chords, as shown, thus completing the figure. 8 M Fig. 1~>8. To Inscribe a Regular Octa- gon within a Oin-n Circle. Fig. 159. To Inscribe a Regular Nuna- gon within a Given Circle. Fig. 160. To Inscribe a Regular Deca- gon within a Given Circle. 30. To Inscribe a Regular Decagon within a Given Circle. In Fig. 100, let D B E A be any given circle in which a decagon is to be drawn. Draw any two diameters through the circle at right angles to each other, as shown by B A and D E. Bisect B C, as shown at-^F, and draw F D. With F as center, and F D as radius, describe the are D G, cutting B A in the point G. Draw the chord D G. "With D as center, and D G as radius, strike the are G II, cutting the cir- cumference in the point II. Connect D and ]I, as shown. Bisect D II and draw the line C It, cutting the circumference in the point I. Draw the lines II I and I D, which will then be two sides of the required figure. Set the dividers to the distance II I and space oil' the circumference of the circle, as shown, and draw the connecting lines D K, K M, M L, L P, P E, E N, N O and. O II, thus completing the ligniv. 31. To Inscribe a Regular Undecagon within a Given 32. To Inscribe a Regular Dodecagon within a Given Circle. In Fig. 102, let M F A I be any given circle in which a dodecagon is to be drawn. From anv point in the circumference, as A. with a radius equal to the radius of the circle, describe the arc C B, cut- ting the circumference in the point B. Draw the chord A P>, which bisect as shown, and draw the line C, cutting the circumference in the point D. Draw A D, which will then be one side of the given figure. With the dividers set to this space step off in the cir- cumference the points B, I, X, II, M, G, L, F, K and F, and draw the several chords, as shown, thus com- pleting the Iignre. 33. General Rule for Inscribing any Regular Polygon in a Given Circle. Through the given circle draw any diameter. At right angles to this diameter draw a radius. Divide that radius into four equal parts, ami prolong it outside the circle to a distance equal to three < i, nun triful Problt in*. 45 of those parts. Divide the diameter of the circle into the same, number of equal parts as the polygon is to have sides. Then 1 mm 1 lie end of the radius prolonged, as above described, through the second division in the diameter, draw a line cutting the ci renni ferenee. Con- neet this point, in the circumference and the nearest end of the diameter. The line thus drawn will lie one side of the required ligure. Set the dividers to this space and step oil on the circumference of the eirclo outside ihe circle to the extent, of three of those' parts, as sho\vn 1^\ abe, thus obtaining the point <:. From r, through the second division in the diameter, draw the line c II, cutting tlie circumference in the point II. Connect II and K. Then JI K will be one side of the required ligure. Set the dividers to the dis- tance 11 Hand step oil the circumference, as shown, thus obtaining ihe points for the other sides, and drw the connecting arcs, all us illustrated in the ligurc. f,V, 161. To Iiixciibe a. Regular Un- decagnn within a Given Circle. Fig. 162. To Inscribe a Regular Dodec- agon within a Given Circle. Fig. ItiJ.To Inscribe a Regular Undec- agun within a Given Circle by Ihe General Rule. Q Fig. M./. Ujiun a (liom Side lo Construct an Equilateral Tri- angle. Fig. Via. To Construct a Triangle, Ihe Length of the Three Sides being (liren. Fig. 166. I 'pan a Givm Side to draw n Regular Pentagon, the remaining number of sides and draw connecting lines, which will complete the ligurc. 34. To Inscribe a Regular Polygon of Eleven Sides (Undecagon) within a Given Circle by the General Rule. Through the given circle, E I) F G in Fig. 13, draw any diameter, as K K. which divide into the same number of equal parts as the ligurc is to have sides, as shown by the small figures. At right, angles to the diameter just drawn draw the radius D K, which divide into four equal parts. Prolong the radius D K 35. Upon a Given Side to Construct an Equilateral Triangle. In Fig. lf.4, let A B represent the length of the given side. Draw any line, as C D, making it equal to A B. Take the length A B in the dividers, and placing one foot upon the point C, describe the are K F. Then from D as center, with the same radius, describe the are (i II, intersecting the first arc- in the point K. Draw K C and K D. Then C D K will be the required triangle. 36. To Construct a Triangle, the Length of the Three Tlte Xew Metal \\~orker Patlf. i Sides being Given. In Fig. 165, let A B. c D and K F be the given sides from which it is required to con- struct a triangle. Draw any straight line, G II, mak- ing it in length equal to one of the sides, E F. Take the length of one of the other sides, as A B, in the compasses, and from one end of the line just drawn, as G, for center describe an arc, as indicated by L M. Then set the compasses to the length of third side, C 1), and from the opposite end of the line first drawn, H, describe a second arc, as I K, intersect- ing the first in the point 0. Connect G and H. Then G II will be the required triangle. 37. Upon a Given Side to Draw a Regular Pentagon. In Fig. 166, let A B represent the given side upon which a regular pentagon is to be constructed. With B as center and B A as radius, draw the semicircle of the circle, obtaining the points M ami L. Draw A M, M L and L D, which will com- plete the figure. 38. Upon a Given Side to Draw a Regular Hexagon. In Fig. 167, let A B be the given side upon which a regular hexagon is to be erected. From A as center, and with A B as radius, describe the arc B C. From B as center, and with the same radius, describe the arc A C, intersecting the first arc in the point C. C will then be the center of the circle which will cir- cumscribe the required hexagon. With C as cent IT. and C B as radius, strike the circle, as shown. Set the dividers to the space A B and step off the circum- ference, as shown, obtaining the points E, G, F and D. Draw the chords A E, E G, G F, F D and D B, thus completing the required figure. Fig. 167. Upon a (liven Side to Draw a Regular Hexagon. Fig. 16$.l'pon a Giivji Side to Draw a lieyular Heptagon. Fig. KO.Upon a Given Side to llmw a Regular Octagon. A D E. Produce A B to E. Bisect the given side A J3, as shown at the point F, and erect a perpendic- ular, as shown by F C. Also erect a perpendicular at the point B, as shown by G H. With B as center, and F B as radius, strike the arc F G, cutting the per- pendicular H G in the point G. Draw G E. With G as center, and G E as radius, strike the arc E II, cutting the perpendicular in the point H. With E as center, and E H as radius, strike the arc II D, cutting the semicircle A D E in the point D. Draw D B, which will be the second side of the pentagon. Bisect D B, as shown, at the point K, and erect a perpendic- ular, which produce until it intersects the perpendic- ular F C, erected upon the center of the given side in point F. Then C is the center of the circle which circumscribes the required pentagon. From C as cen- ter, and with C B as radius, strike the circle, as shown. Set the dividers to the distance A B and step off the 39. Upon a Given Side to Draw a Regular Heptagon. In Fig. 168, A B represents the given side upon which a regular heptagon is to be drawn. From B as center, and with B A as radius, strike the semicircle A E D. Produce A B to D. From A as center, ami with A B as radius, strike the arc B F, cutting the semicircle in the point F. Through F draw F G per- pendicular to A B, which extend in the direction of C. From D as center, and with radius G F, cut the semi- circle in the point E. Draw the line E B, which is another side of the required heptagon. Bisect E B, and upon its middle point, H, erect a perpendicular, which produce until it meets the perpendicular erected upon the center of the given side A B, in the point C. Then is the center of the circle which will circumscribe the required heptagon. From C as center, and with C B as radius, strike the circle. Set the dividers to the distance A B and step off the circumference, as Geometrical Problems. 47 shown, obtaining the points K, N, M and L. Draw the connecting arcs A K, K N, N M, M L and L-E, thus completing the figure. 40. Upon a Given Side to Draw a Regular Octagon. In Fig. 10!>, let A 1> represent the given side upon which a regular octagon is to bo constructed. Produce A 13 indefinitely in the direction of D. From B as center, and with A H as radius, describe the semicircle A E D. At the point B erect a perpendicular to A B, as shown, cutting the circumference of the semicircle in the point E. Bisect the arc E D, obtaining the point F. Draw F 15, which is another side of the re- quired octagon. Bisect the two sides now obtained and erect perpendiculars to their middle points, G and H, which produce tmtil thev intersect at the point C. (J "then is the center of the circle that will circumscribe which produce until they intersect at the point C. Then C is the center of the circle which will circum- scribe the required nonagon. From C as center, and with C B as radius, strike the circle B P A. Set the dividers to the space A B and step oil the circle, as shown, obtaining the points N, P, M, E, and L. Draw the connecting chords, A N, N P, P M, M R, R 0, L and L E, thus completing the figure. 42. Upon a Given Side to Draw a Regular Decagon. In Fig. 171, A B is the given side upon which a regular decagon is to be drawn. Produce A B indefi- nitely in the direction of D. From B as center, and with B A as radius, strike the semicircle A H D. Bisect the given side A B, obtaining the point F. Through the point B draw the line H B G, perpen- dicular to A B. From B as center, and with B F as Fig. 170. Upon a Given Side to Draw a Regular Nonagon. Fig. 171 I 'pan a Given Sid<: to Draw a Regular Decagon. Fig. 172. Upon a Given Side to Draw a Regtilar Undecagon. the octagon. From C as center, and with C B as radius, strike the circle, as shown. Set the dividers to the space A B and step off the circumference, obtaining the points L, K, M, and N. Draw the connecting arcs A L, L K, K M, M 0, O N and N" F, thus com- pleting the required figure. 41. Upon a Given Side to Draw a Regular Nonagon. In Fig. 170, A B is any given side upon which it is required to draw a regular nonagon. Produce A B in- definitely in the direction of D. From B as center, and with B A as radius, strike the semicircle A F D. At the point B erect a perpendicular to A B, cutting the semicircle in the point F. Draw the chord F D, which bisect., obtaining the point G. From D as cen- ter, and with D '/r Mi'lal Worker Pattern Bwk. A well-known and easily demonstrated geometrical principle is that the sum of the three interior angles of a triangle is equal to two right angles, or in other words, as a right angle is one of 90 degrees, if the three angles of any triangle be added together their sum will equal 180 degrees. Hence, if one of the angles of a set- square be fixed at 90 degrees (which is done for con- venience in drawing perpendicular lines) the sum of the two remaining angles must also be 90 degrees, and, if then the two other angles be made equal, each will be 45 degrees, which is the half of 90 degrees. If, however, one of the other angles is fixed at 30 (one-third of 90 degrees), the remaining angle must be 60 degrees, as 30 + 60 = 90. By means, then, of the 45-degree and the 30 X 60-degree triangles, the draftsman has at his command of the 45-degree triangle, as A E, is placed against the blade of the T-square, and the vertical division of the circle is drawn along the other short side C E. In Fig. 178 the vertical and horizontal divisions of the circle, A B and C D, are drawn as before, after which one of the shorter sides of the 45 -degree triangle is placed against the T-square, and the long or oblique side E F is brought to the center of the circle and another division G I is drawn. By reversing the posi- tion of the triangle the last division II K is drawn, thus dividing the circle into eight equal parts. In Fig. 179, after drawing the divisions A B and C D as before, the 30 X 60-degree triangle is placed in the position shown at A E F, and the division E N is drawn along its hypothenuse or oblique side. Bv re- versing the position of the triangle, still keeping the Fig. 177. Circle Divided into Four Equal Parts by the Use of a Triangle or Set-Square. Fig. 178. Circle Divided into Eight Equal Parts by the Use of a 45-degree Triangle. Fig. 170. Circle Divided into Twelve, Equal Parts by the Use of a 80 x 60- degree Triangle. the means of drawing lines at angles of 90, 60, 45 and 30 degrees, and by combination 75 degrees (45 -)- 30) and 15 degrees (90 75). With the 45-degree angle he can bisect a right angle, and with the 30 and 60-degree angles he can trisect it. The pattern draftsman sometimes finds it con- venient to have a set-square in which the sharpest angle is one of 22 degrees (one-half of 45) for use in drawing the octagon in a certain position which will be referred to later. In Figs. 177, 178, 179 and 180 are illustrated the application of the foregoing, in which the circle is divided, by the use of the triangles above described, into four, eight and twelve equal parts. In Fig. 177 the horizontal division A B of the circle is drawn by means of the T-square placed against the side of tin- drawing board, after which one of the shorter sides side A F against the blade of the T-square, the divi- sion J K may be drawn. Changing the position of the triangle now so that its shortest side comes against the blade of the T-square, as shown dotted at G H F, the division G M is drawn, and again reversing its posi- tion, still keeping its shortest side against the T-square, the last division I L may be drawn, thus dividing the circle into twelve equal parts. In Fig. 180 the circle is divided into eight equal parts, but differing from that shown above in this respect that, while in Fig. 178 two of the divisions lie parallel with the sides of the drawing board, in the latter case none of the divisions are parallel with t lie- sides of the board or can be drawn with the f-square; but if this method is used in drawing an octagon, as shown dotted in Fig. 180, four of the sides of the oc- tagon can be drawn with the T-square illR l tue other Geometrical 51 four with the 45-degree triangle. The position of the -2$ X 67-i-degree triangle in drawing tlie divisions of the circle is shown at A 15 (..' and I.)E C, while the posi- tion of the 45-degree triangle in drawing the oblique sides of an octagon figure is shown at F. It will thus be seen that the 22 X tiT^-degree triangle is available in drawing accurately the miter line for all octagon miters. As a triangle in whatever form it may be con- structed is intended to be used by sliding it against, the blade of the T-^quare. "11 the angles above men- tioned are calculated with, reference to the lines drawn by the T-square. In practical use it will be found in- convenient in drawing such lines to actually bring the point of a set-square to the center of a circle. A better method, and one which makes use of the same principles, is shown in Fig. 181. The blade of the Fig. ISO. Circle Divided into Eight Equal Parts by the Use of a 1S&% x Vt^-degree Triangle, T-square is placed tangent to or near the circle, as shown by A B. One side of a 45-degree triangle is placed against it, as shown, its side C F being brought against the center. The line C F is then drawn. By reversing the trrangle, as shown by the dotted lines, the line E D is drawn at right angles to C F, thus dividing the circle into quarters. A similar use of the 30 X 60-degree triangle is shown in Fig. 182, by which a circle is divided into six equal parts. Bring the blade of the T-square tangent to or near the circle, as shown by A B. Then place the set-square as shown by G B M, bringing the side G B against the center of the circle, drawing the line D L. Then place it as shown by the dotted lines, bringing the side A H against the center, scribing the line F E. Then, by reversing the set-square, placing the side G M against 'the straight-edge, erect the per- pendicular C I, completing the division. The follow- ing are a few of the problems to which these principles may be advantageously applied. 48. To Inscribe an Equilateral Triangle within a Given Circle. In Fig. 183, let D be the center of the given circle. Set the side E F of a 30-degree set-square against the T-square, as shown, and move it along D F Fig. 181. Proper Method of Using a 45-degree Triangle. until the side E G touches D. Mark the point B upon the circumference of the circle. Eeverse the set-square so that the point E will come to the right of the side F G and move it along in the reversed position until the side E G again meets the point D, and mark the point C. Now move the T-square upward until it touches the point D, and mark the point A. Then A B and C are points which divide the circle into three equal parts. The triangle may be easily completed from this stage by drawing lines connecting A B, B C and C A, with any straight-edge or rule, but greater accuracy is obtained by the further use of the set- si [uare, as follows : Place the side F G of the set- square against the T-square, as shown in Fig. 184, Fig. 182 Method of Using a 30 x 60-degree Triangle in Dividing the Circle. and move it along until the side E G touches the points A and C, as shown. Draw A C, which will be one side of the required triangle. Set the side E F of the set-square against the T-square, and move it along until the side F G coincides with the points C and B. Then draw C B, which will be the second side of the triangle. 52 The New Metal Worker Pattern Book. Place the side F G of the set-square against the T- square, with the side E F to the right, and move it along until the side E G coincides with the points A and B. Then draw A B, thus completing the figure. The same results may be accomplished with less work by first establishing the point A by bringing the T-square against the center, and then using the set- square, as shown in Fig. 184. The different methods are here given in order to more clearly illustrate the use of the tools employed. Fig. 183. Fig. 184. To Inscribe an Equilateral Triangle within a Given Circle. 49. To Inscribe a Square within a Given Circle - Let D, in Fig. 185, be the center of the given circle. Place the side E F of a 45-degree set-square against the T-square, as shown, and move it along until the side E G meets the point D. Mark the points A and B. Eeverse the set-square, and in a similar manner mark the points C and H. The points A, H, B and C are corners of the required square. Move the J- square upward until it coincides with the points A and H and draw A H, as shown in Fig. 186. In like man- ner draw C B. With the side E F of the set-square against the T-square, move it along until the side G F coincides with the points B and H, and draw B H. In a similar manner draw C A, thus completing the figure. 50. To Inscribe a Hexagon within a Given Circle. In Fig. 187, let be the center of the given circle. Place the side E F of a 30-degree set-square against the T-square, as shown. Move the set-square along until the side E G meets the point 0. Mark the points A and B. Reverse the set-squaiv, and in like manner mark the points C and D. With the side F G of the set-square against the 1 -square, move it along until Fig. 185. Fig. 186. To Inscribe a Square within a Given Circle. the side E F meets the point 0, and mark I and H. Then A, H, D, B, I and C represent the angles of the proposed hexagon. From this stage the figure may be readily finished by drawing the sides by means of these points, using a simple straight-edge; but greater ac- curacy is attained in completing the figure by the further use of the set-square, as shown in Fig. 188. With the side E F of the set-square against the f -square, as shown, draw the line II D, and by mov- ing the T-square upward draw the side C I. Reversing the set-square so that the point F is to the left of the point E, draw the side A H, and also, by shifting the Geometrical Problems. 53 T-square, the side I B. With the edge E F of the set-square against the T-square, move it up until the side G F coincides with the points B and D, and draw the side B D. In like manner draw A C, thus com- pleting the figure. In this figure, as with the triangle, the same results may be reached by establishing the points H and I, by means of a diameter drawn at right angles to the T-square, as shown in the engravings, and, using it as a base, employing the set-square, as shown in Fig. 188. The first method shown is, how- Fig. 18?. Fig. 188. To Inscribe a Regular Hexagon within a Given Circle. ever, to be preferred in many instances, on account of its great accuracy. 51. To Inscribe an Octagon within a Given Circle. In Fig. 189, let K be the center of the given circle. Place a 45-degree set-square as shown in the engraving, bringing its long side in contact with the center, and mark the points E and A. Keeping it in the same po- sition, move it along until its vertical side is in contact with K and mark the points D and II. Reverse the set-square from the position shown in the engraving, and mark the points C and G. Move the T-square up- ward until it touches the point K, and mark the points B and F. Then A, H, G, F, E, D, C and B are cor- ners of the octagon. The figure may now be readily completed by drawing the sides, by means of these points, using any rule or straight-edge for the purpose, Fig. 189. To Inscribe a Regular Octagon within a Given Circle. all as shown by A H, H G, G F, F E, E D, D C, C B and B A, or by means of a 22 X 67^-degree set- square. 52. To Draw an Equilateral Triangle upon a Given Side. In Fig. 190, let A B be the given side. First bring the line A B at right angles to the blade of the T-square. Then set the edge C B of a 30-degree set- square against the T-square, and move it along until the edge B D meets the point B, and draw the line B F. Reverse the set-square, still keeping the side C B against the T-square, and move it along until the Fig. 190. To Draw an Equilateral Triangle upon a Given Side. side B D meets the point A, and draw the line A F, thus completing the figure. 53. To Draw a Square upon a Given Side. In Fig. 191, let A B be the given side drawn at right angles 54 Tin \' : /C M'lttl II //''/ Pntt'-ni ttouk. to the blade of the T-square. Set the edge E F of a 45-degree set-square against the T-square, as shown, and move it along until the side E G meets the point B, and draw B I indefinitely. Reverse the set-square, and, bringing the side E G against the point A, draw A Fig. 191. To Draw a Square -upon a Given Side. F indefinitely. Bring the T-square against the point B and draw B F, producing it until ,it meets the line A F in the point F. In like manner draw A I, meet- ing the line B I in the point I. Then, with the set- square placed as shown in the engraving, connect I and F, thus completing the required figure. 54. To Draw a Regular Hexagon upon a Given Side. In Fig. 192, let A B be the given side in a vertical position. Set the edge G H of a 30-degree set-square against the T-square, as shown, and move it along Fig. 192. To Draw a Regular Hexagon upon a Given Side. until the edge I G coincides with the point A, and draw the line A D indefinitely. Reverse the set- square, still keeping the edge G H against the T-square, and move it along until the side I G coin- cides with the point B, and draw B E indefinitely. These lines will intersect in the point 0, which will be the center of the required figure. Still keeping the edge G H of the set-square against the T-square, move it along until the perpendicular edge I II meets the point O, and through O draw F C indefinitely. With the set-square in the position shown in the engraving slide it along until the edge I G meets the point B, and draw B C, producing it until it meets the line F C in the point C. Reverse the set-square, still keeping the edge G H against the T-square, and draw the line C D, producing it until it meets the line A D in the point D. Slide the set-square along until the side I H meets the point D, and draw the line D E, meeting the line B E in the point E. Move the set- square along until the edge I G meets the point A, and draw the line A F, meeting the line C F in the point F. Now bring the set-square to its first position and Fig. 193. To Draw a Regular Octagon upon a Given Side. slide it along until the edge I G meets the points F and E, and draw F E, thus completing the required figure. 55. To Draw a Regular Octagon upon a Given Side. In Fig. 193, let C D be the given side, drawn perpen- dicular to the blade of the T-square. Place one of the short sides of a 45-degree set-square against the T-square, as shown in the engraving. Move the set- square along until its long side coincides with the point C. Draw the line C B, and make it in length equal to C D. With the T-square draw the line A B, also in length equal to C D. Reverse the set-square, and bring the edge against the point A. Draw A II in length the same as C D. Still keeping a short side of the set-square against the T-square, slide it along until the other short side meets the point H, and draw H G, also of the same length. Then, using the long side of the set-square, draw G F of corresponding length. By means of the T-square draw F E, and by reversing the set-square draw E D, both in length Geometrical equal to the original side, C D, joining it in tin; point D, thus completing the required octagon. 56. To Draw an Equilateral Triangle about a Given Circle. In Fig. 194, let () lie the center of the given circle. Place the edge E F of a 30-dcgree set-square against the T-square, !IS shown, and move it along un- til the edge F G meets the center O, and mark the point A upon the circumference of the circle. Reverse the set-square, still keeping the edge E F against the T-sqnare, and in like manner mark tin- point B. Move Fig. 191. Fig. 195. To Draw an E. Hence, the point of the protractor representing 22 degrees was placed upon the line A B, which represents the center of one side of the required figure. Having thus established the position of one side, the other sides of the figure are located by marking points in the circumference of the circle opposite points in the protractor at regular intervals of 45 degrees. 64. To Inscribe a Dodecagon within a Given Circle. In Fig. 205, let O be the center of the given circle. Fig. 04. To Inscribe an Octagon within a Given Circle. Through draw the diameter A O B, at right angles to which one of the sides of the polygon is required to be. Set the protractor so that the center point of it coincides with the center O, and revolve it until the point marking 15 degrees falls upon the line A B. With the protractor in this position, mark points in the circumference of the circle opposite the points in the protractor representing 0, 30, 60, 90, 120, 150 and 180 degrees, as shown by E, F, G, H, I, K and L. Then these points will represent angles of the required polygon. The remaining angles may be obtained by placing the protractor in like position in the opposite half of the semicircle, or they may be determined by drawing lines from the points F,. G, H, I and K through the center O, producing them until they cut the circumference in the points M, N, P, R and S, which are the remaining angles. The figure is now to be completed by drawing the sides, as shown. In a do- decagon, or twelve-sided figure, each side must occupy a space represented by one-twelfth of 360 degrees, or 30 degrees of the protractor. As the side F E was re- quired to be located in equal parts upon opposite sides of A B, the middle of one division of the protractor representing a side (that is, 15 degrees, or one-half of 30 degrees) was placed upon the line A O B. Having thus established the position of one side, the others are measured off in manner above described. p R Fig. t05. To Inscribe a Dodecagon within a Given Circle. In making use of the protractor to erect a regular polygon upon a given side, the exterior angle, or angle formed by an adjacent side with the given side ex- tended, as E B D in Figs. 168, 170 and 171, is found by dividing 360 degrees by the number of sides in the required polygon ; while the interior angle, or angle between any two adjacent sides on the inside of the polygon, as E B A in the same diagrams, is the supplement of that angle, or, in other words, is found by subtracting the exterior angle from 180 degrees. Thus to find the exterior angle by means of which to construct a regular decagon, divide 360 degrees by 10, which gives 36 degrees; while the interior angle is equal to 180 degrees less 36 degrees, which is 144 de- grees. THE ELLIPSE. For a definition of the ellipse the reader is referred to Chapter I, definitions 78 and 113. It may also be described as a curve drawn with a constantly increas- ing or diminishing radius, or as similar to a circle, but having one diameter longer than another, the diameters referred to being at right angles to each other. If, upon one of two lines intersecting each other at right angles, half of the long diameter be set off each way from their intersection (A A, Fig. 206) and upon the other line half of the sh'ort diameter be set off each way from the intersection (B B, Fig. 206), four prin- cipal points in the circumference of the ellipse will thus 60 T/ie New Metal Worker Pattern Book. be established; and through these four points only one perfect ellipse can be drawn, one-quarter of which is shown by the solid line from A to B in the illustra- tion. It is true that other curves having the appear- ance of an ellipse can be drawn through these points, as shown by the dotted lines, but, as stated above, there is only one curved line existing between those points which can be correctly termed an ellipse. There are several methods of producing a correct ellipse, as by a string and pencil, by a trammel con- structed for the purpose and by projection from an oblique section of a cylinder or of a cone, each of which will be considered in turn. The ellipse is properly generated from two points upon its major axis, called the foci, and its circumference is so drawn that if from any point therein two lines be drawn to the two foci, their sum shall be equal to the sum of Fig. 206. Defining an Ellipse. two lines drawn from any other point in the circum- ference to the foci. 65. To Draw an Ellipse to Specified Dimensions with a String and Pencil. In Fig. 207, let it be re- quired to draw an ellipse, the length of which shall be equal to the line A B, and the width of which shall be equal to the line D C. Lay off A B and D C at right angles to each other,, intersecting at their middle points, as shown at E. Set the compasses to one- half the length of the required figure, as A E, and from either D or C as center, strike an arc, cutting A B in the points F and G. These points, F and G, then are the two foci, into which drive pins, as shown. Drive a third pin at C. Then pass the string around the three points F, G and C and tie it. Eemove the pin C and substituting for it a pencil, pass the same around, as shown at P, keeping the string taut. If the combined lengths from F and G to the several points in the boundary line be set off upon a straight line, their sums will be found equal. For example, the sums of P F and P G, A F and A G, C F and C G, B F and B G, are all the same. Although correct so far as theory is concerned, this method is liable to error on account of the stretch- ing of the string. The same result can be obtained by means of a trammel constructed for the purpose, which is shown in Fig. 208. E is a section through the arms, showing the groove in which the heads of the bolts move. H and G are the bolts or pins by which the movement is controlled and regulated. In the engraving the bar K is shown with holes at fixed dis- tances, through which the governing pins are passed. An improvement upon this plan of construction con- sists of a device that will clamp the pins firmly to the bar at any point, thus providing for an adjustment of the most minute variations. Fig. SOT. To Draw an Ellipse by Means of a String and Pencil. 66. To Draw an Ellipse to Given Dimensions by Means of a Trammel. In Fig. 208, let it be required to de- scribe an ellipse, the length of which shall be equal to A B and the breadth of which shall be C D. Draw A B and C D at right angles, intersecting at their mid- dle points. Place the trammel, as shown in the en- graving, so that the center of the arms shall come directly over the lines. First place the rod along the line A B, so that the pencil or point I shall coincide with either A or B. Then place the pin G directly over the intersection of A B and C D. Next place the rod along the line C D, bringing the pencil or point I to either C or D, and put the pin H over the intersection of A B and C D. The instrument is then ready for use, and the curve is described by the pencil I moved by the hand, and controlled by the pins work- ing in the grooves. When a trammel is not convenient, a very fair substitute is afforded by the use of a common steel Geometrical Problems. Gl square and a thin strip of wood, like a lath. This method of drawing an ellipse is useful under ordinary circumstances when only a part of the figure is re- quired, as in the shape of the top of a window frame to which a cap is to be fitted, in which half of the figure would be employed, or in the shaping of a member of a molding in which a quarter, or less than a quarter, of the figure would .be used. 67. To Draw an Ellipse of Given Dimensions by Means of a Square and a Strip of Wood. In Fig. 209, set off the length of the figure, and at right angles to it, through its middle point, draw a line representing the width of the figure. Place a square, as shown by AEG, its inner edge corresponding to the lines. Lay the strip of wood as shown by F E, putting a pencil at the point F, corresponding to one end of the figure, placed upon a board, and a line drawn around it, the resulting figure will be a circle. If now the pipe be cut obliquely, as in making an elbow at any angle, and the end thus cut be placed upon a board and a line drawn around it, as mentioned in the first case, the figure drawn will be an ellipse. What has thus been roughly done by mechanical means may be also accomplished upon the drawing board in a very simple and ex- peditious manner. The demonstration which follows is of especial interest to the pattern cutter, because the principles involved in it lie at the root of many practical operations which he is called upon to perform. For example, the shape to cut a piece to stop up the end of a pipe or'tube which is not cut square across, the shape to cut a flange to fit a pipe passing through the slope of a roof, and other similar requirements of Fig. SOS. To Draw an Ellipse by Means of a Trammel. Fig. 209. Fig. 210. To Draw an Ellipse by Means of a Square and a Strip of Wood. and a pin at E, corresponding to the inner angle of the square. Then place the stick across the figure, as shown in Fig. 210, making the pencil, F, correspond with one side of the figure, and put a pin at G, corre- sponding with the inner angle of the square. Now move the stick from one position to the other, letting the points E and G slide, one against the tongue and the other against the blade of the square. The pencil point will then describe the required curve. In draw- ing the figure the square must be changed in position for each quarter of the curve. As shown in the en- gravings, it is correct for the quarter of the curve rep- resented by F D, Fig. 209. It must be changed for each of the other sections, its inner edge being brought against the lines each time, as shown. One definition of an ellipse is " a figure bounded by a regular curve, which corresponds to an oblique section of a cylinder." This can be practically illustrated by assuming a piece of stove pipe as tlie representative of the cylin- der. If the piece of pipe is cut square across, the end almost daily occurrence, depend entirely upon the principles here explained. 68. To Describe the Form or Shape of an Oblique Secti n of a Cylinder, or to Draw an Ellipse as the Oblique Projection of a Circle. The two propositions which are stated above are virtually one and the same so far as concers the pattern cutter, and they may be made quite the same so far as a demonstration is con- cerned. The explanation of the engraving is confined to the idea of the cylinder, believing it in that shape to be of more practical service to the readers of this book than in any other. In Fig. 211, let G E F H represent any cylinder, and A B C D the plan of the same. Let I K represent the plane of any oblique cut to be made through the cylinder. It is required to draw the shape of the section as it would appear if the cylinder were cut in two by the plane I K, and either piece placed with the end I K flat upon paper and a line scribed around it. Divide one-half of the plan ABC into any convenient number of equal parts, as shown by the figures 1, 2, 3, 4, etc. Through Tin' \t:ir Metal \Vbr/,-i. r litjok. these points and tit right angles to the diameter A C draw lines as shown, cutting the opposite side of the circle. Also continue these lines upward until they cut the oblique line I K, as shown by I 1 , 2', 3', etc. Draw I' K', making it parallel to I K for con- venience in transferring spaces. With the J-square set at right angles to I K, and brought successively against the points in it, draw lines through I' K 1 , as stown by 1", 2", 3 a , etc. With the dividers take the distance across the plan A B C 1) on each of Fig. 211. The Ellipse as on Oblique Section of a Cylinder. the several lines drawn through it, and set the same distance off on corresponding lines drawn through I 1 K 1 . In other words, taking A C as the base for measurement in the one case and I 1 K 1 as the base of measurement in the other, set off from the latter, on each side, the same length as the several lines measure on each side of A C. Make 2" equal to 2, and 3" equal to 3, and so on. Through the points thus ob- tained trace a line, as shown by I' M K 1 and the opposite side, thus completing the figure. To make this problem of practical use it is neces- sary that the diameter of the cylinder shall be equal to the short diameter of the required ellipse, and that UK- lino 1 K be drawn at such an angle that the dis- tance 1' 'J' shall be equal to its long diameter. Another definition of the ellipse is that "it is a liinire bounded by a regular curve, corresponding to an oblique section of a cone through its opposite sides." It is this definition of the ellipse that classes it among what are known as conic sections. It is generally a matter of surprise to students to find that an oblique section of a cylinder, and an oblique section of a cone through its opposite sides, produce the same figure, but such is the case. The method of drawing an ellipse upon this definition of it is given in the follow- H H Fig. tlS.The Ellipse as an Oblique Section of a Cone. ing demonstration. The principles upon which this rule is based, no less than those referred to in the last demonstration, are of especial interest to the pattern cutter, because so many of the shapes with which he has to deal owe their origin to the cone. 69. To Describe the Shape of an Oblique Section of a Cone through its Opposite Sides, or to Draw an Ellipse as a Section of a Cone. In Fig. 212, let B A C represent a cone, of which E D G F is the plan at the base. Let H I represent any oblique cut through its opposite sides. Then it is required to draw the shape of tin- section represented l>v II I, which will be an ellipse. At any convenient place outside of the figure draw a duplicate of IF I parallel to it, upon which to construct the figure sought, as II' I 1 . Divide one-half of the; Geometrica / plan, as E D G, into any convenient number of equal parts, as shown by 1, 2, 3, 4, etc. From the center of the plan M draw radial lines to these points. From each of the points also erect a perpendicular line, which produce until it cuts the base line B C of the cone. From the base lino of the cone continue each of these lines toward the apex A, cutting the oblique line H I. Through the points thus obtained in H I, and at right angles to the axis A.D of the cone, draw lines, as shown by 1', 2', 3', 4', etc., cutting the opposite sides of the cone. From the same points in H I, at right angles to it, draw lines cutting II' I 1 , as shown by r, 2", 3 a , 4 = , etc., thus transferring to it the same divisions as have been given to other parts of the fig- ure. After having obtained these several sets of lines, the first step is to obtain a plan view of the oblique cut, for which proceed as follows : With the di- viders take -the distance from the axial line A D to one side of the cone, on each of the lines I 1 , 2 1 , 3', 4', etc., and set off like distance from the center of the plan M on the corresponding radial lines 1, 2, 3, 4, etc. A line traced through the points thus obtained will give the plan view of the oblique cut, as shown by the inner line in the plan. This result may be verified by dropping lines vertically from the points in H I across the plan, in- tersecting them with the radial lines in the plan of corresponding number. Thus a line dropped from point 4 on H I should intersect the radial line M 4 at K -s-G -sB measurement, with the dividers take the distance on each of the several cross lines 2 J , 3 s , 4 s , 5 3 , etc., from E G to one side of the plan of the obi:'que cut just de- scribed, and set off the same distance on each side of II' I 1 on the corresponding lines. A line traced through the points thus obtained will be an ellipse. '/7r XV" 5 , - . - ' V^-" \ V s - 7~~ \Ax ^ \s'^-.. Fig. SIS.- To Construct an Ellipse from Two Circles by Intersecting Lines. the same point (4 s ) established upon it by measuring the distance upon line 4' from A D to A B. Having thus obtained the shape of the oblique cut as it would appear in plan, the next step is to set off upon the lines previously drawn through II' I' the width of the oblique cut in plan as measured upon lines of corre- sponding number. Therefore, with E G as a basis of j Fig. 214. To Draw an Ellipse within a Oiven Rectangle by Means of Intersecting Lines. 70. To Construct an Ellipse to Given Dimensions by the Use of Two Circles and Intersecting Lines. In Fig. 213, let it be required to construct an ellipse, the length of which shall equal A B and the width of which shall equal H F. Draw A B and H F at right angles, intersecting at their middle points, K. From K as center, and with one-half of the length A B as radius, describe the circle A C B D. From K as center, and with one-half of the width II F as radius, describe the circle E F G II. Divide the larger circle into any convenient number of equal parts, as shown bv the small figures 1, 2, 3, 4, etc. Divide the smaller circle into the same number of equal and correspond- ing parts, as also shown by figures. By means of the T-square, from the points in the outer circle draw vertical lines, and from points in the inner circle draw horizontal lines, as shown, producing them until they intersect the lines first drawn. A line traced through these points of intersection will be an ellipse. 11. To Draw an Ellipse within a Given Rectangle by Means of Intersecting Lines. In Fig. 214, let E D B A be any rectangle within which it is required to con- struct an ellipse. Bisect the end A E, obtaining the point F, from which erect the perpendicular F G, dividing the rectangle horizontally into two equal por- tions. Bisect the side A B, obtaining the point H, and draw the perpendicular II I, dividing the rectangle vertically into two equal portions. The lines F G and II I are then the axes of the ellipse. F G represents the major axis, and II I the minor axis. Divide the spaces F E, F A, G D and G B into any convenient number of equal parts, as shown by the figures 1, 2, 3. From these points in V K and G 1) draw lines to I, and from the points in F A and G B draw lines to the 11, > New M< />//. \Vn,-ki'r I'lttfcnt Book. point H. Divide F C and G C also into the same number of equal parts, as shown by the figures, and from H and I through each of these points draw lines, continuing them till they intersect lines of correspond- ing number in the other set, as indicated. A line traced through the several points of intersection be- tween the two sets of lines, as shown in the engraving, will be an ellipse. Besides the above methods for drawing correct ellipses there are several methods for drawing figures approximating ellipses more or less closely, but com- Fig. 215.-Firat Method. Fig. 216.-Second Method. To Draw an Apprcximate Ellipse with, the Campasteg, the Length only Being Given. posed of arcs of circles, which it is sometimes necessary to substitute for true ellipses for constructive reasons. The ellipse has been described above as a curve drawn with a constantly changing radius. If, instead of using an infinite number of radii, some finite number be assumed, it will appear that the greater the number assumed the more nearly will it approach a perfect ellipse. Thus, a curve very much like an ellipse can be drawn, each quarter of which is composed of arcs drawn from two centers. If the number of centers be increased to three, the curve comes much nearer a true ellipse, and with four or five centers to each quarter, the curve thus produced can scarcely be distinguished from the perfect ellipse. 72. To Draw an Approximate Ellipse with the Com- passes, the Length only being Given. In Fig. 215, let A be any lepgth to which it is desired to draw an elliptical figure. Divide A C into four equal parts. From 3 as center, and with 3 1 as radius, strike the arc BID, and from 1 as center, and the same radius, strike the arc B 3 D, intersecting the arc first struck in the points B and D. From B, through the points 1 and 3, draw the lines B E and B F indefinitely, and from D, in like manner, draw the lines D G and D II. From the point 1 as center, and with 1 A as radius, strike the arc E G, and from 3 as center, with the same radius, or its equivalent, 3 C, strike the arc II F. From D as center, with radius D G, strike the arc G II, and from B as center, with the same radius, or its equivalent, B E, strike the arc E F, thus completing the figure. A figure of different proportions may be drawn in the same general manner as follows : Divide the length A C into four equal parts, as indicated in Fig. 216. From 2 as center, and with 2 1 as radius, strike the circle 1 E 3 F. Bisect the given length A C by the line B D, as shown, cutting the circle in the points E and F. From E, through the points 1 and 3, draw the lines E G and E H indefinitely, and from F, through the same points, draw similar lines, F I and F K. From 1 as center, and with 1 A as radius, strike the arc I A G, and from 3 as center, with equal radius, strike the arc K C H. From E as center, and with radius E G, strike the arc G D H, and from F as center, with corresponding radius, strike the arc I B K, thus completing the figure. 73. To Draw an Approximate Ellipse with the Com- passes to Given Dimensions, Using Two Sets of Centers. First Method. In Fig. 217, let A B represent the length of the required figure and D E its width. Draw A B and D E at right angles to each other, and intersecting at their middle points. At the point A erect the perpendicular A F, and in length make it equal to C D. Bisect A F, obtaining the point N. Draw N D. From F draw a line to E, as shown, cut- ting N D in the point G. Bisect the line G D by the line H I, perpendicular to G D and meeting D E in the point I. In the same manner draw lines correspond- ino' to G I, as shown by L I, M O and li 0. From I and as centers, and with I G as radius, strike the arcs G D L and M E R, and from K and P as centers, Problems, 65 with K (! as radius, strike tin- arcs R, thus completing the liguiv. 74. To Draw an Approximate Ellipse with the Com- passes to Given Dimensions, Using; two Sets of Centers. Second Method. In Fig. 21s, lot C D represent the length of a ro<|iiiro the width. Lay oil' these two dimensions at right angles to (>aeh other, as shown. On C D lay off a space ei|iial to the width of the required figure, as shown by 1) E. Divide the remainder of I) C, or the space V. (', into three equal parts, as shown in the cut. With a radius equal to t\vo of these parts, and from \\ as center, strike the circle G S F T. Then with V and G as centers, and V G as radius, strike the arcs, as shown, intersecting upon A 1? prolonged at () and P. From 0, through the points G ami F, draw O L and M, and likewise from (enter, with F L as radius, describe a circle, as shown, thus establishing the points M, N and O, which, with L, are the centers from which the ellipse is to be struck. From M, draw M L Q and M N S indefinitely, and in a similar manner () L P and N R. With as center, and O D as radius, strike the arc P D R, cutting O P and O R, as shown. In a similar manner, and with the same radius (or which is the same, with M E as radius) and M as center, describe the arc Q E S. With L and N as centers, and with L 15 or N C as radius, strike the arcs Q B P and R C S, thus com- pleting the figure. The above methods of drawing approximate el- lipses are only available within certain limits of pro- portion, as will be discovered if an attempt is made to draw them very much donga-ted, the limit being e~ f E Fig. 217. -First Method. Fig. 218. Second Method. Fig. 219. Third Method. To Draw an Approximate Ellipse with the Compasses, Using Two Sets of Centers. P, through the same points, draw P K and P N. From O as center, with O A as radius, strike the arc L M, and with the same radius, and P as center, strike the arc K N. From F and G as centers, and with F D and G C as radii, strike the arcs N M and K L re- spectively, thus completing the figure. 15. To Draw an Approximate Ellipse with the Com- passes to Given Dimensions, Using Two Sets of Centers. Third Method. In Fig. 219, let B C represent the length of the required figure and D E its width. B C and D E are drawn at right angles to each other, intersecting at their middle points at F. The next step in describing the figure is to obtain the di (Terence in length between the axes F D and F B, which can be done as indicated by the arc D G. This difference, G B, is to be set off ( 'rom the center F on F B and F D, as shown by F H, F. J, then draw II J and set off half of H J to L, as indicated by the arc K L. The object of the operation so far has been to secure the point L. From F as reached when the long diameter is about equal to two times the shorter diameter. Beyond this limit in the first two methods, if the final arc be drawn with the radius G K (Figs. 217 and 218), it will not reach the end of the long diameter, but will strike it at a point inside of A or C. By the third method, if the long diameter be increased until it is about 2f times the shorter, the point L (Fig. 219) will fall at the extreme limit of the long diameter (B), thus completely cutting out the small arc P Q. It must, therefore, in extreme cases be left to the judgment of the draftsman to adjust or vary the lengths of the radii of the two arcs so as to produce the result which will look the best. 76. To Draw an Approximate Ellipse with the Com- passes to Given Dimensions, Using: Three Sets of Centers. In Fig. 220, let A B represent the length of the re- quired figure and D E the width. Draw A B and D E at right angles to each other, intersecting at their middle points, as shown at C. From the point A draw 66 The Xew Metal Worker Pattern A F v perpendicular to A B, and in length equal to C D. Join the points F and D, as shown. Divide A F into three equal parts, thus obtaining the points Z and I, and draw the lines Z D and I D. Divide A C into three equal parts, as shown by Y and G, and draw E G and E Y, prolonging them until they intersect with Z D and I D respectively, in the points J and II. Bisect J D, and draw K L perpendicular to its central point, intersecting D E prolonged in the point L. Draw J L and H J. Bisect H J, and draw M N perpen- dicular to its central point, meeting J L in N. Draw N II, cutting A B in the point 0. L then is the center of the arc J D P, N is the center of the arc II J, and is the center of the arc H A K. The points S and U, corresponding to N and 0, from which to of X (), draw P R, perpendicular to X O and parallel to K M. Then N (.) and P K are tho axes of the ellipse. 78. In a Given Ellipse, to Find Centers by which an Approximate Figure may be Constructed In Fig. -2-2-2, let A E B D be any ellipse, in which it is required to find centers by which an approximate Jigure may be drawn with the compasses. Draw the axes A B and E D. From the point A draw A F, perpendicular to A B, and make it equal to C E. Join F and E. Divide A F into as many equal parts as it is desired to have sets of centers for the figure. In this in- stance four. Therefore, A F is divided into four equal parts, as shown by P<) and (i. Divide A (' into the same number of equal parts, as shown by R BP Fig. 220. To Draw an Approximate Ellipse with the Compasses, Using Three Sets of Centers. Fiij. %%1. To Find the True Axes of a Given Ellipse. Fit/. 222. In a Given Ellipse, to Find Centers by which mi Approximate Fiyure may be Constructed. strike the remainder of the upper part of the figure, may be obtained by measurement, as indicated. Hav- ing drawn so much of the figure as can be struck from these centers, set the dividers to the distance L P or L J, and placing one point at E, the remaining center will be found at the other point of the dividers, in the line E D prolonged, as shown by X. 11. To Find the True Axes of a. Given Ellipse. In Fig. 221, let N P R be any ellipse, of which it is re- quired to find the two axes. Through the ellipse draw any lines, A B and D E, parallel to each other. Bisect these two lines and draw F G, prolonging it until it meets the sides of the ellipse in the points II and I. Bisect the line H I, obtaining the point C. From C as center, with any convenient radius, describe the arc K L M, cutting the sides of the ellipse at the points K and M. Join K and M by a straight line, as shown. Bisect M K by the line N 0, perpendicular to it. Through C, which will also be found to be the center S T. From the points of division in A F draw lines to E. From D draw lines passing through the divi- sions in A C, prolonging them until they intersect the lines drawn from A F to E, as shown by D U, D V and D W. Draw the chords U V, V W and W E, and from the center of each erect a perpendicular, which prolong until they intersect as follows : The line perpendicular to W E intersects the center line E D in the point D. Now draw D W and prolong the perpendicular to V W till it intersects D W in K, and draw K V. Prolong the perpendicular to U V till it cuts K V in L and draw L I', cutting A C in the point S. Then D is the center of the arc E W, K is the center of the arc W V, L is the center of the an; V U and S is the center of the arc U N. By these centers it will lie seen that one-quarter of the figure (A to E) may be struck. By measurement, corre- sponding points may be located in other portions of the figure. If correctly done the points U, V and W dreamt /na.tl Problems. 67 will be found to fall upon the ellipse. consequently the arcs drawn between those points from the centers obtained cannot deviate much from the correct ellipse. 79. To Draw the Joint Lines of an Elliptical Arch. First Method. In a circular arch the lines representing the joints between the stones forming the arch, or the voussoirs as they are properly called, are drawn radially from the center of the semicircle of the arch. In an' elliptical arch this operation is somewhat more dillicuit, as the true ellipse possesses no such single B C Fig 223. First Method. To Draw the Joint Lines of an Elliptical Arch. point, but, instead, two foci, as has been explained. Therefore, the following course must be pursued : From any point upon the ellipse at which it is desired to locate a joint, as A, Fig. 223, draw a line to each of the foci, as A B and A C. Risect the angle BAG (Prob. 12 in this chapter), as shown at D, and extend the line D A outside the ellipse, which will be the joint line required. 80. To Draw the Joint Lines of an Elliptical Arch. Second Method. In Fig. 224, A R is one-half the curve of the arch, A C its center line and C B its springing line. Draw A I) parallel to C B, and D B parallel to A C, and draw the diagonals A B and C D. From each of the points 1, 2, 3, etc., representing the joints, drop lines vertically, cutting C D. From their inter- sections with C D carry them at right angles to A B, cutting the springing line C B, as shown by the small figures 1", 2", 3% etc. From the points in C B draw 3' 4" 5' B Fig. 224,-Second Method. To Draw the Joint Lines of an Elliptical Arch. lines through corresponding points in the arch A B, as 1' 1, 2" 2, 3' 3, etc., and continue them through the face of the arch which will be the joint lines sought. In the case of an elliptical curve made up of arcs of circles, the joint lines would be drawn radially from the centers of the arcs in which they occur. THE VOLUTE. The volute is an architectural figure of a geo- metrical nature based upon the spiral, and is of quite frequent occurrence in one form or another, conse- quently some remarks upon the different methods of drawing it will not be out of place. 81. To Draw a Simple Volute. Let D A, in Fig. 225, be the width of a scroll or other member for which it is desired to draw a volute termination. Draw the line D 1, in length equal to three times D A, as shown by D A, A R and B 1. From the point 1 draw 1 2 at right angles to D 1, and in length equal to two-thirds the width of the scroll that is, to two- thirds of D A. From 2 draw the line 2 3 perpen- dicular to 1 2, and in length equal to three-quarters of 1 2. Draw the diagonal line 1 3. From 2 draw a line perpendicular to 1 3, as shown by 2 4, indefi- nitely. From 3 draw a line perpendicular to 2 3, pro- ducing it until it cuts the line 2 4 in the point 4. From 4 draw a line perpendicular to 3 4, producing it until it meets the line 1 3 in the point 5. In like manner draw 5 6 and 6 7. The points 1, 2, 3, 4, etc., thus obtained are the centers by which the curve of the volute is struck. From 1 as center, and with 1 D as radius, describe the quarter circle D C. Then from 2 as center, and 2 C as radius, describe the quarter circle C F, and so continue using the centers in their numerical order until the curve intersects with the other curve beginning at A and struck from the same centers, thus completing the figure, as shown. 82. To Draw an Ionic Volute. Draw the line A B, Fig. 226, equal to the hight of the required volute, and divide it into seven equal parts. From the third division draw the line 3 C, and from a point on this line at any convenient distance from A B describe :i 68 The Xcw Mdal Worker I'allcnt Book. circle, the diameter of which shall equal one of the seven divisions of the line A B. This circle forms the eye of the volute. In order to show its dimen- sions, etc., it is enlarged in Fig. 227. A square, D E F G, is constructed, and the diagonals G E and F D are drawn. F E is bisected at the point 1, and the line 1 2 is drawn parallel to G E. The line 2 3 is then 'drawn indefinitely from 2 parallel to F D, cut- ting G E in the point II. The distance from H to the center of the circle is divided into three equal parts, as shown by H a b O. The triangle 2 1 is formed. On the line H set off a point, as c, at a distance from equal to one-half of one of the three equal parts into which H has been divided. From c draw the line c 3 parallel to 1 O, producing it until it cuts 2 3 in the point 3. From 3 draw the line 3 4 parallel to G E indefinitely. From the point c draw a line c 4 parallel to 2 0, cutting the line 3 4 -in the point 4, completing the triangle c 3 4. From 4 draw the line 4 5 parallel to F D, meeting 1 in the point 5. From 5 draw the line 5 6 parallel to G E, meeting the line 2 in the point 6. From 6 draw the line, 6 7 parallel to F D, meeting the line c 3 in the point 7. Fig. 225. To Draw a Simple Volute. Proceed in this manner, obtaining the remaining points, 8, 9, 10, 11 and 12. These points form the centers by which the outer line of the volute proper is drawn. From 1 as center, and with radius 1 F, Fig. 226, de- scribe the quarter circle F G. Then from 2 as center, and with radius 2 G describe the quarter circle G D, and so continue striking a quarter circle from each of the centers above described until the last arc meets the circle first drawn. To obtain the centers by which the inner line of the volute is struck, and which gradually approaches the outer line throughout its course, proceed as follows : Produce the line 3 c, Fig. 227, until it intersects 1 2 in the point I 1 , which Fig. S26.To Draw an Ionic Volute. mark. This operation gives also the points ',)' and 5* of intersection with the lines parallel to 1 2, whidi also mark. In like manner produce 4 c, 1 O and 2 O, as shown by the dotted lines, and mark the several points of intersection formed with the cross lines. Then the points I 1 , 2', 3', 4', etc., thus obtained are the centers for the inner line of the volute, which use in the same manner as described for producing the outer line. 83. To Draw a Spiral from Centers with Compasses. Divide the circumference of the primary some- times called the eye of the spiral into any number of equal parts; the larger the number of parts the more regular will be the spiral. Fig. 228 shows the primary divided into six equal parts. Fig. 22! is an enlarged view of this portion of the preceding figure. Construct the polygon by drawing the lines 1 2, 2 3, 3 4, etc.. producing them outside of the primary, as shown by A, B, D, F, and E. From 2 as center, with 2 1 as radius, describe the arc A B. From 3 as center, and 3 B as radius, describe the arc B D; and (li'iinn C,!) with 4 as center, witli radius 4 D, describe the arc D F. In this manner the spiral may be continued ;( iiv nuiuber of revolutions. Jn the resulting ligure the various revolutions will be parallel. 84. To Draw a Spiral by Means of a Spool and Thread. Set the spool as shown bv A D B in Fig. Fig. 227. Eye. of the Volute in Fig. 226 Enlarged. 230 and wind a thread around it. Make a loop, E, in the end of the thread, in which plaee a pencil, as shown. Hold the spool firmly and move the pencil around it, unwinding the thread. A curve will be de- scribed, as shown in the dotted lines of the engrav- ing. It is evident that the proportions of the figure toj), K A at the bottom and A B at the side, the length of A B, which determines the width of the scroll, being given. Bisect A B, obtaining the point C. Let the distance between the beginning and ending of the first revolution of the scroll, shown by a c. l)o established at pleasure. Having determined Fig. 231. To Draw a Scroll to a Specified Width. Fig. 233. The Center of Fig. SSI Enlarged. this distance, take one-eighth of it and set it off up- ward from C on the line A B, thus obtaining the point b. From l> draw a horizontal line of any convenient length, as shown by b h. With one point of the com- passes set at b, and with b A as radius, describe an arc cutting the line b h in the point 1. In like manner, from the same center, with radius b B, describe an arc cutting the line b h in the point 2. Upon 1 2 as a base erect a square, as shown by 1 2 3 4. Then from 1 as Fig. 228. To Draw a Spiral from Centers. Fig. 129. Enlarged View of the Eye of the Spiral in Fig. 228. Fig. SSO.To Draw a Spiral by Means of a Spool and Thread. are determined by the size of the spool. Hence a larger or smaller spool is to be used, as circumstances require. 85. To Draw a Scroll to a Specified Width, as for a Bracket or Modillion. In Fig. 231, let it be required to construct a scroll which shall touch the line D B at the center, with 1 a as radius, describe an arc, ab; and from 2 as center, with 2 b as radius, describe the arc b c. From 3 as center, with radius 3 e, describe the arc c d. From 4 as center, with radius 4 d, describe the arc d e. If the curve were continued from e, being struck from the same centers, it would run parallel to 70 Tin' \rir MI ti/l \Vnrl-fl- 1'iilli'i'n linn/.-. itself; but as the inner line of the scroll runs parallel to the outer line, its width may be set off at pleasure, as shown by a a', and the inner line may be drawn by the same centers as already used for the outer, and con- tinued until it is intersected by the outer curve. To find the centers from which to complete the outer curve, construct upon the line of the last radius above used (4 e) a smaller square within the larger one, as shown by 5678. Thi? is better illustrated by the larger diagram, Fig. 232, in which like figures repre- sent the same points. Make the distance from 5 to 8 equal to one-half of the space from 4 to 1, making 4 to 8 equal the distance of 5 to 1. Make 5 to 6 equal the distance from 8 to 5. After obtaining the points 5, (!, 7, etc., in this manner, so many of them arc i<> be used as are necessary to make 1 the outer curve inter- sect the inner one, as shown at the? explanation of such principles in order that the reasons for the steps taken in the demonstrations following mav be readily understood. Underlying the entire range of problems peculiar to sheet metal work are certain fundamental principles, which, when thoroughly understood, make plain and simple that which otherwise would appear arbitrary, if not actually mysterious. So true is this that noth- ing is risked in asserting that any one who thoroughly comprehends all the steps in connection with cutting a simple square miter is able to cut any miter what- soever. Since almost any one can cut a square miter, the question at once arises, in view of this statement, why is it that he cannot cut a raking miter, or a pin- nacle miter, or any other equally difficult form? The answer is, because he does not understand how he cuts the square miter. He may perform the operation just as he has been taught, and produce results entirely satisfactory from a mechanical standpoint, without be- ing intelligent as to all that he has done. He does not -comprehend the w\\y and wherefore of the steps taken. Hence it is that when he undertakes some other miter he finds himself deficient. There is a wide difference between the skill that produces a pattern by rote by a mere effort of the memorv and that which reasons out the successive steps. One is worth but very little, while the other renders its possessor independent. It is with a desire to put the student in possession of this latter kind of skill, to render him intelligent as to every operation to be performed, that the present chapter is written. The forms with which the pattern cutter has to deal may be divided, for convenience of deseription, into three general classes : I. The first of these embraces moldings, pipes and r/'ijnhu- continuous forms, and may be called forms of parallel lines, or as a shorter and more convenient name to use, parallel forms. II. The second, which will be called regular taper- ing forms, comprehends all shapes derived from cones or pyramids, or from solids having any of the regular geometric figures as a base and which terminate in an apex. III. The third class will be called irregular forms, and will include everything not classified under either of the two previous heads. Many of these might be properly called transition pieces that is, pieces which have figures of various outlines placed at various angles as their bases, and have figures with differing outlines variously placed, as their upper terminations, thus forming transitions, or connecting pieces between the form which lies next them at one end and the adjacent form on the other end. While pieces of metal of any shape necessary to form the covering of a solid of any shape may prop- erly be called patterns, the shapes of pieces necessary to form the joints between moldings meeting at an angle are known distinctively as miters. This name applies equally well in sheet metal work if the two arms of the moldiag are not of the same profile, or to a single arm coming against any plain or irregular sur- face. These forms comprise the first class referred to above and, so far as principle is concerned, come under the same general rules, which will be subsequently given. Conical forms, with very little taper, coming against other forms are also said to miter with them. In fact, the word miter has come into such general use that it is often applied to any joint between pieces of metal ; but the term can scarcely be considered as cor- rect when the forms have very much taper. The principle involved in the development of such patterns, however, is the same as that applied to the develop- ment of the surfaces of all other regular tapering forms, Tin: .\i-n- Mitnl \\'i>/-/,-rr I'atl'fii I in, I,. referred to above as the second class, whose character- istics will be considered in their proper chapter. The method employed for developing the patterns for forms of the third class has been termed triinii/nlii- tion, and is adopted on account of its simplicity, as it does away with the reduction or subdivision of an irregular form into a number of smaller regular forms, each one of which would have to be treated separately and perhaps by a different method. In fact, there are some shapes which have arisen from force of circum- stances which it would be impossible to separate into regular parts, and even if they could be so separated such a course would result in tedious and complicated operations. After principles have been thoroughly explained the problems in this work will follow in three sections or departments of the final chapter, arranged according to the above classification. This is one of the instances in which the pattern cut- ter is required to be something of an architectural draftsman, and to this end a chapter ofi Linear Draw- ing (Chap. Ill) has been introduced, in which atten- tion is given to this phase of the work, and to which the student is referred. The arrangement of the problems in each of the sections of the succeeding chapter will be made with reference to these two conditions, the simpler ones being placed before those in which preliminary draw- ing is required. Parallel Forms. (MITER CUTTING.) Since in sheet metal work a molding is made by bending the sheet until it fits a given stay, a molding may be defined mechanically as a succession of paral- Fig. 23S. Profile of a Molding. Fig. XS4.A Stay. Fig. 235. A Reverse Stay. Two conditions exist in regard to the work of developing patterns of all forms, no matter to which of the three classes above defined they may belong : yifxt In very many cases a simple elevation or plan of the intersecting parts, together with their profiles, is all that is necessary to begin the work of developing the pattern that is, the plan or elevation, as the case may be, shows the line (either straight or curved) which represents the surface against which another part is to be fitted ; in other words, the much sought for " miter line." X'miul In numerous other instances, however, no view can be drawn either in elevation, oblique or otherwise, or in plan, in which the miter or junction of the parts will appear as a simple straight or curved line against which the points can be dropped. In such cases it becomes necessary to do some preliminary work in order to prepare the way to the actual work of laying out the pattern. A view of the joint must be developed by means of intersections of lines which will show it as it appears in connection with the eleva- tion or plan to be used in developing the pattern. lei forms or bends to a given stay, and, so far as the mechanic is concerned, any continuous form or ar- rangement of parallel continuous forms, made for any purpose whatever, may be considered a molding and treated under the same rules in all the operations of pattern cutting. Keeping this fact in mind all paral- lel forms will be considered as moldings and that word will be used in the demonstrations, remembering that a difference in name simply means a difference of profile, but not a difference in treatment or principle. A molding may be defined theoretically as a form or surface generated by a profile passed in a straight or curved line from one point to another, this profile being the shape that would be seen when looking at its end if the molding were cut off square. A prac- tical illustration of this may be given as follows: In Fig. 233, let the form shown be the profile of some molding. If this shape be cut out of tin plate or sheet iron, as shown in Fig. 234, it is called a stay. For the purpose of this illustration, as will appear fur- ther on, a stay, the reverse of the one shown in Fig. 234, or, in other words, the piece cut from the face or i if Pull, -i-il Cull! inf. outside of the shape represented in that figure, as shown in Fig. !''>>, will !> required. Having made a reverse stav, or " outside stay, as it is sometimes called, Fig. :>:>;>, take some plastic material as potters' day and, placing it against any smooth surface, as of a hoard, place the stay against the board near one end in such a position that its ver- tical lines are parallel with the ends of the board, and move this reverse stav in a straight line along the faee of the board until a continuous form is obtained in the dav corresponding to the profile of the stay, all as illustrated in Fig. -!'!<>. Bv this operation will be produced a molding in accordance with the second definition above given. The purpose in introducing this illustration is to show more dearly than is other- wise possible the principles upon which the different Firj. 236. Generating a Molding in Plastic Material by Means of a Reverse Stay parts of a molding are measured in the process of pat- tern cutting. Suppose that the form produced as illustrated in Fig. 23 (5 be completed, and that both ends of the molding be cut off square. It is evident, upon in- spection, that the length of a piece of sheet metal necessary to form a covering to this molding will be the length of the molding itself, and that the width of the piece will be equal to the distance obtained In- measuring around the outline of the stay which was used in giving shape to the molding. Now with a thin-bladed knife, or by means of a piece of fine wire stretched tight, let one end of the clay molding just constructed be cut off at any angle. By inspection of the form when thus cut, as clearly shown in the upper part of Fig. 237, it is evident that the end of a pattern to form a covering' of this model must have such a shape as will make it when formed up conform to the oblique end of the molding or model. To cut such a pattern by means of a straight line drawn from a point corresponding to the end of the longer side of the mold, to a point corresponding to the end of the shorter side of it, would not be right, evidently, because certain parts of the covering, when formed up, fold down into the angles of the molding, and therefore would require to be either longer or shorter, as the case might be, than if cut as above de- scribed. It is plain, then, that some plan must be devised by which measurements can be taken in all these angles or bends, and at as many intermediate points as may be necessary, in order to obtain the right length at all points throughout its width. This can be done quite simply as follows : Divide the curved parts of the stay into any con- venient number of equal parts, and at each division cut a notch, or affix a point to it. Replace the stay in the position it occupied in producing the molding shown in Fig. 236 and pass it again over the entire length of the model. The points fastened to the stay will then leave tracks or lines upon the surface of the molding. Now, by means of measurement upon the different lines thus produced, the length of the molding at all of the several points established in the stay may be obtained. All this is clearly illustrated in Fig. 237. In the upper right hand corner of the illus- tration is shown the stay prepared with points, by moving which as above described lines are left upon the face of the molding, as shown to the left Now, upon a sheet of paper fastened to a draw- ing board, draw a vertical line, as shown by A B in Fig. 237, and upon that line set off with the dividers the width of each space or part of the profile or stay that is, make the space 1 2 in the line A B equal to the space 1 2 in the stay, and 2 3 in the line A B equal to 2 3 of the stay, and so continue until all the spaces are transferred and from the points thus ob- tained in A B draw lines at right angles to it indefi- nitely, as shown to the left. The lines an dspaces upon the paper will then correspond to the lines and spaces upon the clay molding made by the points fastened to the stay. Next, measure with the dividers the length of the molding upon each of the lines drawn upon it, and set off the same lengths upon the corresponding T/IC l \\'41, by which the miter line is located to a certainty, and in connection with which it is a simple matter to correctly place the profile, it is not uncommon to attempt the operation by drawing the miter line only, placing it either above, below or at one side of the profile. The mistake is made bv hav- ing the line at the side of the profile when it should be either above or below it, and vice versa. Fig. -24-2 illustrates a ease in point. The engraving was made L . B , Fig. 24%- A. Squure Face Miter Produced Where a Square Return Miter ivas Intended, from the drawing of a person who attempted to cut a square return miter by the rule, using a miter line only. By placing the miter line E F at the side in- stead of below the profile, a square face miter such as would be used in the molding running around a panel or a picture frame was produced in place of what was desired. In order to avoid such errors the reader is recom- mended to a careful pei'usal of the chapter on Linear Drawing (Chapter III), where the relation existing be- tween plans, elevations and sections or profiles is thor- ough- explained. It is better to draw a complete plan. as :-hown in Fig. 24-1, thus demonstrating to a eer- Principles uf Pattern Cuttinij. tainty the correct relationship of the parts, than to save a little labor and run the risk of error. As remarked in the earlier part of this chapter, some labor is often neeessarv before the requirements mentioned above in connection with the rule can be fulfilled. Sometimes a miter line must first be de- veloped, and sometimes the profile of a molding must undergo a change of profile known as raking. It is believed that the principles underlying these opera- tions arc made sufficiently dear in connection with the problems in which they are involved not to need es- pecial explanation in this connection. Suffice it to say that, in many instances, half the work is done in the getting ready. (FLARING WORK.) This subject embraces" a large variety of forms of frequent occurrence in sheet metal work, and the de- velopment of their surfaces comes under an altogether different set of rules than those applied to parallel forms. Before entering into the details of these methods it will be best to first define accurately what is here included by the use of the term. These forms include only such solid figures as have for a base the circle or any of the regular polygons, as the square, triangle, Fig. . A Right Cone Generated by the Revolution of a Right- Angled Triangle about its Perpendicular. hexagon, etc. ; also figures though of unequal sides that can be inscribed within a circle, and all of which terminate in an apex located directly over the center of the base. While the treatment of these forms has been said to be altogether different from that of parallel forms there are some points of similarity to which the stu- dent's attention is called that may serve to fix the methods of work in his memory. Whereas in parallel forms the -distances of the various [mints in a miter are measured from a straight line drawn through the mold near the miter for that purpose, as C D, Fig. 23.8, the distances of all points in the surfaces of tapering solids produced by the intersection of some other surface are measured from the apex upon lines radiating therefrom ; and Fig. 244. A Bight Cone with Thread Fastened at the Apex to which are Attached Points Marking the Upper and Lower Bases of a Frustum. whereas the distance across parallel forms (the stretch- out) is measured upon the profile, the distance across tapering forms is measured upon the perimeter of the base. Patterns are more frequently required for por- tions of frustums of these figures than for the com- plete figures themselves and the methods of obtaining the pattern of coverings of said frustums is simply to develop the surface of the entire cone or pyramid and by a system of measurements take out such parts as are required. As the apex of a cone is situated in a perpendic- ular line erected upon the center of its base, it must of necessity be equidistant from all points in the cir- cumference of the base. In works upon solid geometry the cone is de- scribed as a solid generated by the revolution of a right-angle triangle about >ts vertical side as an axis. This operation is illustrated in Fig. 243, in which it will be seen that the base E 1) of the triangle C K 1) is the radius which generates the circle forming the base of the cone, and that the hypothenuse C D in like manner generates its covering or envelope. 80 The Xcw Metal Wurkvr I'aW_/-n Book. If a plane bo passed through u 00110 parallel to the base and at some distance above it, the line which it produces by cutting the surface of the cone must also be a circle, because it, like the base, is perpen- dicular to the axis. The portion cut away is simply another perfect cone of less dimensions than the first. Fig. S45. Frustum of a Right Cone, the Dotted Lines Showing the Portion of the Cone Removed to Produce the Frustum. while the portion remaining is called a frustum of a cone. AFC, Fig. 244, is a cone, and B D E C, Fig. 245, is a frustum. The line B E, Fig. 244, shows where the cone is cut to produce the frustum. If, having a solid cone of any convenient material, as wood, a pin be fastened at tlie apex C of the same, as shown in Fig. 244, and a piece of thread be tied Fig. 246. Envelope of the Cone and Frustum, Described by the Pin and Thread in Fig. 244. thereto, to which are fastened points B and A, corre- sponding in distance from the apex to the upper and lower bases of the frustum, and the thread, being drawn straight, be passed around the cone close to its surface, the points upon the thread will follow the lines of the liases of the frustum throughout its course. If then. taking the thread and pin from the cone, and fastening the pin as a center upon a sheet of paper, as shown in Fig. 24(5, the thread be carried around the pin, keep- ing it stretched all the time, the track of the points fastened to the thread will describe upon the paper the shape of the envelope of the frustum, as shown by (J D E F. By omitting the line produced bv the upper of the two points, the envelope of the complete cone G C F will be described. The length of the arc /ic Dm winy. have both its upper and lower faces oblique to the axis of the cone a level base can be assumed at a con- venient distance below the lower face of the frus- tum from which the circumference can lie obtained and then both the upper and lower faces of the frus- tum can be developed by the method just described. A right cone having an elliptical base might seem to belong in the same class with regular tapering forms, but as the distance from its apex to the various points in the perimeter of its base is constantly varying, it is therefore placed in the class with irregular forms in Principles :> is shown the usual method of draw- ing the plan and elevation of an elliptical flaring ar- ticle, the outer curve of the plan A C B D being the shape at M N of the elevation, while the inner curve I G A' J is the plan at the top K L. As many centers may he employed in drawing the curves of the plan of such an article as desired, all of which is ex- plained in the chapter on Geometrical Problems (Chap. I V. ), Problems 73, 70 and 78. To simplify matters only two sets of centers have been employed in the present drawing, all as indicated by the dotted lines drawn from the various centers and separating the different arcs of circles. Keference to the plan now shows that that portion of the article included between the points E Fi'j. 253. Usual Method of Drawing an Elliptical Flaring Article. C W V G II is a position of the envelope of a cone the radius of whose base is 1) C and whose apex is located at a point somewhere above D ; and likewise that that portion included between the points X A B II I J is part of a cone the radius of whose base is A F and whose apex is somewhere upon a line erected at F. Thus four sectors cut from cones of two differ- ent sizes go to make up the entire solid of which the Article shown in Fig. 253 is a frustum. To determine the dimensions, then, of such cones it is necessary to construct a diagram such as that shown in Fig. 254, which is in reality a section upon the line E D of the plan, in which P and P E are respectively equal to E D and E F of the plan. At points and R, Fig. 254, erect perpendicular lines J and K Z indefinitely. Upon O J set off OS equal to the straight hight O K of the frustum, Fig. 253, and Fiy. 2J4.Diayram Constructed to Determine the Dimensions of the Cones, Portions of which are Combined to Make up the Article Shown in Fig. 253. draw S U parallel to O P, which make equal in length to D H. A line drawn through the points P and U will then represent the slant or taper of the frustum, as shown at M K of the elevation, and if continued till it intersects with the perpendiculars from O and P will determine the respective hights of the two cones, as shown by Z and J. Then P JO is the triangle which, if revolved about its vertical side J 0, will generate the cone from which so much of the figure as is struck from the centers C and D in Fig. 253 is cut ; and P Z R is the triangle which if revolved about its vertical side Z R will generate the cone from which the end pieces of the article are taken. To present this before the reader in a more forcible manner, several pictorial illus- trations are here introduced in which the foregoing operations are more clearly shown. In Fig. 255 is shown a view of the plan of the base A C B D of Fig. 253 in perspective, in which the reference letters are the same as at corresponding parts of that plan, and upon which is represented, in its correct position, a sector of the larger cone from which the side portions The Xvw Mini \Vnrkcr Pattern Buok. of the frustum are taken. Tims the triangular sur- faces. F D E and F D AV, being sections of the cone through its axis, correspond to the triangle J () P of the diagram, Fig. 254. In Fig. 256 two additional sectors from the smaller cone previously referred to are represented as standing upon the adjacent portions of the plan from which their dimensions were derived. Thus C F and II D, the center lines of their bases, correspond respectively to A F and Y B of the plan, Fig. 253, and the triangles L F G and M H K, being- radial sections of the cones, correspond with the tri- angle Z R P of the diagram. In Fig. 257 is presented the opposite view of the combination seen in Fig. 250, Fig. 255. Perspective View of the Plan in Fig. 253, with a Sector of the Larger Cone in Position, showing at C B and D A the joining of their outer surfaces or envelopes. As previously remarked, two sets of centers only were employed in constructing the plan, Fig. 253, for the sake of simplicity. Had a third set of centers been made use of the arrangement of sectors of cones shown in Figs. 256 and 257 would have been supple- mented by a pair of sectors, cut from a cone of inter- mediate size, which would have been placed on either side of the large sector between it and the smaller ones, all being joined together upon the same general principle as before explained. Reference to Fig. 22<) in the chapter on Geometrical Problems shows at .1 L P, P S W and AV IT Y what the relative position of their bases would be. If it be desired to complete the solid, which has been begun in Fig. 256, it will first be .necessary to cut away that portion of the middle sector which stands over the space F H B, Fig. 256. Such a cut might be begun upon the line F H, and passing vertically through the points L and M would finish through the curved surface of the further or curved side of the sector. The cut thus made between the points L and M is shown at D C in the other view, Fig. 257, and is by virtue of the conditions a hyper bola. (See Def. 113, Chap. I.) The piece necessary to complete the solid would then be a duplicate of the shape remaining after making the above described cut, the outer surface of which is shown by A B C D of Fig. 257. The complete solid would then have the appearance shown in Fig. 258. Fig. 256. The Same Plan Showimj Two Sectors of the Smaller Cone in Position Joining the Larger One. By thus resolving the solid from which the ordi- nary elliptical flaring article is cut into its component elements the process of developing its pattern may be more readily understood. This process may now lie easily explained by returning to the string and pin method which was made use of in connection with the simple cone in the earlier part of this section. In Fig. 257 is shown a line some distance above the base representing the top of the frustum shown by K L in the original elevation, Fig. 253. It also shows a pin fastened at the apex of the middle conical sector to which is attached a thread carrying points G and II representing the upper and lower surfaces of the frustum. Now, if the siring be drawn tight and passed along the side of the larger sector of the cone from A to B the points will follow the upper and lower bases of the frustum. AVhen the point B is Principles /' /'"// that the distances between such lines at one end of the article' (provided it does not reach to the vertex) bear a regular proportion to the distances between them at the other end. Hence, in the development of the pattern of an irregular form it becomes necessary to drop all previously described systems and simply pro- ceed to measure up its surfaces, portion by portion, adding one portion to another till the entire surface has been covered. To accomplish this end one of the most simple of all geometrical problems is made use of, to which the reader is referred (Chap. IV., Problem 36) viz. : To construct a triangle, the lengtlts of the three sides //ll:-i I!/' I'l/llri-ll Cl/tlilll/. 87 A 15 (' D of the plan, is a perfect circle lying in a horizontal plane. K II ( t lie eli!\ ation. Its upper sur- face, however. N <> I' <^ of the plan, is elliptical in shape and besides being placed at one side of the center is also in an inclined position, as shown by I' 1 <1 of tlin elevation. T<> the right of this plan is another drawing of tin- same, A' IV C' I)', turned one-quarter around from which, and the elevation, is projected another view, J' K' L M, whieh may be called the perimeter at the top. As F (!. the distance across the top, is greater than X P (its apparent width in the plan), the curve X ( ) P Q evidently does not give the correct distance around the top, and therefore a correct view of the top must 1 btained. The method of accomplishing this docs not dill'er from manv similar operations described in connection with parallel forms and is clearly shown in the drawing. Considering N <) I* of the bottom curve with 4' of the top than in crossing the I line from -i of the bottom to 5' of the top, for the rea- ; son that in the former case the triangles produced would be less scalene or acute. The next step is to devise a means of determining the true lengths which these lines represent or, in other words, their real length as they could be measured if a. full size model of the article were cut from a block of wood or clay upon which these lines had been marked, as shown upon the drawing. The lines upon the plan, of course, "only show the horizontal distances between the points which they connect. The vertical hight above the base of any of the points in the upper curve can easily be found by measuring from its position upon the line F G of the elevation perpendicularly to the base E H. Therefore, having both the vertical and the horizontal distance given between any two points, it is only necessary to construct with these dimensions a right angle triangle, and the hypothenuse will give their true distance apart. Thus in Fig. 262 a b is equal to the line 4 4' of the plan, while a c is made equal to 4 4' of the elevation. Consequently c b represents the true distance between the points 4' of the top and 4 of the base. Therefore, to obtain all of these hypothenuses in the simplest possible manner, it will be necessary to construct one or two diagrams of triangles. To avoid confusion it is better to make two ; one for obtaining the distances represented by the full or solid lines drawn across the plan and the other for those of the dotted lines. To do this extend the base line E II of the elevation, as shown at the left, at any convenient points, in which, as R and S, erect two perpendicular lines. Project lines horizontally from all the points in F G, cutting these two lines as shown, and number the points of intersection. (Some of the figures are omitted in the drawing for lack of space.) From R set off on the base line distances equal to the lengths of the solid lines of the plan 1 1', 2 2', 3 3', etc., numbering the points!, 2, 3, etc., as shown, and connect points of similar number upon the base with those upon the perpendicular. From S set off on the base line dis- tances equal to the lengths of the dotted lines of the plan 1 0', 2 1', 3 2', etc., and number them to corre- spond with figure upon the line of the base ABC. Thus make 1 S equal to 1 0' of the plan, 2 S equal to 2 1' of the plan, 3 S equal to 3 2', etc., and connect Pri. of Puiii m < 'nit! in/. eacli point in tlie base with the point of next lower number upon the perpendicular by ;i dotted line, ;is 1 <>n tin- base with ii nu the perpendicular, 2 with 1, 3 with 2, etc. The entire surface of the piece for which Fig. 63. Top, Back and Bottom for a Model of One-half the Article Shown in Fig. 201. a pattern is required has thus been cut up into two sets of triangles, one set having the spaces upon the base line ABC, which arc all equal, for their bases, and the other set having the spaces in the curve N" 0" P" of the top, also equal to each other, as their bases, and each separate triangle having one solid line and one dotted line as its sides. In all of this work the student's powers of mental conception are called into play. The shape of the sur- face, which is yet to be developed, has been spoken of as if it really existed in fact, it must exist in the mind or imagination of the operator in order to make him intelligent as to what he is doing. If this fails him, he can resort to a model which can easily be con- structed (full size or to scale, according to convenience) as follows : Describe upon a piece of cardboard or metal the shape E F G II, Fig. 261, to which add on its lower side, E H, one-half of the plan of the bottom, ABC, with the curve N P and the solid lines con- necting it with the outside curve traced thereon. Also add on its upper side, F G, one-half the shape of top, N" 0" P", marking the points 1, 2, 3, etc., upon its edge. Now cut out the entire shape in one piece, as shown in Fig. 263, and bend the same at right angles, on the lines F G and E H. Small triangles of the shape and size of each of the triangles shown in the diagram of solid lines, Fig. 261, as K, 1 1 K, 2 2 E, etc., can now be cut out and placed upon the portion representing the bottom, each with its base upon the solid line which it represents, at the same bringing the apex of each to the corresponding number on the top. These can be fastened in place by bits of sealing wax, or if cut from metal the whole can be soldered together. The hypothenuses of the various triangles will thus represent the true distances across the pattern upon the solid lines of the plan, while the distances upon the dotted lines can be represented by pieces of thread or wire, placed so that each will reach from the point at the base of one of the triangles to the point at the top of the one next it. If constructed of metal two or three triangles will suffice to give the model sufficient rigidity, and the remaining points can be connected by pieces of wire, using a different kind of wire to represent the distances on the dotted lines. In Fig. 264, is shown a pictorial representation of a model constructed, as above described, from the draw- ings shown in Fig. 261. In the illustration the tri- angles 2, 5 and 8 only are shown in position, their hypothenuses connecting points of similar number in the upper and lower bases. The other points are rep- resented as being connected by wires or threads repre- senting both the solid and the dotted hypothenuses in the diagrams of triangles in Fig. 261. Such a model if constructed will give a general idea of the shape of the entire covering, and at the same time >f the small pieces, or triangles, of which the covering is composed, with all the dimensions of each. If all of the spaces formed upon this skeleton surface could be filled in Fig. 264. Perspective View of Cardboard Model of One-half the Article Shown in Fig. 261. Avith pieces of cardboard or metal just the size of each and the whole removed together and flattened out (each piece being fastened to its neighbor at the sides), it would constitute the required pattern, the same as will be subsequently obtained by measurements taken from the drawing, and as shown in Fig. 265. 90 Tlir Xrtr Mi-tal HW.vr 1'ntln-n Ifovl: Having by means of the diagrams of triangles in Fig. 261 obtained the lengths of all the sides it is now only necessary to construct successively eaeh triangle in the manner described in Chapter IV. Problem 36, remembering that the last long side of each triangle used is also the first long side of the next one to be constructed. Therefore, at any convenient place draw any straight line, A N of Fig. 265, which make equal to the real distance from A to N, Fig. 261, which has been found to be the distance of the diagram of solid lines. To conduct this operation with the great- est economy and ease it is necessary to have two pairs of dividers, which shall remain set, one to the spaces upon the plan of the base ABC, and the other to tin- spaces upon ~S" O" P", and a third pair for use in taking varying measurements. From A of Fig. 265 as a center, with a radius equal to 1 of the plan, Fig. 261, describe a small arc, and from N as a center, with a radius equal to the true distance from N to 1 of the plan, which has been found to be 1 of the diagram of dotted lines, describe another arc, cutting the first one as shown at the point 1, Fig. 265. The triangle thus constructed represents the true dimensions of one indicated by the same figures of the plan. Next from N of the pattern as a center, with a radius equal to N" 1 of true profile of top, Fig. 261, describe a small arc, which cut with one struck from point 1 of pat- tern as a center, with a radius equal to 1 1 of the dia- gram of solid lines, thus locating point 1' of pattern. This triangle is, in turn, succeeded by another whose sides are next in numerical order, that is 1 2 of the base and 1 2 of the diagram of dotted lines. Thus the operation is continued, always letting the spaces of the circumference of base succeed one another at one side of the pattern, and the spaces upon the true'profile of top succeed one another at the other side of the pat- tern, until all the triangles have been laid out as shown by A N P C, Fig. 265, which will complete one-half the entire pattern. It is not necessary to draw all of the dotted or solid lines across the pattern, as the points where the small arcs intersect are all that are really needed in obtaining the outlines of the pattern, but it is often advisable to draw them as well as to number each new point as obtained, in order to avoid confusion and insure the order of succession. In dividing the curves of top and bottom into spaces, such a number of points should be taken as will insure the greatest accuracy, as in the case of dividing a profile. Thus too few would give too short a stretch-out, while if the spaces were too small error in transferring their lengths might result, which would be increased as many times as there were spaces. I nder the head of transition pieces may be in- cluded a large number of forms having various shaped polygonal or curved figures as their upper and lower surfaces, placed at various angles to each other, some- times centrally located as they appear upon the plan and sometimes otherwise. It often happens that one surface or termination is entirely outside the other in that view, forming an offset between pipes of differing sixes and shapes. Sometimes such an offset takes a curved form, constituting a curved elbow of varying section throughout its length, in which case it consists of a number of pieces, each with a different shape at either end. With such forms may be classed the ship Fig. 265. One-half Pattern of Side of Article Shmvn in Fig. 261. ventilator, whose lower end is usually round and hori- zontal and whose upper end is enlarged and elliptical and stands in a vertical position, the whole being com-: posed of five or six pieces. In such cases, when the shape and position of the two terminating surfaces onlv are given, it becomes necessary to assume or draw as many intermediate surfaces as there are joints required, each of such a shape that the whole series will form a suitable transition between two extreme shapes. It, may be remarked, that what have been spoken of here as "surfaces" do not necessarily mean surfaces of metal forming solid ends to the pieces describe) I. but simply outlines upon paper to work to, as more often the "surface" is really an opening. Still another class of forms demanding treatment by triangulation result, from the construction of arches Princijlles of Pntli-rn Cnttinrj. 91 cut through ciirvwd walls, as when an arch of either round or elliptical form, as a door or window head, is placed in a circular wall in such a manlier 1 hut its sides or jambs are radial, or tend toward the center of the curve of the wall. It will l>e seen that the sofht of such an arch is similar in shape to the sides of a transi- tion piece, having what might be called its upper and lower surfaces curved and placed vertically. In such cases it. is best to consider the horizontal plain; passing through the springing, line of the arch as the base from which to measure the hights of all points assumed in the outer and inner curves. It is believed that a sufficient number of this gen- Fig. S66. Elevations and Plan of an Elliptical Cone. eral class of problems will be found in the third section of the chapter on Pattern Problems to enable the care- ful student to apply the principles here explained to any new forms that might present themselves for his consideration, remembering that any form may be so turned as to bring any desired side into a horizontal position to be used as a base, or that an upper hori- zontal surface can be used as a base as well as a lower. The operations of triangulation undoubtedly re- quire more care for the sake of accuracy than those of any other method of pattern cutting, for the reason that there is no opportunity of stepping off a continuous stretchout, at once, upon any line, either straight or curved. It is therefore not to be recommended if the subject in hand admits of treatment by any regular method without too much subdivision. Triangulation is not introduced as an alternate method, but as a last resort, when nothing else will do. Besides the various forms of transition pieces, an- other class of forms is to be treated under this head, which might almost be considered as regular tapering articles. They include shapes, or frustums cut from shapes, which terminate in an apex, but whose bases cannot be inscribed in a circle, as irregular polygons, figures composed of irregular curves as well as the per- fect ellipse. A solid whose base is a perfect ellipse and whose apex is located directly over the center of its base (in other words, an elliptical cone) is perhaps the best typical representative of this class of figures. If the base of such a cone be divided into quarters by its major and minor axes, it will be seen at once that all of the points in the perimeter of any one quarter will be at different distances from the apex of the cone, because they are at different distances from the center of base or the intersection of the two axes. This is clearly shown in Fig. 266, in which are shown the two elevations and the plan of an elliptical cone. The side elevation shows K E to be the distance of the apex from the point P in the plan of the base, while the end elevation shows K' D to be the distance of the apex from the point D of the base, or the true distance rep- resented by X D of the plan. If one-quarter of the plan of the base, as D P, be divided into any convenient number of equal spaces and lines be drawn to the center X, as shown, each line will represent the horizontal distance of a point in the perimeter from the apex ; and if a section of the cone be constructed upon any one of these lines, as, for in- stance, line 4 X, or, in other words, if a right angle triangle be drawn, of which 4 X is the base and E K the altitude, the hypothenuse will be the true distance of the point 4 from the apex. Therefore, to ascertain the distances from the apex to the various points in the circumference of the base construct a simple diagram of triangles, as shown in Fig. 2G7, viz. : Erect any perpendicular line, as X M, equal in hight to E K of the elevation; from X, on a horizontal line X P, as a base, set off the various distances of the plan, X 1, X 2, X 3, etc., numbering each point, and from each point draw a line to M. These hypothenuses will then represent the distances of the various points in the perimeter of the base from the apex of the cone; or, in other words, the sides of a number of triangles forming the envelope of the cone, the bases of which triangles 92 Tlic \i'n- f'ufti-rn Book. will be the spaces 1 2, 2 3, etc., upon the plan. As all of these triangles terminate at a common apex or center, instead of laying out each one separately to form a pattern, as in the case of an article of the type shown in Fig. 261, the simplest method is as follows : Fig. S67. Diagram of Sections on the Radio] Lines of the Plan in /'(', the upper sur- face A B being shown in Fig. 266 by the lines G II and N O. If the envelope of such a frustum is desired the cut which its upper surface would make through the envelope of the entire cone could lie obtained in exactly the same manner as that of its lower base, be- cause the upper surface of the frustum is in reality the base of the cone, which remains above after the lower part has been cut away. But as part of the operation has already been performed in obtaining the cut at the base, it is most easily accomplished as follows : First draw radial lines from the point M of the diagram of Fig. S6S. Elevation of the Frustum of an Elliptical Cone. triangles, Fig. 267, to each of the points previously obtained in the cut at the bottom of the envelope, between A and D; also draw a horizontal line at a hight above the base X P equal to R L, Fig. 266, cut- ting the hypothennses M 1, M 2, etc., as shown by G H. Now place one foot of the dividers at the point M, and bringing the other foot successively to the various points of intersection of the liiie G II with the various hypothenuses, describe arcs cutting the radial lines in the envelope of corresponding number. A line traced through the points of intersection, as B C, will give the cut at the top of the envelope of the frustum, of which A D is the bottom cut. If the cut at the top of the frustum is to be oblique instead of horizontal, a means must be devised for of Pattern Cutting. measuring the distance from the apex at which the oblique plane cuts each of the hypothennses, or in otlier words, eacli of the lines drawn from the apex of the cone to the various points in its liase. In Fig. 269, K S T F is the elevation of an oblique frustum of an elliptical cone, whose apex is at K, and whose base is the same and lias been divided in the same manner as that shown in Fiji. 266. Erect lines from each of the points iu the curve of one-half the plan P 1) A to the base line K F of the elevation, thence carrv them toward the apex K, cut- ting the line S T ; the vertical liight of the points upon S T can then most easilv lie measured 1>\- carrying them horizontally, cutting the center line R K of the cone, where to avoid confusion they should be numbered to Fig. 270. Diagram of Sections on the Radial Lines of the Plan in Fig. 269, with the Pattern of One-half the Envelope. correspond with the points of the plan from which each was derived. These points may now be transferred in a body by any convenient means to the vertical line X' M' of the diagram of triangles, Fig. 270, seeing that each point is placed at the same distance from M' that it is from the point K of Fig. 269. A horizontal line from any one of the points on the line X' M' extended to the hvpothenuse of corresponding number will then give the correct distance of that point from the apex cf the cone. The diagram M' X' I)' is a duplicate of M X P of Fig. 267, and the lower outline of the en- velope is the same as that shown in Fig. 267. It will be noted, however, that half the stretchout of the base is iiecessarv in this case to give all the essentials of the pattern of the envelope, while one-quarter was suiiicient for the previous operations. When all the points in the uppei line of the frustum have been obtained in the diagram they may be transferred to the various radial lines in the envelope, from M' as a center, by the use of the compasses as before, all as shown in the draw- ing. If the apex of the cone were not located directly over the crossing of the two axes of the ellipse that is, if the cone were scalene or oblique instead of right the method of obtaining its "envelope, or parts of the same, would not differ from the foregoing. Lines drawn from the points of division in the circumference of the base to the point representing the position of the apex in the plan will lie the horizontal distances used in constructing a diagram of triangles, which dis- tances can be used in connection with the vertical hight of the cone, as before, in obtaining the various hypothenuses. If the apex of a scalene cone be located over the line of either axis of the ellipse, either within the perimeter of the base or upon one of those lines continued outside the base, one-half the pattern of the entire envelope will have to be obtained at one operation ; but if the apex is not located upon either of those lines in the plan, then the entire envelope must be obtained at one operation, as no two quarters or halves of the cone will be exactly alike. The method of obtaining the envelope of any scalene cone, even though its base be a perfect circle, is governed by the same principles as those employed in the above dernonsti-ations. It will be well to remember that any horizontal section of a scalene cone is the same shape as its base, which fact can be used to ad vantage in determining the best method to be employed in obtaining the envelope of any irregular flaring surface that may be presented. If, for instance, the plan of any article, whose upper and lower surfaces are horizontal, shows each to con- sist of two circles or parts of circles of different diame- ters not concentric, it is evident that the portion of the envelope indicated by the circles of the plan is part of the envelope of a scalene cone. An illustration of this is given in Fig. 271, which shows a portion of an article having rounded corners and flaring sides and ends, but with more flare at the end than at the side. The plan shows the curve of the bottom corner A B to be a quarter circle with its center at X, and that of the top C D to be a quarter circle with its center at Y. The rounded cornet A B I) C is then a portion of the envelope of a frustrum of a scalene cone, and the method of finding the dimensions of the complete cone is quite simple and is as follows: First draw a line, Z N, through the centers of the two circles in the plan, T/ie Xew Metal Worker Pattern Book. at right angles to which project an oblique elevation, as shown below, making the distance between the two lines E F and G H equal to the hight of the article. Lines from X arid M of the plan of the bottom fall upon G H, locating the points X' and H, while lines ! from Y and N of the top locate the points Y' and F in the upper line of the oblique elevation. A line drawn through Y' and X', the centers of the circles, will then represent the axis of the cone in elevation, which can be continued to meet a line drawn through the points F and H, representing the side of the cone, thus locat- ing the apex Z' of the scalene cone. The point Z' can then be carried back to the plan, as shown at Z, thus locating the apex in that view. As the line N Z rep- resents the horizontal distance between the point F and the apex Z' of the cone, so lines drawn from Z to any number of points assumed in the curve of the base C D will give the horizontal distances between those points and the apex, to be used as the bases in a dia- gram of triangles similar to that shown in Fig. 267, while V Z' gives their hight. Having drawn a dia- gram of triangles the pattern follows in the manner there shown. For greater accuracy in the case of a very tapering cone, the circles of the. plan can be completed, as shown dotted, and their points of intersections with the line Z N can be dropped into oblique elevation, as seen at S arid T, through which a line can be drawn to meet a line through F and H with greater accuracy than one through Y' and X', as the angle in the former case is twice as great. In the above methods of obtaining the envelopes of what may be termed irregular conical forms, it will be clearly seen that the operation of dividing the curve of the base into a great number of spaces really re solves the conical figure into a many sided pyramid, and that the lines connecting the apex with the points in the base, which have been referred to as hypothe- n uses, are really the angles or hips of the pyramid. It is therefore self evident that any method of de- velopment which is applicable to a many sided pyra- mid is equally applicable to one whose sides are fewer in number, with the only difference, however, that the lines representing the angles or hips in the case of a pyramidal figure mean angles or sharp bends in the pattern of the envelope, while in the case of the conical envelope the bends arc so slight as to mean only a con- tinuous form or curve. It is believed that the foregoing elucidation of the principles governing the development of the surfaces of irregular shaped figures is sufficiently clear to make the demonstrations of this class of problems, given in Chap. VI, Section 3, easily understood by the student, as well as to enable him to apply them to any new forms that may present themselves for solution. This chapter is intended to present, under its three different heads, all the principles neccssarv to guide the student in the solution of any problem that may Fi(j. ~'71. Elevations and Plan of an Article the Corner of which is a Portion of a Scalene Cone. arise. Its aim is to teach principles rather than rules, and the student is to be cautioned against arbitrary rules and methods for which he cannot clearly under- stand the reason. His good sense must govern him in the employment of principles and in the choice of methods. There is hardly a pattern to be cut which cannot be obtained in more than one way. Under some conditions one method is best, and under other conditions another, and careful thought before the of Pattern Outliu*/. 95 drawing is begun will show which is best for the pur- pose in hand. The list of problems and demonstrations "in the rhaptt-r which follows is believed to be so comprehen- sive that therein will be found a parallel to almost any- thing that may be required of the pattern cutter, and it is believed that he will have no difficulty in applying them to his wants. CHAPTER VI. Every effort has been put forth in the preceding chapters of this book to prepare the student for the all important work which is to follow viz., the solution of pattern problems. It is always advisable in the study of any subject to be well grounded in its funda- mental principles. For this reason a chapter on Linear Drawing has been prepared to meet the requirements of the student in pattern cutting, which is preceded by a description of drawing materials and followed bv a solution of the geometrical problems of most frequent occurrence in his work. But the most important chapter is the one immediately preceding this, in which the theory of pattern cutting is explained, and which, if thoroughly understood, will render easy the solution of any problem tlic student may chance to meet. The selection of problems here presented is made sufficiently large and varied in character to anticipate, so far as possible, the entire wants of the pattern cut- ter, and the problems are so arranged as to be con- venient for reference by those who make use of this part of the book without previous study of the other chapters. Tn the demonstrations, onlv the scientific phase of the subject will be considered; consequently, all al- lowances for seams, joints, etc., as well as determining where joints shall be made, arc at the discretion of the workman. In some of the problems it has been necessary to assume a place for a joint, but if the joint is required at a place other than where shown, the method of procedure would be slightly varied while the principle involved would remain the same. Each demonstration will be complete in itself. Al- though references to other problems, principles, etc., will be made where such references will be of.advnn- tage to the student. As stated in the preceding chapter, the problems will be classed under three different heads according to the forms which thev embodv viz. : First, Parallel Forms; Second. Regular Tapering Forms, and Third, Irregular Forms. SECTION 1. (MITER CUTTING). The problems given in this section are such as occur in joining moldings, pipes and all regular con- tinuous forms at. any angle and against any other form or surface, and in fact include everything that may legitimately be termed Miter Cutting. In the problems of this class two conditions exist, which depend upon the nature of the work. Accord- ing to t\\e first, a simple elevation or plan of the inter- secting parts shows the miter line in connection with Uie profile, which is all that is necessary to begin at once with the work of laving out the patterns. It frequently happens, however, that moldings are brought obliquely against sloping or curved surfaces in such a manner that no view can be drawn in which the miter line will appear as a simple straight line. Hence it becomes necessary to produce by the inter- section of lines a correct elevation of the intersections of the various members of the molding, which when done results in the much sought miter line. Or it may be necessary to develop a correct profile of some oblique member or molding in order to effect a perfect miter. Thus some preliminary drawing must be done before the work of laying out the miter patterns can be prop- erly begun, which constitutes the wr\- lines dropped from the points. An illustration of this occurs in Fig. 280, where it will be seen that no point obtained by the dividing of the profile A' B 1 strikes the point X of the miter line, which is absolutely necessary to the shape of the pattern. Therefore, after spacing the profile, a line is drawn from X back to A 1 B', foring the point No. 6. In turn this point is transferred to the stretchout O P, also marked 6, from which a measur- ing line is drawn in the same manner as through the other points in the stretchout, upon which a point from X is dropped, as shown by X'. In actual practice such expedients as this must be resorted to in almost every case, because usually there is less correspondence be- tween the members of dissimilar profiles, between which a miter is required, than in the illustration here given. By this means profiles, however unlike, can be joined. PROBLEM 8. A Butt Miter Against an Irregular or Molded Surface. Let B A in Fig. 281 be the profile of a cornice, against which a molding of the profile, shown by G H, is to miter, the latter meeting it at an inclination, as indi- Fig. 81- -4 Butt Miter arjainst an Irregular or Molded Surface. cated by C D. Construct an elevation of the oblique molding, as shown by C D F E, in line with which draw the profile G II. Divide & H in the usual man- ner into any convenient number of parts, and through the points draw lines parallel to the lines of the in- clined molding, cutting the profile B A, all as indicated by the dotted lines. At right angles to the lines of the molding, of which a pattern is sought, lay off a stretch- out, M N, in the usual manner, through the points in which draw measuring lines. Place the T-square at right angles to the lines of the inclined molding, or, what is the same, parallel to the stretchout line, and, bringing it against the points of intersection formed by the lines drawn from the profile G II across the profile B A, cut the corresponding measuring lines. In the event of any angles or points occurring in the profile B A which are not met by lines drawn from the points in G H, additional lines from these points must be drawn, cutting the profile G H, in order to establish corresponding points in the stretchout. Thus the points 3 and 13 in the profile G II arc inserted after spacing the profile, as described in Problem 7, because the points with which they correspond in the profile B E are angles which must be clearly indicated in the pattern to be cut. Having thus cut the measuring lines corresponding to the points in the profile B A, draw a line through the points of intersection, as shown by O P. Then O P will be the shape of the pattern of the incline cornice to miter against the profile A B. 104 77;e New Metal Worker Pattern Book. PROBLEM 9. The Pattern of a Rectangular Flaring: Article. In Fig. 282, let C A B E be the side elevation of the article, of which F I K M is the plan at the base and G II L N the plan at the top. Let it be required to produce the pattern in one piece, the top included. Make H 1 L 1 N 1 G 1 in all respects equal to II L N G of the plan. Through the center of it likewise draw E P indefinitely, and through the center in the opposite direction draw S indefinitely. From the lines H 1 L 1 and G 1 N 1 set off T and W S respectively, each in length equal to the slant hight of the article, as shown by C A or E B of the elevation. Through O and S respectively draw I 1 K 1 and F 1 M 1 , parallel to H 1 L 1 and G 1 N 1 , and in length equal to the corresponding sides in the plan I K and F M, placing one-half that length each way from the points and S. In like manner set off V P and U E, also equal to C A, and draw through E and P the lines F" I' and K'' M", parallel to the ends of the pattern of the top part as already drawn, and in length equal to I F and K M of the plan. Draw I' H 1 , K 1 L 1 , K' L', M 3 N 1 , M 1 N', F 1 G 1 , F 1 G 1 and I' H 1 , thus completing the pattern sought. In the same general way the pattern may be described, including the bottom instead of the top, if it be required that way. Considering this problem in the light of miter cut- ting proper, I H G F and F G N M may be regarded as the plan of two similar moldings of which A C is the profile, I H, G F and N M being the miter lines. T is the stretchout line, drawn at right angles to F M, while I 1 K 1 and H 1 L' are the measuring lines representing respectively the points C and A of the profile. The points F and G are then dropped into their respective measuring lines, thus locating the points I' LLE AT ON. PATTERN. Fig. S82. The Pattern of a Rectangular Flaring Article. and H' at one end of the pattern, while points K 1 and L 1 are derived from M and N at the other end. PROBLEM 10. Patterns of the Face and Side of a Plain Tapering Keystone. Let A B D C in Fig. 283 be the elevation of the face of a keystone, and G E' F* K of Fig. 284 a sec- tion of the same on its center line. Sometimes problems occur which are so simple that it is not apparent that their solution is an ex- emplification of any rule. That this, with others in which plain surfaces form the largest factors, may be so designated, will be sufficient excuse for a brief ref- erence to first principles. This problem is generally referred to as finding the " true face " of the keystone, because, the face being inclined, the elevation ABC D does not represent the ' ' true face " or " true ' ' dimensions of the face. To state the case, then, in con- formity with the rule, A B and C D are the upper and Pattern Probkms. 105 lower lines of a molding, of which E a F* of Fig. 284 is the profile, and A G and B D are the surfaces against Fig. 284. Patterns of the Face and Side of a Plain Tapering Keystone. which it miters, or the miter lines. Therefore, to lay out the pattern, draw any line, as E' F 1 , at right angles to A B for a stretchout line, upon which lay off the stretchout taken from the profile E' F', Fig. 284, which in this case consists of only one space, as shown by E' F' ; through the points E' and F 1 draw the hori- zontal lines A' B' and C' D', which are none other than the measuring- lines. Then, with the T-square placed parallel with the stretchout line, drop the points from the miter lines A C and B D into lines of correspond- ing letter, which connect, as shown by A 1 C 1 and B 1 D 1 , which completes the pattern. In developing the pattern for the side, W G and F 3 K are the lines of the molding, B D of Fig. 283 its profile and E" F a the miter line. Hence upon any vertical line, as L K', lay off the stretchout of profile B D, locating the points M' and H*, all as shown by L M H 1 K 1 , through which points draw the measuring lines; then, with the T-square placed parallel to L K', drop the points E 1 and F J into lines of corresponding letter, as shown by E 3 F'. As the vertical lines at Gr- and K represent the position of surfaces against which the side is required to fit at the back, bring the T-square against each, thus locating them in the pattern at G 1 and K', as shown. As the side must also fit over the molding of the arch an opening must be cut in it corresponding in shape to the profile of the arch molding N, which is given in the sectional view. It is therefore only necessary to transfer this profile to the pattern, placing the top at the measuring line M and the bottom at the measuring line H', all as shown at N 1 . PROBLEM II. Patterns for the Corner Piece of a Mansard Roof, Embodying: the Principles Upon Which All Mansard Finishes are Developed. One of the first steps in developing the patterns for trimming the angles of a mansard roof is to obtain a representation of the true face of the roof. In other words, inasmuch as the surface of the roof has a slant equal to that shown in the profile of the return, the length of the hip is other than is shown in the eleva- tion, and this difference in dimensions extends in a pro- portionate degree to the lines of the various parts form- ing the finish. Not only are the vertical and oblique dimensions different, but, as the result of this, the angle at A is different from that shown in a normal ele- vation. Hence, it is of the greatest importance to ob- tain a " true face " or elevation of the roof as it would appear if swung into a vertical position, which may be accomplished as follows : In Fig. 285, let A E F C be the elevation of a mansard roof as ordinarily drawn, and let A 1 G be the profile showing the pitch drawn in line with the eleva- tion. Set the dividers to the length A 1 G, and from A 1 as center, strike the arc G G 1 , letting G' fall in a vertical line from A 1 . From G' draw a line parallel to the face of the elevation, as shown by G 1 C 1 , and from the several points in the hip finish, as shown by C and K, drop lines vertically, cutting G 1 C 1 in the points C 1 and K', as shown. From these points carry lines to corresponding points in the upper line of the 106 The New Metal Worker Pattern Book. elevation, as shown by C' A and K' h. Then A C' F 1 E represents the pattern of the surface shown by A C F E of the elevation. In cases where the whole hight of the roof cannot be put into the drawing for use, as B' draw the horizontal line, as shown by B 1 B 3 , and from B drop a vertical line cutting this line, as shown, in the point B s . By inspection of the engraving it will be seen that the point B 3 falls in the line A C' obtained Elevation Fig. $85. The Plain Surfaces of a Mansard Roof Devetoped. above described, the same result may be accomplished by assuming any point as far from A as the size of the drawing will permit, as B, and treating the part between A and B as though it were the whole. That is, from A, in a vertical line, set off AB 1 , equal to A B. From in the previous operation, thus demonstrating that the latter method of obtaining the angle by which to pro- portion the several parts results the same as the method first described, and therefore may be used when more convenient. PROBLEM 12. A Face Miter at Right Angles, as in the Molding Around a Panel. In Fig. 286, let A B D C represent any panel, around which a molding is to be carried of the profile at E and E'. The miters required in this case are of the nature commonly called " face " miters, to dis- tinguish them from other square miters, which can only be shown in a plan view. A correct elevation of the panel A B D C, with the lines of the molding carried around the same, determines the miter lines A F and G C, which, in connection with the profiles at E and E' are all that is necessary to the development of the pat- Pattern Problems. 107 tern. The two profiles are here drawn, thus constitut- ing an entire section of tin- panel. Keeause it is usual, for constructive reasons, to cut the two moldings with the intervening panel in one piece where the width of V.-M+H 1 -X4-I-+ Fig. 286. A Face Miter at Right Angles, us in the Molding Around a Panel. the metal will permit it. Divide the two profiles in the usual manner into the same number of parts, from which points draw lines parallel to the lines of the molding, cutting the miter lines, as shown. For the pattern o[ tin- side corresponding to A B lay off a stretchout at right angles to it, as shown by II K, through which draw measuring lines in the usual man- ner. Place the T-square at right angles to A B, or, what is the same, parallel to the stretchout line H K, and, bringing it successively against the several points in the miter line A F, cut measuring lines of corre- sponding number. Then a line traced through these points, as shown by L M, will be the pattern sought. The other pattern is developed in like manner. It is usual to draw the stretchout lines, K II and K 1 H 1 across the lines of the moldings which they represent, beginning the stretchouts at the inner lines of the mold- ing, thus : Point 10 of profile E would be located at V, while point 10 of profde E 1 would be at W. While this is apt to produce some ^confusion of lines in actual practice, it gives the entire profile in one continuous stretchout for the purpose alluded to above that of cutting the entire width of the panel in one piece. Should it be desired to make one of the moldings separate from the rest, an additional point for the pur- pose of a lap is assumed at one of the moldings, as 11 of profile E'. The pattern for the end piece, A C, may be derived without drawing an additional profile, as its profile and stretchout are necessarily the same as that of the other two arms ; therefore reproduce H K on a line at right angles to A C, as shown by N O, through the points in which draw measuring lines in the usual manner, producing them sufficiently far in each direction to intercept lines dropped from the points in the two miter lines. Place the "["-square at right angles to A C, and, bringing it successively against points already in A F and C G, cut measuring lines of corresponding numbers. Then lines traced through the intersections thus formed, as shown by P R and S T, will be the shape of the pattern of the end piece. It may be noticed in the last operation that drop- ping the points from either of the miter lines, as A F, into the measuring lines is, in fact, only continuing in the same direction the lines previously drawn from the profile E to the line A F; and that in reality the shape of the cut at P R is developed without the assistance of the miter line, thus giving another instance of the fact that any square miter can be cut by the short method when the relation of the parts is understood. 108 The New Metal Worker Pattern Book. PROBLEM 13. The Patterns of the Moldings Bounding; a Panel Triangular in Shape. In Fig. 287, let D E F be the elevation of a triangular panel or other article, surrounding which is a molding of the profile, shown at G andG 1 . Construct an elevation of the panel molds, as shown by ABC, and draw the miter lines A D, B E and C F. For the patterns of the several sides proceed as follows : Draw a profile, Or, placing it in correct relative position to the side D F, as shown. Divide it into any con- of the three sides, at convenient points, draw stretch- out lines, as shown by H I, H 1 I 1 and IP P, through the points in which draw the usual measuring lines. With the T-square parallel to each of the several stretchout lines, or, what is the same, at right angles to the respective sides, bringing the blade successively against the points in the several miter lines, cut the corresponding measuring lines, all as indicated by the Fig. tS7.The Patterns of the Moldings Bounding a Triangular Panel. venient number of parts in the usual manner, and through these points draw lines, as shown, cutting the miter lines F C and A D. In like manner place the profile G' in a corresponding position relative to the side E F. Divide it into the same number of parts, and draw lines intersecting those drawn from the first profile in the line F C, also cutting the line E B. By this operation points are obtained in the three miter lines A D, E B, F C, from which to lay off the patterns in the usual manner. At right angles to each dotted lines. Then lines traced through the points of intersection thus obtained will describe the patterns required. A' C' F' D 1 will be the pattern for the side A D F C of the elevation, and likewise C' B a E' F 3 is the pattern for the side described by similar letters. Placing another profile in the molding A B D E would, if divided the same as the others, only result in another set of intersections at the points already existing on- the lines A D and B E, as occurred on Pattern Problems. 109 the line F C, hence to save labor one profile in this case is all that is really necessary, the points being carried around from that and dropped into the three stretchouts respectively. PROBLEM 14. The Patterns of a Molding Mitering Around an Irregular Four-Sided Figure. In Fig. 288, let A B C D be the elevation of an ir- regular four-sided figure, to which a molding is to be fitted of the profile shown by K. Place a duplicate profile against the side opposite, as shown, from which in these several stretchouts draw measuring lines in the usual manner, producing them until they are equal in length to the respective sides, the pattern of which is to be cut. Placing the T-square at right angles to the Fig. tSS.The Patterns of a Molding Mitering Around an Irregular Four-Sided Figure. project the lines necessary to complete the elevation of the molding as it would appear when finished, all as shown by B F G H. Draw the several miter lines B F C G, D H and A E. Divide the two profiles into the same number of parts in the usual njanner, through the points in which draw lines parallel to the lines of the molding in which they occur, cutting the miter lines, as shown. At right angles to each of the several sides lav off a stretchout from the profile, as shown by L M, L' M', L' M', L 3 M 3 . Through the several points lines of the several sides, or, what is the same, parallel to the stretchout lines, bring it against the points in the miter lines, cutting the corresponding measuring lines, all as indicated by the dotted lines. Then the lines traced through these points of intersection will give the several patterns required. Thus E 1 II' D' A 1 will be the pattern of the side E H I) A of the eleva- tion; H 3 D" C 1 G' will be the pattern ot the side H D C G ; G a F 1 B' C' that of F B C G ; and F' B J A' E 3 that of the remaining side. 110 The Xcir \\'<>rkur Pattern Book. PROBLEM 15. The Patterns of Simple Gable Miters. In Fig. 289, let 'A B K and 15 K B be the angles of the miters at the foot and peak of a gable. Draw profiles of the required molding in correct relation to both the horizontal and inclined moldings, as shown at H and H', through the angles of which draw the other parallel lines necessary to complete the ele- vation. Their intersection at the base of the gable produces the miter line B C, while the miter line at the top of the gable is a vertical line, because the two sides of the gable, K B and K R, are of the same pitch. The profile II is so placed as also to repre- sent the return at the side at its proper distance from B. Divide the profile 11 in the usual manner into any convenient nuin- A ber of equal parts. Place the T-square parellel to the lines in the horizontal molding, and, bringing it successively against the points in the profile, cut the miter line B C, as shown. At right angles to the lines of the hori/.oiital cornice draw the stretchout E K, through the points in which draw the usual measuring lines, as shown. Reveoe the ^-square, letting the blade lie parallel to the stretchout line E F, and, bringing it against tile several points of the profile H, cut the corresponding measuring lines. Then a line traced through these points of intersection, as shown from G to V, will be the pattern of the end of the horizontal cornice inhering with the return. In like manner, with the T-square in the same position, bring it against the points in the miter line B C, and cut corresponding measuring lines drawn tli rough the same stretchout. Then a line traced through the points of intersection thus obtained, as shown by T U, will be the pattern of the end of the horizontal cornice inhering against the inclined cornice. Divide the pro- file H' into any convenient number of equal parts, all as indicated by the small figures. Through these points draw lines cutting the miter line B C, and also the miter line K L at the top. At right angles to the lines of the raking cornice place a stretchout, E' F 1 , of the profile H 1 , through the points in which draw the usual measuring lines, as shown. Place the f-dqu&ro parallel to this stretchout line, and, bringing it suc- i 144.4 1 Fig. 289. The Patterns of Simple Gable Miters. cessively against the points in B C and K L. cut the corresponding measuring lines, all as indicated by the dotted lines. Through the points thus obtained trace lines, as indicated by M N and O P. Then M N will be the pattern for the bottom of the raking cornice inhering against the horizontal, and O P will be the pattern for the top of the same. The pattern shown at G V will also be the pattern for the return miter- ing with A D of the elevation, it being necessary only to reverse it and to .establish its length. Pattern Problems. PROBLEM 16. ill The Pattern for a Pedestal of Which the Plan is an Equilateral Triangle. should be drawn so as to show one side in profile and the plan placed to correspond with it. Draw the miter lines E and G 0. Divide the profile B D into spaces of convenient size in the usual manner, and number them as shown in the diagram. From the points thus obtained drop lines, cutting E and G O. as shown. Lay off the stretchout N P at right angles to the side E G, and through the points in it draw measuring lines. Place the T-square at right angles to E G, and, bring- ing it successively against the points in the miter lines E and G 0, cut the corresponding measuring lines. A line traced through these points will be the pattern, as shown by H L M K. The principle involved in this and several follow- ing problems is exactlv the same as that of the preced- ing regular and irregular shaped panels. In this case the shape of the article is shown in plan instead of ele- Fig. 290. The Pattern for a Pedestal of which the Plan is an. Equilateral Triangle. Let A B D C in Fig. 290 be the elevation of a pedestal or other article of which the plan is an equi- lateral triangle, as shown by F E G. This elevation vation, and the profile is too large to permit of its being drawn within the plan, as were the profiles of the panel moldings in their elevations. 112 The Xew Metal Worker Pattern Book. PROBLEM 17. The Pattern for a Pedestal Square in Plan. In Fig. 291, let A B D C be the elevation of a pedestal the four sides of which are alike, being in plan as shown by E H G F, Fig. 292. Since the plan is a rectangular figure the miters involved are square miters, or miters forming a joint at 90 degrees. A square miter admits of certain abbreviations, the rea- sons for which are explained in Problem 3, as well as in Chapter V, under the head of Parallel Forms. The abbreviated method which is here illustrated is always used. The plan is introduced only to show the shape of the article, and is not employed directly in cutting Fig. S92.The Plan of Square Pedestal. the pattern. Space the profiles, shown in the eleva- tion by A C and B D, in the usual manner, numbering the points as shown. Set off a stretchout line, L R, at right angles to the base line C D of the pedestal, through the points in which draw measuring lines. Place the T-square parallel to the stretchout line, and, bringing it successively against the points in the two profiles, cut the corresponding lines drawn through the stretchout. A line traced through these points, as shown by L M N K, will be the pattern of a side. Fiy. 291. The Pattern for a Pedestal, Square in Plan. PROBLEM 18. The Patterns for a Vase, the Plan of Which is a Pentagon. In Fig. 293, let S C K T be the elevation of a vase, the plan of which is a pentagon, as shown O C' C a B P. The elevation must be drawn in such a man- ner that one of the sides will be shown in Draw the plan in line and in correspondence with it. Divide the profile into spaces of convenient size in the Pattern Problems. 113 usual manner and number them. Draw the miter lines C' II 1 and C* IF in the plan, and, bringing the T-square successively against the points in the profile, drop lines across these miter lines, as shown by the dotted lines in the engraving. Lay off the stretchout M 1ST at right angles to the piece in the plan which the case of a complicated profile, or one of many dif- ferent members, to drop all the points across one sec- tion of the plan C' H' IF C a would result in confusion. Therefore it is customary, in practice, to treat the pattern in sections, describing each of the several pieces of which it is composed independently of the Fig. 294. Pattern for the Base. Fig. 293. Pattern for the Upper Part. The Patterns for a Vase, the Plan of which is a Pentagon. corresponds to the side shown in profile in the eleva- tion. Through the points in it draw the usual measur- ing lines. Place the T-square parallel to the stretch- out line, and, bringing it against the several points in the miter lines which were dropped from the elevation upon them, cut the corresponding measuring lines drawn through the stretchout. A line traced through the points thus obtained will describe the pattern. In others. In the illustration given the pattern has been divided at the point II, the upper portion being developed from the profile and plan, as above, while the lower part is redrawn in connection with a section of the plan, as shown in Kig. 294. Corresponding letters in each of the views represent the same parts, so that the reader will have no trouble in perceiving just what has been done. Instead oi redrawing a pov- 114 Worker Pattern tion of the elevation and plan, as has been done in this case, sometimes it is considered best to work from one profile rather than to redraw a portion of it, as tha 1 always results in more or less inaccuracy. Therefore, after using the plan and describing a part of the pattern, as shown in the operation explained above. a piece of clean paper is pinned on the board, cover- ing this plan and pattern, upon which a duplicate plan is drawn, from which the second section of the pattern is obtained. (J reat care, however, is neces- sary in redrawing portions of the plan to insure accuracy. PROBLEM 19. The Pattern for a Pedestal, the Plan of Which is a Hexagon. sides, drawn so that one of the sides will be shown in profile. Place the plan below it and corresponding with it. Divide the profile shown in the elevation into any convenient number of spaces in the usual manner, and, to facilitate reference to them, number them as shown. Bring the T-square against the points in the profile and drop lines across one section of the plan, as shown bv II X M. At right angles to this section of the plan layoff the stretchout line .N < >. through the points in which draw the usual meas- uring lines. Place the T' sc l uare parallel to the stretchout line, and, bringing it successively against the points in the miter lines II X and M X, cut the corresponding measuring lines, as indicated by the dotted lines. Then a line traced through the points Fig. 205. The Pattern for a Pedestal, the Plan of which is a Hexagon. In Fig. 295, let C D F E be the elevation of a thus obtained will be the required pattern, as shown by pedestal which it is desired to construct of six equal P S T E. Pattern 115 PROBLEM 20. The Pattern for a Vase, the Plan of Which is a Heptagon. sides will be shown in profile. In line with it draw the plan, placing it so that it shall correspond with the elevation. Spaee the profile L P in the usual manner, and from the points in it drop lines crossing one sec- tion of ihe plan, cutting the miter lines It S and II V, as shown. Lay oil' a, stretchout, A H, at riirht anules to the side of the plan corresponding to the side of the vase shown in profile in the elevation. Through the, points in it draw tin; usual measuring lines. Place the "|"-s(|iiare parallel to tins stretchout line, and, bringing it successive] v against the points in the miter lines, cut the corresponding measuring lines, as shown. A line traced through these points, as shown by K O Fig. 296. The Pattern fur a Vase, the Plan nf which is a Heptagon. In Fig. 296, let E L P G be the elevation of the W U, will be the pattern of one of the sides of the vase, constructed in sneh a manner that one of its ; vase. PROBLEM 21. The Patterns for an Octagonal Pedestal. Let K II G W L in Fig. 21)7 be the elevation of a pedestal octagon in plan, of which the pattern of a section is required. This elevation should be drawn in such a manner that one side of it will appear iu profile. Place the plan so as to correspond in all respects with it. Divide the profile G W, from which the plan of the side desired is projected, in the usual manner, and from the points in it drop points upon each of the miter lines F T. and P U in the plan. Lay off a stretchout, B E, at right angles to the side of the plan corresponding to the side of the article shown in profile in the elevation, and through the points in it draw the usual measuring lines. Place the T-square parallel to the stretchout line, and, bringing it -suc- cessively against the points dropped upon the miter lines from the elevation, cut the corresponding measur- ing lines. A line traced through the points thus obtained will describe the pattern of one of the sides of which the article is composed. In eases where the profile is complicated, consisting of many members, and where it is verv long, confusion will arise if all the points are dropped across one section of the plan, as above described. It is also quite desirable in many cases to construct the pattern in several pieces. In 116 The New Mdal Worker Pattern Book, such cases methods which are described in connection the pattern is cut 'by means of a part of the plan with Problem 18 may be used with advantage. ! redrawn above the elevation, thus allowing the use In the present case the pattern is constructed of two of the same profile for both. The same letters refer t Fig. 297. The Patterns for an Octagonal Pedestal. pieces, being divided at the point 8 of the profile. The lower part of the pattern is cut from the plan drawn below the elevation, while the upper part of to similar parts, so that the reader will have no diffi- culty in tracing out the relationship between the dif- ferent views. l'nttt'1-n PnJ'li nis. PROBLEM 22. The Patterns for a Newel Post, the Plan of Which is a Decagon. it; In Fig. 298, let \V US I 1 <> H T V be the eleva- tion of a newel post which is required to be constructed in ten parts. Draw the plan below the elevation, as shown. The elevation must show one of the sections w the plan, as shown by e the elevation of an urn to 1)0 constructed in twelve pieces. The elevation must l>e drawn so as to sliow one side in profile. Con- struct the plan, as shown, 1o correspond with it and the several points in the miter lines X X and OX. cut the corresponding measuring lines. A line traced through the points thus obtained will describe ihe pattern sought. In this illustration is shown a method sometimes resorted to by pattern cutters to avoid the confusion resulting from dropping all tin- points across one section of the plan. The points from 1:5 to 20 inclusive arc dropped upon the line OX. The stretchout C D is drawn in exactly the middle of the pattern that is, it is drawn from X, the central point of the plan. Points are transferred liv the T- s<{uare from () X to the measuring lines on one side of V Plan Fig. SOO.T'he I'ritlrrns for an Urn, the Plan of wltii-h is a Dodecagon. draw the miter lines. ]>ivide the prolile A S (! into spaces in the usual manner, and from the points thus obtained drop lines across one section, X X 0, of the plan. Lav off the stretchout C D at right angles to the side N O of the plan. Place the T-square parallel to the stretchout, and, bringing it successively against the stretchout, the points on the other side being ob- tained by duplicating distances from C Don the several lines. The points 1 to 13 are dropped on N X only. The stretchout E F is laid off at right angles to the side MNfrom the point X, and, the J-^piare being set parallel to E F, the points are transferred to the 119 measuring lines on one side of E F, while the distances scribed in the lirst instance. This plan will be found ad- on the opposite side are set off by measurement, as de- vaiitagcons in complicated and very extended profiles. PROBLEM 24. The Pattern for a Drop Upon the Face of a Bracket. In Figs. I'.ol and '.}o-2, methods of obtaining tlie II K. as shown bv O P, and on P lay off a stretch- return strip lifting around a. drop and mitering against out, through the points in which draw the usual tin 1 face of a bracket an,' shown. Similar letters in measuring lines. From the points in the profile F G the two liguivs represent similar parts, and the follow- I carry lines in the direction of the molding that is, The Pattern for a Drop upon the face of a hracket. ing demonstration may be considered as applying to both. Let A 15 I) C be the elevation of a part of the face of the bracket, and II Iv L a portion of the side, showing the connection between the side strip of the dr.>p E F (1 aiul the face of the bracket. To state the case simply, F G is the profile and 1ST M the miter line, because N M is the outline of the surface against which the side strip miters. Then, following the rule, divide F G into any convenient number of parts in the usual manner, as shown by the small figures. Produce parallel to K M intersecting the face of the bracket N M. .Reverse the T- s quare, placing the blade parallel to the stretchout line O P, and, bringing it suc- cessively against the points in N M, cut the cor- responding measuring lines, as indicated bv the dotted lines. Then a line traced through these several points of intersection, as shown by O R P, will be the pattern of the strip fitting around E F G and mitering against the irregular surface N M of the bracket face. 120 The New Metal Worker Pattern Book. PROBLEM 25. The Pattern of a Boss Fitting: Over a Miter in a Molding". Let A B C in Fig. 303 be the part elevation of a pediment, as in a cornice or window cap, over the miter and against the molding and fascia in which a boss, F K G H, is required to be fitted, all as shown by N E D of the side view. The outline F K G H of the boss is to be con- sidered as the profile of a molding running in the direc- tion shown by D E in the side view, and mitering against the surface of the cornice shown by N E. For the patterns proceed as follows: Divide so much of the profile of the boss K F H G as comes against the cornice, shown from K to F, into any convenient number of parts, and from these points draw lines parallel to D E that is, to the direction of the mold- ing under consideration until they intersect the miter line N O E, which in this case is the profile of the cornice molding. As the boss is so placed over the angle in the cornice molding that the distance from K to F is the same as that from K to G, the part of the boss K G will be an exact duplicate of K F and may be duplicated from the pattern of K F without another side view drawn especially for it, which would have to be done if the boss was otherwise placed. Therefore, extend the line N D upon which to lay off a stretchout of K F H G, dividing the portion K 1 F 1 into the spaces shown at K F of the. profile, through which draw the usual measuring lines. Make the por- tion F 1 G 1 equal in length to the part FUG and, lastly, the portion G 1 K" a duplicate of F 1 K' reversed, as shown. Place the T-square parallel to the stretch- out line K' K", and, bringing it against the several points in N 0, cut corresponding measuring lines, as Pig. SOS. The Pattern of a Boss Fitting Over a Miter in a Molding. shown. Then lines traced through these points of in- tersection, as shown by K 1 L M K s , will be the re- quired pattern. PROBLEM 26. The Patterns for a Keystone Having: a Molded Face With Sink. In Fig. 304, let E A B F be the front elevation of a keystone, as for a window cap, of which K L M P S R is a sectional view, giving the profile of the mold- ing M N O P, over which it is required to fit. The sink in the face extends throughout its entire length, and is shown by G H D C, its depth being shown by the line K T of the section. E F H G and A B D C thus become moldings, of which E A and F B are the parallel lines, E F, G H, C D and A B the miter lines, and K R the profile. Likewise C G H D becomes a molding, of which G H and C D are the miter lines and K T the profile. Therefore, to obtain the pattern of the face pieces, divide the profile of the face K R into any convenient number of spaces, and from the points thus obtained carry lines across the face of the keystone, as shown. At right angles in the top of the keystone lay off a stretchout of K R, as shown by K" R 1 , through which draw the usual measuring lines. Placing the T-square parallel to the stretchout line, and bringing it successively against the points in the lines C D and A B bounding the face strip, cut the corresponding measuring lines. Then a line traced through these points, as shown by C" A" B" D 1 , will be the pattern for this part. In developing the pattern for the sink the usual method would be to divide K T into equal spaces, carrying lines across the face, and thence into the stretchout; but since this would result in confusion of Pattern Problems. 121 lines, the same points as were established in K R have been used, which arc quite as convenient as the others mentioned, save that the points in K T must be ob- tained from the points in K R, bv oiirrviiig lines back to K T, as shown, and in laying off the stretchout each individual space must be measured bv the dividers. K* K S' Fig. $04. The Patterns for a Keystone Havinij a Molded Face with Sink. At right angles to the line II D of the keystone lay off a stretchout of K T, as shown by K 1 T', through the points in which draw the usual measuring lines. Place the T-square at right angles to the lines across the face of the keystone, and, bringing it successively against the points in the lines G H and C D, forming the sides of the sink, cut the corresponding measuring lines drawn through K' T'. Then lines traced through these points, as indicated by G 1 H 1 and C" I) 1 , will form the pattern of the required sink piece. For the pattern of the piece forming the sides of the sink in the face of the keystone, K R T becomes the elevation of a molding running in the direction of R T, of which K R and K Tare the miter lines and C 1) the profile. Hence, at any convenient place above or below the sectional view, lay off the stretchout of the line C D, as determined by the lines drawn across it, in the first operation, all as indicated by C a D". Through the points in ('" D" draw measuring lines in the usual man- ner. The next operation, in course, would be to drop lines from the points in the profile to the miter lines; but as this has already been done by the lines of the first operation, it is only necessary to place the T-- square at right angles to the measuring lines, and bring it successively against the several points in the lines K R and K T, and cut the corresponding measur- ing lines, as shown. Then a line traced through these points, as indicated by K 3 R' and K 3 T', will be the pattern of the piece required. For the side of the keystone, K L S R becomes the face of a molding, of which A B is the profile and K R the miter line at one side, and L M and P S the miter lines at the other. From this point forward the problem is, in principle, the same as Problem 10. For convenience, and to avoid confusion, it is best to again make use of the same set of lines instituted in the first part of the demonstration. Therefore, lay off the stretchout A' B' equal to A B, putting into it all the points occurring in A B, through which draw measur- ing lines in the usual manner. Place the T-square at right angles to these measuring lines, and, bringing it successively against the points in the line K R, and likewise against L M and P S of the back, cut corre- sponding measuring lines, as shown. Then a line traced through these points of intersection, as shown by N' M 1 L 1 K 4 R 2 S 1 P' 0', will be the outline of the required pattern, with the exception of that part lying between N' and O', which make a duplicate of N O. By examination of the points in A 1 B 1 and the lines drawn through the same and making comparison with the points in A B, it will be seen that in order to locate accurately the position of the profile of the window cap molding M N O P, two additional points, as shown by a; 1 and '/', have been introduced, corresponding to x and y, the points of intersection between the extreme lines of the cap molding itself and the side of the keystone A B, as shown in the elevation by the curved lines of that molding. In practice it is frequently necessary to introduce extra points in operations of this character. 122 The New Mdul Worker Pattern Book. PROBLEM 27. The Pattern of a Square Shaft to Fit Against a Sphere. In Fig. 305, let H A A 1 K be the elevation of a square shaft, one end of which is required to fit against the ball D F E. Draw the center line FL, upon which locate the center of the ball G. Continue the sides of the shaft across the line of the circumference of the ball indefinitely. From the points of intersection be- tween the sides of the shaft and the circumference of the ball, A or A 1 , draw a line at right angles to the sides of the shaft, across the ball, cutting the center line, as shown at B. Set the dividers to G B as radius, and from G as center, describe the arc C C 1 , cutting the sides in the points C and C 1 . Then II C C 1 K will be the pattern of one side of a square shaft to fit against the given ball. Fi(j. SOS. The Pattern of a Square Shaft to Fit Aijainst a Sphere. PROBLEM 28. To Describe the Pattern of an Octagon Shaft to Fit Against a Ball. Let H F K in Fig. 300 be the given ball, of which G is the center. Let D' C" C 3 D" E represent a plan of the octagon shaft which is required to fit against the ball. Draw this plan in line with the center of the ball, as indicated by F E. From the angles of tlte plan project lines upward, cutting the circle and con- stituting the elevation of the shaft. From the point A or A 1 , where the side in profile cuts the circle, draw a line at right angles to the center line of the ball F E, cutting it in the point B, as shown. Through B, from the center G by which the circle of the ball was struck, describe an arc, cutting the two lines drawn from the inner angles C 2 C 3 of the plan, as shown at C and C 1 . Then M C C 1 N will be the pattern of one side of ;.a octagon shaft inhering against the given ball H F K. If it be desired to complete the eleva- tion of the shaft meeting the ball, it may be done by carrying lines from C and C' horizontally until they meet the outer line of the shaft in the points I);md D 1 . Connect C 1 and D 1 , also C and D, by a curved line, the lowest point in which shall touch the hori- zontal line drawn through B. Then the broken line D C C 1 D 1 will be the miter line in elevation formed by the junction of the octagonal shaft with the ball. Fig. S06.The Pattern of n.i (M.iij.-, Shaft to Fit Against n Ball. 123 PROBLEM 29. The Patterns of an Octagonal Shaft, the Profile of Which is Curved, Fitting over the Ridge of a Rool. In Fig. 3(>S. The shaft is octagon throughout, anil if it were designed to stand upon a level surface, the method of obtaining its patterns would be the same in all respects as that de- -M 1 Pig. 3(17. Plan, Elevation and Patterns. The Patterns of an Octar/imtil Shaft, Curved in Profile, Fitting nri'r tt Rtdyp. scribed in Problem '1\ . As shown by the line K / K, however, its lower end is designed to tit over the ridge of a roof or gable, to obtain the patterns of which pro- ceed as follows : Construct a plan of the shaft at its largest sec- tion, as shown bv A B I) K K. from the center of which draw miter lines, as shown by <1 B'and G F. Divide the profile of the shaft .1 L. corresponding to I' 'i K of the plan, into any number of parts in the usual manner, and from these points carrv lines verti- cally crossing the miter lines (! K ami and K I 1 ' in the plan. Two others, indicated in the plan by C 1) and II 1, will be cut to lit over the ridge of the roof, as shown in the elevation by // /// a. The remaining four pieces, shown in plan bv B 0, D E, F I and A II, will be cut obliquely to fit against the pitch of the roof, as shown bv n o in the elevation. For the sides D and II I, shown in the center of the elevation, it will be seen that the line drawn from 4 touches the ridge in the point m, while the line drawn from:! corresponds to the point at which the side terminates against the pitch of the roof. Therefore, in the pattern draw a line from the center of it, on the measuring line 4, to the sides of it, on the measuring line :>, all as shown by m' n' and m' n'. Then these are the lines of cut in the pattern corre- sponding to in a and /// //of the elevation. By further inspection of the' elevation, it will be seen that for the remaining four sides it is necessary to make a cut in the pattern from one side, in a point corresponding to :? of the profile, to the other, in a point corresponding to 1 of the profile, all as shown by ////. Taking corre- sponding points, therefore, in the measuring lines of the pattern, draw the lines it,' o', as shown. Then the original pattern, modified by cutting upon these lines, will constitute the pattern for the four octagon sides. 124 'ihe New Metal Worker Pattern Book. PROBLEM 30. To Construct a Ball in any Number of Pieces, of the Shape of Gores. Draw a circle of a size corresponding to the re- quired ball, as shown in Fig. 309, which divide, by any of the usual methods employed in the construction of polygons, into the number of parts of which it is desired to construct the ball, in this case twelve, all as shown by E, F, Gr, H, etc. From the center draw radial lines, R E and R F, etc., representing the joints between the gores, or otherwise the miter lines. If the polygon is inscribed, as shown in the illustration, it will be observed that the joint or miter lines will lie in the surface of the sphere and that therefore the middle of the pieces, as shown at W, C and ', will fall inside the surface of the sphere a greater or less distance according to the number of gores into which the. sphere has been divided, and that therefore it becomes necessary to construct a section through the middle of one of the sides for use as a profile from which to obtain a stretchout. It will be well to distinguish here between absolute accuracy and something that will do practically just as well and save much labor. This profile, if made complete, would have for its width the distance W u\ while its hight or distance through from R to a point opposite would be equal to the diameter of the circle, or twice the distance R U. As one-quarter of this section will answer every pur- pose, it may be constructed with sufficient accuracy as follows : Supposing R E F to be the piece under con- sideration, draw a line parallel to its center line R C conveniently near, as A V 1 , upon which locate the points A and V by projection from C and R, as shown by the dotted lines. From the point V erect the line B V perpendicular to V A, and make B V equal to the radius of the circle, or R V ; then an arc of a circle cutting the points B and A will complete the section. This can be done by taking the radius R U between the points of the compasses and describ- ing an arc from the point V, whose distance from V is equal to the distance n' U. To develop the pat- tern divide B A into any convenient number of equal parts, and from the divisions thus obtained carry lines across the section P] R F at right angles to a line drawn through its center, and cutting its miter lines, all as shown in R E and R F. Prolong the center line R C, as shown by S T, and on it lay off a stretchout obtained from B A, through the points in which draw measuring lines in the usual manner. Place the T-square parallel to the stretchout line, and, bringing it successively against ;lie points in the miter lines R E and R F, cut the corresponding measuring lines, as shown. A line traced through these points will give the pattern of a section. If, on laying out the plan of the ball, the polygon had been drawn about the circle, instead of inscribed, as shown in the engraving, it is quite evident that a quarter of Fig. 309. To Construct a Ball in any Number of Pieces, of the Shape of Gores. the circle would have answered the purpose of a profile. These points, with reference to the profile, are to be observed in determining the size of the ball. In the illustration presented, the ball produced will corre- spond in its miter lines to the diameter of the circle laid down, while if measured on lines drawn through the center of its sections it will be smaller than the circle. The patterns for a ball made up of zones or strips having parallel sides will be found in Section "2 of this chapter (Regular Tapering Forms). Pattern Problems. PROBLEM 31. The Pattern of a Round Pipe to Fit Against a Roof of One Inclination. 125 /TT^K ' ^ D 7 \ Is"" J^ i~ 1 E 1 1 | j 1 1 i * i 5 ( ; ' \ 1 1 1 1 1 1 A | ^ N ^~i L Ir^tf-i t-t \ rX y s 1 |_ 1 L/"3 \ / / \/ ^ \ z. 1 7T /*4 \ " T~"yr ^ ^^ -^ '" - -R y "** fv 7, 6 In Fig. 310, let A B be the pitch of the roof and C F D E the profile of the pipe which is to miter against it. Let G P H be the elevation of the pipe as required. Draw the profile in line with the eleva- tion, as shown by C F D E, and divide it into any con- venient number of equal parts. Lay off a stretchout in the usual manner, at right angles to and opposite the end of the pipe, as shown by I K, and draw the measuring Tines. Place the T-square parallel to the sides of the pipe, and, bringing it successively against the divisions of the profile, cut the pitch line, as shown by A B. Reverse the T-square, placing it at right angles to the pipe, and, bringing it successively against the points in A B, cut the corresponding measuring lines. A line traced through the points thus obtained, as shown by L M N, will finish the pattern. Fiij. 310 The Pattern of a Round Pipe to fit. Against a Roof of ' One Inclination. PROBLEM 32. The Pattern of an Elliptical Pipe to Fit Against a Roof of One Inclination. In Fig. :.l 1, let N C D O be the elevation of an elliptical pipe litting against a roof, represented by A 15. Let K F G Q be the section or profile of the pipe. Draw the profile in convenient proximity to the elevation, as shown, and divide it into any convenient number of equal parts. Place the T-square parallel to the sides of the pipe, and, bringing it against the points in the profile, drop lines cutting the roof line A B, as through the points in it draw measuring lines in the usual manner. Reverse the T- S( l uare j placing it at Fig. 311. The Pattern of an Elliptical Pipe to Fit Against a Roof of One Inclination. shown. Opposite to the end of the pipe, and at right right angles to the pipe, and, bringing it successively angles to it, lay off a stretchout, as shown by H I, and ; against the points in A B, cut the corresponding 126 Tin- Ne 1'iiti'i-n Ik HI/.-. measuring lines, as indicated. A line traced through tin- sliurl diameter Q I'' crossing tin- roof. In such ca.-e these points, as shown by l\ I. M . will lie ilie rei|iiirc(l ilie prolile should lie turned sn that Q F is across the pattern. In the illustration the long diameter of the roof, or jiarallel to D, and the elevation duly pro- ellipse, or E G, is shown as crossing the roof. The jeete.l from it. The pipe migbl with equal facility be method of procedure would be exactly the same if the placed that the long diameter should lie at an pipe were placed iu the opposite position that is. with oblique angle desired. PROBLEM 33. The Pattern of an Octagon Shaft Fitting Over the Ridge of a Roof. In Fig. 312, let A B C be the section and I) II G. representing the ridge of the roof, cut the corre- G I E the elevation of an octagon shaft mitering sponding measuring lines. Then a line traced through against a roof, represented by the lines F f the in the engraving, will be the lower end of the pattern angle lines in the elevation, and also against the point ' required. /'illti I'll I'rnlil, ///.v. 127 PROBLEM 34 The Pattern of a Round Pipe to Fit Over the Ridge of a Roof. Let A I> (J iii Kin'. :i:; lie \\ section of tin 1 roof and 1> S 15 T 1'] an elevation of tin.' pipe. 1'raw a profile of the pipe 111 hue, as shown hv !' (j \\. Since, both iiielinations of the roof are. to the same angle, -\K I Kit I fore againsl one slope of the roof, and lay off the stretch- out of the same upon the stretchout line I K, drawn at right angles to the lines of the pipe, which may lie duplicated in a reverse order for the other half, as .shown. Draw measuring lines through these points in the usual manner. Place the T-square parallel to the sides of the pipe, and, bringing it against the points in the prolile, cut the roof line, as shown from B to T. Reverse the T-square, placing it at right angles to the lines of the pipe, and, bringing it successively against the points dropped upon the roof line, cut the corresponding measuring /Y,/. Jl3.-The Pattern ../ .1 /.'..ii.irf Pipe t,, Fit Oner the J}i,(,,e ,./ .1 l!,,of. through the points, as shown by L M N O halves of the pattern will be the same. There- i P, will form that end of the pattern which meets space oil' the half of the profile which miters the roof. PROBLEM 35 An Octagon Shaft Mitering Upon the Ridge and Hips of a Roof. In Fig. :> 14 are shown the front and side elevations of a hipped roof, below which are placed plans, each turned so as to correspond with the elevation above it. Before the pattern of the shaft can be developed it will be necessary to obtain a correct elevation of its inter- section with the roof. Therefore, considering the plan l ~s~ E 2 3 45C 7R!) 10111 1 \ p^ ,TTE R 4 i \ y ~T x / L \ / x / 0-7 /\ 37 9~!~ uf- 1 f. ' S S- ">< y r FRONT ELEVATIpN ' Fir. 314. The Pattern of an Octagon Shaft Fitting Ortr the Ridije and Hips nf a Rnnf. Aii octagonal shaft is required to be mitered down upon this roof, so that its center line or axis shall inter- sect ihe apex of the roof C, as shown upon the plans. of the shaft as the prolile of a molding, number 'all its points in both plans, beginning. For convenience, at the ridge of the roof, and including the points where the 128 The New Metal Worker Pattern Book. oblique sides cross the hips of the roof, as shown by the small figures 1 to 11. The next step is to project lines upward into the elevations from each of these points, continuing them till they intersect the lines of the roof, as shown by the vertical dotted lines. From each of these intersections in either view lines can be projected horizontally to the other view till they in- tersect with lines of corresponding number. Thus the points 9 and 10 cut the line of the hip in the front ele- vation at the point B, which, being carried across to the side view and intersected with lines from points 9 and 10 from the plan below it, give the correct posi- tion of those points in the side view. In like manner the intersection of lines from points tj and 7 in the side view, with the hip line at D, give the correct hight of those points in the front view. Points 5 and 8, being upon the hips, must appear in the elevations at points where the vertical lines from them cut the hip lines in the elevations. Lines connecting these points (o, 6, 7 and 8, and 8, 9, 10 and 11) will complete the eleva- tions. In case all sides of the roof have the same pitch and the shaft is a regular octagon, all the angles of the shaft except ~2 and 11 will intersect the roof at the same hight, in which case it will only be necessary to draw the front view. But should the slope of the front of the roof be different from that of the sides, it will be necessary to follow the course above described. To develop the pattern, draw any horizontal line, as E F, upon which place the stretchout of the octagon shaft obtained from the plan, as shown bv the small figures, through which draw the usual measuring lines at right angles to it, and intersect the measuring lines with lines of corresponding numbers drawn horizontally from the intersections in the elevation. A line traced through these intersections, as shown by X Y Z, will be the desired pattern. PROBLEM 36. The Pattern of a Flange to Fit Around a Pipe and Over the Ridge of a Roof. In Fig. 315, let A B C be the section of the roof against which the flange is to fit, and let O P S B. be the elevation of the pipe required to pass through the flange. Let the flange in size be required to extend from A to C over the ridge B. Since both sides of the roof are of the same pitch, both halves of the opening from the point B will be the same. Therefore, for convenience in obtaining both halves of the pattern at one operation, the line B C may be continued across the pipe toward A 1 , and used in place of B>A, the dis- tance from B on either line to the side O li being the same. Under these conditions it will be seen that the process of describing the pattern is identical with that in the previous problem. Make B A 1 equal to B A, and proceed in the manner described in the problem just referred to. Divide the profile D E F G into any number of equal parts in the usual manner, and from the points so obtained carry lines vertically to the line A' C, and thence, at right angles to it, indefinitely. Also carry lines in a similar manner from the points A' and C. Draw II L parallel to A' C. Make II I the width of the required flange, and draw I K parallel to H L. Through that part of the flange in which the center of the required opening is desired to be draw the line A* C', crossing the lines drawn from the pro- file. From each side of this line, on the several Fig. SIS. The Pattern of a Plunge to Fit Around a Pipe and Over the Ridge of a Roof. Pattern Problems. measuring lines, set oil the same distance as shown upon the corresponding lines between D F and I) E F, as shown. A line traced through the points thus obtained, as shown by D' E 1 F' G', will be the required opening to fit the pipe. Through the center, across the flange, draw the line N M, which represents the line of bend corresponding to the ridge B of the section of the roof. PROBLEM 37. The Pattern of a Flange to Fit Around a Pipe and Against a Roof of One Inclination. Let L M, Fig. 316, be the inclination of the roof and P R T S.an elevation of the pipe passing through it. N then represents the length of the opening which is to be cut in the flange, the width of which will be the same as the diameter of the pipe. Let A B D C be the size of the flange desired, as it would appear if viewed in plan. Immediately in line with the pipe draw the profile G H I K, putting it in the center of the plan of the flange A B D C, or otherwise, as required. Divide one-half of the profile in the usual manner, and carry lines vertically to the line L M, representing the pitch of the roof, and thence, at right angles to it, indefinitely. Carry points in the same manner from A and B. Draw C' D 1 parallel to L M. Make C 1 A 1 equal to A C, or the width of the required flange, and draw A 1 B 1 parallel to C 1 D 1 . Then C 1 A 1 B 1 D 1 will be the pattern of the "required flange. Draw E 1 F 1 through it at a point corresponding to E F of the plan, crossing the lines drawn from the profile. From E 1 F' set off on each side, on each of the measuring lines crossing it, the width of opening, as measured on corre- sponding lines of the plan, measuring from E F in the plan to the profile. Through the points thus obtained draw a line, which will give the shape of the opening to be cut, all as shown by G 1 H 1 I 1 K 1 . Fig. 316 The Pattern of a Flange to Fit Around a Pipe and Against a Roof of One Inclination. 130 Tlie New Metal Worker Pattern Book. PROBLEM 38. The Pattern for a Two-Piece Elbow. In Fig. 317, let A C B D be the profile of the pipe in which the elbow is to be made. Draw an elevation of the elbow with the two arms at right angles to each other, one of which is projected directly from the pro- file, as shown by E G I H K F. Draw the diagonal line G K, which represents the joint to be made. Divide the profile into any convenient number of equal parts. Place the T-square parallel to the lines of the arm of the elbow, opposite the end of which the profile has been drawn, and, bringing the blade successively against the several pojnts in the profile, drop corre- sponding points on the miter or joint line K G, as shown by the dotted lines. Opposite the end of the same arm, and at right angles to it, lay off a stretchout line, M N, divided in the usual manner, and through the divisions draw measuring lines, as shown. Place the blade of the T-square at right angles to the same arm of the elbow, or, what is the same, parallel to the stretchout line, and, bringing it successively against the points in K G, cut the corresponding measuring lines, as shown. A line traced through these points, as indicated by K P O, together with M N, will form the required pattern. Fig. 317. The Pattern for a Two-Piece Elbow, PROBLEM 39. The Patterns for a Two-Piece Elbow in an Elliptical Pipe. Two Cases. The only difference to be observed in cutting the patterns for elbows in elliptical pipes, as compared with the same operations in connection with round pipes, lies with the profile or section. The section is to be placed in the same position as shown in the rules for cutting elbows in round pipe, but it is to be turned broad or narrow side to the view, as the requirements of the case may be. In Figs. 318 and 319 are shown elevations and profiles of two right angled two-piece elbows in elliptical pipes. In Fig. 318 the broad side of the ellipse is presented to view, while Fig. 319 shows the narrow side, as indicated by the respective positions of the profiles. Although the results in the two cases are different in consequence of the position of the profiles, the method of procedure is exactly the same. Similar parts in the two drawings have been given the same reference letters and figures, so that the following demonstrations will apply equally well to either : Let A C E F D B be the elevation of the elbow and II G K I its section. Draw C D, the miter line. Divide the profile i:; the usual manner, as indicated by the small figures, and by means of the T-square placed parallel lo the "arm, drop points upon the miter line, as shown. Opposite the, Pattern Problems. 131 end of the arm lay oil' a stretchout, M N, and through tin- points in it draw the usual measuring lines. points in the miter line, cut the corresponding meas- uring lines. A line traced through these points, Fig. 318. Fig. 319. A Two-Piece Elbow in Elliptical Pipe. liVverse the T-square, placing it at right angles to the arm, and, bringing it in contact with the several as shown by L P O, will constitute the required miter. PROBLEM 40. The Patterns for a Three-Piece Elbow. In Fig. 320, let E M L I H K N F be the ele- vation of a three-piece elbow. The drawing of a three-piece elbow, at any angle whatever, should be so constructed that the middle section or portion bears the same angle with reference to the two arms. Since the two arms in the present instance are at right angles (90 degrees) to each other, the middle section must therefore be drawn at an angle of 45 degrees to both. Make its diameter the same as that of the two .arms, and draw the miter lines M N and L K. Draw the profile A B C in line with the arm from which the pattern is to be taken, as shown, and divide it into any convenient number of equal parts. Place the blade of the T-square parallel to this arm of the elbow, and, bringing it against the points in the profile, drop cor- responding points upon the miter line L K. At right 132 The Xew Metal Worker Pattern Book. angles to L I draw a stretchout, as R S, through the divisions in which draw measuring lines in the usual manner. Placing the T-square at right angles to L I, and bringing it successively against the points in the miter line L K, cut the corresponding measuring lines, drop like divisions upon M X. At right angles 1<> 1. M lay off a stretchout of the profile A 15 0, as shown by P 0, through the points in which draw measuring lines in the usual manner. Reversing the position of the set-square so that its long side shall come at right Fig. 310. A Three-Plece Elbow. as shown. Then the line T U V, traced through the points thus obtained, forms in connection with S R the pattern of an end section. Place the 45-degree set square against the blade of the T-square so that its oblique or long side shall coincide with the lines of the middle section of the elbow, and, bringing it against the points in L K, angles to M L, or, what is the same, parallel to the stretchout line, bringing it successively against the several points in the miter lines M N and L K, and cut the corresponding measuring lines. Then lines traced through these points, as shown by D X Y and G W Z, will be the pattern of the middle section. PROBLEM 41. The Patterns for a Four-Piece Elbow. In constructing the elevation of a four-piece elbow, first draw the profile A B C, from which pro- ject one of the arras of the elbow, as shown by the lines A F and C G, Fig. 321, At right angles to this lay off the other arm of the elbow. M T. X T. continuing the lines of each until they intersect. Through the points of intersection draw the diagonal line a d. Establish the point a on this diagonal line at con- Pattern Problems. 133 venienee, and from it. draw the lines f e r, as shown, which divide into three equal parts, thus obtaining the points- /and e. Through /' w Fig. SSI. A Four-Piece Elbow. and e, to the center a, draw the lines f a and e , which will represent the centers of the middle sec- tions of the elbow, at right angles to which the sides of the same are to be drawn. Through/, and at right angles to /a, draw L K, meeting M L in the point L, and stopping on the line Mi'tnl \Yoi-L-'' i- I'altfrii JSouk. center lines made use of in dividing the quarter circle I c represents the number of pieces in the elbow. Therefore, to draw an elevation of an elbow in any number of pieces, construct the quadrant ale as above described, then divide b c into such a number of parts that the number of lines drawn to a (including a b and a c) shall equal the number of pieces required. Thus the five lines a 6, a/, a c, a d and a c are the center lines of the five pieces of which the elbow shown in Fig. 322 is constructed. Although the two extreme lines a b and a c are not, strictly speaking, center lines, their relation to the adjacent miter lines is the same as that of the other lines radiating from a. Through/, and at right angles to fa, draw V S, joining the side of the arm E S in the point S, and joining a corre- sponding line drawn through e in the point V. In like manner draw the line T E, representing the inner side of the same section. The remaining sections are to be obtained in the same way. As but one section is necessary for use in cutting the patterns, the others may or may not be drawn, all at the option of the pattern cutter. Draw the miter or joint lines S E, V T, etc. Divide the profile (or one-half of it) in the usual manner. Place the T-square parallel to the lines of the arm, and, bringing the blade against the several points in the profile, drop corresponding points upon the miter line S E. Shift the T-square so that the blade shall be parallel to the part V S E T, and trans- fer the points in S E to V T, as shown. For the pat- tern of the arm, at right angles to it lay off a stretch- out of A B C, as shown by F G, through the points in which draw the usual measuring lines. Place the T-square at right angles to the arm, and, bringing it against the points in E S, cut the corresponding meas- uring lines, as shown. Then a line traced through these points, as shown from H to I, will be the pattern. For the pattern of the piece S V T E prolong the line a/, as shown by L K, upon which lay off a stretchout, through the points in which draw the measuring lines in the usual manner. Placing the T-square at right, angles to S T, or, what is the same, parallel to the stretchout line, bring it against the several points in Fig. SSS A Five-Piece Elbow. the lines E S and T V, and cut the corresponding measuring lines. Then lines traced tli rough the points thus obtained, all as shown by N P M, will be the pattern sought. PROBLEM 43. The Patterns for a Pipe Carried Around a Semicircle by Means of Cross Joints. In Fig. 323, let F E D be the semicircle around which a pipe, of which A C B is a section, is to be carried by means of any suitable number of cross joints, in this instance ten. Divide the semicircle F E D into the same number of equal parts as there are to be joints, which, as just stated, is to be ten, all as shown by D, 0, P, E, S, E, etc., and draw lines from each point to Z. As there are to be ten joints there must necessarily be eleven pieces, therefore, according to the directions given in the previous problem, the semicircle must be divided into such a number of equal parts that the number of lines radiating from Z Pntlvnt 135 shall be eleven, all as shown, each line serving as the center line <>f a piece. From L> toward the center Z set off the diameter of the pipe A B. as shown l>v tlie point A'. From Z as eenter, with the radius Z A 1 , draw the dotted line representing the inner line of the pipe, and cutting the radial lines previously drawn in the points 0', P 1 , etc. Through and 0' draw lines at right angles to Z and continue them in either direction till they intersect with the lines drawn through P and P 1 on the one side and through D and A' on the other. Each pair of lines is to be drawn at corresponding to the full sections composing the body of the pipe. For the pattern of the end section pro- ceed as follows : Divide the profile A C B in the usual manner into anv convenient number of equal parts, and from the points thus obtained carry lines upward at right angles to Z D. cutting T 1 T. Prolong the line Z D. and upon it place a stretchout from the profile A C B, perpendicular to which draw measur- ing lines in the usual manner. With the T-square placed parallel to Z D, and brought successively against the points in T' T, cut the measuring lines of Fig. StS.A Pipe Carried Around a Semicircle by Means of Cross Joints. right angles to its respective radial or center line. Through the points of intersection draw the lines T T 1 , U U', etc., which will represent the lines of the joints or miters. It will appear by inspection that the point U is equidistant from P and O, and that U' is also equidis- tant from P' and 0', and that therefore the lines U U', T T 1 , etc., if continued inward must arrive at the center Z. Thus the joint lines, like the center lines, must radiate from the center of the semicircle. Draw the profile of the pipe A C B directly below and in line with one end of the pipe, all as shown in the engraving. As may be seen by inspection of the diagram, two patterns are required, one corresponding to the half section occurring at the end, and the other corresponding numbers. Then a line traced through the points of intersection thus obtained, as shown by I K L, will be the shape of the miter cut, and G I K L H will be the complete pattern for one of the end pieces. For the pattern of one of the large pieces, as U V V U', lay off a stretchout of A C B upon its center line extended, as shown by M N, and through, the points in it draw measuring lines in the usual man- ner. Place the T-square parallel to U V and, bringing it against the points in U U 1 , cut the line V V. Next place the T' S( l uare parallel to the stretchout line, and, bringing it against the several points in the miter lines U 1 U and V V, cut the corresponding meas- uring lines, all as shown, thus completing the pat- tern. 136 The New Metal Worker Pattern Book. PROBLEM 44. The Patterns for an Elbow at Any Angle. LetDFHKLIGE in Fig. 324 be the eleva- tion, of a pipe in which elbows are required at special angles. In convenient proximity to and in line with Fig. Sij.An Elbow at Any Angle. one end of the pipe draw a profile, as shown by A B C, which divide in the usual manner. Placing the T-square parallel to the first section of the pipe, and, bringing it against the several points in the profile, drop corresponding points upon F G. Shift the J-square, placing it parallel to the second section, and, bringing it against the several points in F G, drop them upon H I. At right angles to the first section lay off a stretchout of A B C, as shown by T U, through the points in which draw the customary measuring lines. Placing the T-square at right angles to this section of the pipe, and bringing it against the several points in F G, cut the corresponding measuring lines. Then the line ELS traced through these points will, with the line T U, be the pattern sought. The pattern for the opposite end is to be obtained in like manner, all as shown by M N P, and therefore need not be described in detail. For the pattern of the middle section lay off a stretchout, W V, at right angles to it, with the cus- tomary, measuring lines. Placing the T-square at r i gh t. angles to the section, bring it successively against the points in G F and I H, and cut the correspond- ing measuring lines, as shown. Then lines traced through these points, as shown by Y X Z Q, will be the pattern sought. The positions of the longi- tudinal joints in the several sections of this elbow, as well as those of all others, are determined by the order in which the measuring lines drawn through the stretchout are numbered. In the present instance the joints are allowed to come on the back of the pipe, or, in other words, upon D F II K, which corresponds to the point 1 in the profile. Hence, in numbering the measuring lines in the several stretch- outs, point 1 is placed at the commencement and ending, while if it were desired to have the joint in either piece come on the opposite side, or at a point corresponding to 9 of the profile, the stretchout would have commenced and ended with that figure, the Pattern Problems. 137 figure 1 in that case coming, in regular order, where 9 now occurs. The effect of such u change upon anv of the patterns here given would be the same as if they were cut in two upon the line 9 and the two halves were transposed. PROBLEM 45. The Patterns for a Bifurcated Pipe, the Two Arms Being the Same Diameter as the Main Pipe, and Leaving: It at the Same Angle. In Fig. 325 is shown an elevation of a bifurcated pipe, all arms being of the same diameter. In this problem, as in many others, it becomes necessary to first make a correct drawing of the intersection of the parts showing the miter lines correctly ; after which of the miter lines will at once be determined. Thus the intersection of the three bisecting lines at E gives the point at which the miter lines starting from the points P, E and K must meet. In line with the upper end of the pipe draw a Fig. MS. The Pattern for a Bifurcated Pipe. the method of laying out of the miter patterns is the same as that employed in several other problems immediately preceding this. If. in this case, each arm of the pipe be divided longitudinally into two equal parts, as shown by the center lines, and each half be considered as a separate molding the correct position profile of it, as shown by A C B. A profile will also be needed in one of the oblique arms, a half only being shown at A' C' B' on account of the limited space. For the pattern of the upper portion of the pipe, divide the profile A C B into any number of equal spaces, and place the stretchout of the same on 138 The New Metal Worker Pattern Book. any line drawn at right angles to S P, as shown by the continuation of S D to the left, and draw the usual measuring lines. Next drop the points from the profile A C B parallel with S P till they cut the miter line PEE; then placing the T-square at right angles to S P, drop the points from the miter line P E into measuring lines of corresponding num- ber. A line traced through these points of intersection, as shown from E" to P', will give the miter cut on the lower end of the pipe S D E P, one-half of which only is shown in the engraving. The pattern for the piece E F J K is obtained in exactly the same manner, and might be obtained, so far as the half indicated by C' B' on profile is concerned, from the original profile, by simply continuing the lines through to the miter line J F, as shown. For simplicity, therefore, the profile A' C' B' is divided into the same number of equal parts as the original profile, and a stretchout of it is placed upon any line, as T U. drawn at right angles to E F. The points are then dropped from the profile both ways, cutting the miter lines K R E and J F, after which, with the T-square placed parallel to T U, they can be dropped into the measuring lines of the stretchout. Lines traced through the points of inter- section will constitute the required pattern, as shown by K' R' E' R" K" X W V. PROBLEM 46. The Patterns for the Top and Bottom of a " Common " Skylight Bar. In Fig. 326, A B represents a portion of the pro- file of the ridge bar, or of the ventilator forming the top finish of a skylight, against which the upper end of a "common" bar is required to miter; and C D represents the profile of the curb or finish against which the lower end of the bar miters. The parallel oblique lines connecting the two show the side eleva- tion of the bar whose profile is shown at E F F 1 . As the profile consists of two symmetrical halves, either half, as E F or E F 1 , may be chosen to work from, and as it contains no curved portions it is simply necessary to number all of its points or angles, and then to place a complete stretchout of the same upon any line drawn at right angles to the lines of the molding, as Gr H, and to draw the usual measuring lines, all as shown. As a properly drawn elevation shows the intersection of the points of the profile with the two miter lines A B and C D 1 , it is only neces- sary to place the T-square parallel to the stretchout lines G H, and bring it successively against the points in A B and C D 1 , and cut corresponding measuring lines, as shown at I J and K L. Straight lines con- necting the points of intersection will complete the pattern, as shown at I J L K. The length of the pattern, which is here shown indefinite, must be determined by a detail drawing, in which the rise M B and the run M D 1 are correctly given. The patterns for the "jack" bar and for the "hip " bar will be given later among those problems in which the development of the miter line and the Fig. 32fi.The Patterns for a " Common " Skylight Bar. raking of the profile are necessary, with which they are properly classed. 1'atkrn I'rMems. PROBLEM 47. 139 The Patterns for a T-Joint Between Pipes of the Same Diameters. Let D F G II M F K ft in Fig. 327 be the eleva- tion of two ]>i])cs of the same size meeting at right Fig. 327. A T- Joint between Pipes of the Same Diameter. angles and forming a T, of which A B C D and A' B' (J 1 D' are profiles drawn in line with either piece. As the two profiles are alike, and as the end of one piece (D E F K) comes against the side of the other piece (G I M H), both halves of D E F K, B A D and B C D, will miter with one-half, B' C 1 D', of the piece G I M H. By projecting the points B and B' from the profiles through their respective elevations the point L is found, which being connected with the points F and K gives the miter lines. Space the profile A B C D into any number of equal parts and lay off the stretch- out 1ST O ut right angles to the pipe of which ABC D is the profile, as shown, through the points in which draw the usual measuring lines. Set the T-square at right angles to this pipe, and, bring- ing the blade against the several points on the rniter lines, cut the corresponding measuring lines drawn through the stretchout, as indicated by the dotted lines. Then N F 1 U V W O will be the pattern for the upper piece. As both halves of this piece (dividing now upon the line A C) will be alike only one-half of the profile (A B C) has been divided, but the stretchout is made complete. For the pattern of the other piece, divide its profile into any convenient number of equal parts and lay off the stretchout on the line R T, drawn at right angles to the pipe. Placing the T-square parallel with the pipe drop points upon the miter lines from that portion of the profile (B 1 C' D') which comes in line with them ; then place the blade of the T-square at right angles to the pipe, and, bringing it against the several points in the miter lines, cut the corresponding measuring lines, as shown by the dotted lines. A line, X Y Q Z, traced through these points will bound the opening to be cut in the pattern for the lower pipe. From the points 1 in the stretchout draw the lines R P and T S, in length equal to the. length of the pipe. Connect P S. Then P R T S will be the required pattern. The seam iji the pipe may be located as shown in the engraving, or at some other point, at pleasure. 140 The New Metal Worker Pattern Book. PROBLEM 48. The Patterns for a Square Pipe Describing a Twist or Compound Curve. As problems of this nature frequently occur in connection with hot air pipes, grain chutes, etc., this problem is given as embodying principles which can often be made use of. The upper opening of the pipe in this case is required to be in a horizontal plane, while the lower opening is in a vertical position and placed at a given distance below and to one side of the top, the pipe describing a quarter turn when viewed from either the top or the front. To more fully illustrate the nature of the prob- lem, a perspective view of it is shown in Fig. 328, in which the pipe is represented as being contained within a cubical shaped solid. The solid, of which the pipe is represented as forming a part, is shown in outline, the pipe itself being shaded to show its form, while upon the front and lower side of the solid are shown in dotted lines the front elevation and plan of the pipe. Thus G F T C represents the front view of a solid just large enough to contain the pipe, in which A B C D shows the position of the lower opening, and A B E FD C shows the curve of the pipe as seen from the front. G II S B 1 is the top of tin- solid in which the upper opening N P S R is situated. The curve of the pipe in plan has been projected to the lower face of the solid by vertical lines, R L, and others not shown, and is shown by C J K L M D. To state the case simply, then, ABE is the profile of the piece of metal forming the top of the pipe, while 1) M L and C J K are the two miter lines, or the plans of the intersecting surfaces, and CDF is the profile of the lower side of the pipe intersecting the same miter lines. The top and bottom pieces being developed, it is only necessary to reverse the operation and con- sider the lines of the plan I) M L and C J K as the profiles of the front and back pieces respectively, while ABE and CDF become the miter lines or elevations of the intersecting surfaces. A part of these operations are carried out in detail in Fig. 339, where the elevation and the plan are drawn directly in line with each other; the various points being represented by the same letters in the two illustrations. For the pattern of the top piece divide its profile A B E by any convenient number of points (1, 2, 3, etc.), from which drop lines ver- tically cutting the two miter lines D' M and C' J of the plan, as shown (the figures of the plan 2 to 11 have no reference to this part of the operation). Upon R 8, drawn at right angles to the direction of the mold, lay off the stretchout of ABE, through which draw the usual measuring lines. With the T-square placed parallel to R 8 and brought against the several points in the two miter lines cut lines of corresponding num- ber; lines traced through the points of intersection, as shown by R T and U S, will give the pattern of the top piece. It will be noticed that owing to the contrary relation of the two curves it is necessary to have the points of the profile occur more frequently near B than E, as otherwise they would intersect the miter Fig. SS8. Perspective View of a Pipe Describing a Twist or Compound Curve. line D' M too far apart near D', while they would occur more frequently than is necessary near M. As there is no curve from A to B of the profile, that part of the pattern from R to S will be a duplicate of the plan view, consequently the curve from R to the measuring line drawn from S may be traced from the plan. The development of the pattern for the lower /'a ttcni Problems. HI Mile of the pipe is not given, but it would be accom- plished in exactly the same manner as that of the top piece, using C D F as the profile instead of A B E. For the pattern of the front piece of the pipe, divide its profile L M D' by any convenient number the T-square placed parallel to Q 1 and brought against the various points' in B E and D F cut corresponding measuring lines. Lines traced through the points <>f intersection, as shown by Q P and N, will give the required pattern. / - 512 3 t 5 6 1 Pig. iS9. Patterns for a Pipe Describing a Compound Curve. of points 1, 2, 3, 4, etc., from which drop lines ver- tically, cutting the two miter lines 1) F and B E, as shown. Upon Q 1, drawn at right angles to the direc- tion of the mold, place the stretchout of L M D', The pattern of the back piece not given in the illustration can be developed in exactly the same man- ner as that of the front by using C' ,T K as the profile and proceeding otherwise the same as in the fore- through which draw the usual measuring lines. With ' going. T/ic yew Metal IFcr/ar Patient Jlwk. PROBLEM 49. The Construction of a Volute for a Capital. It is sometimes desirable in designing capitals of large size to construct the volutes of the same of strips of metal cut and soldered together. The principal characteristic entering into the design of the volute, and that which distinguishes it from an ordinary scroll, consists in a pulling out or raising up of each succes- sive revolution of the scroll beyond the former, thus producing a ram's horn effect. This feature of its de- sign is also frequently embodied in the construction of scrolls used to finish the sides of large brackets or head blocks, such as may be seen by reference to Fig. 87 on page 12. As all volutes, except those of the Ionic order, always occur under the corners of the abacus and project diagonally from the bell of the capital, their forms can only be correctly delineated in a diagonal elevation. In Fig. 330 is shown a diagonal elevation of a portion of the bell and abacus of a capital with the volute. Immediately below the same, D A C B shows one-quarter of the plan of the capital, turned to corre- spond with the elevation, in which the various curves of the volute have been carefully projected from the elevation, as shown by the dotted lines. As the pat- tern cutter is dependent upon the drawing of the plan for his miter lines, considerable care must be given to this part of the work. On account of the small scale necessary in drawing Fig. 330, an enlarged view of the plan of the helix of the volute, as seen from below, is shown in Fig. 331, in which the various curves can be followed throughout their course. The volute as here given consists of two side pieces or scrolls, an outside cover or face strip, an in- side cover and two narrow strips to rill the space where the second curve of the scroll projects beyond the first. The outside cover or face strip extends from F of tin- elevation to G, where it is met by the inside face strip, which begins at H. To obtain the pattern for the in- side cover, divide the profile from II to G into any convenient number of equal spaces, and lay off a stretchout of the same upon the center line of the volute in plan, A B, extended toward K, as shown bv the nine spaces on the upper side of the line. Drop lines vertically from each of these points intersecting the upper line of the side of the scroll in plan. Place the J-square parallel to the stretchout line B K, and bringing it successively against the points in the plan, drop lines cutting corresponding lines of the stretch- out. Then a line traced through the points of inter- section, as shown from I to J, will give the shape of the side of the strip to cover the space between the points H and G of the elevation. A similar course is to be pursued in obtaining the outside cover or strip extending from F to G. This stretchout consists of fourteen spaces, and is shown on the lower side of the center line A K, the pattern being shown from L to M. The pattern for the remaining strip consists of a stretchout of seven pieces taken from the profile be- tween G and the termination of the scroll line. Points from this part of the profile are intersected with two miter lines in the plan, one forming the outer line of the strip, or its finish against the more projecting part of the scroll, and the other forming its finish against the lower scroll or inner edge of its first or outer curve. In Fig. 331 the lines showing the projection of the inner part of the volute beyond the outer curve are clearly seen. In the lower half the lines correspond- ing to the points 1 to 7 of the profile are shown by cor- responding numbers. Lines dropped from the points on both these lines to corresponding lines of the stretchout will give the pattern as shown from M to N. By inspection of the drawing it will be seen that the outline of the volute, as given in the elevation, does not represent exactly the "true face" of the scroll. As the variations in the angle of the side of the central part or helix of the scroll are only such as can be produced by the springing of the metal necessary to bring it into shape, no allowance need be made for such variation in cutting the pattern directly from the eleva- tion. Careful measurements of the stern or lower part of the volute, as shown in the plan, however, show that, the distances from point 9 to points a and i, if laid olf on a line parallel to A B, would reach to points a' Pattern 143 and b\ These points projected back into the elevation locate them in that view at a and V. Therefore the outline of the back of the stem will have to be ex- tended as shown by the dotted line from F to a'. This and need not be repeated here. The correct outline, from G to b' is omitted to avoid confusion with the figures. To avoid confusion of lines in dropping the points from the different parts of the profile to the miter liues Fig. S3 1, Enlarged, View of Helity PATTERNS Fig. 330. The Construction of a Volute for a Capital. outline can be accurately obtained, if deemed necessary, by the raking process described in connection with a number of other problems in this section of this chapter, and thence to the stretchout, only the first and last of each series or stretchout have been shown by dotted lines in the drawing. 144 Tin- Sen- Miiul Wvrhr I'ntim, /;<,/,; PROBLEM 50. The Pattern for a Pyramidal Flange to Fit Against the Sides of a Round Pipe Which Passes Through Its Apex. A pictorial illustration of the flange fitting against the sides of the pipe, as stated above, is shown in Fig. 332. In Fig. 333 K L M represents the elevation of Fig. 333. Pattern for a Pyramidal Flange to Fit Against a Round Pipe. pyramid, and P K T S elevation of the pipe that is to pass through it, A B D C being plan of pipe and pyra- mid. As the pyramid has four sides, each side will miter or fit against one-quarter of the profile oj the pipe, as will be seen by reference to the plan. Again, as each side consists of two symmetrical halves, as shown by the dotted line dividing the side B D, one- eighth of the profile of the pipe (as G I) is all that need be used in obtaining the pattern. Therefore, divide G I into any C&nvenient number of parts and carry ver- tical lines to L M, which represents one side of the pyramid, and then, from these points and the points L and M carry lines at right angles to it indefinitely, as shown. L M in the elevation represents the complete length of one side of the pyramid, as it would be if not cut by the pipe. Lav olT on the line from M the length of one side of the base of the pyramid, as B D in the plan, as shown by M M 1 . Biseet M M 1 at F, from which point draw F E parallel to L M, cutting the line from point L at E. The lines from K to the points M and M 1 would give the pattern of one side of the pyramid if it were not to be cut by the pipe. It simply remains now to measure the width of the pat tern at the various points of the curved portion, which Fig. 332. Perspective View of Pyramidal Flange. can be done by measuring the distance of each point in the profile G I, from the center line of the side B D in plan, and setting off these distances upon lines of corresponding number drawn through the pattern from the line L M, measuring each time from the center line E F. Thus the distance of point 4 from the center line in plan is set off from the center line of the pat- tern each way upon lino 4, and coincides with this point as previously established by the lines drawn from E to M and M'. The distance of the point 3 frcm the center line in the plan is set off from the center line of pattern each wavMipon line 3 of the pattern. Point '1 is established in the same manner. A line traced through the points 4321234 completes the pat- tern. 145 PROBLEM 51. The Patterns for a Square Pyramid to Fit Against the Sides of an Elliptical Pipe Which Pisses Through Its Center. Fa Fig. :>:-i, A B C D shows the plan of a square pyramid, whose ;i]ic.\, if completed, would be at E. F II 1 J shows the horizontal section of an elliptical pipe, against, the sides of which the sides of the pyra- mid are required to be fitted. From the side A B (or patterns will be necessary, one for A B G S to fit against the broad side of the pipe, and another for B G T C to iit against what might be termed the edge or narrow side of the pipe.- To obtain the pattern of the side of the pyramid shown by B G T 1 C D C C 1 Fig. SS4. The Pattern! for a Square Pyramid to Fit Against the Hides of an Elliptical Pipe. D C)of the plan is projected a front elevation, in which K L M N shows the broad side of the pipe. To the right another or side elevation is projected, in which the narrow view of the pipe is shown by P Q li. An inspection of the plan will show at once that two (or B E C, if the pyramid were complete), first divide that portion of the profile of the pipe from G to H by any convenient number of points, as shown by the small figures, from which, together with B and E, project lines vertically to the elevation above, cut- 146 The New Metal Worker I'utl'-rn lion!.-. ting that side in profile as shown from E' to B 1 . At right angles to E' B 1 carry lines from each of the points indefinitely, as shown. At any convenient distance away, cut these lines by any line, as E 3 V, drawn parallel to E 1 B'. Upon each of the lines drawn from the points in E 1 B', set off from E 3 V 3 the distances upon lines of corresponding number in the plan meas- ured from E V. Thus upon line 5 of the pattern set off either way from its intersection with the line E 1 V 3 a length equal to the distance of point 5 of the plan from the line E V. Upon line 4 of the pattern set off distances equal to that of point 4 from E V of the plan, ct.-. Also make V 3 C 3 and V s B 3 equal to V C and V B of the plan. Lines drawn from C 3 and B 3 toward E 3 will meet the points previously set off on line 5 of the pattern, indicated by T" and G 1 , and will constitute the sides or hips of the pattern, and a line traced through the points set off on lines 1 and 5 inclusive will give the shape of that portion of the pattern to fit against the pipe.' An exactly similar course is to be pursued in ob- taining the pattern of the side of the pyramid A S G B (or A K B of tin 1 complete pyramid'), whose profile is shown bv B 3 E J of the side elevation, showing the nar- J row view of the pipe. The pattern is shown at A 4 B" G 2 S*, and the operation is clearly indicated by the lines of projection. If it is desired to complete the elevations bv show- ing the lines of intersection of the sides of the pipe with the sides of the pyramid shown respectively in each elevation, as from c to b and <-. toy', it can be ac- complished as follows : To obtain the line c i, erect lines vertically from points ti. T and S (not shown), passing through the space between c and b in the front elevation, upon each of which set off the hight of each point as measured upon lines of corresponding number from B 2 C" to B 3 E 3 , as shown from R toward in the front, elevation; then a line traced through the points will give the line ef. PROBLEM 52. The Patterns for a Rectangular Pipe Intersecting a Cylinder Obliquely. In Fig. 335, let A B C represent the plan of a drum or cylinder, and B E D C the plan of rectangular pipe, the profile of which is shown by F G II I. In the elevation, J K L M represents the drum, N O P Q the rectangular pipe, and K n O the angle at which they are to intersect. Draw the end view, or plan, of circular drum in line with the elevation, as shown. Also extend n and P q so a line dropped from point C of plan will cut them, as shown by N and Q. Then n N" Q q is the joint between the drum and pipe, as shown in elevation. For the pattern of rectangular pipe proceed as follows: Divide B C of plan into anv convenient number of equal parts, and from these points carry lines horizontally cutting E D. Also from the points in C B drop lines vertically cutting Q P and N O. On P extended lay off a stretchout of profile F G II I, as shown by 1 1'. transferring the spaces in E D to H G and F I', and through the points in it draw the usual measuring lines, as shown. Place the "T-square parallel with the stretchout line 1 1', and, bringing it successively against the points in the miter lines N n Q q, cut the corresponding measuring lines, as indicated by the dotted lines. Lines traced through the points thus obtained, as indicated by ihgfi', will give the desired pattern. It will be observed that 1 II li i is a duplicate of O P Q N, and that G F/// is also a dupli- cate, only in a reversed position. The points in h >/ of pattern arc derived from Q y, as the points in f i' of pattern are derived from N //. If the size of tin- work- is such as to render it inconvenient to drop points from the elevation to the pattern by means of the T- square, the stretchout line I I' can be drawn where convenient, the usual measuring lines erected and the distances from O P to points in N / and (.) */ transferred by means of the dividers to lines of similar number drawn from the stretchout line. For the pattern or shppe of opening in drum, pro- ceed as follows: On L M extended, as R U, la}- off a fit//' /;. I'rnlilems. 147 stretchout of BO of plan, and from the points thusob- tainc"! erect the usual measuring lines, as shown. Place in.tr lines of corresponding number. Through the points thus obtained trace the lines V W and YX; PLAN i Fig. 335. Patterns for Rectangular Pipe Intersecting a Cylinder Obliquely. the T-square parallel with M L, and, bringing it suc- cessively against the points N n and Q q, cut measur- then V W X Y will be the shape of the required open- ing in the side of the drum. PROBLEM 53. The Pattern lor the Intermediate Piece of a Double Elbow Joining: Two Other Pieces Not Lying in the Same Plane. In Fig. 33fi is shown a front and side view of a somewhat complicated arrangement of elbows such as sometimes occurs when pipes have to be carried around beams or through limited openings. An inspection of the drawing will show that once the correct angle of the different elbows is ascertained the development of the miters will be quite simple, and is the same as those occurring in several of the problems preceding 148 Tin- Sf.w Metal Worker Pattern B<>k. this. The lower section of the pipe rises vertically to the first elbow, B, from which it must be carried up- ward a distance equal to C M, to the left a distance equal to B M, as shown in the front view, and back a distance equal to o C, as shown by the side view. methods employed in drawing the two views shown in Fig. 336 will be of assistance to the pattern cutter. According to the principles of projection each indi- vidual point must appear at the same hight in both ele- vations, and at the same distance right or left and for- 2. s FRONT VIEW SIDE VIEW Fig. 836. Elevations of Double Elbow. Fig. a38. Diagram Used in Obtaining Correct Side View of Upper Elbow. Fig. 337. Correct Side View of Lower Elbow. 1 2 Fig. 840. Method of Obtaining the Pattern of Middle Portion in One Piece. Fig. 339. Correct Side View of Upper Elbow. The Pattern for the Intermediate Piece of a Double Elbow Joining Two Other Pieces Nut Lying in the Same Plane. From the elbow C it then rises vertically, as seen in front, but really toward the observer as shown by the side view. The problem then really consists in find- ing the correct angles of the elbows, and becomes a question of draftsmanship rather than of pattern cut- ting. Some suggestions then with regard to the ward or back, with reference to the center lines of the plan. As front and side views are here required, begin by first placing the given plan in two positions, turning those sides of it to the bottom which corre- spond to the sides required in the elevations, and proceed by erecting the center lines of the differ- Pattern Problems. 149 ent pieces in their proper positions and building the pipe around them, so to speak. The plan being a circle, the different sides can only lie indicated by numbering the points, as will be seen by referring to the plans, point 2 appearing in front in the front elevation, and point 3 appearing in front in the side elevation. The plans having been so arranged and corresponding parts in both given the same number, proceed now to erect the center line of the lower sec- tion, making the bight of the first bend, B, the same in both views, as indicated by the dotted horizontal line. From this point the center line is continued in both views, giving it its proper inclination to the left in the front view, and to the right in the side view, all accord- ing to the specified requirements, thus establishing the point C, making it agree in bight in both views. From this point the pipe appears inclined only in the side view, which means that it leans toward the observer in the front view. Next draw the outlines of the pipe at equal distances from the center line and on either side of it throughout the entire course of the pipe in both views, deriving them from the points of plans 1 and 3 in the front view and 2 and 4 in the side view. Their intersection in the front view will give definitely the positions in the miter of points 1', 1", and 3', 3", and in the side view of points 2', 2', and 4', 4 J . As point 3' has been established in the front view, if a line be carried horizontally across till it intersects the line from point 3 of the side view, it will give the hight of point 3' in the miter, as shown in the front view. In the same manner a horizontal line from 1' in front, inter- secting the perpendicular f?om point 1 in plan of side, will give the true hight of point 1' in the side view. A careful inspection of the dotted lines of Fig. 336 will make the subsequent operations necessary to the com- pletion of the elevations clear to the reader without further explanation. Since neither of the views gives a true side view of the intermediate piece, one must be constructed from the facts now known, so as to get the true angle of the elbow B. By dropping a vertical line from the point C of the front view into the plan it will appear that the horizontal distance between the points C and B would be measured by the line K P of the plan; but bv further reference to the side eleva- tion the position of the point C is found to be to the right of its center line by a distance equal to B C' of the plan ; therefore, if this distance be set off on the vertical line from the point E in the plan below the front view, which is indicated by E C, the point C will determine the true position in the plan of the point C of the elevations, and the distance C P will be its horizontal distance from B. Since, now, its vertical distance cap easily be obtained from either front or side elevation, a new diagram can now be easily constructed which shall contain the proper dimensions to obtain a correct side view of this elbow. Proceed, then, to construct diagrams shown in Fig. 337, making C M equal to C M, Fig. 336, M B equal to C P of the plan, Fig. 336 ; a line connecting the points C and B will represent the center line of the intermediate portion of the pipe and give its true relation to the vertical por- tion whose center line is represented by B H, Fig. 337. By drawing the outlines of the pipe at the re- quired distance on either side of the center lines B H and B C, a correct side view of the miter is obtained. Since, as has been referred to above, the upper portion of the pipe appears vertical in one view and inclined in the other (see Fig. 336), a correct side view of the upper elbow is more difficult to be obtained. While different methods may be devised for obtaining it, the following is perhaps the simplest : As the upper sec- tion of the pipe, as shown by Fig. 336, is of indefinite length, any point may be assumed, as D, from which to take measurement for obtaining the angle of the upper elbow. Since the true length of the line C B of either elevation has already been obtained and given in Fig. 337, and since the true length of the part C D can be derived from the side view of Fig. 336, it is necessary only to obtain the true distance between the points D and B of the elevations to obtain the proper angle at the point C. By dropping a vertical line from the point D to a horizontal line drawn from the point C in the side view, Fig. 336, the horizontal distance be- tween C and D may be obtained. By tranf erring this distance, o C, to the plan of the front view, and locat- ing its distance from C, as indicated by D. this point will give the true position of the point D in the plan, and the line 1) P will give the true horizontal distance between the points D and B. In Fig. 338 let the distance C be equal to the line D P of Fig. 336. At point O erect a perpendicular, O D, making the dis- tance O D equal to o D of the side elevation, Fig. 336. From the point C drop a perpendicular, C B, making that distance equal to the vertical hight between the points C and B, as measured on line C M of the front view ; a diagonal line connecting the points D and B will readily be seen to give the true distance between the points bearing those letters in Fig. 336. Proceed now to construct the triangle shown in Fig. 339, mak- ing C B equal to C B of Fig. 337. From C as a center, 150 Xf/r .!/<'/"/ Worker Puttcm IlooL: with a radius equal to C D as obtained from the side view in Fig. 336, draw a small arc. which intersect with the arc drawn from the point R, with a radius equal to B D as obtained in P'ig. 338 ; this will give the correct angle of the upper elbow at C. A complete view of the miter may be obtained by further adding outlines of the pipe at equal distances on either side of the center lines, and connecting their angles, as shown by the line bf. Having now obtained two correct side views of the two elbows, the problem of obtaining the patterns for the same can be solved by the regular method. To obtain the pattern for the middle portion in one piece further calculations, however, will be re- quired. This, of course, could be obviated by making a slip joint in the middle portion of the pipe, by means of which the two elbows could be made separate, and then simply turned upon each other till the required angle is obtained. But as it might be desirable to make the pattern of the middle portion in one piece some means must be employed of ascertaining just how far one elbow would have to be turned upon the other were they made separately. As the seam in pipe con- taining elbows is usually made at either the shortest or the longest point of the miter, it may be easily seen, by an inspection of Fig. 336, that a line from the shortest point, or throat, b of the upper miter of the piece in question, would not meet the longest point, or point a, Fig. 337, in the miter of the other end, and some means must be devised for obtaining the real position of these points, of which the following is perhaps the simplest : From either of the points D or B, Fig. 339, draw a line through the point i, continuing it to the further side of the triangle, as indicated by the line B X. Lay off the distance D X upon the line D C of the side view, Fig. 336, thereby locating the position of the point x in that view. A line connecting this point with point B must intersect the miter line 2", 4:', in this view at the same point which it does in Fig. 339, thereby locating its position just as much as in Fig. 339. This point having been obtained, its equivalent upon the lower miter may be found by means of a line drawn parallel to the center line of the middle portion, intersecting it at, the point ij, from which point it can be carried vertically to the plan, as shown by Z, where its distance from other points can be measured with accuracy. The position of the point a in Fig. 337 will readily be seen to be at point // in the plan of the front view, Fig. 336. By transferring the point Z from the plan of the side view to the plan of the front view, which can be done by measuring its distance from either of the points 2 or 3, the relative position of the points h and Z upon the same circle will be apparent. Fig. 340 shows a diagram, in which a correct side view of the two elbows is shown, giving , the proper distance between the points B and C. Considering the lower one to be in its proper and lixed position, the profile is constructed and divided into points for the purpose of obtaining a stretchout and the miter pattern according to the usual method, the | stretchout being shown upon the line E F in the j profile and point 8 will readily be seen to correspond with point h in the plan of the front view. The posi- tion of the point Z in the same plan can be obtained by measuring its distance from point h and transferring it to Fig. 340, as indicated by M. As the point I of the upper elbow is in relation to the highest point, or , of the lower elbow as the point M is to the point 8 in the profile, it becomes necessary to place the point 8 in the stretchout of the upper elbow as far from the point 8 on the stretchout of the lower one as the dis- tance from 8 to M in the profile, which is shown by m in the stretchout. The stretchout of the upper elbow is thus moved, as it-were, in its relation to the stretchout of the lower elbow, that portion of it which extends beyond the point 1 at the left end being added to the other so as to make the seam continuous. The points are then dropped from the profile to the two rniter lines, and thence into measuring lines of corre- sponding number in the stretchout. Lines traced through the points of intersection, as shown by Y I* R X, will be the required pattern. The miters for the upper and lower sections would, of course, be inverted duplicates of the adjacent ends of the middle piece. Pattern ProU<-inx. PROBLEM 54. A Joint Between Two Pipes of the Same Diameter at Other Than Right Angles. 151 Let L F D E K I II _\l of Fig. 341 represent the elevation of two pipes of the same diameter meeting at the angle M H I, for which patterns are required. Draw the profile or section A' B' C' in line with the branch pipe, and the section A B C in line with the main pipe. As both pipes are of the same diameter, and the end of one piece comes against the side of the other piece, both halves of the branch pipe (dividing at the point B) will miter with one-half, B D, of the main pipe. By projecting lines through the elevations of each piece from the points B or 4 of their respective profiles the point G is obtained, which, being connected with points F and H, gives the required miter line. Space both the profiles into the same number of equal divisions, com- mencing at the same point in each. For the pattern of the arm proceed as follows : Lay off the stretchout O N opposite the end of the arm and draw the usual meas- uring lines at right angles through it, as shown. Place the T-square at right angles with the arm, or, what is the same, parallel with the stretchout line, and, bringing the blade successively against the points in the miter line F G H, cut the corresponding measuring lines. Through the points thus obtained trace the line PEST, which will form the end of the pattern required. For the pattern of the main pipe proceed as follows : Opposite one end lay off the stretchout, as shown by V Y, and opposite the other end lay off a corresponding line, as shown by U X. Connect U V and X Y. From so many of the points in the stretchout line V Y as represent points in the half of the profile BAD draw the usual measuring lines. With the T-square placed parallel to the mold- ing D I, drop the points from the profile onto the miter line F G H; then, placing it at right angles to the molding, drop lines from the points in the miter line intersecting the corresponding measuring lines. A line traced through these points of intersection, as F' Z H 1 W, will describe the shape required. The position of the seam in both the arm and the main pipe is determined by the manner of numbering the spaces in the stretchout. In the illustration the seam in the arm is located in the shortest part, or at a point corre- sponding to 1 of the profile. Accordingly, in number- lY Fig. S41 A Joint between Two Pipes of the Same Diameter at Other than Right Angles. ing the divisions of the stretchout, that number is placed first. In like manner the seam in the main pipe is located at a point opposite the arm. Therefore, in 152 The New Metal Worker Pattern Book. numbering the spaces in the stretchout commence at 1, which, as will be seen by the profile, represents the part named. If it were desirable to make the seam come on the opposite side of the main pipe from where it has been located that is, come directly through the opening made to receive the arm the numbering of the stretchout would have been begun with 7. In that case the opening F 1 W IT Z would appear in two halves, and the shape of the pattern would be as though the pres- ent pattern were cut in two on the line 7 and the two pieces were joined together on lines 1. By this expla- nation it will be seen that the seams may be located during the operation of describing the pattern wherever desired. It is not necessary, as prescribed at the out- set of this problem, that both profiles should be spaced off exactly alike. Any set of spaces will answer quite as well, provided there be points in each exactly half way between A and B of either profile that is, where points 4 are now located. They are spaced alike in this case to show that lines dropped from points of the same number in each profile arrive at the same point on the miter line, and that therefore when both pipes are the same diameter and their axes intersect, one profile may be used for the entire operation. 910 9-10 Note. In the nineteen problems immediately following, the conditions are such that it will be necessary to obtain the miter line from the data given by the operation of raking before the straight- forward work of laying out the patterns can be begun. However, as certain parts of the work of raking the miter line and of laying out the pattern are common to both operations, the two are usually carried along together, and therefore such points and spices should be assumed upon the profiles at the out- set as will be required in the final stretchout. PROBLEM 55. To Obtain the Miter Line and Pattern for a Straight Molding Meeting a Curved Molding of Same Profile. In Fig. 342, let F G J K represent a piece of straight molding joining a curved mold, G H I J, the profiles of the straight and curved molding being the same. To obtain the miter line or line of joint, Gr J, proceed as follows : Draw the profile in line with the straight molding, as shown by C D E, and divide into any convenient number of parts. From the divi- sions in the profile draw lines parallel to F G in the direction of the miter indefinitely, and also in the opposite direction, cutting the vertical line C E of the profile, as shown by the small figures, which corre- spond in number to the divisions on the profile. From B, the center from which the curved molding is struck, draw the line B A through the molding, as shown. Transfer the hights of the various points of the profile as obtained on the line C E to the line A B, placing the point E at the point o of the intersection of the lower line of the curved molding with the line A B, all as shown by X o. Then, with B as a center, draw arcs from the divisions on the line X o, intersecting lines of corresponding numbers drawn from the profile parallel to the lines of the straight molding. A line traced through these intersections, as shown by Gr J, will be the required miter line, and, as will be seen, is not a straight line. To obtain the pattern for the K/ I | : N i ! 1 -fi\ ! ii n \ n \ 1 I ij I , \ i J ! i V i N PATTERN Fig. Sj.The Miter Line between a Curved and a Straight Molding of the Same Profile. Pattern /'/-otilems. 153 straight molding, draw the line L N at right angles to it, upon which place the stretchout of the profile C D K, as shown by the small ligures. At right angles to the stretchout line L X. and through the points in it, draw the usual measuring lines. With the T-square placed at right angles to K J, bring it successively against the points forming the miter line G ,], and cut lines of corresponding number in the stretchout. Then a line traced through these points of intersection will form the miter end of the pattern shown by L M N O. The methods employed in obtaining the patterns for the curved portions are treated in Section 2 of this chapter. PROBLEM 56. A T-Joint Between Pipes of Different Diameters. In Fig. 343 it is required to make a joint at right angles between tin smaller pipe D F G E and the larger used in both operations the following course will be most economical. At a convenient distance from the end of the smaller pipe in each view draw a section of it. Space these sections into any suitable number of equal parts, commencing at corresponding points in each, and set- ting off the same number of spaces, all as shown by A B C and A 1 B 1 C 1 . From the points in A B C draw lines down- ward through the body of the large pipe indefinitely. From the points in A 1 B' C' drop point onto the profile of the large pipe, as shown by the dotted lines. For the pattern of the smaller pipe the requirements are its profile A B' C 1 and the line F 1 G', which - b - 6 - -7 - e - 9 Fiij. 3(3. A 1-Joint between Pipes of Different Diameters. pipe H K L I. For this purpose both a side and an end view are necessary. As the two pieces forming the J are of different sections this problem really consists of two separate operations, but as certain steps can be is the outline of the surface against which it miters, and therefore its miter line. Therefore, take the stretchout of A' B 1 C 1 and lay it off at right angles opposite the end of the pipe, as shown by V W. Draw the measuring lines, as shown. Then, with the T-square set parallel to the stretchout line, and brought succes- sively against the points between F' and G' upon the profile of the large pipe, cut corresponding measuring lines, as shown. Then a line traced through these points, as shown from X to Y, will form the end of the pattern. For the pattern of the larger pipe the stretchout is taken from the profile view F 1 G' L' and laid off at right angles to the pipe opposite one end, as shown by "N P. A corresponding line, M O, is drawn opposite the other end, and the connecting lines M N and P are. drawn, thus completing the boundary of the piece through which an opening must be cut to meet or miter with the end of the smaller pipe. According to the rule given in Chapter V, a profile and a rniter line are necessary. The profile F' G 1 L' has already been stated, but no line has yet been drawn in the elevation 154 The Neiv Metal \Vorker Pattern Book. of the larger pipe which shows its connection with the smaller pipe. This can only be found l>v projecting line's from the points dropped upon K' (}' through the elevation till they intersect with lines previously drawn from the 'profile A B C, as shown between F and G. F G then constitutes the miter line. For economy's sake, then, the spaces 1 to 4 previously obtained in the profile are duplicated upon the stretchout, as shown, to which are added as many more (4 to 10) as are neces- sary. As the points 1 to 4 have already been dropped upon the miter line in its development it is now only necessary to drop them parallel to the stretchout line into measuring lines of corresponding number, when a line traced through the points of intersection, as shown by R S T U, will give the pattern of the opening required. It may be noticed that the development of the miter line F G is not really necessary in this case, as the points are really dropped from the profile ABC right, through the elevation till they intersect the measuring lines. This happens in consequence of the arm or smaller pipe being at right angles to the larger one. Different conditions are shown in Problems 57 and 58 following. PROBLEM 57. The Joint Between Two Pipes of Different Diameters Intersecting at Other Than Right Angles. Let ABC, Fig. 344, be the size of the smaller pipe, and Y N 1 Z the size of the larger pipe, and let II L M be the angle at which they are to meet. Draw an elevation of the pipes, as shown by G K 1 N M L H, placing the profile of the smaller pipe above and in line with it, as shown, also placing a profile of the larger pipe in line with its elevation, as shown. In this problem the profiles of the moldings or pipes are given, but the line representing their junction must be ob- tained before going ahead. To obtain this miter line, first place a duplicate of the profile of the smaller pipe in position above the end view of the larger pipe, as shown by A 1 B' C 1 , the cen- ters of both being on the same vertical line, C' N 1 . Divide both profiles of the small pipe into the same number of spaces, commencing at the same point in each. From the points in A B' C project lines indefinitely through the elevation of the arm, as shown. From the points in A' B' C 1 drop lines on to the profile of the large pipe, and. from the points there obtained carry lines across to the left, producing them until they intersect corresponding lines in the elevation. A line traced through these several points of intersection gives the miter line K L, from which the points in the two patterns are to be obtained. For the pattern of the small pipe proceed as follows : Opposite the end lay off a stretchout, at right angles to it, as shown by E F. Through the points in it draw the usual measuring lines, as shown. In the developing of the line K L the points have already been dropped upon the miter line. It therefore only remains to carry them into the stretchout, which is done by placing the -square at 1 1 1 1 1 II II II 1 H | i 1 i| i ii i| i 1 i | Y6 ;( Fig. 344. The Joint between Two Pipes of Different Diameters Intersecting at Other than Right Angles. Pattern Problems. 155 right angles with the pipe, and, bringing it successively against tin 1 ]iDints iii the miter line K L, cut the corre- sponding measuring lines, as shown liy the dotted lines. A line traced through the points thus obtained will give the pattern of the end of the ami, as indi- cated. For the pattern of the large pipe proceed as fol- lows: Opposite one end, and at right angles to it, lay off a stretchout line, as shown bv K S In spacing off this stretchout u is l>est to transfer the spaces from 4 to 4 as they exist, as by so doing measuring lines will result which will correspond with points already exist- ing in the miter line K L, thereby saving labor, as in the case of the smaller pipe, and also avoiding con- fusion. The other points in the profile arc taken at convenience, simply for stretchout purposes. Draw a corresponding line, P T, opposite the other end, and connect P R and T S. In laying off the stretchout R S, that number is placed first which represents the point at which it is desired the seam shall come. For the shape of the opening in the pattern, draw measur- ing lines from the points 4, 3, 2, 1, 2, 3, 4, as shown, and intersect them by lines dropped from corresponding points in the miter line. Through the points thus ob- tained trace the line U V W X, which will represent the shape of the opening required. PROBLEM 58. The Joint Between an Elliptical Pipe and a Round Pipe of Larger Diameter at Other Than Right Angles. Two Cases. In Fig. 345 J K L M is the side elevation of the round pipe and E F G II that of the elliptical pipe joining the larger pipe at the angle F G J. In the WJ ioQ Fig. 345 The, Joint between an Elliptical Pipe and a Round Pipe of Larger Diameter at Other than Right Angles. First Case. The Major Axis of the Elliptical Pipe Crossing the Round Pipe. end elevation T S I shows the profile of the round pipe and U R S T the intersection of the elliptical pipe whose profile is shown at A B C D and N P Q, respectively, in the side and end views. From an in- spection of the drawings it will be seen that the side elevation shows the narrow view of the elliptical pipe, while the end elevation shows its broad view, or in other words, that the profile of the elliptical pipe is so placed that its major axis crosses the round or larger pipe. In Fig. 346 the elevations show the same pipes intersecting at the same angle,, but with the difference that the profile of the elliptical pipe is so placed that its minor axis crosses the round pipe. The reference letters and figures are the same in the two drawings and the following demonstration will apply equally well to either: By way of getting ready to lay out the miter, it will first be neeessarv to obtain a correct elevation of the miter line or intersection between the two pipes, as sliown, from H to G. To do this divide the two profiles A B C D and N O P Q into the same number of equal parts, commencing at the same points in each. Draw lines from the points in N O P Q, parallel with U T, cutting T S. In a similar manner draw lines in- definitely from the points in A B C D, parallel with 156 The Nno Mni, but the whole of the profiles must be used instead of the halves, because the two axes or center lines of the pipes do not intersect. In Fig. 347, let A B C be the size of the small pipe and F 1 H 1 M 1 be the size of the large pipe, between which a right-angled joint is to be made, the smaller i pipe being set to one side of the axis of the large pipe, as indicated in the end view. Draw an elevation, as shown by D F I L M K G E. Place a profile of the small pipe above each, as shown by ABC and A' B' C 1 , both of which divide into the same number of equal parts, commencing at the same point in each. Place the T-square parallel to the small pipe, and, bringing it successively against the points in the profile A 1 B' Pattern Problems. 157 C', drop lines cutting tin- profile of the large pipe, as shown, from F 1 to II'; and in like manner drop lines from the points in the profile A B C, continuing them through the elevation of the larger pipe indefinitely. For the pattern of the small pipe set off a stretchout Ai - -12 13 R Fig. .Itf.A t-Joint between Pipes of Different Diameters, the Axis of the Smaller ripe Passing to One Side of that of the Larger. line, V W, at right angles to and opposite the end of the pipe, and draw the measuring lines, as shown. These measuring lines are to be numbered to corre- spond to the spaces in the profile, but the place of beginning determines the position of the seam in tin- pipe. In the illustration giv. . the seam lias been located at the shortest part of the pipe, or, in other words, at the line corresponding to the point 10 in the section. Therefore commence numbering the stretchout lines with 10. Place the T-square at right angles to the small pipe, and, bringing the blade suc- cessively against the points in the pro- file of the large pipe from F 1 toH', cut the corresponding measuring lines, as shown. A line traced through the points thus obtained, as shown by X Y Z, will form the end of the required pattern. For the pattern of the large pipe, lay off a stretchout from the profile shown in the end view, beginning the same at whatever point it is desired to locate the seam, which in the present instance will be assumed on a line corresponding to point 13 in the pro- file. After laying off the stretch- out opposite one end of the pipe, as shown on R, draw a corresponding line opposite the other, as shown by N P, and connect N and P R, thus completing the outline of the pattern, through which an opening must be cut to miter with the end of the smaller pipe. In spacing the profile of the large pipe, the spaces in that portion against which the small pipe fits are made to correspond to the points obtained by dropping lines from the profile of the small pipe upon it, as shown by 1 to 7 inclusive. This is done in order to furnish points in the stretchout correspond- ing to the lines dropped from the profile A B C, as shown. No other measuring lines than those which represent the portion of the pipe which the small pipe fits against are required in the stretchout. Accord- ingly the lines 1 to 7 inclusive are drawn from R, as shown, and are cut by corresponding lines dropped from ABC. A line traced through the several points of intersection gives the shape S T U, which is the opening in the large pipe. If it be necessary for any purpose to show a correct elevation of the junction be- tween two pipes, the miter line F H G is obtained by intersecting the lines dropped from ABC with cor- responding lines carried across from the .same points obtained on the profile F 1 IF, by dropping from A' B' C', explained in Problems ;">(>, ;">7 and 5.S, and all as shown by the dotted lines. As remarked in Problem .">'., this line is not abso- lutely necessary, but is of great advantage in illustrat- ing the nature and principles of the work to be done. 158 TIte Xcw Metal \\'u/-L-,'f I \iltrn, !',<>,, I. PROBLEM 60. A Joint at Other Than Right Angles Between Two Pipes of Different Diameters, the Axis of the Smaller Pipe Being Placed to One Side of That of the Larger One. In Fig. 348, let C' B 1 A' be the size of the smaller pipe, and D' E 1 I 1 the size of the larger pipe, between which a joint is required at an angle represented In \V F K, the smaller pipe -to be placed to the side of the larger. Draw an eleva- tion of the pipes joined, as shown by V D G H I K F W. As in the preced- ing problems, the miter line or line giving a correct elevation of the junction of the pipes must be developed before the actual work of laying out the miter pat- terns can be begun, therefore place a profile or section of the arm in line with it, as shown by C 1 B' A 1 , and opposite and in line with the end of the main pipe draw a section of it, as shown by D' E 1 I 1 . Directly above this section draw a second profile of the small pipe, as shown by C B A, placing the center of it in the required position relative to the center of the profile of the large pipe. Divide the two profiles of the small pipe into the same number of equal spaces, commencing at the same point in each. From the divisions in C' B' A' drop lines parallel to the lines of the arm indefinitely. From the divisions in C B A drop lines until they cut the profile of the large pipe, as shown by the points in the arc D' E 1 . From these points carry lines horizontally to the left, producing them until they intersect the corresponding lines from C' B 1 A 1 . A line traced through these points of in- tersection, as shown by D E F, will be the miter line between the two pipes. For the pattern of the arm proceed as follows : Lay off a stretchout at right angles to and opposite the end of the arm, as shown by R P, and through the points in it draw the usual measuring lines. Place the T-square at right angles to the arm, and, bringing it successively against the points already in the miter line, cut the correspond- ing measuring lines. A line traced through these points, as shown by UTS, will form the required pattern. For the pattern of the main pipe draw a stretch- out line Opposite one end of it, as shown by M 0, numbering the divisions in it with reference to locat- ing the seam, which can be placed at any point desired. The spaces of the profile between D' and K' should be transferred to the stretchout point by point as they occur, as by so doing measuring lines will be Fig. 348 A Jnint at other than Right Angles between Two Pipes of Different Diameters, the Axin of the Smaller Pipe being Placed lt> One Side of that of the Lartjtr One. obtained which will correspond to the points already in the miter line. Draw a line corresponding to the Pattern I'roliterns. 159 stretchout line opposite the other end of the pipe, as shown by L N, and connect L M and N 0. Through the points in the stretchout line corresponding to the points between D 1 and E' of the profile draw measuring lines, as shown by 10, 11, 12, 13, 14, 15 and 1. Place the T-square at right angles to the main pipe, and, bring- ing the blade against the points in 1) K K successively, cut the measuring lines of corresponding number, all as shown by the dotted lines. A line traced through these points of intersection, as shown near the middle of M L N 0, will give the shape of the opening to be cut in the pattern of the main pipe. PROBLEM 61. The Patterns for a Pipe Intersecting a Four-Piece Elbow Through One of the Miters. In Fig. 34!), lot A B C 1) E E 1 U 1 C' 13' A' repre- sent the four-pieced elbow in elevation, F G II J its Fig. S49. Pattern for the Pipe Intersecting an Elbow Tliroiigh Our. of the Miters. prolile, and K L M X the elevation of the pipe which intersects the elbow through a miter joint. In line with the pipe draw the profile of sftme, as indicated by P R S. Extend F H of the profile of elbow, upon which as a center line draw another profile of the small pipe, as shown by 0' P' R' S 1 . Divide both profiles of the small pipe into the same number of parts, commencing at the same points in each, as S and S 1 . Now parallel to F P 1 of the profile draw lines from the points in O' P 1 R' S 1 intersecting the profile F G II J. as shown. A profile should properly be drawn in its correct relation to the part of which it is the section. As the part C D D 1 C' is about to be considered first, the pro- file should be pjaced with its center line F H at right angles to C D ; but as in a regular elbow of any num- ber of pieces the miter lines all bear the same angle with the sides of the adjacent pieces, the profile may for convenience be placed in proper relation to one of the end pieces, after which lines may be carried from it parallel to the side it represents to the miter line, thence from one miter line to another, always keeping parallel to the side, continuing this throughout the entire elbow if necessary. Therefore parallel to D E of the elevation draw lines from the points in G H of the profile, cutting the miter line D 1 D, and continue these lines parallel to D C and C B. From the points in the profile P R S draw lines parallel with L K intersecting lines of corresponding numbers drawn from G II. A line traced through these intersections will give the miter line K Z N. From the point Z in the miter line carry a line back to the profile of the pipe, as indicated by Z a. This gives, upon the pro- file of the pipe, the point at which the rniter line K N crosses the rniter line of the elbow C C 1 , so that it can be located upon the stretchout line, where it is marked '. For the pattern of the pipe K L M N proceed as follows: At right, angles to K L draw the line M' M", upon which lav oil' the stretchout of () I' 1! S. as shown bv the smiill figures, through the points in which, and at right angles to it, draw the 100 The \'t'tn Mini \\~vrkcr Pattern liuuk. usual measuring lines, which intersect with lines of corresponding numbers drawn at right angles to the line of the pipe L K from the intersections on the miter line K Z N. A line traced through the inter- sections thus obtained, as shown by M 1 N 1 K 1 N 2 M", will be the required pattern for the intersecting pipe. To avoid confusion of lines in developing the patterns of the intersected pieces of the elbows a duplicate of those parts, as shown by B C D D' C 1 B 1 , is given in Fig. 350, in which the miter line K Z N is also shown. The profiles F G H J and O' P' K' S' are presented merely to show the relationship of parts, as the patterns are obtained from the miter line K Z N, in connection with the stretchout of as much of the profile as is covered by the intersection. It is not necessary to include in this operation the entire elbow pattern, therefore only such a part of the pattern will be developed as is contained in profile from V to H. For the pattern for that portion of elbow shown in elevation by U Z N or V II of profile, proceed as follows : At right angles to C D of elevation draw the line R S, upon which lay off the stretchout of V H of the profile, as indicated by the small figures, through which draw the usual measuring lines at right angles to it, which intersect with lines of corresponding num- bers drawn from the intersections on the miter line Z N at right angles to C D. Trace a line through the intersections thus obtained, as shown by U 3 Z 3 N 1 . Then will U 3 Z s represent the pattern for that part of elbow shown in elevation by U Z, and Z 3 N 1 be the pattern for the cut on the miter line Z N. The pat- tern for the other half of opening shown by N 1 X V S is simply a duplicate of the half just obtained re- versed. Then X N' Z" shows the shape to be cut out of what would otherwise be a regular elbow pattern. The point a in the profiles O' S' and V II is so near the line drawn from the point 4 that separate lines are not shown, and on this account when obtaining the shape of K 1 Z' the points 4 and a are shown on the same line. In order to show that the pattern is produced by the regular method -that is, by the intersection of points from the miter lino into lines of corresponding number in the stretchout it should be noted that the profile and Stretchout of the piece alivadv developed is' properly designated by the figures 1, '2, 3, , 5, (i, while that of the piece next to be considered is properly designated by the figures 1, 2, 3, -ia, 5, 6, the point 4 not occurring in the first piece at all, while the points i and a both fall upon the same line in the stretch- out of the second piece, all of which is clearly shown. The pattern of the cut on the miter line K Z is obtained in the same manner as for Z N. At right angles to B C draw the line R' S', upon which place the stretchout of V H of the profile, as shown. s 1 B 1 D ELEVATION Fig. S50. Patterns for the Pieces of an Elbow Intersected by the Pip*. Through the points in the stretchout and at right angles to same draw the usual measuring lines, which intersect with lines of corresponding numbers drawn from the intersections on the miter line K Z, at right angles to B C. Trace a line through the intersections thus obtained, as shown by K 1 Z' U'. Then will Z' U' represent the pattern for that part of elbow shown in elevation by Z U, and Z 1 K' be the pattern for the cut on the miter line Z K. The pattern for the other half of opening shown by K' X' V S' can be obtained by duplication. Then will X 1 K' Z' repre- sent the shape to be cut out of the regular clb<>\\- pattern. Pattern Problems. 161 PROBLEM 62. The Pattern for a Gable Molding: Mitering Against a Molded Pilaster. Let N X V E in Fig. 351 be the elevation of a \- N G E 2 , will be the required pattern. Although the roof strip A B of the gable mold- ing is perfectly straight, points will have to be intro- duced between A and B for the purpose of ascertaining the shape of the cut from N to F, its intersection with the side of pilaster. The simplest method of obtain- ing these points is to derive them from the points be- tween B' and C 1 , as shown by 0' to 5' in the plan. They can then be transferred to their proper place in the stretchout, as shown, between A 3 and B". By so doinir points of like number fall in the same place on the pro- lile H I, and the vertical lines dropped therefrom ran !>< intersected with F X for the pattern of the roof strip and with the other lines from F to E for the pattern of the face of the mold, all of which is clearly shown. PROBLEM 63. The Patterns for an Anvil. It frequently occurs that sheet metal reproductions of various emblems or tools are desired for use as orna- ments or signs. In the following problem is shown how the various pieces necessary to form an anvil may be obtained. The description, of course, only applies to the several sides, as a representation of the horn can only be obtained by hammering or otherwise stretch- ing the metal. In Fig. 352 is shown a side and end elevation and two plans of the anvil, exclusive of the horn, the plans being duplicates and so placed as to correspond re- spectively with the side and the end views. Before the pattern of the side piece J N B G can be developed the line P Q R S, which is the result of the miter- ing together of the two forms shown by U V W of the plan and Z X T of the end view, must be obtained. Therefore divide the curved portion Z X T of the profile of the side into any convenient number of equal spaces, as shown by the small figures, and from the points thus obtained drop lines vertically cutting W V U 1 , the profile of the gore piece. Transfer the points thus obtained on W V U' to W V U of the other plan, and from these points erect lines vertically through the elevation of the side, and finally intersect them with lines of corresponding number drawn from the points originally assumed in Z T. Then a line traced through the points of intersection, as shown by ' P Q R S, will be the required miter line. To obtain the pattern of the side, first lay off a stretchout of the profile Z T, as shown, upon v q, through the points in which draw the usual measuring lines. K J N M \ ^ N 1 M' I~\K' B E c B' E' ft? 1 l\< 2 3 2 l 1 I | ELEV/ TION 1 | \ \ ELEV TION | | | j 1 l! 1 i i l l l i i b B /> 1 1 i! 1 1 1 i 1 1 i ! 1 1 \ i 1 1 1 1 1 1 ( II n 1 1 1 1 1 1 \ c ' m, m 1 ' / PATTEF N FOR \ 3 \ 1 TATTEF y N FOR \ | / / FRONT f N Xi // BACK , EN X " D Fig. S5S. P.itterns for the End Pieces of an Anvil. With the T-square placed parallel with J G drop lines from the points in the profile Z T cutting the outlines. Pattern Problems. 163 of the side from J to a and (i to //, and also cutting the miter lines of the gore piece O Q S (which last opera- tion has really been done in the raking operation above described). Placing the T-square parallel with v q, bring it successively against the points- in the several miter lines of the side elevation and cut corresponding measuring lines; then lines traced through the points the raking operation, lay off a stretchout of the same upon any line running at right angles to the form of this piece, as shown upon U" W. As the points have already been dropped from the profile to the miter lines in the operation of obtaining them, it only remains to place the I -square parallel to IP W* and bring it suc- cessively against the points in Q and Q S. cutting || I (ELEVATION jSIDE | Fig. 352. Patterns for the Side, Gore Piece dud Bottom of an Anvil. of intersection, as shown, from y to e, o to X, X to s and g toe?, will give the pattern for the lower portion of the side. As that part of the side from Y to Z of the pro- file is straight and vertical, that portion of the pattern shown on the stretchout line from X to I I -4-- +44 L 1 1 1 i i i n i r (. i 1 f ,._ 12 1/ Fig. 357, Kan and Sectional View of a Level Molding Mitering Obliquely Against Another Livel Molding of Different Profile. Fig. S58. Method of Obtaining the Pattern of the Lintel Molding Shown in Fig. S57. miter line, as shown in plan by G H C F. To obtain remembered that as nights are all to be compared the this miter line proceed as follows : Opposite to and in vertical lines of each profile must be placed parallel I 'ill III' II I 'fill ill' nix. Hi!) ;iinl their ii]i|xT ends turned in tin- same direction. Therefore, the back or line 1 \'.\ of the profile / is placed parallel to P> (', which represents a vertical line with reference to the profile V, and the point 12 is placed exactly opposite the point .1. according to the requirements of the side view. Fig. '~>7. Divide the profiles X and / into the same number of parts, as in- dicated l>v the small figures in each. From the points thus obtained in profile / carry lines at right angles to P> ('. cutting K J of profile Y, as shown. With the T-square placed parallel with the line B (J of the plan of the window carry lines from the points on the pro- file K J in the direction of (i and F; also draw lines from the points in the profile X parallel to Gr K, cutting the lines of corresponding munlier drawn from the pro- file Y. A line traced through points of intersection, as shown by (r H (< F, will give the miter line, as shown in the plan. While the curved portions of the profiles X and / have been divided into such a number of spaces as will answer the purpose, of an ordinary miter, it will be noticed that the plane surfaces between points 2 and ."> and 1.1 and 12 intercept so much of the curve of K .1 as to produce a curve between those points in the pattern. Therefore, for accuracy it is necessary to subdivide those spaces on K J, as indicated by a l> and c d e there shown. These points must be dropped back to the profile Z, and the spaces thus produced transferred to the stretchout line L M, all as indicated. The lines indicated by the small letters in K J have only been drawn partway in the engraving to avoid confusion, and the measuring Hues produced by these points in the stretchout have been shown dotted for the sake of distinction. These points are then intersected with the surfaces to which they belong in the miter line G H C, as shown between 2 and 3 and 11 and 12. For the pattern of the lintel molding first draw a line at right angles to it, as shown by L M, on which line lay off a stretchout of the profile X, as indicated by the small figures. Through the points thus ob- tained draw the usual measuring lines. With the T-square placed parallel with the stretchout line L M j carry lines from the points in G H C F to measuring lines of -corresponding number, when a line drawn through the points of intersection, as shown by N P, will complete the pattern. PROBLEM 68. The Patterns for a Square Shaft of Curved Profile Mitering Over the Peak of a Gable Coping Having a Double Wash. Let A B C in Fig 359 be the front elevation and D E F G H be the side elevation of a coping to sur- mount a gable, the profile of the top of which is shown at D K II in the side elevation. Also let M N P be the elevation of a square shaft or base, as of a finial, having a curved profile, as shown, which it is required to miter down upon the top of the coping. As the matter of drawing the elevations of the shaft in correct position upon the washes of the coping is attended with some difficulty, the method of obtaining these will be briefly described first : Through the lowermost point on either side of the front elevation, as P, draw a line' at the correct angle of the pitch of the coping or gable, as shown by V W, and extend the same to the right far enough to permit a section of the coping to be con- structed upon it. Upon this line set off the distance X W, equal to P L (half the width of the shaft at its base), and through the point W draw a line perpendicu- lar to V W ; next through the point X draw a line, making the same angle with Y W that D K, of the given profile of the coping, does with the horizontal line D L 1 , and extend this line to the right till it meets the line from W at K', and to the left, making D 1 K 1 equal to D K. This gives one-half the profile of the wash, which is all that is necessary in obtaining the elevation. Now from the points in the profile D' K 1 project lines parallel to Y W till they meet the center line of the front elevation, and duplicate them on the other side of the center line, which will complete the front elevation of the gable. From the points in the profile D K H erect vertical lines indefinitely, which may be intersected with lines projected horizontally from the points on center line I L to complete the side elevation. Thus a line from point B intersected with 170 Tlie New Metal Worker Pattern Book. line from K will give the apex of coping, and a line from point Z intersected with lines from D and H will give the points E and G, front and back of the washes at the apex. As the shaft is exactly square, the side elevation point U ; while the crossing of the side P with the top line of coping B C (marked 4) is projected upon the center line of the side view, thus giving the point Y. This completes the elevations with the exception of the lines M U P and M 1 Y P 1 . If the profile of the of it, M 1 N' 0' P 1 , is in all respects the same as that of | shaft P were a straight line, either slanting or vertical, U4-U 1 i i i M--M- FRONT ECEVATION Fig. S59.The Patterns for a Square Shaft of Curved Profile Mitering Over the Peak of a Gable. the front, and is drawn in line with it, as shown by the horizontal lines of projection connecting them. This having been completed, the point at which its side M 1 N' crosses the line E F (marked 4) is projected back to the front elevation, as shown, thus locating the the lines U P and M 1 Y would be straight lines, because they would represent the intersection of two plain sur- faces ; but as the profile is a curved line it will be necessary to obtain correct elevations of these two lines, as they are essential in obtaining the patterns to follow. I'n n,, - H I'nJdems. 171 The sliaft being square, the miters at its angles are plain square miters and are developed by the ordinary method, as explained in several problems in the earlier part of this chapter. The peculiarity of this problem, then, consists in obtaining the miter lines U P and M 1 Y and the part of the pattern corresponding to the same, which can all be done at one operation, as follows : Divide the profiles P and M' N' into the same number of equal parts, and place the stretchout of the same upon the center lines extended, as shown at I J and I 1 J, 1 through which draw the usual measuring lines for subsequent use. From the points in M 1 N' from 4 down drop lines vertically upon the profile of coping D K, as shown by the dotted lines, and trans- fer the points thus obtained to the profile D' K', from which points draw lines parallel to V W, as shown by the dotted lines. Intersect these with lines of corre- sponding number (2, 3 and 4) drawn horizontally from either profile, as shown at 2', 3' and 4', thus obtaining the miter line U P. After the points 1 to 11 of the profile have been dropped into the measuring lines of corresponding number of the stretchout the points 2', 3', 4' and 4^' are also dropped into the measuring lines of corresponding number, thus giving the cut U' T at the bottom of the pattern, which can be duplicated on the other side of the center line, thus completing the pattern. Q R S T U 1 of the front of the sliaft. From the points 1 to 4 of the profile P draw lines parallel to B C cutting the profile D 1 K', as shown by the solid lines, and transfer the same to the profile of coping D K, and from these points erect perpendicu- lar lines (also shown solid) indefinitely, as shown, which intersect with lines drawn horizontally from points of corresponding number in either profile, as shown at 2," 3" and 4". This will give the correct miter line M 1 Y. The miters at the sides of piece M 1 N 1 O 1 P 1 are of course the same as those of the front piece, therefore after they have been obtained the points 2" and 3" are dropped into measuring lines 2 and 3 of the stretchout I 1 J 1 , which when duplicated on the other side of the center line complete the line Q' Y 1 T 1 , which is the bottom cut of the side piece. It has been remarked that in obtaining the inter- sections between U and P and M' and Y horizontal lines may be drawn from points in either profile. The reason for this is simply that the two profiles O P and N 1 M' are identical and have been divided into the same number of equal parts. If a case should occur in which the side and face should be dissimilar it must be borne in mind that N' M' is the profile of the face piece and its points must be used in obtaining the intersec- tions between U and P, while O P is the profile of the side piece, and its points must be used in obtaining the intersections between M 1 and Y. PROBLEM 69. The Patterns of a Cylinder Mitering with the Peak of a Gable Coping; Having a Double Wash. Let A B C in Fig. 360 be the elevation of a coping to surmount a gable, the profile of which is D E F E 1 D 1 , which, as will be seen, shows a double wash, E Y and F E 1 . Let M P N be the elevation of a pipe or shaft which is required to miter over this double wash at the peak of the gable. Before any patterns can be developed it will be necessary to first obtain a correct elevation of the miter line or intersection of the shaft with the coping. To accomplish this proceed as fol- lows : In line with the pipe or shaft construct a profile of the same, as shown by G 1 L 1 K 1 H 1 , which divide into any convenient number of equal parts, and from the points thus obtained drop lines vertically through the elevation. Draw a corresponding profile, as shown by II G L K, directly over the profile of the coping, all as shown, which divide into the same number of equal parts, beginning to number at a corresponding place in the profile, and from the points in it drop lines on to the profile of the coping, cutting the washes E F and F E 1 , and thence carry the lines parallel to the lines of the coping, producing them until they intersect the lines dropped from the profile G 1 L 1 K' II'. Through the points of intersection thus obtained trace a line, as shown from to P, then B 1 P will be the miter line in elevation. As both halves of the shaft are alike (dividing on the line II L in one profile, and on H' L' in the other), it is really only necessary to use one-half of the profile, 172 Tin- New Worker /'W-. MS to use both halves as in the diagram requires the additional work of carrying the points from E F E' to the center line B B' for one-half and then down the other side of the gable for the other side. For the pattern of the shaft proceed as follows : In line with the end M N of the shaft, and at right angles to it, lay off a stretchout of the profile G' H 1 K 1 L 1 , as shown by R S ; in ihc usual manner, through the points in which draw measuring lines. Commence numbering these unifies to the lines of one side of the coping, as A B, lay oil' a stretchout of the wash of the coping K I 1 ' K', all as shown by K' J F' E 3 . In this stretchout line set oil' points corresponding to the points in E F E 1 , ob- tained by the lines previously dropped from the profile G H K L. Place the J-square at right, angles to A B, and, bringing it against the points in the miter line O B 1 , cut lines of corresponding numbers drawn through the stretchout E" E 3 , all as indicated by the dotted L'9 Fig S60. The Patterns of a Cylinder Mitering with the Peck of a Oable Coping Having a Double Wnxh. measuring lines with, the figure corresponding to the point at which the seam is desired to be, in this case 1. Place the J square at right angles to the shaft, and, bringing it against the points in the miter line B 1 P cut the corresponding measuring lines. Then a line traced through these points of intersection, as shown by T U V W X, will be the miter required. In case it should be desired to miter the coping against the base of the Shaft, the pattern for it may be obtained from the same lines in the following manner: At right lines. Then a line traced through these points of in- tersection, as shown by Z 1 Y, will be the pattern of the wash for the side of the gable A B required to miter against the base of the shaft. Incase the design should call for a shaft octagonal in shape, the same general rules would apply. Less divisions, however, will be required in the profile, it. only being necessary to drop points from each of the angles of the octagon, as in the ease <>f Problem 33, previously given. 1'iitlrni. /'/<>/:/> ,,/.-,; 173 PROBLEM 70. A Butt Miter of a Molding; Inclined in Elevation Against a Plain Surface Oblique in Plan. Let A B in Fig. 3(>1 be the profile <>l' a given cornice, and let E D represent the rake or incline of the cornice as seen in elevation. Let G H represent the angle of the intersecting surface in plan. The lirst step in developing the pattern will be to obtain the miter line in the elevation, as shown by E F. For this purpose draw the profile A B in connection with the raking cornice, which space in the usual manner, as indicated by the small figures. Draw a duplicate of this profile, as shown by A 1 B', placing it in proper position with reference to the lines of the plan. Space the profile A 1 B 1 into the same number of parts as A B, and through the points thus obtained carry lines parallel to the lines of the cornice, as seen in plan, cutting the line G II, as shown. In like man- ner draw lines through the points in A B, carrying them parallel to the lines of the raking cornice in the direction of E F indefinitely, as shown. Place the T-square at right angles to the lines of the cornice, as shown in plan, and, bringing it against the points of intersection in the line G H, carry lines vertically, cutting corresponding lines in the inclined cornice drawn from the profile A B. Through the points of intersection thus obtained trace a line, as shown from E to F. Then E F will be the miter line in elevation, formed by an inclined cornice of the profile A B meet- ing a surface in the angle shown by G H in the plan. At right angles to the raking cornice lay off a stretchout of A B upon any line, as K L, and through the points draw the usual measuring lines, all as shown. Place the T-square at right angles to the lines of the raking cornice, and, bringing it against the several points in the miter line E F, cut corresponding meas- uring lines drawn through the stretchout K L. A line traced through these points of intersection, as shown from M to N, will be the pattern required. B' /'!/. ,161. A Butt Miter of a Molding Inclined in Elevation Against a. Plain Surface Oblique in Plan. PROBLEM 71. Patterns for the Moldings and Roof Pieces in the Gables of a Square Pinnacle. Fig. 302 shows the elevation of one of four sim- in a square pinnacle. The profile ilar gables occurring of the molding is shown at P. The first step is to obtain the miter line or elevation of the miter shown at K, from which to derive the pattern. Draw the pnrtile P in the molding, as shown, placing it so that its members will correspond with the lines of the molding. Draw a second profile, P', in the side view 174 The New Metal Worker Pattern Book. of the gable, placing it, as shown in the engraving, so that its members will coincide with the lines of the side view. Space both of these profiles into the same number of parts in the usual manner, and through the points thus obtained draw lines parallel to each of the moldings respectively, as shown, until they inter- sect, and trace a line through the points of intersec- tion, as shown at K. Then K is the line in elevation upon which the moldings will miter. Draw the center line O M, which represents the miter at the top of the gable. For the pattern of the molding lay off a stretch- out of the profile upon any line, as G H, drawn at right angles to the line of the gable in elevation, as shown by the small figures. Through these points draw measuring lines, as shown. Place the T-square parallel to the stretchout line, or, what is the same, at right angles to the line of the gable, and, bringing it successively against the several points in the miter lines M and K, cut the corresponding measuring lines, as shown, and trace lines through the intersections. This completes the pattern of the molding, to which the piece forming the roof may be added as follows : Make L D 1 equal to E D of the side view of the gable and set it off at right angles to L B'. In like manner, at right angles to the same line, set off A' B 1 equal to A B- of the side view, and draw the lines D' A' and A 1 F, as shown. Fitj. $62. Patterns for the Moldings and Roof Pieces in the Gables of a Square Pinnacle. PROBLEM 72. Pattern for the Moldings and Roof Pieces in the Gables of an Octagon Pinnacle. Fig. 363 shows a partial elevation and a portion o,f the plan of an octagon pinnacle having equal gables on all sides. The first step in developing the patterns is to obtain a miter line at the foot of the gable, as shown by L. To do this proceed as follows : Draw the profile K, as shown, placing it so that it shall cor- respond in all its parts with the lines of the molding in elevation. Divide into spaces and number in the usual manner, and through the points draw lines par- allel to the lines of the gable toward L, as shown. Draw a duplicate profile in the plan, K 1 , so placed as to correspond with the lines of the molding in plan. Divide it into the same number of spaces, and through the points in it draw lines parallel to the lines of the plan, cutting the line D F, representing the plan of the miter. From the points in D F thus obtained carry lines vertically, intersecting corresponding lines dniwn from the profile in the elevation. A line traced through the several points of intersection, as shown by L, will be the line of miter in elevation between the moldings of the adjacent gables. The center line N forms the miter line for the top of the gable. For the pattern proceed as follows : Upon any line, as E E, drawn at right angles to the lines of the gable, lay off a stretchout of the profile, as shown by the small figures. Through the points of the stretch- out draw the usual measuring lines. Place the T- square at right angles to the lines of the gable, and, Pattern Problems. 175 bringing the blade successively against the points in the two miter lines above described, cut the corre- be added by setting off A' B 1 at right angles to A' C 1 , equal in length to A B of the side view. In like man- F C Fig. S6S. Pattern for the Moldings and Roof Pieces in the Gables of an Octagon Pinnacle. sponding measuring lines, as shown. Lines traced through the points of intersection thus obtained will give the pattern of the molding. The roof piece may ner, upon the line from M set off D 1 C 1 equal to C D of the side view. Then draw F' D 1 B', thus completing the pattern. 176 Ttie Xcw Metal Worker Pattern Bouk. PROBLEM 73. The Pattern for the Miter Between the Moldings of Adjacent Gables Upon a Square Shaft, Formed by Means of a Ball. In Fig. 364, let A C be one of the gables in pro- file and B D the other in elevation, the moldings form- ing a joint against a ball, the center of which is at E. The first operation necessary will be that of obtaining the miter line, or, in other words, the appearance in elevation of the intersection of the molding with the ball. Place the profile of the mold in each gable, as shown at F and H. Divide each of these profiles into the same number of equal parts, as indicated by the small figures. From the points thus obtained in F drop lines vertically, meeting the profile of the ball, as shown from C to J. From the center E of the ball erect a vertical line, as shown by E J. From the points in C J already obtained carry lines horizontally, cutting E J, as shown, and thence continue them, by arcs struck from E as center, until they meet lines of corresponding number dropped from points in the pro- file H parallel to the gable in elevation. Through the intersections thus obtained trace a line, as indicated by D G M. Then D G M will be the miter line in eleva- tion. To develop the pattern for the molding, first lay off at right angles to the gable a stretchout of the pro- file, as shown by P R, through the points in which draw the usual measuring lines. Place the T-square parallel to the stretchout line, or, what is the same, at right angles to the lines of the gable, and, bringing it successively against the points in the miter line D M, cut the corresponding measuring lines. A line traced through the points of intersection from 2 to 7 (that is, from U to V) will give the pattern for the curved por- tion of the profile. As any section of a sphere is a perfect circle whose length of radius depends upon the proximity of the cutting plane to the center of the sphere, the curves S to U and V to T of the pattern, representing the plain surfaces 1 2 and 7 8 of the profile, must be arcs of circles, whose lengths of radius can be determined from the elevation. As the pattern for the plain surface 1 2 is simply a duplicate of the cut from D to G of the elevation, set the dividers to the radius E G of the ele- vation, and from S and U respectively as centers strike arcs, which will be found to intersect at N. Then N is tin; center by which to describe the arc S U. To find the radius for the curve from V to T continue the line 8 M through the sphere, cutting its opposite sides at M and L ; then M L will be the diameter of the Fig. S64. The Pattern for the Miter Kftween the Moldings of Adjacent Gables Upon a Square Shaft, Formed bij Means of a Ball. circle of which the arc 7 8 or V T is a part. There- fore with K M (one-half of M L) as a radius, and V and T respectively as centers, strike arcs, which will inter- sect in the point 0. From 0, with the same radius, describe the arc V T. Then S U V T will be the pat- tern, of the molding to miter against the ball. Note. The remaining problems in this section of the chapter involve the necessity of " raking " or de- veloping a new profile from the given or normal profile before the pattern for the required part can be ob- tained. One of the principal characteristics of this work is that, as the normal profiles are usually spaced into equal parts for convenience in beginning the work, the resulting or raked profiles must by force of cir- cumstances be made np of a number of unequal spaces : in consequence of which their stretchouts must be transferred to given straight lines, space by space, as they occur upon the new profiles. 177 PROBLEM 74. The Pattern of a Flaring: Article of which the Base is an Oblong: and the Top Square. Let A B D E of Fig. 365 be the elevation of the article, and F N I the plan at the base, K M P L being the plan at the top. If the sides are to be de- veloped in connection with the top (supposing the bottom to be open) proceed for the pattern as follows : Draw K' M' 1" L', Fig. 366, equal in all respects to K M P L of the plan. Through the center of it, and at right angles to each other, draw lines V U and S T indefinitely. While the elevation in Fig. 365 shows the slant hight of the ends it does not give the slant and L' P', making them in length equal to F N and I O of the plan, letting the points V and U come mid- way of their lengths respectively. Draw K' F', M' N 1 , L' I' and P 1 0', thus completing the pattern for the sides. Upon S T set off Z T from M 1 P 1 , and Y S from K 1 L', in length equal to A B or D E of the ele- vation, and through the points S and T draw F' P and N' 0' parallel to K 1 L 1 and M 1 P 1 , and in length equal to F I and N of the plan, letting the points S and T- fall midway of their lengths respectively. Draw F J K 1 Fig. 365. Elevation and Plan. Fig. 366. Pattern. The Pattern of a Flaring Article uf Which the Base is Oblonij and the Top Square. hight or profile of the sides, therefore through the ele- vation, and perpendicular to the base and top, draw the line C G, which will measure the straight hight of the article. From G set off G II, in length equal, to M R of the plan. Draw H C. Then H C will be the profile of the article through the side, and therefore the width of the pattern of that portion. Upon V U of the pattern, from K' M 1 , set off "W V, and from L 1 I" set off X U, in length equal to H C of the eleva- tion. Through U and V draw lines parallel to K' M' P L 1 , N 2 M 1 and O ! P 1 , which will complete the pattern of the ends. If it is required to produce the pattern with the sides joined to the bottom, supposing the top to be open, lay out first a duplicate of F N I 0, through the center of which draw the stretchout lines V U and S T as before, and proceed in the same general manner as described above to obtain the sides, placing then- wider ends against corresponding sides of the base or bottom. 178 The New Metal Worker Pattern Book. PROBLEM 75. The Envelope of the Frustum of a Pyramid which is Diamond Shape in Plan. In Fig. 367, let A B D E be the elevation and K G I the plan of the pyramid at the base. Project the points B and D into the plan, as shown, locating the points M and P, and draw the sides of the plan at top, each parallel to the corresponding line of the plan at the base. By projection from G or O of the plan draw C F of the elevation, representing R of the plan and also the straight hight of the frustum. Before the slant hight or stretchout of a side can be obtained it will be necessary to construct a section on any line crossing the plan of the side at right angles as S T. Therefore extend the top and bottom lines of the elevation, as shown dotted at the right, cutting the vertical line S 1 S 5 , thus making 'S 1 S" equal to the straight hight of the frustum. Upon the base line extended set of from S' the distance S 1 T 1 , equal to S T of the plan, and draw S 3 T 1 . Then will S" T 1 be the true profile or slant hight of the frustum. At right angles to M R of the plan draw S W, making its length equal to the slant hight of the frus- tum, as shown by S* T 1 of the section. Througli W draw N II indefinitely, parallel to K 0. At right angles to K 0, through the points K and 0, draw lines K N and H, cutting N II in the points N and II, thus establishing its length. Connect M N and R II. Then M R II N will be the pattern of one of the four sides composing the article. PATTERN. Fig. 367. The Envelope of the Frustum of a Pyramid Which is Diamond Shape in Plan, PROBLEM 76. The Pattern of the Flaring: End of an Oblong Tub. In Fig. 368, A B D C shows the elevation and N P O R the plan of a vessel having straight sides and semicircular ends, one end of which is slanting. First draw a correct plan and elevation of the article, seeing that each point of the elevation is carefully projected from its corresponding point in the plan. Divide half of the boundary line of the top into any number of equal spaces, commencing at O, all as shown by the small figures 1, 2, 3, etc., in the plan. From the points thus obtained carry lines vertically until they cut the top of the elevation, as shown in the points between B and L ; also continue the lines downward until they meet the line T 0, all as shown. From the points between L and B thus obtained draw lines parallel to B D, producing them upward indefi- nitely, and continue them downward until they meet the bottom line of the elevation F D, as shown. At right angles to the lines thus drawn, and at any con- venient distance from the elevation, draw G II. With the dividers, set off from the line G H, on each of the lines drawn through it, the distance from T O, on the lines of corresponding number, to the curved line P 0. In other words, make G K equal to T 6 of the plan. Set off spaces on the other lines corresponding to the distance on like lines in the plan. Through the points thus obtained trace a line, as shown by K H. Then G H K will be the half profile of the end of the vessel on any line, as L M, drawn at right angles to Pattern Problems. 179 the line D B. The stretchout of the pattern is to be taken from the profile thus constructed. At right angles to D II, and at any convenient distance from it, draw U V, upon which lay off twice the stretchout of K II, numbering each way from the central point 1, as shown. From the points in the stretchout thus obtained draw measuring lines at right spending measuring lines, as shown, for the top of the pattern. If it is desired to make a joint upon the line E F of the elevation, the triangular shaped piece ELF may be added to the pattern as follows : With the di- viders take the distance E F of the elevation as radius, and from the point Z of the pattern as center describe an arc. In like manner, with E L of the elevation as PLAN R Fig. 368. Pattern of the Flaring End of an Oblong Tub. angles to it indefinitely. With the blade of the T-square set at right angles with D B, and brought successively against the points in F D, cut lines of corresponding number drawn through the stretchout. Then a line traced through the points of intersection, as shown by Y Z, will be the pattern of the bottom end of the piece. In like manner bring the blade of the T-square against the points in L B and cut corre- radius, and point R in the upper line of the pattern as center, describe a second arc, cutting the first arc in the point W. Connect W with R and also with Z. The triangular piece at the opposite end terminating in point X is added in a similar manner, thus com- pleting the entire pattern of that portion of the vessel from the line E F to the right. PROBLEM 77. Pattern for the Flaring Section of a Locomotive Boiler. While the pattern here described is especially adapted to the tapering section or " taper course " of a locomotive boiler its principles are equally applicable to tanks, cans or pipes whose shapes are governed by the spaces or positions which they are to occupy. The section of the boiler at A F, Fig. 369, is round, as shown by I N L M. The lower half of the circle I M L is the profile from L F to G D, but the upper half is raked or slanted from B K to C II, retaining its semi- circular character at C H. The line H G is a vertical 180 New Metal Wwkvr /'//> m liwk. line, as shown by S L of the sectional view, and the surface H K G being vertical is simply a flat triangular surface, exactly as shown in the elevation. Fig. 369. Pattern for the Flaring Section of a Locomotive Boiler. Since the part B K H C is semicircular when cut upon any vertical line, the first step will be to obtain a section of it as it would appear if cut upon a line at right angles to B C. This section must be derived from the normal section of the level part, and may be done as follows : Assume any line, as U ~W, drawn at right angles to B C at any convenient position outside of the elevation, us the vertical center lino of the new section. Divide one-half, the normal section, as N L, into any convenient number of spaces, as shown bv the figures, and from the points thus; obtained draw lines parallel to A B, cutting B K, as shown, also extend- ing them back to the center line N M. From 15 K carry them parallel to B C, cutting the line C II, and extend them indefinitely, cutting also the line U AY. AVith the dividers measure the horizontal distance of the various points in the normal profile K L from the center line N M, and transfer these distances to lines of corresponding number, measuring each time from U AY. Thus make AA" 7 equal to O 7; the distance from U W to the point 6 equal to P l> ; and so con- tinue till all the distances have been measured. A line traced through these points will constitute a profile of the raking portion on a line at right angles to its direction, undBK and C il will be the miter lines. To develop the pattern first layoff the stretch- out of the profile T U A r upon any line drawn at right angles to B C, as A' B 1 . As the points in U T have already been dropped upon the miter lines in the previous process it is now only necessary to place the T-square parallel to A 1 B 1 and, bringing it successively against the points in C II and B K, drop lines cutting the measuring lines of corresponding number. A line traced through the points of intersection, as shown 1>\- X X Y Y, will be the pattern of the raking portion B K H C. To this may be added the flat triangular piece K H G, as shown by X Y Z. From the points X and Z lines may be drawn at right angles to X /, as shown by Z J and X Q, extending them sufficiently to complete the lower portion of this part of the boiler, shown by K E D G of the elevation. PROBLEM 78. The Pattern for a Blower for a Grate. In Fig. 370 D F K H E shows a front view, P L M a side view, and A C B a plan of a blower. The conditions which determine the course to be pursued in arriving at the pattern are that its upper outline shall conform to the semicircle F K IT of the eleva- tion, and that it shall slant from K to.(i at an angle indicated by the line L M of the profile. Therefore, the first step will be to determine a true section of its Pattern Problems. 181 upper portion or hood upon a line at right angles to ]j M from the puiut N of the side view; after which, Fig. 370 Pattern for a Blower for a Orate. with N M and N L as miter lines, the pattern can be developed in the usual manner. To obtain a true pro- file of the hood, divide one-half of the semicircle F K H into any number of equal spaces, and carry lines from each of the several points to the vertical line L N of the side view. From the points thus obtained in L N" carry lines parallel with L M indefinitely, which intersect at right angles by the line T S, located at any convenient point outside of the diagram. With the dividers take the horizontal distances between the points in the arc F K to the line K G, and set them off on the lines of corresponding number, measuring from the line T S. Then a line drawn through the points thus obtained, and as indicated by T R, will be a correct section through the inclined portion of the blower. Take the stretchout of the profile T R point by point and place the spaces on the line U V, which is drawn at right angles to L M. Through the points in U V draw the usual measuring lines at right angles to it. As the points from the profile T R have already been dropped upon both the miter lines M N and N L, it is only necessary to carry them, at right angles to L M, on to the measuring lines of corresponding numbers. Then a line traced through the points thus obtained, and as indicated by I" K' H', will be the desired pattern. PROBLEM 79. Pattern for a Can Boss to Fit Around a Faucet. In Fig. 371 is shown a top and side view of a boss whoso sides are in part parallel and just suffi- ciently apart to allow the faucet to fit between them. L N represents the diameter of the opening at the top. K L M X represents the general shape of the boss where it joins the can and is the result of the condi- tions existing in the side view, but is not made use of in the process of obtaining the pattern. The essen- tial points are the curve of the can body, D A B E, the diameter of boss at top, L N, the distance be- tween 1) and E and the distance X C, all of which are shown in the side view. Divide one-quarter of the plan of the top, as indi- cated by N, into any convenient number of spaces, as indicated by 1, 2, 3, etc. From the points thus established drop lines vertically, cutting the line repre- senting the top in the side view, as shown from F to C. From the points thus established in F C carry lines parallel to the side F D, producing them until they cut the curved line D A B E, as shown between D and A. The next step to be taken is to obtain the profile which would be shown by a section taken through the article at right angles to the line D F. For this purpose at any convenient point draw a line through D F and at right angles to it, as shown by P R. From the points established in the plan of the top, as shown from O to N, carry lines vertically until they meet the horizontal line K M passing through the cen- 182 Neiv Metal Worker Pattern Book. ter of the top, as shown. Taking the length of each of the distances thus obtained in the dividers, set it off from either side of P R on the lines of correspond- ing numbers, and through the points thus obtained trace the curve, as shown. Then this curve will repre- sent the required section from which the stretchout of the. envelope may be obtained. On the line E P, pro- duced sufficiently outside of the side view for the pur- pose, lay off the stretchout of one-half of this curve, as shown, and through the points thus established draw measuring lines parallel to D F. Then, with the T-square placed parallel to P R, or, what is the same, at right angles to D F, and brought successively against the points in the profile of the can body be- tween D and A, cut the measuring lines of correspond- ing numbers. In like manner bring the J-square against the points in the top of the article shown from F to C and cut the measuring lines of corresponding numbers. Then lines traced through the points thus obtained, as shown by D' A' and F' C', will be one- half of the pattern of one of the ends. As that por- tion of the boss lying between points A, B and C is simply a flat triangular piece it is only necessary to add a duplicate of its shape to that part of the pattern just obtained, bringing one of its straight sides against the line 5, all as shown. To the other straight side C' B' must be added a duplicate of the first part of the pattern reversed, as shown by B' C' G' E'; the result- ing shape will then constitute the pattern of one-half the boss. Fig. 371. Pattern for a Can Boss to Fit Around a Faucet. PROBLEM 80. The Patterns for a Molded Base in which the Projection of the Sides is Different from that of the Ends. Let A B C D, in Fig. 372, represent the side view and E F G H the plan of a base in which the projec- tion of the sides, as shown at O P, is less than that of the ends, as shown at M C. B C and A D show the profile of the ends of the base. As the projection through the sides of the base is less than that of the ends, a profile must be obtained through the side P in plan, from which to obtain the stretchout in pro- ducing the pattern of sides. To obtain the pattern for the end proceed as follows : Divide the profile B C into an equal number of parts, as shown by the small figures, and from the points obtained drop lines at right angles to A B until they intersect the miter lines J P and L Gr in plan. At right angles to F G draw the line B" C", upon which place the stretchout of the profile B C, as shown by the small figures, through which draw the usual measuring lines, which intersect with lines of corresponding numbers drawn from the miter lines at right angles to F G. A line traced through these intersections, as shown by J' L' G' F', will be the re- quired pattern for the end of the base. To obtain the profile through the side proceed as follows : From B in elevation draw the vertical line B M, as shown, and from the divisions on the profile B C draw lines par- Pattern Problenifi. 183 allel to the lines of the moldings until they intersect the line B M, as shown l>v tlie small figures. From the intersections obtained on the miter line L G in plan, as before explained, draw lines parallel to H G and ex- tend them indefinitely, cutting the miter line K H, as shown. Upon K L of the plan extended set off IV M', the side of the base. To obtain the pattern for the side proceed as follows : At right angles to II G draw the line B* N", upon which place the stretchout of the profile B' N', as shown by the small figures. It will be noticed that the spaces in the profile B' N' are unequal, and therefore each must be separately placed on the line N' W PROFILE THROUGH 0. P. IN" PLAN, F' , Fig. H' N 2 G Patterns for a Molded Base in Which the Projection of the Sides {s Different from that of the Ends. in hight equal to B M of elevation, and transfer the spaces from B M to B' M', as shown. At right angles to B' M' and from the points on same draw lines, as shown, intersecting lines of corresponding numbers drawn from the rniter line L G. A line traced through these intersections will be the desired profile through B 8 N J . Through the points in this stretchout line draw the usual measuring lines, as shown, which intersect with lines of corresponding numbers drawn at right angles to H G from the miter lines K II and L G. A line traced through these intersections, as shown bv K' H' G' L', will be the desired pattern for the side. PROBLEM 81. The Patterns for an Elliptical Vase Constructed in Twelve Pieces. The first essential in beginning the work is an ellipse, which may be drawn by whatever rule is most convenient, and which must be of the length and breadth which the vase is required to have. Draw the plan of the sides of the vase about the curve, as shown in Fig. 373, in such a manner that all the points X, Y, Z, etc., shall have the same projection beyond the curve. Complete at least one-fourth of the plan by drawing miter lines, as shown by P C, M C, C, U C and N C. Above the plan construct an elevation of 184 Tin 1 \/'/r Mini II '<;//>; I'lilli-ni the article, as shown bv II L K G. Only the profile considered before the article is constructed. As the H V W L of the elevation is needed for the purpose of projection of cadi of lhe sides upon the plan is differ- Fig. 37S.The Patterns for an Elliptical Vase Constructed in Twelve Puves. pattern cutting, but the other lines are desirable in process of designing, in order that the effect may be ent when measured from the center C on lines at right angles to the lines of the sides, it will be necessary Pattern 185 lirst to develop the profiles of each of the van ing sides from the normal prolile II V \V L. Therefore. di\ idc -II Y \\' Ij in the usual manner, and from the several points in it drop lines across its corresponding section (.No. 1) of the plan. Across the second section in the plan, from the points already obtained in U C, draw lines parallel to <) U, the side of it cutting O C, and produce them until thcv meet A C, which is drawn from C at right angles to 1' <) produced. Then the points in AC give the projections from which to obtain a profile of the section numbered '2. In like manner continue the points from C O across the third section in the plan, parallel to O M, the side of it cutting M C, and pro- duce them until they cut C 15, which is drawn from C at right angles to O M produced. Then C B contains the points requisite in obtaining a profile of the third section. Continue the points in C M across the fourth section, cutting its other miter line C P. From C draw C D at right angles to the side P M of the sec- tion, cutting the lines drawn across section 4. Then upon C D will be found the points necessary to determine the profile of the fourth pattern. Pro- duce the line of the base of the elevation indefinitely, as shown by C' C" C :l . and also the line of the top A' W I)'. From the several points in the profile H Y "W L draw lines indefinitely, parallel to the lines just de- scribed and as shown in the diagram. From C 1 , upon the base line produced, set off points corresponding to the points in C A of the plan, making the distance from C' in each instance the same as the distance from C in the plan. NumVr the points to correspond with the numbers given to the points in the profile II V W L, from which they were derived. In like manner from C* set off points corresponding to the points in C B of the plan, numbering them as above described. From C 3 set off points corresponding to those in CD of the plan, likewise identifying them by figures in order to facilitate the next operation. From C 1 erect the perpendicular C' A 1 ; likewise from C" and C 3 erect the perpendiculars C 2 B and C 3 D 3 . From each of the points laid off from C 1 , and also from each of those laid off from C' and C 3 , erect a perpendicular, producing it until it meets the horizontal line drawn from the profile H V W L of corresponding number. Then lines traced through these several intersections will complete the profiles, as shown. Perpendicular to the side of each section in the plan lay off a stretchout taken from the profile corresponding to it, just described, and through the points in the stretchouts draw measuring lines in the usual manner, all as shown by E F, E' F', E J F J and E 3 F 3 . Place the T-square parallel to each of these stretchout lines in turn, and, bringing it against the several points in' the miter lines bounding the sections of the plan to which they correspond, cut the measur- ing lines in the usual manner. Then lines tracad through the points of intersection thus obtained, all as shown in the diagram, will complete the patterns. PROBLEM 82. The Patterns for a Finial, the Plan of which is an Irnegfular Polygon. In the central portion of Fig. 374 is shown the plan B C D E F G II, upon which it is required to construct a fiuial, the only other view given being a section through one of the sides, being that numbered 1 on the plan. The section of side ABC, or No. 1, is shown above and in line with the plan of the same, and is marked Profile No. 1, and is a section on the line A M, which is drawn at right angles to B C of the plan. To obtain the pattern of A B C of plan, or No. 1, divide the profile K L in the usual manner, and, with the T-square placed parallel with C B of plan and brought successively against the several points in profile K L, drop ; lines cutting the miter lines A B and A C. On A M extended, as Al Ml, lay off a stretchout .of K L of profile, through the points in which draw the customary measuring 'lines. Place the T-square parallel to the stretchout line Al Ml, and, bringing it against the several points in the miter lines A B and A C, cut corresponding measuring lines. Tracing lines through the points thus obtained, as shown by 186 The New Metal Worker Pattern Book. Al Cl Bl, will give the pattern of part of article shown on plan by A B C. at right angles to eacli that is, since A N, A and A P, drawn at right angles respectively to C D, D E PROFILE NO.4 PROFILE NO. 3 PROFILE NO.1 K2 PROFILE NO.2 cij D3 02 C2 Fig. 374. Pattern for a F inial, the Plan of Which is an Irregular Polygon. Since the point A in the plan is not equidistant from all the sides of the same, when measured on lines and E F, differ in length from each other and from A M a correct section must be obtained for each Pattern Problems, 187 of the other sides before their patterns can be developed. These different sections can be most conveniently obtained at one operation in the following manner: With the T-square placed parallel with D C, and brought successively against the points in A C, draw lines cutting A D and A N. Then the points in A N can be used to obtain a profile of section No. 2. Also continue the points from A D across the third section of the plan, and parallel with D E, and produce them until they cut A E, also A 0, which is drawn from A at right angles to D produced. Then A contains the points necessary in obtaining a profile of the third section. Continue the points from A E across the fourth section of the plan, and parallel with E F, cut- ting A P and A F. Then the points in A P can be used to obtain a profile of section No. 4. While the projection of the several points in each of the new profiles can be obtained respectively from the lines A N, A and A P, the hights of the several points must be the same in all and must be derived from the normal profile K L. Therefore continue J L, which represents a horizontal line of profile No. 1, in either direction, as shown by L4 L2. From the several points in profile No. 1 draw lines parallel with L4 L2, extending them indefinitely in either direction. At any convenient position on L4 L2 set off points corresponding to the points in A N of the plan, as shown at J2 L2, numbering them to correspond with the points in A N and in the normal profile. From each of the points in J2 L2 erect lines perpendicular to the same, intersecting lines of corresponding number drawn from K L. Then a line traced through the points of intersection, as shown from K2 to L2, will be the cor- rect section on A N of the plan, from which to obtain a stretchout of piece No. 2. The sections of pieces No. 3 and No. 4 are ob- tained in a similar manner from A O and A P. J3 L3 is a duplicate of A and J4 34 of A P. Perpen- diculars are erected from each of the points cutting horizontal lines of corresponding number, thus devel- oping K3 L3 and K4 L4. To obtain the pattern of A D C (No. 2) continue A N downward indefinitely, upon which lay off a stretchout of K2 L2, as shown by A2 N2, through the points in which draw the usual measuring lines. Place the "[-square parallel with A2 N2, and, bringing it against the several points in A C and A D, cut measuring lines of corresponding numbers. Lines traced through these points, as shown by A2 C2 and by A2 D2, will be the pattern sought. The stretchouts for pieces Nos. 3 and 4 are taken respectively from profiles 3 and 4. A3 03 is the stretchout of K3 L3 and is laid off on a continuation of A O, while A4 P4 is taken from K4 L4 and is set off on a continuation of A P. The remaining opera- tions are the same as those employed in obtaining the other pieces. PROBLEM 83. Pattern for a Three-Piece Elbow, the Middle Piece Being: a Gore. Let A B C D E F Gr in Fig. 375 be the elevation of a three-piece elbow to any given angle, as G F E, the middle piece of which, B C II, forms a gore extend- ing around one-half the diameter. The lines H B and II C are drawn parallel respectively to the ends of the two outer pieces, therefore the patterns for the end pieces will be straight from H to C and H to B and mitered from H to F. To obtain the pattern for one of the ends, as F H ODE, divide N K L into any convenient number of equal parts. With the 7-square at right angles to N L, carry lines from the points in N K L, cutting the miter line F H C, as shown. (X any line, as E D extended, lay off a stretchout of M K L, as shown by e d, through the points in which draw the usual measuring lines, as shown. With the T-square brought successively against the points in F II C, cut corresponding measuring lines, as shown. A line traced through the points of intersection, as shown by fh" c, will be the half pattern of end, as represented in elevation by F H C D E, or in profile by N K L. The other half of the pattern can be obtained by dupli- cation. 188 The New Metal HW.vr Pattern Bool'. Since II C is drawn parallel to E D, the distance II J is less than one-half the diameter of the normal profile, or less than L, therefore it will lie necessary before obtaining the pattern of B II C to obtain a cor- rect section on line H J of elevation. To do this, first place tlie T-square parallel with B C and carry lines from the points in H C, cutting II J and H B. Next draw any line, as K' M' of the section, and erect the perpendicular H' J'. From II', on H' J', set off the spaces in H J of elevation, transferring them point by point. Through the points thus obtained draw lines parallel with K' M', as shown. With the dividers take the distance across K O L or L O M, on the several lines drawn parallel with K M, and set off the same distance on lines of corresponding number drawn through H' J'. Thus H' M' and H' K' are the same as K and M. A. line traced through the points thus obtained, as shown by K' J' M', will be the section desired. For the pattern of II B C, lay off on H J extended, as h h', a stretchout of K' J' M' of section, through which draw the usual measuring lines. With the T-square placed parallel with II J, and brought successively against the points in B II and C H, cut corresponding measuring lines drawn through k h', as indicated by the dotted lines. Lines drawn through these points of intersection, as shown by 1> 1> h' r, will be the pattern of the part shown in. elevation by B II 0. K' H' SECTION. Fig. 37S.lttltei-n /../ a Three-Piece Elbuw, tlie Middli- Piece lie in ij a (Jure. PROBLEM 84. The Patterns of a Tapering Article which is Square at the Base and Octagonal at the Top. A B D C in Fig. 376 shows the plan of the article at the base, IKLMHGrFE represents the shape at the top, E 3 H' J D" C 3 is an elevation of one side. In order to obtain the slant hight of the octagonal sides it will be necessary to construct a diagonal section or ele- vation. Therefore extend the lines of the base C 3 1) 2 and top E 3 ir, as shown, to the left, through which draw a vertical line, as R 1 C 2 . Upon the line of the base set off each way from C' a distance equal to A R or R 1) of the plan. In like manner upon the line of the top, set off from R' each way a distance equal to one-half I G. Draw I 1 A' and G' D 1 , thus completing a diagonal section. If it is desired to complete a diagonal eleva- tion, set off E s F" equal to E F of the plan and draw lines to C ! , as shown. To obtain the pattern of one of the smaller sides, produce the diagonal line R C, upon which set off the length or stretchout of I' A 1 , as shown by N C', and draw the measuring line E' F 1 . By means of the T-square, as indicated by the dotted lines, set off E 1 K 1 equal to E F of the plan and draw C' E 1 and C' F'. Then E' C' F' is the pattern of one of the smaller sides 1'iitii i-ii Problems. 189 Fig. 377. Pattern in One Piece, of the article. For the pattern of one of the larger sides, draw 11 P perpendicular to the side A C, upon which set oil' P, in length equal to E 3 C' of the ele- vation, at right angles to which through and P draw measuring lines. By means of the T-square, as shown by the dotted lines, make A 2 C 4 equal to A C of the plan. In like manner make F E' equal to I E of the plan. Connect A 2 P and C 4 E 4 . Then A' T E 4 C 4 will be the pattern of one of the larger sides of the article. If for any reason the pattern is desired to be all in one piece the , shapes of the different sides may be laid off adjacent in c:ich other, the large and small sides alternating, all as indicated by i a a 1 , Fig. 377. Pig. 376. Elevations, Plan and Patterns. A Taperinij Article Which is Square at the Base and Octagonal at the Top. PROBLEM 85 The Patterns of a Finial, the Plan of which is Octagon with Alternate Long: and Short Sides. In Fig. 378, let A L M N P K S T be the ele- vation of the iinial corresponding to the plan which is shown immediately below it. The elevation is so drawn as to show the profile of one of the long sides, for the pattern of which proceed in the usual manner. Divide the profile A L M N O P into any number of convenient spaces, as shown by the small figures, and from the points thus obtained drop lines across tin; corresponding section in the plan, cutting the miter lines DEC and D 2 F C, as shown. A duplicate of the part E C F of the plan is shown below by K 1 C 2 F', and in the demonstration C 2 K 1 and 0" K' are to be consid- ered the same in all respects as C K and C F. The same mav be said of the two parts of the stretchout line bearing like letters; the division having been made on account of the extreme length which the pat- tern would have if made in one piece. Perpendicular to D D 2 lay off a stretchout, as shown by G H, through the points in which draw measuring lines in the usual manner. Place the y-square parallel to the stretchout line, and, bringing it against each of the several points in D E C and D 2 F C, cut the corresponding measuring lines. Then a line traced through these points of intersection will be the pattern sought. For the pattern of the short sides a somewhat dif- ferent course is to be pursued. As the distance from C to the line K E is greater than that from C to E F a profile of the piece as it would appear if cut on the line C 1) must first be obtained. To do this proceed as follows : From the points in C E, dropped from the 190 The Sew Metal Worker Pattern Bwk. 1 31+7 4 '6 B 1 5 17 Fig. 378. Elevation, Plan and Patterns. Fig. 379. Diagonal Section. The Patterns of a Finial, the Plan of Which is Octagon with Alternate Loruj and Short Sides. profile, carry lines parallel to E K across C D, cutting C K, as shown. At any convenient place lay off B' JP 1 , Fig. 379, in length equal to C D of the plan, on which lay off points corresponding to the points ob- tained in C D, and for convenience in the succeeding operations number them to correspond with the num- Pattern Problems. 191 Vrs in the profile from whirh they are derived. At IV erect the perpendicular B' A', equal to B A of the elevation. From the several points in the profile of the elevation draw horizontal lines cutting the 'central vertical line A B, as shown. Set off points in A' B 1 in Fig. 379 to correspond, and through these points draw horizontal lines, which number for convenience of iden- tification in the following steps. From the several points in B 1 P' carry lines vertically, intersecting correspond- ing horizontal lines. Then a line traced through these points, as shown by A' L' M' N' O 1 P', will be the profile of the short side on the line C D of the plan. After obtaining the profile as here described, for the pattern of the short side proceed as follows : Perpen- dicular to K E of the short side^ or on C D extended, lay off a stretchout of the diagonal section A 1 O' of Fig. 379, as shown by C' D 1 , through the points in which draw measuring lines in the usual manner. Place the T-square parallel to the stretchout line, and, bring- ing it against the several points in the miter lines D K C and DEC bounding the short side in the plan, cut the corresponding measuring lines. Then a line traced through these points, as shown in the diagram, will be the required pattern. PROBLEM 86. The Pattern for a Gore Piece Forming a Transition from an Octagon to a Square, as at the End of a Chamfer. In Fig. 380, let F F F F represent the plan of the square portion of a shaft and A A A A that of the Pattern Fig. 380. The Pattern for a Gore Piece Forming the Transition from an Octayon to a Square. octagon portion. Let D P C be the elevation of the gore piece which is required to form the transition be- tween the two shapes. The outline C D, which repre- sents the intersection of the gore piece with the side of the shaft, may be of any contour whatever at the pleasure of the designer, the method of laying out the pattern being the same no matter what its outline. By reference to the plan it will be seen that the lines of the molding, of which C D shows only the termination, run octogonally, or in the direction of A A. There- fore, before a stretchout of the piece can be obtained a correct profile must be developed on a line at right angles to its lines that is, on the line E F. To do this proceed as follows : Divide the line C D, as it appears in the elevation, into any convenient number of spaces, as shown by 1, 2, 3, 4, etc. From the points thus obtained drop lines down upon the side of the plan A F, which should be placed in line below the ele- vation. Continue the lines from the side A F across the corner, as shown, all parallel to A A of the octagon, crossing E F, and number the lines to correspond with the numbers of the points in the elevation from which they were derived. Draw the vertical line G n at a convenient distance from P D, and cut G n by lines drawn at right angles to it from the points in C D, as shown by the connecting dotted lines. G II then may be considered to represent the point F in the plan, or 11 of the numbers on the line E F. From G II, on each of the several lines drawn through it, lay off a distance equal to the space from E to the corre- sponding number in the same plan. Thus lay off from G H on line 1 a distance equal to 11 1 on E F, and on line 2 a distance equal to 11 2 of E F, and so on for each of the lines through G H. Then a line traced ' through these points, as shown by I H, will be the 192 The New Metal ll'w/ar Pattern Book. profile of the gore piece, or the shape of its section when cut by the line E F. Prolong E F, as shown by K L, and lay off on the latter a stretchout of the profile I H, the spaces of which must be taken from point to point as they occur, so as to have points in the stretchout corresponding to the points on the miter lines A F, previously derived from C D. Through the points thus obtained draw the usual measuring lines, as shown. Place the T-square at right angles to the measuring lines, or, what is the same, parallel to E F, and, bringing it against the points in A F and F A, cut the corresponding lines drawn through the stretchout. Lines traced through these points, as shown, will constitute the pattern. PROBLEM 87. The Pattern for a Gore Piece in a Molded Article, Forming: a Transition from a Square to an Octagon. In Fig. 381, let A B D C represent the elevation of an article of which G H I J is the half plan at the base and K L M N O P the half plan at the top. A C of the elevation is the normal profile or profile of one of the square sides, and L H and M II of the plan show the miter lines between the square sides and the gore piece. C E and D F, the elevations of the miter lines II M and I N, are shown as part of the design, but are not necessary in cutting the pattern. As only the normal profile, which would be used in cutting the pattern of one of the square sides, is shown in the elevation, the first step will be to obtain from this a profile of the gore piece, or in other words, a section upon its center line, R H of the plan. Divide the profile A C into any convenient number of parts, and from the points obtained drop lines at right angles to A B, cutting the miter line L H in plan, as shown. From the intersections obtained on the miter line L II draw lines parallel to L M, as shown, cutting the other miter line H M, and continue them indefinitely. At any convenient position outside the plan draw the line A 1 A" parallel to H R, and draw a duplicate of the profile A C in the same relative position to A 1 A 2 that A C holds to A B, and divide the same into the same spaces as A C, all as shown by A 1 C'. From the points in A 1 C 1 draw lines parallel to A 1 A", cutting lines of corresponding number drawn through the plan of the gore piece. A line traced through these intersections, as shown from A" to C", will be the profile of the transition piece from which to obtain the stretchout for the pattern. To obtain the pattern proceed as follows : Upon K' R 2 , a continuation of II R, place the stretchout of the profile A' C a , as shown by the small figures, through which draw the measuring lines, as shown. These are n<>\v to be intersected bv lines drawn from points of corresponding number upon the miter lines H L and II M. Lines traced through the points of intersection, Fig. X81.The Pattern for a Gore Piece in a Molded Article, Forniinrj shown by R T and R S, will give the desired pattern. a Transition from a Square to an Octa,j,, lt . PROFILE OF GORE PIECE. f '/!//< r n lit 3 PROBLEM 88. The Patterns for a Raking: Bracket. Tliis is one 1 of the many instances which calls for special draftsmanship on the part of the pattern cutter. Frequently the architect's drawings give only a detail of a bracket for the level cornice of a building, while the scale elevations show one- or more of the gables to be finished with raking brackets. In such cases the detail of the " level " bracket, and the pitch of the roof are the only available facts from which to produce the required bracket. In Fig. :;s-J, let M X or <) I' be drawn at the re- quired angle, with reference to any horizontal line, to represent the pitch of the gable cornice. The first step is to redraw the normal side elevation of the level bracket so that its vertical lines shall be at right angles to the lines of the rake, all as shown at L Q P. Next, at any convenient distance from this draw two rertical lines, as M ( ) and N' 1", the horizontal distance between which shall be the required face width of the bracket. Lines projected parallel to the rake from the various j angles in the profile between these vertical lines will complete the front elevation of the raking bracket. The additional lines E ',-i< Tiook, by N 1 U 1 K M, being u Hat surface, as indicated in the side view N U, is obtained by pricking directly from the face view of the bracket, no development of it being necessary. To avoid confusion of lines, the sink piece E FH prolilc, as shown by ' d\ is laid oil at right angles to the lines of the rake, and through the points in it the usual measuring lines are drawn. The " -square is then placed at right angles to the lines <>f the rake, and, be- ing brought successively against the points in the sides VP \^.\\\\$^\\Y c in the side view, being vertical, is obtained by pricking directly from its elevation in llie face view of the bracket, A B D' C being the shape. For the other straight strip bounding this panel, shown in the side view by a b, the length is laid off equal to a b, while the width is taken from the face view, equal to the space indicated by A B. For the strip representing the irregular part a to c proceed as follows : Divide the profile a d c into any convenient number of parts, from the points in which carry lines crossing the face view of the same part, as indicated by A B 1)' C. At right angles to the lines of the rake lay off a stretchout of the profile just named, as indicated by a' c', through the points in which draw the usual measuring lines. Place the 1 -square at right angles to the lines of rake, and, bringing it against the several points in the line A C and B D', cut the corresponding measuring lines drawn through the stretchout. Then lines traced through the several points of intersection thus formed, as indi- cated by A' G" and B 1 D', will be the pattern of the curved strip forming part of the boundary of the panel in the side view of the bracket. Of the three pieces of molding forming the head of the bracket, the profile of the piece across the face is normal, as shown at L N, while that of the two side pieces, or returns, requires to be modified or raked before a square miter with the face piece can be eH'ec.tcd. These principles will be further explained in Problems 91 to 94. The first step will be to draw a correct elevation of the head, which includes raking the profiles of the upper and lower returns. Divide the normal profile L N into convenient spaces, and from the points thus obtained carry lines indefinitely parallel to the rake across the top of the I'.K e view of the bracket. Draw duplicates of the nor- mal profile, placing them in a vertical position directly above where the new sides are required to be, as shown b\- it I and k m. Divide these two profiles into the same number of parts employed in dividing the normal profile, and from these points drop lines vertically, intersecting those drawn from L N. Then lines traced through these points of intersection, as shown by L' N 1 and K M, will be respectively the profiles of the moldings on the upper side and on the lower side of the bracket. Lay off a stretchout of the profile L N at right angles to the line of the rake and through the points in it draw the usual measuring lines. "With the blade of the T-square at right angles to the lines of the rake, and brought successively against the several points in the profile N' L' and K M, cut the measuring lines drawn through the stretchout. Then lines traced through the points of intersection thus obtained, as shown by L 3 N' and K 1 M', will be the shape of the ends of the molding forming the front of the bracket head. Before laying out the pattern for the return mold- ing forming the upper side of the bracket head a cor- rect side elevation of it must be drawn. A duplicate of the profile L' N 1 is transferred to any convenient place, as shown at L 3 1ST* in Fig. 383, and parallel lines from its angles are extended to the right, as shown, making L' Q 1 equal to L Q of the side view of the bracket. At Q 1 repeat the outline L" N 4 , which represents the intersection of the bracket head with the bed mold of the cornice. L' N 1 of Fig. 382 is then the cor- rect profile and the lines L' N 4 and Q 1 X" are the miter lines of this return ; however, as both the miter lines are identical with the profile, the stretchout q x may be taken from either one, the other being divided into the same number of spaces as the first, which is easier than dropping the points from one to the other. The T-square may then be placed at right angles to the lines in the molding and brought successively against the points in the lines L 3 N 4 and Q' X*, and the cor- responding measuring lines intersected. Then lines traced through these points, as shown .by L 4 N* and Q 3 X 3 , will form the pattern. The pattern for the return molding of the head occurring on the lower side of the bracket is obtained in the same manner. A duplicate of the profile K M of the face view of the bracket is drawn at any con- venient place, as shown by K 3 M 2 in Fig. 384. The proper length is given to the molding by measuring upon the side view of the bracket, and a duplicate of the profile is drawn at the opposite end. Space the profile K" W into any convenient number of parts, as indicated by the small figures, and in like manner di- vide the profile K 3 M 3 into the same number of parts. At right angles to the line of the molding lay off a stretchout of these profiles, as shown by k' m 1 , through which draw the usual measuring lines. With the blade of the T- sr l uarc a t right angles to the lines of the molding, and brought successively against the several points in the profiles K 3 M 3 and K 3 M 3 , cut the corre- sponding measuring lines. Then a line traced through these points of intersection, as shown by K 5 M 5 and K 4 M 4 , will constitute the pattern of the return mold- ing, or the lower side of the bracket. 196 Tlie New Metal Worker Pattern Book. PROBLEM 89. The Pattern for a Raised Panel on the Face of a Raking Bracket. In the solution of the problem stated above, and which is given in Fig. 385, the first requisite is tin- design or outline of the side of the normal bracket, as such an outline is really a section through the raking bracket upon a line at right angles to the rake. N S T shows the side view of a normal bracket, or the bracket as it would appear in a level cornice, the part from G to H being molded as shown by the shaded profile, which profile, being a section on line a b of the normal bracket, is given complete at J and called the nor- mal profile. The first step is to derive from these factors a front elevation of the molded panel upon tin- face of the raking bracket. To accomplish this first divide the profile of the panel molding into any con- venient number of equal parts, as shown in the section shaded in the side of the normal bracket, and through these points draw lines parallel to the face of the bracket, producing them iintil they cut the upper sur- face against which the panel terminates, and in the opposite direction until they meet the vertical surface in the lower part of the bracket against which the panel terminates at the bottom. From the points thus ob- tained in the horizontal surface near the top of the bracket and in the vertical surface near the bottom of the bracket draw lines at right angles to the face, thus transferring the points to the line representing the outer face of the panel, as shown from G to II. These points will be used a little later in develop- ing the view of the panel at rjght angles to the face. Next, from the points already obtained in the line rep- resenting the vertical surface near the bottom of the bracket carry lines parallel with the rake, extending them across the front elevation of the bracket. In the diagram, to avoid confusion, these lines terminate at the intersections shown from A to B, but in actual work they would be extended across the front eleva- tion, thereby making also the intersections shown from C to D. At any convenient place in line with the front elevation of the raking bracket draw the normal profile, as shown below the elevation, and divide it into spaces corresponding to the spaces used in dividing the profile in the side view. From the points thus obtained carry lines vertically, intersecting those just drawn from the side of the normal bracket across the front elevation. A line traced through the points of intersection gives the outlines shown at A B and C D. These outlines constitute a front, elevation of the lower end of the molded panel, or the view as seen from u point exactly in front of the face <>f the raking bracket when finished and in its proper or linal position. The outline or shape of the upper end of the panel \vouhl appear as a simple straight line in this view because it miters against a surface which is horizontal from front to back. ABC I) F K shows the entire front view of the molded panel. This view furnishes the means for the next step, which is to obtain a view at right angles to the face G H, and at the same at right angles to the lines of the rake N 0. To do this, first continue the lines from the normal profile of panel in their vertical course till they intersect the upper line of the panel E F. These lines are omitted through the face of the bracket, the points only being indicated on the line K F. From the points thus established in K K. and from the points derived in the outlines A B and C D, carry lines at right angles to the raking cornice, producing them indefinitely, as shown. At right angles to the raking cornice, at any convenient place, draw the line H' and G 1 , setting off on it spaces corresponding to those established in II G, already described. Through the points in H 1 G 1 draw lines at right angles to it to the left, producing them until they intersect lines al- ivadv drawn from the outlines A B and C D and the points in the line E F. Through the points of inter- section thus obtained, a.s indicated by 1 T in the lower left hand corner, 8 14 in the lower right hand corner, s, 9, 10, etc., in the upper right hand corner, and 1, 2, 3, 4, etc., in the upper left hand corner, trace lines, thus completing a view of the panel piece at right angles to its face. The next step to be taken is to develop a true profile of this panel, or in other words, a section at right angles to its lines, from which to obtain a stretchout for the required pattern. To do this, first assume any line, as P 0, at right angles to the lines of the view just obtained as the surface of the panel in the new profile. Upon this line extended,, as at K, draw a duplicate of normal profile so that the points 7 and S shall lie in it. Divide the profile K into the same number of spaces as in previous instances, and from these points carrv lines through the face view in- tersecting them with lines of corresponding number, as Pattern Problems. 197 14 NORMAL PROFILE, a Fig. 385. The Pattern for a Raised Panel on the Face of a Raking Bracket. 198 Tlic New Metal Worker Pattern Book. shown at L P and Q R. Then L P Q K will be the true profile of the moldings along the face of the raking bracket. The student will observe that only half the profile is shown at K, as both halves are alike, one- half will answer all purposes if it be kept in mind while making the intersections by number that the points 1-7 in one profile are 14-8 in the other. At any con- venient place lay off the stretchout of the true profile, as shown to the left by the line L M. Through the points in this line draw the usual measuring lines, as shown. Then, with the blade of the T-square placed parallel with the stretchout line and brought against the several points of intersection at the corners of the " View at Right Angles to the Face," cut correspond- ing measuring lines. Lines traced through the points thus obtained will produce the pattern shape, as shown. PROBLEM 90. The Patterns for a Diagonal Bracket Under Cornice of a Hipped Roof. In Fig. 386 is shown a constructive section of the cornice of a hipped roof, under which the bracket L fits against the planceer and over the bed molding C. Fig. 387 shows an inverted plan of the angle of such a cornice, including two normal brackets B and C, and the diagonal bracket D, of which the patterns are re- quired. At A, in line with one arm of the cornice in plan, is also shown a duplicate of the profile of the normal bracket. E F represents the miter line of the planceer over which the diagonal bracket is required to fit. Two distinct operations are necessary in obtaining the patterns of the bracket D, one for the face pieces and the other for the sides. As the bracket is placed exactly over (or more properly speaking under) the miter in the cornice, one-half its width must be drawn on either side of the miter line, as shown in Fig. 387. Each half of its face thus becomes a continuation of the moldings forming the faces of the course of normal brackets of which it is a part. Therefore the normal profile X 8 of the bracket A is the profile to be used, and I G and J F form the miter lines for one half the face. The usual method in obtaining the pattern for the face piece would be to divide the profile of A into any convenient number of spaces and lay off a stretchout of the same upon any line drawn at right angles to the direction of the mold that is, at right angles to I Jor G F after which lines should be dropped from the profile upon the rniter lines and thence into-the stretch- out. However, as the miter is a square miter, the short method is available ; hence the stretchout line is drawn at right angles to the horizontal line of the ele- vation X X, as shown at II G. The usual measuring lines are drawn and intersected with lines from points of corresponding number on the profile. Lines traced through the points of intersection, as shown by K M and L N", will give the pattern for half the face. The operation of obtaining or " raking " the pat- tern of the side is exactly similar to that employed in Problem 88, with the difference that while in Problem 88 the side is elongated vertically, in the present in- stance (the cornice remaining horizontal, and the bracket being placed obliquely) it is elongated laterally or horizontally. The operation is also complicated by the addition of a profile at the back edge of the bracket Fig. S80. Sectional View of the Cornice of be divided into a convenient number of spaces, as shown by the small figures, which must also be dropped upon E F, as shown, and numbered correspondingly. H -r transfer the points and spaces from E F. Low from each point in the line E' F', erect lines vertically, in- tersecting lines of corresponding number previously drawn to the right from the elevation. Thus, lines drawn upward from the intersections 1, 2, 3, 4, etc., on the line E' F' intersect with horizontal lines 1, 2, 3, 4, etc., while lines drawn upward from the intersec- tions 1', 2', 3', 4', etc., on the line E' F' intersect with horizontal lines 1', 2', 3', 4', etc. Lines traced through the points of intersection, as shown by R S P, will be the required pattern of the side. If it be desirable to ascertain the exact angle to which to bend the edges or flanges of the bracket to fit against the planceer it may be accomplished in the following manner : Extend P of the pattern of the side till it intersects the line from X of the side eleva- Fig. S87. Inverted Flan of Cornice and Method of Obtaining Patterns. From each of the points in the profile of the elevation carry lines indefinitely to the right, as shown. At any convenient point at the right of the plan, draw another plan of the diagonal bracket, so placed that its sides shall be parallel with the horizontal line X X of the elevation, all as shown, and upon its center line E' F' tion, as shown at Y. Upon the solid line X X in diagonal elevation establish any point, as T. Through the point T and at right angles to Y P draw a line in- tersecting the line Y P at U. As the angle of the plan shown in Fig. 387 is a right angle, construct a right angle, ABC, Fig. 388, 200 Tlie New Metal Worker Pattern Book. and bisect it, obtaining the miter line B L. Now take the distance from Y to T in diagonal elevation, and place it 'on the miter line B L in Fig. 388, from B to D. At right angles to B L draw a line through the point D, intersecting the sides of the right angle A B Fig. S88. Diagram for Obtaining Angles for Sending the Flanges. C at E and F. Now take the distance T U in diag- onal elevation and set it off from D toward B, locating the point H. Connect the points E, II and F; then will the profile E H F in Fig. 388 represent a section across the hip at right angle to its rake and will also be the angle to be used in putting the straight parts of the face together, as shown by E' H' F' in Fig. 389. Tin- anisic which the sides of the bracket make with the planceer will be the complement of the angle II K D of Fig. 388 and may be obtained as follows: Paral- lel to B L, in Fig. 388, and through the point E, draw Fig. S89. Perspective View of Finished Bracket. I K, representing the vertical side of the bracket ; then will the angle J E I represent the profile required IW bending the flanges on the side of the bracket, the pro- file being shown in position by I' E' J' in Fig. 389. In Fig. 389 is shown a perspective view of the finished bracket as seen from below. PROBLEM 91. To Obtain the Profile of a Horizontal Return, at the Foot of a Gable, Necessary to Miter at Right Angles in Plan With an Inclined Molding of Normal Profile, and the Miter Patterns of Both. In the elevation B C E D, and plan F G H K L I, of Fig. 390 is presented a set of conditions which necessitate a change of profile in either the horizontal or raking molding, in order to accomplish a miter joint at I II in the plan. In other words, the condi- tions are such that with a given profile, as shown by A 1 , in the raking molding, the profile of the horizontal molding forming the return will require to be modi- fied, as shown by the profile A*, in order to form a miter upon the line I H in the plan. The reason for this is easily found. If a vertical line be erected from point 9 in profile A it will be seen that each line emanating from a point in the nor- mal profile A 1 becomes depressed after passing this vertical line, more or less, according as its distance away from this line increases, all in proportion to the amount of rake or incline of the face molding, as shown by the dotted lines. If, on the contrary, the profile A" be considered as the normal profile, the pro- file A' will have to be changed or "raked," in this case increased in hight, in proportion to the inclina- tion. (These conditions are treated in the succeeding problem.) The vertical hight of the profile of the re- turn may be measured in the side elevation and com- pared with that of the inclined molding by measuring across the latter at right angle's to the line B C. In this problem it is assumed that the profile as well as the pitch, or rake, of the cornice B C are established and that the profile of the horizontal re- turn is to be modified, or "raked," to suit it. To obtain this profile, first draw the normal profile in the raking cornice, as shown by A 1 , placing it to corre- spond to the lines of the cornice, as shown. Draw another profile corresponding to it in all parts, directly above or below the foot of the raking cornice, in line with the face of the new profile to be constructed, placing this profile A so that its vertical lines shall cor- respond with the vertical lines of the horizontal cor- Pattern Problems. 201 nice. Divide the profiles A and A' into the same number of parts, and through the points thus obtained draw lines, those from A 1 being parallel to the lines of the raking cornice, and those from A intersecting them vertically. Through these points of intersection of like numbers trace a line, which gives the modified profile, as shown by A 2 . Then A 3 is the profile of the horizontal return, indi- cated byG II 1 F in the plan. It is also the elevation of the miter line I II of the plan. Therefore at any convenient point at right angles to the lines of the raking cornice lay off the stretchout M N of the profile A 1 , through the points in which draw measuring lines in the usual manner. Place the J-^quare at right angles to the lines of the raking cornice, and, bringing it suc- cessivelv against the points in the profile A 2 , cut the corresponding measuring lines just described. Through the points of intersection trace a line, as shown by O P II. Then OPE will be the shape of the lower end of the raking cornice mitering against the return. For the pattern of the return proceed as follows: Construct a side elevation of the return, as shown by S V U T, mak- ing the profile V U the same as the profile A' of the elevation. Let the length of the return correspond to the return, as shown in the plan by F I. In the pro- file V.U set off points corresponding to the points in the profile A 2 as shown from B to D. At right angles to the elevation of the return lay off a stretchout of V U, or, what is the same, of the profile^A. 2 , as shown by W X, through the points in which draw measuring lines in the usual manner. Placing the T-square parallel to this stretchout line, and bringing it success- ively against the points in V U, cut tne corresponding measuring lines. Then a line traced through, these Z 1 Fig. 390. To OotzLi tte Pr.-f.le cf c, Il.-rizontzl return at the Foot of a Gable, Necessary to Miter at night Angles in Plan with an Inclined Molding of Normal Profile, and the Patterns of Both. points of intersection, as usual, from Y to Z, will be the pattern of the horizontal return. PROBLEM 92. To Obtain the Profile of an Inclined Molding Necessary to Miter at Right Angles in Plan with a Given Horizontal Return, and the Miter Patterns of Both. The conditions shown in this problem are similar to those in the one just demonstrated. In this, how- ever, the normal profile is given to the horizontal re- turn, and the profile or the raking cornice is modified to correspond with it. To obtain the new profile pro- ceed as follows: Divide the normal profile A 1 , Fig. 391, into any convenient number of parts in the usual manner, and from these points carry lines parallel to 202 Tlte New Metal Worker Pattern Book. the lines of the raking cornice indefinitely. At any convenient point outside of the raking cornice, and at F' 4 / \ | 1 K 1 ( * 1 2 3 4 [/ 5 F / 7 8 - 7 9 10 / ii 12 / 13 -I 2 4 K i H Plan Fig. 391. To Obtain the Profile nf an Inclined Mold- ing Necessary to Miter at Rir/ht Angles in Plan with a Given Horizontal Return, and the Pat- terns of Both. right angles to its lines, construct a duplicate of the normal profile, as shown by A", which divide into like number of spaces. With the T-square at right angles to the lines of the raking cornice, and brought suen sa- ively against the several points in this profile, cut corresponding lines drawn through the cornice from the profile A 1 . Then a line traced through these points of intersection, as shown by. A", will be the profile of the raking cornice. For the pattern of the foot of the raking cornice mitering against the return, take the stretchout of the profile A 3 and lay it off on any line at right angles to the raking cornice, as shown by P 0. Through the points in this stretchout line draw the usual measuring lines, as shown. With the T-square at right angles to the lines of the raking cor- nice, or parallel to the stretchout line, bring it successively against the points in the pro- file A', which is also an elevation of the miter, and cut the measuring lines drawn through the stretchout P O. Then a line traced through the points of intersection, as shown by B' R 1 , will be the miter pattern of the foot of the raking cornice. For the pattern of the return proceed as follows : Construct an elevation of the return, as shown by F' G' K 1 H', in di- mensions making it correspond to F G K II of the plan. Space the profile A of the re- turn in the same manner as A'. At right angles to the lines in the return cornice draw any straight line, as M N, on. which lay off its stretchout, through the . points in which draw measuring lines in the usual manner. Place the T-square at right angles to the lines of the return cornice, and, bringing it successively against the points in the profile A, cut the correspond- ing measuring lines. Through the points of intersec- tion ti'ace a line, as shown by G 2 K". In like manner draw a line corresponding to F 1 II 1 of the side eleva- tion. Then F 2 G" K s IP will be the pattern of the horizontal return to miter with the raking cornice, as described. PROBLEM 93. To Obtain the Profile of the Horizontal Return at the Top of a Broken Pediment Necessary to Miter with a Given Inclined Molding:, and the Patterns of Both. In Fig. 392, C B D represents a portion of the elevation of what is known as a "broken pediment," the normal profile of whose cornice is shown at A 1 . With these conditions existing it becomes necessary lo obtain new profiles for the returns at both the top And the foot. The method of raking the return at the foot has been described in Problem 91, and the method of raking the return at the top is exactly the same. If, Pattern Problems. . 203 in the designing of the pediment, the normal profile should be placed in the return at the foot, as is some- times necessary, then the profile <>f tin- inclined mold- ing must be first obtained, which in turn must be con- sidered as a normal profile and used as a basis of ob- taining the third profile, that of the return at the top. In Fig. 392, let A' be considered as the normal profile of the inclined molding. Divide A 1 into- any convenient number of parts in the usual manner, and through these points draw lines parallel to the lines of the cornice indefinitely.- At any convenient point out- side of the cornice, and in a vertical line with the point at which the new profile is to be constructed, draw a duplicate of the profile of the raking cornice, as shown lv A, which space into the same number of parts as A', already described. From the points in A draw lines vertically, intersecting lines drawn from A 1 . Then a line traced through these several points of intersec- tion, as shown by A 3 , will constitute the profile of the horizontal return at the top and also the miter line as shown in elevation. If the normal profile were in the horizontal return at the foot of the pediment and the modified profile in the position of A 1 , it would he immaterial whether the normal profile or a duplicate of the modified profile were in the place of A by which to obtain the intersecting lines, as the projection of the points only is to be considered in this opera- tion, and that is the same in both cases. For the pattern of the inclined molding proceed as follows : At right angles to the lines of the raking cornice lay off a stretchout of the profile of the raking cornice A 1 , as shown by F G, through the points in which draw measuring lines in the usual manner. Place the T-square at right angles to the lines of the raking cornice, and, bringing the blade successively against the points in the profile A", which is the miter line in the elevation, cut the corresponding measuring lines, and through these points of intersection trace a line, as. shown by G II. Then G II will be the pat- tern of the top end of the raking cornice to miter against the horixontal return. For the pattern of the' horizontal return the usual method would be to con- struct an elevation of it in a manner similar to that described for the return at the foot of the gable in the preceding demonstrations ; the equivalent of this, how- ever, can be done in a way to save a considerable por- tion of the labor. As the view of the miter line is the same in both the front and the side elevation the pattern may be de- veloped from the front just obtained in the following manner, with the result, however, that the pattern will be reversed : Draw the line K M perpendicular to the Fig. SOS. To Obtain the Profile of the Horizontal Return ' the Top of a Broken Pediment Necessary to Miter with a Given Inclined Molding, and the Patterns of Both. lines of the horizontal return, as it would be if shown in elevation. Upon K M lay off a stretchout of the profile A 1 , all as shown by the small figures, and through the points draw the usual measuring lines. With the T-square parallel to the stretchout line K M 204 Tlie New Metal Worker Pattern Book. bring the blade successively against the points in the profile A% cutting the corresponding measuring lines. Through these points of intersection trace a line, as shown by N L, which will be the pattern of the end of the horizontal return to miter against the trable cornice, as shown. PROBLEM 94. To Obtain the Profile and Patterns of the Returns at the Top and Foot of a Segmental Broken Pediment. The preceding three problems treat of the various miters involved in the construction of angular pedi- ments. In Fig. 393 is shown an elevation of a curved or segmental broken pediment in which the normal profile is placed in the horizontal return at the foot. The profiles for the curved molding and for the return at the top can both be obtained at one operation in the following manner : Divide the normal profile ABC into any convenient number of parts, and from the points thus obtained draw lines at right angles to the horizontal line C F of elevation, as shown. At any convenient point draw G II, at right angles to A Gr, cutting them. With Q, the point from which the curve of the molding was struck, as center, strike arcs from the points in A B C, extending them in the direction of D indefinitely. From any convenient point in the arc A D, as L, draw a line to the center Q. From L draw L M, at right angles to L Q, upon which, beginning at L, set off the distances contained in II G, as shown by the small figures in L M. From the points of intersection where arcs struck from Q cut L Q draw lines at right angles to L Q. From the points in L M, and at right angles to it, drop lines cutting those of similar number drawn at right angles to L Q. A line traced through these points of inter- section, as shown by M K, will be the profile of curved molding. It will be observed that the points for obtaining the profile are where he perpendiculars dropped from L M intersect the lines drawn at right angles to L Q, and not where the perpendiculars dropped from L M intersect the arcs. For the profile D E draw N D, parallel to C J, or at right angles to N 0, and, starting from D, set off on D N the same points as are in G II. Drop perpen- diculars from these points to the arcs of similar num- bers drawn from A B, when a line traced through the points of intersection will form the desired profile, as show by D E. The normal profile is also drawn above G II and N D at X and Z to show that the same result is obtained by using the points in G II to set off on L M and N D as would be obtained by dropping the points from the profiles. The patterns for the returns would be obtained as described in the previous prob- lems. Fig. 393. To Obtain the Profiles and Patterns of the Returns at the Top and Foot of a Segmented Broken Pediment. Problems describing the method of obtaining the pattern for the blank for the curved molding will be found in Section 2 of this chapter. Pattern Problems. 205 PROBLEM 95. From the Profile of a Given Horizontal Molding, to Obtain the Profile of an Inclined Molding Necessary to Miter with it at an Octagon Angle in Plan, and the Patterns for Both Arms of the Miter. Another example wherein is required a change of prolilc in order to produce a miter between the parts is shown in Fig. 394. In this case the angle shown us indicated, and in the corresponding side, as shown in elevation by N OLK, draw a duplicate profile, as shown by A'. S94. From the Profile of a Given Horizontal Moldiiiy to Obtain the Profile of an Inclined Moldiny Necessary to Miter n-ith it at an Octa- gon Angle in Plan, and the Patterns fur Roth Arms of the Miter. iii plan between the abutting members is that of an octagon, as indicated by BCD. To produce the modified, profile and to describe the patterns proceed as follows : In the side B draw the normal profile A, Divide both of these profiles into the same num- ber of parts, and from the points in each carry lines parallel to the lines of mold- ing in the respective views, producing the lines drawn from profile A until they meet the miter line C X. From the points thus obtained in C X erect lines vertically until they meet those drawn from profile A', intersecting as shown from to L. Through these points of intersection draw the line O L, which will be the miter line in elevation corresponding to C X of the plan. From the points in O L carry lines parallel with- the raking molding in the direction of P indefinitely. At any convenient point outside of the raking cornice draw a duplicate of the normal profile, as shown by A 2 , placing its vertical line at right angles to the lines of the raking cornice. Divide the profile A 2 into the same number of spaces as employed in A and A 1 , and from these points carry lines at right angles to the lines of the raking cornice, intersecting those of corresponding number drawn from the points in L. Trace a line through these intersections, as shown from E to S. Then R S will be the required profile of a raking cornice to miter against a level cornice of the profile A at an angle indicated by B C D in the plan, or an octagon angle. For the pattern of the level cornice, at right angles to the arm B C in the plan lay off a stretchout of the profile A, as shown by E F, through the points in which draw the usual measur- ing line. With the T-square at right angles to B C, bringing . the blade suc- cessively against the several points in X C, cut corresponding measuring lines drawn through I 1 ! !'. Then a line traced through these points, as shown from II to G, will be the required pattern of the hori- zontal cornice. In like manner, for the pattern of 206 The Xew Metal Worker Pattern Book. the raking cornice, at right angles to its lines lay off a stretchout of the profile R S, as shown l>y U T, through the points in which draw measuring lines in the usual manner. With the T-square at right angles to the lines of the raking cornice, arid brought success- ively against the points in the miter line O L, as shown in elevation, cut the corresponding measuring lines. Then a line traced through the points thus obtained, us shown by W V, will be the required pattern for the raking cornice. PROBLEM 96. From the Profile of a Given Inclined Molding, to Establish the Profile of a Horizontal Molding to Miter with it at an Octagon Angle ia Plan, and the Patterns for Both Arms. In Fig. 395, let B C D be the angle in plan at which the two moldings are to join, U V the angle in elevation, and A or A 1 the nor- mal profile of the raking mold. To form a miter between moldings meeting under these conditions a VL Fig. SOS. From the Profile of a Given Iiu-lined Moldiwj. to Establish the Profile of a Horizontal Moldimj to Miter with it at, an Octagon Angle in Plan, and the Patterns for Both Arms. change of profile is required. To obtain the modified profile for the horizontal arm and the miter line in elevation proceed as follows: Draw the normal profile A with its vertical side parallel to the lines in the plan of the arm E X D C, corresponding to the front of the elevation. Draw a dupli- cate of the normal profile in correct position in the elevation, as shown by A 1 . Divide both of these profiles into the same number of parts, and through the points in each draw lines parallel with the plan and with the elevation respectively, all as indicate.] by the dotted lines. From the points in the miter line of the plan C E, obtained by the lines drawn from the profile A, carry lines vertically, inter- secting the lines drawn from A 1 . Then a line traced through the inter- sections thus obtained, as shown from N to 0, will be the miter line in ele- vation. From the points in N O carry lines horizontally along the arm of the horizontal molding N O U Y, as shown. At any convenient point outside of this arm, either above or below it, draw a duplicate of the nor- mal profile, as shown by A", which divide into the same number of parts as before, and from the points carry lines vertically intersecting the lines drawn from N 0, just described Then a line traced through tin .-( points of intersection, as shown by T S, will give the required modified profile. For the patterns of the arm Y N O U proceed as follows: At right angles to the same, as shown in plan by W E C B, lay off on any straight line, as G F. a stretchout of the profile T S, all as sh'own by the small Pattern Problems. 207 figures I 3 , 2", 3", etc. Through these points draw measuring lines in the usual manner. With the T-square parallel to the stretchout line, and brought against the points of the miter line E C in plan, cut corresponding measurirg lines, as indicated by the dotted lines, and through these points of intersection trace a line, as shown by K II. Then K II will be the shape of the end of Y N U to miter against the raking molding. It will be easily understood that the points as found iipon the line E C arc just the same as would be obtained there if the newly obtained profile were drawn into the plan of the arm C 13 W E and the points were dropped from it to the line E C according to the rule. For the pattern of the raking molding, at right angles to the arm N Z V O in the elevation lay out a stretch- out, L M, from the profile A'. Through the points in this stretchout draw measuring lines in the usual manner. Place the T-square parallel to the stretchout line, and, bringing it against the several points in the miter line in elevation N 0, cut corresponding measur- ing lines, as indicated by the dotted lines. Then a line traced through these points of intersection, as shown by P R, will be the shape of the cut on the arm N Z V to miter against the horizontal molding. PROBLEM 97. The Miter Between the Moldings of Adjacent Gables of Different Pitches upon a Pinnacle with Rectangular Shaft. The problem presented in Figs. 396 and 397 is one occasionally arising in pinnacle work. The figures represent the side and end elevations of a pinnacle which Firj. 396. Side Elevation of Rectangular Pinnacle, Showing the Miter Between the Moldinijs of Adjacent Gables. is rectangular, but not square. All of its faces are tiuished with gables whose moldings miter with each other at the corners, and which are of the same hight in the line of their ridges, as indicated by L M and L' M 1 . Whatever profile is given to the molding in one face of such a structure, the profile of the gable in the adjacent face will require some modification in order to form a miter. In Fig. 396 let A be the normal profile of the molding placed in the gable of the side elevation. Before the miter patterns can be developed it will first be necessary to obtain the miter line or joint between the moldings of the adjacent gables as it will appear in the elevation, to accomplish which proceed as fol- lows : Draw a duplicate of A, placing it in a vertical position directly below or above the point at which the two moldings are to meet, as shown by A 1 . Divide both of these profiles into the same number of parts, as indicated by the small figures, and through these points draw lines intersecting in the points from H to K, as shown. Then a line traced through these intersections will be the miter line in elevation. For the pattern of the molding of the side gable lay off at right angles to II M a stretchout of the profile A, as shown by B C, through the points of which draw the usual measuring lines. Place the T-square at right angles to the lines of the molding, or, what is the same, parallel to the stretchout line, and, bringing it against the several points in the miter line II K, cut corresponding meas- uring lines. Then a line traced through these points, as shown by D E, will be the shape of the cut at the foot of the side gable to miter against the adjacent gable. The next step is to obtain the correct profile of the molding on the adjacent gable. H K having been established as the correct elevation of the miter, its 208 The Xcw Mdal Wurkvr Pattern Bovk. outline may now be transferred, with its points, to the end elevation of the pinnacle, as shown tit II 1 K', Fig. 397, reversing it, because it appears here at the right side of the gable, whereas it appeared at the left of the other. Draw a duplicate of the normal profile, as shown at A", placing its vertical lines at right angles to the lines of the gable, and divide it into the same spaces as in the first operation. From these points draw lines at right angles across the molding, which intersect with lines drawn parallel to the molding from the points in the miter line IP K'. Then a line traced through these points of intersection will form the required modi- fied profile, as shown by W X. For the pattern of the molding of the end gable proceed as follows : At right angles to the lines of the. raking cornice lay off a stretchout of the profile W X, as shown by P R, through the points in which draw measuring lines in the usual manner. With the T-square at right angles to the lines of the raking cornice, bringing it successively against the points in K 1 H 1 , cut corresponding measuring lines. Then aline traced through these points of intersection, as shown from S to T, will be the pattern required. Fly. S97. End Elevation of Rectangular Pinnacle, Slwn-ing Same Miter as in Fig. 396. PROBLEM 98. The Miter Between the Moldings of Adjacent Gables 01' Different Pitches upon an Octagon Pinnacle. This problem differs from the preceding one in that the angle of the plan is octagonal instead of square, but like it requires a change of profile in one of the gables in order to effect a miter. In Figs. 398 and 399 are shown a quarter plan of pinnacle and the elevations of two adjacent gables of different widths but of similar bights. Let A 1 B' F' O G 1 D' of Fig. 398 be a correct elevation and A B C G be a quarter plan of the structure. In that portion of the plan cor- responding to the part of the elevation shown to the front draw the normal profile E, placing its vertical side parallel to the lines of the plan. Divide it into any convenient number of spaces, and through these points draw lines parallel to the lines of the plan, cut- ting C O', the miter line in plan, as shown. In like manner plaee a duplicate of the normal profile, as shown by E 1 in the elevation. Divide it into the same num- ber of equal parts, aud through the points draw lines parallel to the lines of the raking cornice, which pro- duce in the direction of N O indefinitely. Bring the T-square against the points in C O 1 , and with it erect vertical lines, cutting the lines drawn from E 1 , as shown from N to O. Then a line, NO, traced through these points of intersection will be the miter line in elevation. For the pattern of the miter at the foot of the wide gable or gable shown in elevation proceed as fol- lows : At right angles to the lines of the gable cornice lay off a stretchout of the profile E 1 , as shown by II K, through the points in which draw the usual measuring lines. Placing the T-square at right angles to the lines of the cornice, or, what is the same, parallel to the .stretchout line, and bringing it against the several points in N O, cut corresponding measuring lines. Then a line traced through the points of intersect:. <:. thus obtained, as shown from L to M, will be the put- P/ltf/'/'ll Frnlil, in . 209 tern of the miter at the foot of the gable shown in ele- vation. Fur the mudilied profile of the gable molding / 'HIS. Quarter Plan nnd Elevation of Octayon Pinnacle, iiirj Miter Bctu-een Moldinrjs of Adjacent (tables of Different Pitches. upon the narrow side proceed as follows: Draw n cor- rect elevation of the narrow side, reproducing therein the miter line N () from Fig. 398 (reversing tne same), as shown by K P in Fig. 399, and through the points, also reproduced from N O, carry lines parallel to the lines of the gable cornice indefinitely, as shown. Draw a duplicate of the normal profile at any convenient point outside of the gable cornice, as shown by E', placing its vertical side at right angles to A" 11, or the lines of the cornice. Divide E 3 into the same number of parts as used in the other profiles, and through the points draw lines at right angles to the lines of the cor- nice, intersecting the lines drawn from P R. Through Fig. S'JO. Elevation of Narrow Side of Octagon Pinnacle, Showing Same Miter as in Fig. 398. these points trace a line, as indicated by E 3 , which will be the modified profile. To lay out the pattern take the stretchout of E 3 and lay it off on any straight line drawn at right angles to the lines of the cornice, as S T, and through the points in it draw the usual measuring lines. Place the T-square at right angles to the lines of the gable cornice, and, bringing it against the points in P R, cut the measuring lines, as indicated by the dotted lines. Then a line traced through these points of intersection, as shown by U T, will be the pattern for the molding at the foot of the gable on narrow side. 210 Tlie New Metal Worker Pattern Book. PROBLEM 99. The Patterns for a Cold Air Box in which the Inclined Portion Joins the Level Portion Obliquely in Plan. The conditions of the problem are clearly shown in the plan and side elevation of Fig. 400, in which Z B C is the elevation and X C' D' Y is the plan of the level portion of a cold air passage joining a furnace just above the floor line. The inclined portion of the air passage or box is required to join the level portion at the angle Z A E of the side elevation, and at the angle Y A' E' when viewed in plan. These conditions are in many respects similar to those given in Problem 95, with the difference, however, that in this case the joint or miter between the level and the inclined por- tions does not appear as a straight line in the plan. It may be here remarked that the solution of this prob- lem is more a matter of drawing than of pattern cutting, as nothing can be more simple than the cutting of a miter between two pieces of rectangular .pipe when the required angle between them is known. This problem is capable of two solutions, both of which will be given, leaving the reader to choose which is the more adaptable to his requirements. First Solution. As above intimated, before the pattern can be developed it will be necessary to make tareful drawings, in the preparation of which a knowl- edge of the principles of orthographic projection is necessary. (See Chapter III). To proceed, then, with the drawings, first draw a plan and elevation of as much of the furnace as is necessary to show its connection with the cold air box, placing each part of the plan directly under its corre- sponding part in the elevation, so that as soon as any new point is determined in either of the views its posi- tion can be located in the other by means of a perpen- dicular line dropped from one view to the other. Upon the plan set off the width of the box b and draw parallel lines from the side of the furnace body to the right indefinitely, and upon the elevation set off its hight, a, from the floor line up, and draw A Z. A vertical line from the point X of the plan will give the point Z upon the elevation, or, in other words, show how far the curve of the furnace body cuts into the top and bottom surfaces of the cold air box. Next, upon the elevation locate the point A the required distance from the side of the body according to specification and find its position in the plan by means of a vertical line, as shown. From the point A in both views lines must be drawn to represent the angle or deflection of the pipe as it would appear in those views. Thus the ele ration would show the slant, which is determined by the two dimensions c and d. Therefore from the point A of the elevation erect a perpendicular line equal to the required hight c, from the top of which draw a horizontal line to the right of a length equal to the amount of slant r the elevation, which have been correctly obtained, gives the true length of the inclined piece that is, flic true distance from A to E it will be necessary to obtain still another elevation, in which such distance is correctly shown. As A' E' of the plan gives the liori/.ontal distance between the points A and K, and c represents the vertical distance between them, if a right angled triangle lie constructed with A' K' as a base and the hight c as the perpendic- ular, itshypothenuse will then give the desired measurement. Such a triangle properly forms part of an oblique eleva- tion which may be pro- jected from the plan in the following manner: Parallel to A' E', at any convenient distance away, draw a line to rep- resent the level of the floor, as shown ; above which, at a distance equal to , draw another paral- lel line, X" A 2 , represent- . ing the hight of the hori- zontal arm of the pipe. Above the line X" A 3 , at a hight equal to c, draw still another line, upon which the point E is sub- sequently to be located. Fig. 400. Plan and Elevations of a Cold Air Box iti Which the Inclined Pcrtion Joins the Level Portion Obliquely in Plan. First Solution, 212 Tlie Xt-w Metal Worker Pattern Book, Now drop lines from all the points of the plan at right angles to A' E', intersecting each with its correspond ing line of the new elevation, thus locating each point of the miter in that view As points D' and C' are upon the floor, their position will be found at D 3 and C 2 . Like- wise lines from A' and B' will locate those points in the upper surface of the horizontal pipe, as shown at A* and B a , where they are also shown to be in the side elevation. A line dropped from E' will also locate that point at its proper hight, as shown at E a . A line con- necting A" and E 3 will then be the hypothenuse above alluded to and be the correct length sought. As all edges or corners of the pipe are necessarily parallel, lines drawn from B 2 C' and D' parallel to A' E J will complete this part of the elevation as far as necessary. In these, as in all geometrical drawings, lines showing parts concealed from view by other parts are always shown dotted. Lines from X and Y locate those points in the new elevation and show that, while a correct elevation of the inclined arm of the pipe has been ob- tained, the view of the horizontal portion is oblique, the space between X" and Y" showing the open end to fit against the furnace body. Having now obtained a correct oblique elevation, the next step is to obtain a correct profile upon any line, as F H, drawn at right angles across the pipe, which may be accomplished in the following manner : From each point upon the line of the section F, G, J and H project lines parallel with the direction of the pipe to a convenient point outside the elevation, as shown at the left, across which draw a line, x y, at right angles to them as a base from which to measure distances from front to back. Assuming its crossing with the line from G (point 1) to represent the near angle of the pipe, set off from x on the line from F the horizontal breadth of the pipe 6, thus locating point 4, which corresponds to the point F in the elevation. In like manner on the line from II set off from y the distance o of the plan, locat- ing the point 2, which corresponds to point II of eleva- tion, and draw the lines 1 4 and 1 2. The distance of point 3 from line x y is equal to distance b plus the distance o, or in other words, draw the line 2 3 par- allel to 1 4 and the line 4 3 parallel to 1 2, thus locating the point 3. Having now a profile and a correct elevation of the miter, nothing remains but to lay off a stretchout, as shown, upon the line H K and drop the points in the usual manner from the profile to the miter line A 5 B" C'' D", thence into the measuring lines of the stretchout, all as clearly shown in the drawing. As the plan shows all the dimensions of the hori- zontal arm of the pipe, the pattern for that can be de- veloped in the usual manner. To avoid confusion a duplicate of that part of the plan has been transferred to Fig. 401, where a stretchout of the normal profile is laid off at right angles to the lines of the pipe, into which the points are dropped from the miter line A 1? C D. In the normal profile of course the distances 1 4 and 2 3 are equal to I and the distances 1 2 and 4 3 equal to a of Fig. 400. *> D 12 PLAN Fig. 401. Plan and Pattern of Level Arm of Cold Air Boar. It may be noted here that, as is the case in all raked profiles, the dimensions and shape of the profile obtained from the oblique elevation differ somewhat from .those of the normal profile shown in Fig. 401, and that their stretchouts are therefore necessarily dif- ferent. Second Solution. It may be asked naturally, is there no way of producing a miter without a change of profile, just as a carpenter would saw off the ends of two square sticks of timber of the same section and produce a perfect miter at an oblique angle ? There is, but the method of doing it is not so apparent as the Pattern Problems. 213 one just described. To accomplish this a drawing or view must be obtained, in which the surface of the paper represents a plane commou to both arms of the shown in Fig. 402, in which the plan shown in Fig. 400 has been reproduced, but turned around in such a manner as to facilitate the projection from it of an end Fig. 402.Pfiiterns of Cold Air Box. Second Solution. pipe. As three points determine the position of a plane, it will be seen at once that such a plane passes through the points Z, A and E of the side elevation, Fig. 400. The best means of obtaining this view is elevation, all of which is clearly shown in the drawing. This view shows the offset e and the rise c of the oblique portion of the pipe. The new view, which will give the required conditions, is obtained by look- ing at the pipe in a direction at right angles to A E of the end elevation, and is obtained as follows : Parallel to A E at any convenient distance away draw A' E', which make equal to A E by means of the lines drawn at right angles to A E, as shown. Upon the line E' E set off from E' the slant d as given in the side eleva- tion and plan, Fig. 400, locating the point E", and draw the line E 2 A'. From all points of the profile or end view of the horizontal pipe, 1, 2, 3 and 4, project lines also at right angles to A E, continuing them across the line A' E', and make A' Y* equal to A Y of the plan. Then A' Y" will be the length of the horizontal arm in the new view and A' E" will be the length of the inclined arm, both lying in the same plane, and the angle E" A' Y" will be the angle at which the two arms meet. Under the above conditions, then, a line which bisects that angle, as A' C, will be the t)ie New Metal Worker Pattern Hook. miter line between the two arms. As the two arms of the miter are symmetrical, the view can be completed, if desired, by drawing lines parallel with A' E 2 from the points of intersection with the lines from the end view with the miter line A' C. As 1234 is the profile from which the short arm was projected in the new view, a stretchout may now be taken from it and laid off on any line at right angles to C W and the points dropped in the usual manner, all as shown. If desired. the stretchout may also be laid off at right angles to the inclined arm and the pattern for this piece thus developed from the same miter line, although the miter cut A B C D A is the same in both pieces, one simply being the reverse of the other. PROBLEM 100. The Patterns for the Inclined Portion of a Cold Air Box to Meet the Horizontal Portion Obliquely in Plan. This problem is here introduced on account of the similarity of its conditions with those of the one immediately preceding, although, as its patterns are obtained entirely without the use of profiles, it does not properly belong in this connection. Its solution will serve to show what widely different means may be employed to obtain the same ends. In the preceding case the miter cut was obtained without refer- ence to the miter at the upper end of the oblique arm. In this case the oblique portion is required to join, at its upper end, with another arm like and exactly parallel with the arm join ing its lower termination. Under such conditions it follows that the planes of the upper and lower miters must be parallel, and, therefore, that miter cut at the upper end of either of the faces of the oblique por- tion must be parallel with that at the lower end of the same. Advantage may be taken of these conditions to obtain a very simple solution of the problem, as will be seen below. The first requisite is, of course, a correctly drawn elevation and plan in which all the points in each are duly projected from corresponding points in the other view. In Fig. 403 is shown a plan and elevation of the box, with the lines of projection connecting cor- responding points in each, all of which may be con- structed very much as described in the preceding problem. ' The inclined arm is required to have a rise equal to a of the elevation and a forward projection equal to I of the plan. Corresponding points in the two views are lettered alike. Thus the elevation shows clearly that it is an elevatio:i of the front A B F E of the plan, with the back C I) II G dotted behind, while the plan shows clearly A B D G of the elevation with the bottom E F H Gr clotted below. Fig. JOS. Patterns for the, Inclined Portion of a CoM, Air Pox to Meet the Level Portion Obliquely in Plan. The first important information to be derived from the correctly drawn views is that the front and back are the same, likewise the top and bottom are alike. The patterns of the top and front are given separately, upon the supposition that joints will be made at all of Pattern Problems, SIB the angles ; should they he wanted in one piece they could readily be connected. As all the surfaces of the inclined portion of the pipe are oblique to the given view, only some of their dimensions will be correct as they appear on the paper. An inspection of both elevation and plan will show that the lines A C and 13 D are both horizontal and parallel, and, therefore, correct as they appear in the plan, and may be used as given in the construction of a pattern of the top piece. The shortest distance between these two lines will be represented by a line at right angles to both, as M N. Since the point N in the line B D is higher than the point M of the lino A C, by the distance a of the elevation, it will be necessary to construct the diagram J L K in order to get the correct distance be- tween the points M and N. J K is made equal to the distance M X, as indicated by the dotted lines. K L is equal to the rise given in the elevation; hence the distance J L represents the true distance between the points M and N. Upon the continuation of the line M X of the plan set off the distance J L, as shown at ,1 ' L'. Through each of these points lines are drawn parallel to A and B D of the plan. The line A' C' is made equal to A C, and B' I)' is made equal to B D by means of the dotted lines drawn parallel to M N. This pattern is completed by connecting the point A' with B' and C' with D'. In developing the pattern of the side A B F E the same course might be pursued, beginning with the lines A E and B F, whose lengths are correctly given in the elevation, but for the sake of diversity another method has been employed. Beginning with the known fact that the point B is higher than the point A, as shown by a in the elevation, con- struct a diagram, P K, making O P equal to and parallel with A B of the plan, and E equal to a, thus giving K P as the correct length of the line represented by A B of the plan. From the points E and F draw, at right angles to E F, the lines E S and F T indefinitely. Since the distances A E and B F are the same and are correctly given in the elevation, take that distance between the feet of the dividers, and placing one foot at the point E describe a small arc, cutting the line E S in the point S. By repeating this operation from the point P, the point T is established in the line F and T. Lines connecting the points E S, S T and T P will complete the pat- tern of the front and back. PROBLEM 101. The Pattern of a Hip Molding: upon a Right Angle in a Mansard Roof, Mitering Against the Planceer of a Deck Cornice. Let Z X Y V in Fig. 404 be the elevation of a deck cornice, against the planceer of which a hip mold- ing, shown in elevation by U W Y T, is required to miter. Let the angle of the roof be a right angle, as shown by the plan Q I) A', Fig. 405, D N" represent- ing the plan of the angle over which the hip molding is to be placed. This angle is also shown bv B A of the elevation. As the only view which will show the correct angle at which the hip molding meets the plan- ceer is a view at right angles to the line D N, the first step in the development of the patterns will be to con- struct such a diagonal elevation. Assume any point, as A, in the elevation on any line representing a plain surface in the profile of -the roof, as B A. Through A draw a horizontal line indefinitely, as shown by LAC. From B, the point in which the line A 5 meets the planceer, drop a vertical line, cutting the horizontal line drawn through A at the point C, all as shown by B C. Produce the line of planceer W Y, as shown by W Y'. Draw a duplicate of the plan, Q D A' in Fig. 405, in such a manner that the diagonal line DN shall lie parallel to the horizontal line drawn through A, all as shown by Q 1 D 1 A'. At right angles to the line D' A", at any convenient point, as A 5 , draw the line A" C', in length equal to the distance A C in elevation, and through C' draw a line parallel to D 1 A", as shown by I N 1 , cutting the diagonal line I) 1 N 1 in the point N 1 . Then D 1 N' represents the diagonal plan of that part of the hip from B to A in the elevation. From N 1 erect a perpendicular, N 1 M, which produce until it 216 Tlie New Metal Worker Pattern Book. meets the line carried horizontally from the plancccr in the point B'. In like manner from I) 1 erect a per- pendicular, which produce until it meets the horizontal line L C in the point L. Connect L and B', as shown, which will constitute the desired oblique projection of A B. The next step will be to construct a section of the hip molding u-pon a line at right angles with it, as G the side D' A 1 , from which erect a line perpendicular to D 1 N', as shown by K K, which produce until it meets the horizontal line L C in the point L 1 , and thence carry it upward parallel to L B 1 , cutting G II in the point F 1 . On either side of F 1 lay off a space equal to F E of the diagonal plan, as shown by F' K' and F 1 E". Through these points E' and E" draw lines to K, the intersection of the lines L B' and G II. From Fig. 404. The Pattern of a Hip Molding in a Mansard Roof, Mitering Against, the Planceer of a Deck Cornice. II, assumed at any convenient point. It might be supposed that in such a section the two fascias of the hip molding would be at right angles to each other, as they undoubtedly would appear in the plan Q D A' of Fig. 405 or in a section on any horizontal line, as L M. The object of this part of the demonstration is to show exactly what that angle would be and how to obtain it. Assume any point in the diagonal plan, as K, in K as a center describe the curve of the roll of the re- quired diameter. Upon the lines K E' and K E" ccs off from K a distance sufficient to make the desired width of fascia, thus completing the profile of the hip molding in the diagonal elevation. Space one-half of this profile, as G E J , in the usual manner, through the points in which carry lines par- allel to L B', cutting the line of plam-ccr W V, which Pnaern Problems. 217 is the miter line of the roll. The edges of the fascia will of course miter with the lower edge of the fascia Fig. 405. Plan of the Fascias and Angle of the Mansard Shown in Fig. 404. at the top of the mansard, shown in profile at B E 3 , all as shown by the dotted lines projected from E 3 . At right angles to the line L B' draw the straight line S R, upon which lay off a stretchout of the profile in the usual manner, and through the points draw measuring lines. With the T-square parallel to this stretchout line, or, what is the same, at right angles to the lines of the molding in the diagonal elevation, and, bringing it successively against the points in W 1 Y 1 , cut corre- sponding measuring lines drawn through the stretchout. The measuring lines 7 and 8 are cut from the inter- section of the fascia of the hip with lines projected from E 3 as above explained. Then a line traced through these points, as shown in the engraving, as shown by J 1 P J, will be the pattern of the hip molding inhering against the horizontal planceer. PROBLEM 102. The Pattern for a Hip Molding upon a Right Angle in a Mansard Roof, Mitering Against a Bed Molding at the Top. Let A C B, in Fig. 406, be the section of a por- tion of a mansard roof, the elevation of which is shown to the left, and let P E be any bed molding whose profile does not correspond to or member with the molding used to cover the hips, a section of the hip molding being shown at Z Y C'. The solution of this problem will be accomplished by means of a "true face" of the roof, rather than by means of a diagonal elevation as in the problem immediately preceding this. Therefore, supposing the section A C B to give the correct pitch of the roof, the first step will be to obtain the true face, or elevation of the roof as it would appear if tipped or swung into a vertical position, for the purpose of get- ting the correct angle at A' B 1 F'. To do this reproduce the section of mansard and bed molding as a whole at a convenient point below, but so turned as to bring the faces of the roof into a vertical position, maintaining the same distance be- tween the points A and B as shown by A" and B 2 . Project lines horizontally to the left from this section for the true face, marking the lines from the points A 2 and B 2 . From A of the original section carry a line across intersecting the line A 1 B' at the point A 1 . Next drop line from A 1 and B' vertically intersecting iiiK-s of corresponding letter, as shown by the dotted lines. Then A 3 B 3 F' will be the correct angle upon which to construct the corner piece and develop the miter line between the hip molding and the bed mold- ing of the deck cornice. The next step will be to obtain a correct section of the hip molding upon a line at right angles to the line of the hip. To do this it is necessary to first construct a diagonal section through the hip. At any convenient place lay off a plan of the angle of the roof, as shown by D 1 F D" in Fig. 407, and through this angle draw a plan of the hip, as shown by F K. From D 1 erect a line perpendicular to F D', as D 1 C', in length equal to D C of the section. Through C", parallel to D' F, draw C 2 K, producing it until it cuts the line repre- senting the plan of the hip. From the points F and K in the lines representing the plan of the hip erect per- pendiculars, as shown by F L and K C 3 . Draw L C 3 parallel to F K, as shown at the base line of a diag- onal section. From C 3 erect a perpendicular, C 3 E 1 , in length equal to C E of the original section. Connect E 1 L. Then L C 3 E' will be a diagonal section of a portion of the roof, and L E 1 will be the length of the hip through that portion. At right angles to L E 1 draw M H 1 , upon which to construct a correct section of the hi}) molding. Take any point, as G in the line F D 1 , at convenience, and from it erect a perpendicular 218 The New Metal Worker Pattern Book. to P K, cutting F K in the point H, and produce it ' will be obtained by which the angle contained between also iintil it cuts the base line of the diagonal section the facias of the hip molding mav be determined. L C 3 , as shown, and from this point carry it parallel to the line L E 1 , representing the pitch of the hip, until it Therefore from II' on either side set off the distance H G of the plan, as shown by G 1 and G 2 . Through Fiij. jfOfi.Thr Pattern of a Hip Molding Upon a Right Anyle in a Mansard Rnof, Miteriny Atjm. The section also shows the side elevation of a '' common" bar whose profile is at F. The plan immediately below shows a corner of the skylight with one of the hip bars, H K, the patterns for which are required. It will be neces- sary first to sec that the plan is correctly projected from the elevation, and afterward that a diagonal ele- vation of the hip bar bo obtained from this plan, be- fore the correct or raked profile of the hip bar can be obtained. ' -Draw, a duplicate of -normal profile F with its cen- ter line on the center line of the hip, as shown at F', as a means of obtaining thejmefal projection of all its points, numbering corresponding points in both profiles the same. Number the intersections of all the points in the normal profile F with the top and .bottom of the skylight finish, as shown by the small figures in A B and C G. From each of the points in the profile F carry lines parallel to the center line of the hip in either direction, intersecting lines of corresponding number dropped vertically from-bo-tk 4-ho -miters of the trans- verse section to the plan. Lines traeed through these points of intersection will give the miter lines at top and bottom as they appear in plan. At right angles to the lines of the hip carry lines, as shown, by means. of which to construct the diagonal elevation. Assume anv line, as E' G', as the base or 22C Tlte New Metal Worker Pattern Book. horizontal line of the diagonal elevation representing E G of the section. At E' erect a perpendicular upon the horizontal molding at the top whose profile is shown at A B, it will be found most convenient to Fig. 408. Plan and Section of a Skylight and Patterns for the Hip Bar. which to obtain the hights of the various points in the upper miter. As the hip bar is required to miter with carry all the points of the upper profile to the vertical line A B, as shown by 1, 2, 3', 4, 5 and (5 1 , and after- Pattern Problems. 221 ward to transfer them, as shown, to the line A' B', keeping the perpendicular hight from E 1 to B 1 equal to E B. From all the points in A' B' carry lines hori- zontally that is, parallel to E' G' to the right indefi- nitely, as shown. These lines will then represent a partial elevation of the top molding A B in the diagonal elevation. Lines from each of the points in the plan of the upper miter at II may now be carried parallel to II E' until they intersect with lines of cor- responding number drawn from A' B 1 . Lines connect- ing the points of intersection will give the required miter line at the top of the hip bar. From each of the points obtained in this miter line carry lines parallel to B' G', or the rake of the hip bar, and intersect them with lines projected parallel to II E' from the lower miter in plan at K. Lines con- necting these points of intersection will give the re- quired miter line at the bottom of the hip bar. It now remains only to obtain the correct profile of the hip bar before a stretchout can be obtained. To accomplish this, draw any line cutting the lines of the hip bar in the diagonal elevation at right angles,'as shown at R. Upon this line, and above or below the hip bar, as shown at F", draw a duplicate of the normal profile F, from the points in which carry lines at right angles to the hip bar, cutting lines of corresponding number in the same. Then lines connecting the points ' of intersection will give the raked profile, as shown at R. On account of limited space the important details in Fig. 108 are necessarily small, but great care has been taken in the preparation of the drawing, and all the points in the several views of both miters have been carefully numbered, so that the reader will have no difficulty in following out the various intersections from start to finish. The profile and the two miter lines now being in readiness, the pattern may be de- veloped in the usual manner, as follows : Upon any line drawn at right angles to the hip bar, as L M, lay off a stretchout of the profile R, as shown by the small figures, through which draw the measuring lines. Keeping the blade of the I -square parallel with L M, bring it successively against the points of intersection previously obtained in the upper and lower miters and cut corresponding measuring lines. Then lines traced through the various points of intersection, as shown by N and P Q, will constitute the required patterns. It may be noticed that while most of the points from the normal profile come squarely against the inner beveled surface of the curb G, the points 1 and . 2, representing the vertical portion of the bar, pass over the curb to a point beyond. The line from point 2, therefore, intersects at both 2 and 2', which points are duly carried through the views of this miter at K and G 1 and finally into the pattern, as shown ; from which it may be seen that the miter pattern may be cut as shown by the solid line from P to Q, or that portion from point 2 to 3 may be cut as shown by the dotted line. PROBLEM 104. Pattern for the Top of a Jack Bar in a Skylight. The jack bar in a skylight is the same as the " common " bar in respect to its profile, and the miter at its lower end with the curb. At its upper end, however, it is required to miter against the side of the hip bar. instead of against the upper finish of the sky- light. As the hip bar occupies an oblique position with reference to the jack bar, it is evident that a perfect miter between the two could not be effected without a modification or raking of the profile of the hip bar, all of which has been demonstrated, in the preceding problem. It may be here remarked that the raking of the profile of the hip bar is done not so much to affect a perfect joint with the top finish as to make a perfect j miter with the jack bar, or what is the same thing, that the surfaces indicated by 2 3 of the profile of the hip bar in Fig. 408 shall lie in the same plane with that portion of the profile of the jack bar. However, as the raked hip bar presents exactly the same appear- ance when viewed in plan as a bar of normal profile, it will not be really necessary, so far as the miter cut on the jack bar is concerned, to perform the raking operation. In Fig. 409 is shown a sectional and a plan view of a portion of a skylight containing the miter above referred to. The normal profile of the jack bar shown at F and F' is not exactly the same in its proportions as that of the preceding problem, but possesses the 222 77/e Xvw Mi'lul Worker Pattern Book. same general features. The view of the bar given in the section from A to 15 represents an oblique eleva- tion of that side of ihe hip bar which is toward the jack bar. From B to D the view shows the side of the jack bar, while beyond D is shown a continuation of the full hip bar with its profile correctly placed in po- sition at F". The first step before the pattern can be laid out is to obtain a correct intersection of the points in the plan, as at B 1 , and afterward an elevation of the same, as shown at B. Draw u normal profile of the jack bar in correct position in the plan, as shown at F'. Al>n place a profile of the hip bar in the plan of the same, as shown at F : . As only the lateral projection of the [mints are here made use of a normal profile will answer as well as the raked profile shown, as above intimated. Number all the points in both profiles Correspondingly, and from the points in each carry lines respective! v parallel to their plans, intersecting as shown at B'. From the points of intersection of like numbers erect lines vertically into the sectional view, cutting lines of corresponding number drawn from the points in the profile F parallel to the lines of Uhe rake, as shown near B. It will' be seen that both sides of the profile F' intersect with one side of the profile F". both sets of intersection being numbered alike, as I 1 , ~2', 3', etc. This gives rise to two miter lines at B in the sectional view. The line correspond- ing to the intersections on the upper side of the jack bar are here numbered 1", 2", '.'>', etc., while those points belonging exclusively to the lower intersection are numbered 3 3 , ^ and it ;i . A stretchout of the normal profile F mav now be laid off on anv line, as (i II. drawn at right angles to the elevation of the jack bar, through which the usual measuring lines are drawn. Now place the blade of the T-sqnare parallel to G H, and, bringing it against the various points in the two miter lines above de- scribed, cut corresponding measuring lines, carrying the points from the upper miter line into one side of the pattern and those from the lower one into the other side; then lines connecting the points of intersection, as shown from K to L, will constitute the required miter cut. As it is desirable to cut the miter on the jack bar so as to fit o.ver the hip bar (that is, so as not to cut the hip bar at all) and in order to prevent the surface from 4 to 5 of the jack bar from hipping on to a like por- tion of the hip bar, as shown between the points 4', 5" and x in the plan, the line from point 4 of F 1 is al- lowed to intersect with the line from 5 of F 3 , as shown at x, which point is carried into the sectional view and thence into the pattern, where it intersects with lines 4, as shown by x, so that the cut in the pattern is from x to 5 instead of from 4 to 5. For the same reason, Fig. 409. Section and Plan of Miter at the Top of the Jack Bar in a Skylight and Pattern of the Same. if it is desired to prevent the surfaces 2 3 from over- lapping the line from 2 of F 1 may be intersected with 3 from F 2 , as shown at ?/, and carried into the pattern, as shown, producing the cuts in the pattern shown by the dotted lines 3 y in the place of those shown from 3 to 2. I'lttti'.rn Problems. 223 PROBLEM 105. The Pattern of a Hip Mold Upon an Octagon Angle in a Mansard Roof, Mitering Against a Bed Molding of Corresponding Profile. This problem, like many others pertaining to man- sard roofs, m;iv reach the pattern cutter in drawings either more or less accurate, and in diil'erent stages of completion. Certain facts, however vi/., the profiles of the moldings, the pitch of the roof and the angle in plan must be known before the work can be accom- plished ; but with these given the pattern cutter will have no difficulty in drawing such elevations as are necessary to produce the required patterns. In Fig. 410, let A B C D be the given section of the mansard trimming shown, A C the profile of the bed molding and apron, and B D E the pitch of the roof. According to the statement of the problem above the angle of the plan is octagonal ; it might be a special angle cither greater or less than that of an octagon, but the principle involved and the operation of cutting the patterns would be the same. As in all other problems connected with mansard trimmings, the first requisite is an elevation of the "true face," in order to obtain the correct angle between the bed mold ing and the hip molding. A normal elevation, such as is likely to be met with in the architect's 'drawings, is shown in the engraving at the left of the section, merely for purposes of design. In obtaining the true face, shown below, it is best to use the section and plan onlv. Therefore, redraw the section as shown im- mediately below it, placing the line of the roof in a ver- tical position, all as shown. From all the points of this section lines may now be projected horizontally to the left, as the first step in developing the required true face. Immediately above the space allotted to the elevation draw a plan of the horizontal angle, as shown by I E' K. As it will be impracticable to include the entire profile of the roof in the drawings, some point must be assumed at a convenient distance below' the bed mold, as D, from which to measure hight and pro- jection, which locate also in the section below, as shown at D', making B 1 D' equal to B D. From A draw a line at right angles to the line of the roof, meeting it at B, which point may be assumed for convenience as the upper limit of that part of the roof under consider- ation. Now, from the point B drop a vertical line, *which intersect with one drawn horizontally from D, as shown at E; then 1) E will represent the projection. From the lines I K' aud F' K. upon lines at right angles to each, set oil' the projection D E, as shown at I F and K II; through the points F and 11 draw linos parallel to the first lines, meeting in G; then a line from he points of intersection will with D* G' give the correct elevation of the transition side. For the pattern of this side it will be necessary to first construct a section upon its center line, P R of the plan. At any convenient place outside of the plan draw a duplicate of P R parallel to it, as shown by P 1 A', and from the point A 1 erect a perpendicular, A 1 B', in length equal to A B of the original section. In A' B' set off points corresponding to the points in A B, and through them draw horizontal lines, as shown. Place the "{-square parallel to A' B 1 , and, bringing it against the points in E II previously obtained from the profile 0' B 1 , cut corresponding measuring lines. Then a line traced through these points of intersection, as shown by B' P', will complete the diagonal section corresponding to P R in the plan. From this diagonal section take a stretchout, which lay off on the straight line corresponding to P R produced, all as shown by P 3 B 3 . Through the points in P" B 3 draw the usual measuring lines. With the T-square placed parallel to this stretchout line, and brought successively against the points in R II, cut the measuring lines, as shown. Then a line traced through these points of intersection, as shown by E 1 to II', will be one side of the required pattern. In like manner, having transferred points from E II across to the corresponding line D G, cut the measuring lines from it, which will give the other side of the required pattern. The width of fascias (whose intersection forms a panel in this case) may be obtained as suggested above and as given in Problem 6. In locating the points N 1 and M 1 of this pattern it Pattern Probkras, 227 IV _- _ "nx/*- / u i a :. Bitterns /or the Fascias o/ a Ht_p Molding Finishing a Curved Mansard Roof Which is Square at the Eaves and Octagonal at the Top. 238 Tlte Xew Metal Worker Pattern Bwk. is desirable for the sake of design tliat they be, when finished and in position, at the same vertical distance below the corniee as are the points S and T on the square sides of the roof. To accomplish this it will be necessary to go back to the points S' and T 1 , in the first pattern obtained, and from them carry lines back into the stretchout line O* B 4 , where they are numbered 10 and 11. Their positions may now be transferred by means of the dividers to the normal profile O B, where their vertical bights can be measured on the line A B, as shown, and transferred again to the vertical line A 1 B 1 of the diagonal section. It is only neces- sary now to carry them across, as shown, to the profile P' B', where their distances from adjacent points may be measured by the dividers and placed upon the stretch- out line P 2 B'. By similar means the appearance of this panel both in the plan and in the elevation may be completed if s<> desired, all of \vhichwillbemade clear by inspection of the drawing. In the ease of very large roofs, where the develop- ment of a profile or a pattern to the full size would be impracticable, it is possible to perform the work to a scale of li or :5 inches to the foot; after which full size patterns of parts of convenient size may be ob- tained by multiplying their various dimensions by 8 or 4. As the patterns for the roll, usually finishing the hip, are properly included under the head of Flaring Work, which subject is treated in the fol- lowing section of this chapter, they will not be given here. The radii from which they can be ob- tained, however, may be derived from the diagonal section in the manner described in the following problem. PROBLEM in. To Obtain the Curves for a Molding Covering the Hip of a Curved Mansard Roof. The method of obtaining the pattern of the fascias of a molding covering a curved hip has been given in Problem G. As it is necessary in obtaining the pat- terns of the molded portion or roll, that the curve of the hip should be established, this problem really consists of developing from the normal profile of the roof a profile through the hip, or, in other words, a diagonal section of the mansard. Let A E B iii Fig. 422 represent the plan of a mansard roof or tower, the elevation of which is shown by II K, over the hip of which a molding of any given profile is to be fitted, in this case a three-quarter bead, the diagonal line E F in the plan representing the angle of the hip as it would appear if viewed from the top. At any convenient point parallel to E F, and equal to it, drawK' F', and from F' erect a perpendicu- lar, F 1 K 1 , in length equal to the vertical line in eleva- tion G K. Divide G K and F' K' into the same num- ber of equal spaces. From the points in G K draw lines cutting the profile H K, as shown, and from the points thus obtained in H K drop lines vertically, pro- ducing them until they cut the diagonal line E F of the plan, as shown. Through the points in F 1 K' draw measuring lines in the usual manner, and intersect them by lines erected perpendicularly to E F from the points therein. Then a line traced through these points of intersection, as shown by E 1 K', will be the profile to which the molding covering the hip is to be raised. Inasmuch as in the usual process of mold raising all curves must be considered as segments of circles, to accommodate both the adjustment of the machine used and the describing of the patterns, the curved line E 1 K' just obtained must be so divided that each section or segment will approach as nearly as possible an arc of a circle. In this case the section from E 1 to L will be found to correspond to an arc struck from a center, M, while the section from L to K' corresponds to an arc struck from a center not shown in the engrav- ing, but which will be found by the intersection of the lines L N and K 1 N 1 produced. In the lower part of Fig. 423 is shown an enlarged section of the hip molding, including the fascias, as it would appear at the bottom of the hip, and above it another section taken at the top, which has been de- rived from the normal section or section at the bottom by the method used and explained in Problems 105, I'n Htm Prvbkms. lot! and K>7, previously demonstrated. A dotted the roll require trimming after being raised so that the reproduction of the lines oi the upper suction is j roll may have an equal projection throughout its course. SE.CTIONATTOP Fig. 423. Enlarged Sections Through Hip Finish at Top and Bottom, Showing Change in Flare of Fascias. Fig. 42%. Diagonal Section of a Curved Mansard Roof Obtained for the Purpose of Mold Raising. placed here to show the change in the flare that takes place between fascias in going from the bottom to the top of the hip, thus showing that the outer edges' of Methods of obtaining the patterns of curved mold- ings will be found in the following section of this chapter. 240 The New Metal Worker l*uttcru JJuuk. SECTION 2. (FLARING WORK.) It will be well to place before the reader here a clear statement of the class of problems he may ex- pect to meet with under this head. It will include only the envelopes of such solid figures as have for a base the circle, or any figure of equal or unequal sides which may be inscribed within a circle, and which terminate in an apex located directly over the center of the base. According to the definition of an inscribed poly- gon (l)ef. 66), its angles must all lie in the circum- ference of the same circle. So the angles or hips of a pyramid whose base can be inscribed in a circle must lie in the surface of a cone whose base circumscribes its base and whose altitude is equal to that of the pyramid. Therefore the circle which describes the pattern of the base of the envelope of such a cone will also circumscribe the pattern of the base of the pyra- mid contained within it. The envelopes of such solids, therefore, as scalene cones, scalene pyramids and pyramids whose bases cannot be inscribed within a circle are not adapted to treatment by the methods employed in this section. Even the envelope of an elliptical cone cannot be included with this class of problems because it possesses no circular section upon which its circumference at any fixed distance from the apex can be measured. In this connection it is proper to call attention to the difference between a scalene cone and a right cone whose base is oblique to its axis. According to Ih'li- nition 96, a scalene cone is one whose axis is inclined to the plane of its base, and according to Definition !M the base of a cone is a circle. As any section of a cone taken parallel to its base is the same shape as its base, any section of a scalene cone taken parallel to its base must be a circle, and any section taken at right angles to its axis could not, therefore, be a circle, but would be elliptical. Again, as any section of a right cone (Def. 95) at right angles to its axis is a circle, if its base be cut off obliquely, such base would. according to Definition 113, be an ellipse. There- fore, since its horizontal section is a circle, its envelope may be obtained by methods employed in this section. (See Problem 136.) And since the sec- tion of a scalene cone taken at right angles to its axis is an ellipse, the scalene cone becomes virtually an elliptical cone with an oblique base that is, with a base cutoff at such an angle as to produce a circle and, as stated above, cannot be included in this section. The principles governing the problems of this section are given in Chapter V, beginning on page 79, which the reader will find a great help in explaining anything which he may fail to understand. PROBLEM 112. The Envelope of a Triangular Pyramid. Let A B C of Fig. 424 be the elevation of the the point K erect K H, perpendicular to F K and equal pyramid, and E F Gr of Fig. 425 the plan. From the in length to the hight of the pyfamid, as shown by Fig. 424. -Elevation. Fig. 425. Plan. The Envelope of a Trianr/ular Pyramid. Fig. 426.-Pattern. center K draw the lines E K, F K and G K in the plan, representing the angles or hips of the pyramid. From A 1) of the elevation. Draw the hypothenuse F II, which then represents the length of the corner lines. Pattern Problems. 241 From anv point, as L of Fig. 426, for center, with radius equal in V II, describe the arc M N O I indef- inite! v, and draw L M. From M set off the chord M N, in length equal to the sideF G of the plan. lu like manner set off N and I respectively, equal to G E and E F of the plan. Connect I and L, as shown, and draw L () and L N. Then L I N M is the pattern sought. PROBLEM 113. The Envelope of a Square Pyramid. Let E A C of Fig. 427 be the elevation of the pyramid, and F II K L of Fig. 428 the plan. The diagonal lines F K and L II represent the plan of the angles or hips, and G a point corresponding to the apex A of the elevation. From the apex A drop the or circumscribe the pattern, as shown in the diagram From any center, as M, Fig. 429, with a radius equal to A D, describe an arc, as P R S N, indefinitely and draw M P. From P, on the arc drawn, set off a chord, P R, in length equal to one of the sides of _ N Fig. 437. Elevation. Fig. 428.-Plan. The Envelope of a Square Pyramid. Fig. 429. Pattern. line A B perpendicular to the base E C. Prolong E in the direction of D, making BD equal to G F, one of the angles of the plan. Connect D and A. Then A D wiir be the slant hight of the article on one of the corners, and the radius of an arc which will contain the pyramid shown in the plan. From R set off another chord, R O, in like manner, and repeat the same operation, obtaining S and S N. Draw the lines M N, M S, M O and M R. Then M N S R P will be the required pattern. PROBLEM 114. The Envelope of a Hexagonal Pyramid. Let H G I of Fig. 430 represent the elevation of a hexagonal pyramid, of which D F C L B E of Fig. 431 is the plan. The first step is to construct a section on a line drawn from the center of the figure through oue of its angles in the plan, as A B. From the center A erect A X perpendicular to A B, making it equal to the straight hight of the article, as shown in the eleva- tion by G K. Draw the hypothenuse B X. Then X represents the apex and XB the side of a right cone, the plan of the base of which, if drawn, would circum- scribe the plan of the hexagonal pyramid. From any convenient center, as X 1 of Fig. 432, with X B of 242 Tlie Xew Metal Worker Pattern Book. Fig. 431 as radius, describe an arc indefinitely, as shown by the dotted line. Through one extremity of the arc to the center draw a line, as shown by D' X 1 . B' L' in the arc thus obtained draw lines to the center, as shown by E' X', B 1 X 1 , etc., which will represent the angles of the completed shape, and serve to locate Fig. 430. Elevation. Fig. 431.-Plan. The Envelope of a Hexagonal Pyramid. Fig. 432.-Pattern. With the dividers set to a space equal to any side of the plan, as D E, commencing at D 1 , set off this distance on the bends to be made in process of forming up. Then X 1 D 1 E 1 B 1 L 1 C' F 1 D 3 will be the complete the arc six times, as shown. From the several points E' ; pattern. PROBLEM 115. The Envelope of the Frustum of a Square Pyramid. In Fig. 433, let G H K I be the elevation of the Construct a diagonal section on the lin? A P as fol- article, A E D the plan of the larger end and L M lows : Erect the perpendicular P F, making it equal Fig. 433.-Plan and Elevation. Fig. 434. Pattern. The Envelope of the Frustum of a Square Pyramid. N the plan of the smaller end. Produce the hip lines C L, A M, etc., in the plan to the center P. to the straight hight of the article, as shown by R K of the elevation. Likewise erect the perpendicular Pattern Problems. 243 M B of the same length. Draw F B and A B. Then P A B F is the diagonal section of the article upon the line P A. Produce A B indefinitely in the direc- tion of X, and also produce P F until it meets A B extended in the point X. Then X is the apex of a right cone and X A the side of the same, the base of which, if drawn, would circumscribe the plan C A E D. Therefore, from any convenient center, as X' of Fig. 434, with X A as radius, describe the arc C' D' E' A 1 C", and from the same center, with radius X B, draw the arc L 1 N' 0' M' L 1 , both indefinitely. Draw C' X', cutting the smaller arc in the point L 1 . Make the chord C' D 1 equal in length to one side, being at the intersection of the lines D and C A forming the sides of the pan ; and that A D is the top of the frustum or the base of another cone, A O B. which remains after cutting the frustum from the orig- inal cone. For the pattern then proceed as follows : Pattern Problems. 251 Through the elevation draw a center line, K B, indef- initely. Extend one of the sides of tho- pan, as, for example, D O, until it meets the center line in the point B. Still greater accuracy will be insured by ex- tending the opposite side of the pan also, as shown the three lines meeting in the point B which deter- mines the apex of the cone to a certainty. Then B and B D, respectively, are the radii of the ares which contain the pattern. From B or any other convenient point as center, with B as radius, strike the arc P Q indefinitely, and likewise from the same center, with B D as radius, strike the arc E F indef- initely. From the center B draw a line across these arcs near one end, as P E, which will be an end of the pattern. By inspection and measurement of the plan determine in how many pieces the pan is to be constructed and divide the circumference of the pan into a corresponding number of equal parts, in this case three, as shown by K, M and L. With the dividers or spacers step off the length of one of these parts, as shown from M to L, and set off a corresponding num- ber of spaces on the arc E F, as shown. Through the last division draw a line across the arcs toward the center B, as shown by F Q. Then P Q F E will be the pattern of one of the sections of the pan, as shown in the plan. Fi'j. 451. The Envelope of the Frustum of a Right Cone. PROBLEM 124. To Construct a Ball in any Number of Pieces, of the Shape of Zones. Tn Fig. 452, let A I G II be the elevation of a' ball which it is required to construct in thirteen pieces. Divide the profile into the required sections, as shown by 0, 1, 2, 3, 4, etc., and through the points thus ob- tained draw parallel horizontal lines, as shown. The divisions in the profile are to be obtained by the fol- lowing general rule, applicable in all such cases : Divide the whole circumference of the ball into a num- ber of parts equal to two times one less than the num- ber of pieces of which it is to be composed. In convenient proximity to the elevation, the center being located in the same vertical line A N, draw a plan of the ball, as shown by K M L N. Draw the diameter K L parallel to the lines of division in the elevation, With the -square placed at right angles to this diameter, and brought successively against the points in the elevation, drop corresponding points upon it, as shown by 1, 2, 3, 4, etc. Through each of these points describe circles from the center by which the plan is drawn. Each of these circles becomes the plan of one edge of the belt in the elevation to which it corresponds in number, and is to be used in estab- lishing the length of the arc forming the pattern of the zone of which it is the base. Extend the center line K- A in the direction of indefinitely. Draw chords to the several arcs into which the profile has been divided, which produce until they cut G A 0, as shown by 1 2 E, 2 3 D, 3 4 C, 4 5 B and 5 6 A. Then E 2 and E 1 are the radii of parallel arcs which will describe the pattern of the first division above the cen- 252 The New Metal Worker Pattern Book. ter zone, and D 3 and D 2 are the radii describing the pattern of the second zone, and so on. From E l in Pig. 453 as center, with E 2 and E 1 as radii, strike the arcs 2 2 and 1 1 indefinitely. Step the centers D 1 , Fig. 454; C 1 , Fig. 455; B l , Fig. 456, and A', Fig. 457. The pattern for the smallest section, as indicated by F in the plan, may be struck by a radius equal to F 6 in the plan. The center belt or Fig. 455. Pattern of Zone 3 1. . 453. Pattern of Zone 1 2. Fig. 456. Pattern of Zone 4 i. Pig. 457. Pattern of Zone 5 6. Fig. 452. Plan and Elevation. Fig. 454. Pattern of Zone 2 3. Fig. 458. Pattern of Middle Zone. To Construct a Ball in any Number of Zones. off the length on the corresponding plan line, and make 1 1 equal to the whole of it, or a part, as may be de- sired in this case a half. In like manner describe patterns for the other pieces, as shown, struck from zone, shown in the profile by 1 0, is a flat band, and is therefore bounded by straight parallel lines, the width being 1 in the elevation, and the length meas- ured upon line 1 of the plan, all as shown in Fig. 458. Pattern Problems. PROBLEM 125. The Patterns for a Semicircular Pipe with Longitudinal Seams. 253 By the nature of the problem the pipe resolves it- self, with respect to its seetioii or profile, into some regular polygon. In the illustration presented in Fig. iait an oetagoual form is employed, but any other reg- the profile ABC D F II G E and project the points 15 and C back upon N R and complete the elevation by drawing the semicircles O U and P T. By inspection of the diagram it is evident that the pattern for the sections corresponding to U T P in the elevation may be pricked directly from the drawing as it is now constructed, and that the pat- terns for the sections represented by E A and D F of the profile will be plain straight strips of the width of one side of the figure, as shown by either E A or D F, and in length corresponding to the length of the sweep of the elevation on the lines N L V and R X S, respectively. By virtue of the bevel or flare of the pieces N L V U T and R X S T P, as shown by A B and C D of Fig. 459. Elevation and Section. Fig. 480. Pattern. A Semicircular Pipe with Longitudinal Seams. ular shape may be used, and the patterns for it will be cut by the same rule as here explained. Let N L V be any semicircle around which an octagonal pipe is to be curried. Draw N V, passing through the center W. Through W draw the perpendicular L K indefinitely. Let N R be the required diameter of the octagon. Immediately below and iu line with N R construct the profile, each becomes one-half of the frustum of a right cone, with its apex above or below the point W. Therefore prolong C D of the profile until it cuts the center line L K of the elevation in the point M. Then M D and M C are the radii of the pieces corresponding to P T S R of the elevation. Also prolong the side A B, or, for greater convenience, its equivalent, E G, until it 254 The X'n- M'i'tl Worker Pattern Book. cuts the center line in the point M'. Then M' G and M' E are the radii of the pieces corresponding to N L V U O of the elevation. From M'' in Fig. 460 as center, using each of the several radii in turn, strike arcs in- definitely, as shown by N' V, O' U 1 , P 1 T' and K 1 S'. Step off the length N L V in the elevation, Fig. 457, and makeN 1 V of Fig. 45S equal to it. Draw N' O' and V U' radial to W. Then N 1 V U 1 O 1 will con- stitute the pattern for the pieces N L V U O of the elevation. In like manner establish the length of P' T 1 , and draw P' R 1 and T 1 S 1 , also radial to the center, as shown. Then P 1 T 1 S 1 R 1 will be the pattern for the pieces P T S X R of the elevation. This rule may be employed for carrying any polyg- onal shape around any curve which is the segment of a circle. The essential points to be observed arc the placing of the profile in correct relationship to the ele- vation and to the central line L K, after which prolong tin- oblique sides until they cut the central line, thus establishing the radii by which their patterns may be struck. In the case of elliptical curves, by resolving them into segments of circles and applying this rule to each segment, as though it were to be constructed alone and distinct from the others, no difficulty will be met in describing patterns by the principles here set forth. The several sections may be united so as to produce a pattern in one piece by joining them upon their radial lines. This principle is further explained in the pattern for the curved molding in an elliptical window cap in Problem 128. PROBLEM 126. The Blank for a Curved Molding. As curved moldings necessitate a stretching of the metal in order to accommodate them to both the curve of the elevation or plan and the curve of the profile at machinery designed for that purpose, care being taken to make the widtli of the flaring strip suffi- cient to include the stretchout of the curve of the pro- Fiy. 461. Obtaining the Blank for a Curved Core or Ovolo Molding. Fig. 462. Obtaining the Blank for a Curved Ogee Moldiny. the same time, the patterns for their blanks can only be considered as flaring strips of metal in which the curve of the elevation or plan only is considered. The curve of the profile requires to be forced into them by file. Blanks for curved moldings thus become frus- tums of cones and are cut according to the principles of regular flaring articles, as explained in the preceding problems. The method of determining the exact flare Pattern Problems. 255 necessary to ]>roy tho nature of the profile and tho kind of machinerv to lie used in forming the same. I'sually a line is drawn through the extremities of the profile, as shown at A D in either of the two illustrations here given, Figs. 401 and -t(i2, and is continued until it meets the center line, for length of radius, as shown at F. Therefore, to describe the pattern of the blank from which to make a curved molding corresponding to the elevation AGED, proceed in the same manner as though the side E C were to be straight. Through the center of the article draw the line B F indefinitely, and draw a line through the points C and E of one of the sides, which produce until it meets B F in the point F. Then F E will be the radius of the inside of the pattern. The radius of the outside is to be ob- tained by increasing F C an amount equal to the excess of the curved line E C over the straight line E C, as shown by the distance C S. Then F S is the radius of the outside of the pattern. The length of the pattern can be obtained as in previous problems. PROBLEM 127. The Patterns for Simple Curved Moldings in a Window Cap. In Fig. 4ii.'{ is shown the elevation of a window cap. in the construction of which two curved moldings are required of the same profile, but curved in opposite essary for joining it to the face and roof pieces will be obtained in one piece. The method of developing the pattern for the blank is the same for both curves. The II 10 3 Fig. 464. Blank for Center Piece. B' Fig. 403. Elevation or Window Cap. Fig. 465. -Blank for Side Piece. Thf Patterns for Simple Curved Moldings in a Window Cap. directions. It is advisable to include as much in one piece as can be raised conveniently with the means at hand; therefore, the curved part of the profile with its lillets or straight parts adjacent and the two edges nee- two pieces will raise to the form by the same dies or rolls, it being necessary only to reverse them in the machine. Before the blank for the middle piece can be developed it will be necessary to first construct a 256 file New Metal Worker Pattern Book. section upon the center line, as shown atS K; from all points in the mold and the center of the curve upon the center line project horizontal lines to the right. Draw any vertical line, as H K, to represent the face of the cap in the section and at S draw the profile of the mold, as shown. The principle to be employed in strik- ing the pattern is simply that which would be used in obtaining the envelope of the frustum of a cone of which A D is the axis. The general average of the profile is to be taken in establishing the taper of the cone, or, in other words, a line is passed through its extreme points. Draw a line through the profile in this manner and pro- long it until it intersects A D in the point A, all as shown by C A. Then A is the apex of the cone, of which A C is the side and H D the top of the frustum. Divide the profile S, as in ordinary practice for stretch- outs, into any number of spaces, all as shown by the small figures. Transfer the stretchout of the profile S on to the line A C, commencing at the point 1, as shown, letting the extra width extend in the direction of C. From any convenient center, as A in Fig. 464, with radius A C, describe the pattern, making the length of the arc equal to the length of the correspond- ing 'arc in the elevation, all as shown by the spaces and numbers. From the same center draw arcs correspond- ing to points 9, 10 and 11 of the stretchout, thus com- pleting this pattern. For the pattern of the carved molding forming the end portion of the cap proceed in the same general manner. Upon any line drawn through the center N of the curve, as L M, construct a section of the mold. as shown at R. From N draw the perpendicular X !> indefinitely. Through the average of the profile R, as before explained, draw the line to B, cutting N B in the point B, as shown. Lav off the stretchout of the profile upon this line, commencing at the point 1, in the same manner as explained in the previous operation. From any convenient point, as B' in Fig. 465, as center, with radius B 1, describe the inner curve of the pat- tern, as shown, which in length make equal to the eleva- tion, measuring upon the are 1, all as shown by the small figures, after which add the outer curves, as shown by E 1 E". The straight portion forming the end of this mold- ing, as shown in the elevation, is added by drawing, at right angles to the line E 5 B 1 , a continuation of the lines of the molding of the required length, as shown in the pattern. Upon this end of the pattern a square miter is to be cut by the ordinary rule for such purposes, to join to the return at the end of the cap. PROBLEM 128. The Pattern for the Curved Molding: in an Elliptical Window Cap. In Fig. 466 is shown the elevation and vertical sec- tion of a window cap elliptical in shape, the face of which is molded. In drawing the elevation such centers have been employed as will produce the nearest approach to a true ellipse after the manner described in Problem 76 of Geometrical Problems, page 65. The centers B, D and F, from which the respective segments of the elevation have been described, may then be used in obtaining pat- terns as follows : Through the center F, from which the arc forming the middle part of the cap is drawn, and at right angles to the center line of the cap G H, draw the line I K indefinitely. Project a section on the center line of the cap, as shown by P K at the right, the line P K being used as a common basis of measurement upon which to set off the semi-diameters of the various cones of which the blanks for the moldings form a part. Through the average of the profile, as indicated, draw S R, producing the line until it meets I K. Divide the profile of the molding in the usual manner and lay off the stretchout, as indicated by the small figures. Then R S is the radius of the pattern of the middle segment of the cap. With the dividers, measuring down from the pro- file, lay off on P K distances equal to the length of the radius A B, as shown by the point O, and of C 1), as shown by the point M. Through these (mints () and M, at right angles to P K, draw lines cutting S R in the points T and U. Then U S is the radius for the pattern of the segment C E of the elevation, and T S the radius of the pattern for the segment A C. In order to obtain the correct length of the pattern, not only as regards the whole piece, but also as regards the length of each arc constituting the curve, step off the length of the curved molding with the dividers upon any line of the elevation most convenient, as shown, numbering the spaces as indicated, and setting off a like number of spaces upon a corresponding line of the pattern. As a matter both of convenience ami Pattern Problems. 257 accuracy, the spaces used in measuring the arcs arc greater in the one of longest radius and are diminished in those of shorter radii, as will be noticed by examina- tion of the diagram. To lay off the pattern after the radii are obtained as al>o\-e described, proceed as follows: Draw any straight line, as G' II' in Fig. 467, from any point in which, as F 1 , with radius equal to R S, as shown 1'V F 1 E', describe an arc, as shown by E'G'; and likewise, from the same center, describe other arcs cor- responding to other points in the stretchout of the pro- iile. Make the length of the arc E' G 1 equal to the length of the corresponding arc in the elevation, as de- scribed above. From E' to the center F 1 , by which this arc was struck, draw E' F'. Set the dividers to the distance U S as radius, with which, measuring from length of the corresponding arc in the elevation, all as shown by the small figures. From C 1 draw the line C 1 D 1 to the center by which this arc was struck. Fig. 467.-Blank for the Curved Molding. Fig. 486. Elevation and Section of Window Cap.' The Pattern for the Curved Molding in an Elliptical Window Cap. along the line E' F 1 , establish D' as center, from which describe arcs corresponding to the points in the profile, as shown from E' to C'. Make E' C 1 equal to the Set the dividers to the distance T S in the section, and, measuring from C' along the line C 1 D 1 , establish the point B 1 , from which as a center strike arcs cor- 258 The Xew Metal Worker Pattern B\- means of which the lengths of the nidii to be used in describing the pattern are to l>e obtained. Draw the horizontal line II U indefinitely, and at right angles to it draw II A, indefinitely also. Make II U equal to II U of the plan, Fig. 471. Make II C equal to the vertical hight of the vessel, as shown in the w Fig. 473. Plan and Elevation. N 0. From the points H and L of the arc first drawn draw lines to A, thus intercepting the arc N O and determining its length. In the diagram, Fig. 474, set off from II, on the line II U, the distance R H, making it equal to R II of the plan, Fig. 473. Also, upon the line C G, from the point C, set off C I, equal to R N of the plan. Then, through the points R and I thus established, draw the line R B, which produce until it intersects A H. Then R B will be the radius for those portions of the pattern lying between V and H and L and W of Fig. 474. Diagram of Radii. Fig. 475. Pattern of One Half. The Envelope of the Frustum of a Cone, the Base of Which is an Elliptical Figure. elevation by D X. Draw the line C G parallel to II U, making C G in length equal to U N of the plan. Through the points U and G thus established draw the line U G, which continue until it meets II A in the point A. Then A U will be the radius by which to describe that portion of the pattern which is included between the points H and L of the plan. With A U as radius, and from any convenient point as center as A, Fig. 475 draw the arc II L, which in length make equal to II L of the plan, Fig. 473, as shown by the points 1, 1 2, 3, etc. From the same center, and with the radius A G of Fig. 474, describe the parallel arc i the plan. From the point H, on the line H A, Fig. 475, set off the distance H B, equal to R B of Fig. 474. Then, with B as center, describe the arc E II, and from a corresponding center, C, at the opposite end on pattern, describe the arc L K. 'From the same centers, with B I as radius, describe the arcs N M and O P, all as shown. Make H E and L K in length equal to II E and L K of the plan. From E and K, respectively, draw lines to the centers B and C, inter- cepting the arcs N M and P in the points M and P. Then E K P M will be one-half of the complete pat- terns of the vessel. PROBLEM 133. The Pattern of a Heart-Shaped Flaring Tray. Let E C G 1 F G C 1 of Fig. 476 be the plan of the article, and I N O K the elevation. By inspection of the plan it will be seen that each half of it consists of two arcs, one being struck from D or D 1 as center, and the other from C or C 1 as center, the junction between the two arcs being at G and G', respectively. From C 262 The Neiu Metal Worker Pattern Book. draw C F, and likewise draw C G. Upon the point D 1 erect the perpendicular D 1 C'. To obtain the radii of the pattern construct a dia- Lay off the perpendiculars X U and P $ indefinitely. Upon P 8, from P, set off P R, equal to 1)' C 1 of the plan, and on X U, from X, set off X W, equal to D' c of Fix. 476. Plan and Elevation. Fig. 478. Pattern. The Pattern of a Heart-Shaped Flaring Tray. n R. 477. Diagram of Uadii. gram, shown in Fig. 477, which is in reality a section upon the line C G of the plan. Draw X P in Fig. 477, in length equal to the straight hight of the article. the plan. In like manner make P S equal to C G of the plan, and X U equal to C .7 of the plan. Connect U S and W R. Produce P X indefinitely in the direc- Pattern Problems. 263 tion of Z. Also produce R W until it meets P X in tin- point Y. and in like manner produce S I" until 'it meets P Z in the point Z. Tlien / I" and 'A S are the radii for that portion of the article contained between G and K of the plan, and V \V and Y Rare the radii of that portion shown from (i to E of the plan. To lay out the pattern after the radii are estab- lished, draw anv straight line, as Z' G* in Fig. 47s, in length equal to Z S of the diagram. From Z' as cen- ter, with Z S as radius, describe the arc G* F 1 , in length ei[tial to G F of the plan. In like manner, with radius Z I", from the same center, describe the arc y'f, in length equal to #/ of the elevation. Draw/' F'. Set of]' from G 2 , upon the line G 2 Z', the distance R Y of Fig. 477, as shown at Y', and from Y 1 as center, with the radius R Y, describe the arc G 2 E', which in length make equal to G E of the plan. In like manner, from tin; same center, with radius Y "W, describe the arc .ex of Fit/. 4*4. The Enrelopf of th< Frustum of a Rif/hf Cone Whose Upper Plane is Oblique to its Ajrin. the cone of which C B D E is a frustum. Draw the axis A G, which produce below the lignre, and from a center lying in it draw a half plan of the article, as shown by F G II. Divide this plan into any number of equal parts, and from the points carry lines parallel to the axis until they cut the base line, and from there extend them ^in the direction of the apex until they cut the upper plane B D. Place the T-square at right angles to the axis, and, bringing it against the several points in the line B D, cut the side A E, as shown. From A as center, with A E as radius, describe the arc C' E', on which lay off a stretchout of either a half or the whole of the plan, as may be desired, in this case a half, as shown. From the extremities of this stretchout, C 1 and E', draw lines to the center, as C' A and E 1 A. Through the several points in the stretchout draw similar lines to the center A, as shown. With the point of the compasses set at A, bring the pencil to the point D in the side A E, and with that radius describe an arc, which produce until it cuts the corresponding line in the stretchout, as shown at D 1 . In like manner, bringing the pencil against the several points between D and E in the elevation, describe arcs cut- ting the corresponding measuring lines of the stretchout. Then aline (raced through these intersections will form the upper line of the pattern, the pattern of the entire half being con- tained in C 1 B' D' E 1 . PROBLEM 136. The Envelope of a Right Cone Whose Base Is Oblique to Its Axis. In Fig. 485, let G D H be the elevation of u right cone whose base is oblique to its axis, the pattern of which is required. It will be necessary first to assume any section of the cone at right angles to its axis as a Tlie Xew Metal Worker Pattern Book. base upon which to measure its circumference. This can be taken at any point above or below the ol>li<|iu> base according to convenience. Therefore at right angles to the axis D O, and through the point G, draw the line G F. Extend the axis, n.s shown by D B, and upon it draw a plan of the cone as it would appear when cut upon the line G F, as shown bv ABC. Divide the plan into any convenient number of equal parts, and from the points thus ob- tained drop lines on to G F. From the apex D, through the points in G F, draw lines to the base G II. From D as center, with D G as radius, describe an arc indefinitely, on which lay off a stretchout taken from the plan ABC, all as shown by I M K. From the center D, by which the arc was struck, through the points in the stretchout, draw radial lines indefi- nitely, as shown. Place the blade of the T-square parallel to the line G F, and, bringing it against the several points in the base line, cut the side D H, as shown, from F to II. With one point of the com- passes in D, bring the other successively to the points 1, 2, 3, 4, etc., in F II, and describe arcs, which pro- duce until they cut the corresponding lines drawn through the stretchout, as indicated by the dotted lines. Then a line, ILK, traced through these points of intersection, as shown, will complete the re- quired pattern. Pattern Elevation. ^^ Fig. Jj8S.Thf Envelope of a Cone whose Base is Obliqur to its ATI'S PROBLEM 137. A Conical Flange to Fit Around a Pipe and Against a Roof of One Inclination. In Fig. 486 is shown, by means of elevation and plan, the general requirements of the problem. A B represents the pitch of the roof, G H K I represents the pipe passing through it, and C D F E the required flange fitting around the pipe at the line C D and against the roof at the line E F. The flange, as thus drawn, becomes a portion of the envelope of aright cone. At any convenient distance below the elevation assume a horizontal line as a base of the cone upon which to measure its diameter, and continue the sides downward till they intersect this base line, all as shown at L M. Also continue the sides upward till they in- tersect at W, the apex. Below the elevation is shown a plan, and similar points in both news are connected by the lines of projection. S T represents the pipe and N O the flange. While the pipe is made to pass through the center of the cone, as may be seen by ex- amining the base line L M in the elevation, and also P R of the plan, it does not pass through the center of the oblique cut E F in the elevation, or, what is the same, N of the plan. For the pattern of the flange proceed as shown in Fig. 481, which in the lettering of its parts is made to correspond with Fig. 486, just described. Divide the half plan P X 11 into any convenient number of parts in this case twelve and from each of the points thus established erect perpendiculars to the base of the cone, obtaining the points 1', :>', 3 1 , etc. Fmni these points draw lines to the apex of the cone \V . cutting the oblique line E F and the top of the flange (' 1>. as shown. Inasmuch as C D cuts the cone at right angles to its axis, the line in the pattern corresponding to it will be an arc of a circle; but with E F, which cuts the cone obliquclv to its axis, the ease is different, eaeli point in it being at a different distance from the apex. Accordingly, the several points in E F, obtained by Pattern Problems. 267 the lines from the plan drawn to the apex \V, must In- transferred to one of the sides of the cone, where their distances from \Y can he accurately measured. There- fore from the points 0", I 3 , 2 3 , 3 :i , in K l'\ draw lines ut right angles to the axis of the cone AY X, cutting the side W M, as shown. With W as center, and with \V M as radius, strike the arc 1" 11' indefinitely, and. in the plan P X 11, all as shown hy 0', 1", 2', 3", etc. From these points draw lines to the center \V,us shown. With one point of the dividers set at AY and the other brought successively to the points obtained in AY M by the horizontal lines drawn from E F, cut the correspond- ing lines in the stretchout of the pattern, as indicated hy the curved dotted lines. A line traced through w^i Fig. 46.-Plan and Elevation. Fig. 487. Pattern. A Conical Flange to Fit Around a Pipe and Against a Roof of One Inclination. these points, as E 1 F', will represent the lower side of the pattern. As but one-half of the plan has been used in laying out the stretchout, the pattern C' E 1 F 1 D' thus obtained is but one-half of the piece required. It can be doubled so that the scam can be made to come through the short side at C E, or through the long side at D F, at pleasure. with the same center and with W D as radius, strike the arc C 1 D' indefinitely, which will form the bound- ary of the pattern at the top. At anj- convenient dis- tance from M draw AY P', a portion of the length of which will form the boundary of one end of the pat- tern. On P' K', commencing with P', set off spaces equal in length and the same in number as the divisions PROBLEM 138. The Pattern for a Cracker Boat. LetE F H G, in Fig. 4SS, be the side elevation, A B C D E the end, and I K J L the plan of a dish sometimes called a cracker boat or bread trav. The sides of the dish are parts of the frustum of a right cone. To the plan have been addc] the circles show- ing the complete frustums of which the sides are a part, 268 Tlie New Metal Worker Pattern Book. L and K being the centers, all of which will appear clear from an inspection of the drawing, and below is further shown a side view of this frustum. While in Fig. 488.-Plan and Elevations. Fig. 489. Pattern. The Pattern for a Cracker Boat. the plan the top and bottom of the sides have been shown parallel, in the side view the top appears curved at C, the cut producing which curve being shown by B C of the end view. Extend the sides IT "W and V X of the frustum until they meet at Z, which is the apex of the com- pleted cone. Before the pattern can be described it will be necessary to draw a half elevation of the cone U V Z, showing the end view of the tray in its relation to the same, as in Fig. 480. Draw any center line, as K' Z'. From the point L', as center, strike the arc K' T', being one-fourth of the plan of top, as shown by K Tin Fig. 488. Below the plan of top draw one-half of frustum of cone, as shown by ' V'X' w' ', in which draw the end elevation of boat A' B' C' X' K', letting V X' be one of its sides, and extend the line b' B' through the arc K' T' at B". Divide the part of plan B" T' into any convenient number of parts, and from the points carry lines parallel to the center line or axis until they cut the top line u' V, and from there extend them in the direction of Z' until they cut the line B' C'. Place the T-square at right angles to the axis, and, bringing it against the several points in the line B' C', which represents the shape shown by E C F in eleva- tion of side, cut the side V X'. as shown. From Z' as center, with Z' V as radius, describe the arc I' J', upon which lay off a stretchout of plan. As the part of the plan B" T' corresponds to B' C', which shows one-half of one side of boat, and as this part of plan is divided into three parts, six of these parts are spaced off on the arc I' J and num- bered from 1 to 4, and 4 to 1, 4 being the center line. Through these points in the stretchout draw measuring lines to the cen- ter Z', as shown. AVith one point of the compasses set at Z', bring the pencil point up to the several points between V and C' in the elevation, and describe arcs cuttingmeasuringlines of corresponding numbers in the stretchout; then a line traced through these points of intersection will form the line I' K' J', show- ing the upper line of the pattern for one side of the boat. To obtain the bottom line of the pattern, with Z' as center and radius Z' X', describe the arc M' N'. Divide the plan of bottom of boat, as M T X in Fig. 488, into any convenient number of equal parts, in this case six, three on each side of the center T, and start- ing from the center line 4 of pattern, space off three spaces each way on the arc M' N', thus establish- Pattern Problems. 269 ing the points M' and N' of pattern, corresponding to the points M and N of plan. By drawing the lines M' I' and N' J' the pattern for one side of the boat, shown by E F H G in elevation, is completed. PROBLEM 139. Pattern for the Frustum of a Cone Fitting: Ag-ainst a Surface of Two Inclinations. In Fig. 490, let A B C D represent the frustum of a cone, the base of which is to be so cut as to make it lit against a roof of two inclinations, as indicated by P R D. Continue the lines of the sides of the cone A B and D C upward until they meet in the point X, which is the apex of the complete cone. Through the apex of the cone draw the line X R, representing the axis of the cone, meeting the ridge of the roof in the point R, and continuing downward in the direction of Y, as shown. At any convenient distance below A D draw a horizontal line, G H, as a base, and immediately below it draw a plan of the same, as shown by E SFY. Subdivide this plan into any convenient number of spaces, as indicated by the small figures 0, 1, 2, 3, etc. From the points thus established carry lines vertically until they cut the base of the cone G II, and from this line carry them in the direction of the apex X until they cut the line of the given roof. From the points established in the roof line A R draw lines at right angles to the axis of the cone X Y, continuing them until they strike the side of the cone A B. From X as center, with X G as radius, describe the arc GK, upon which lay off a stretchout of the plan. As the pattern really consists of four equal parts or quarters, the divisions of the plan have been num- bered from to 4 and from 4 to alternating, the points representing the lowest and the points 4 the highest points of each quarter. Therefore in number- ing the points of the stretchout G K, any point can be assumed as a beginning which is deemed the best place for the joint (in this case 4), numbering from 4 to and reversing each time, all as shown. From these points established in the arc G K draw lines to the apex X. Then, with X as center, and with radii corresponding to the points already established in the side B G of the cone, strike arcs as shown by the dotted lines, cutting meas- Fig. 490. Pattern for the Frustum of a Cone Fitting Against a Surface of Two Inclinations. uring lines of corresponding number. Then a line traced through the points of intersection, as shown by L, will 270 The New Metal Worker Pattern Book. be the shape of the pattern at the bottom and N M L will constitute the entire pattern of the frustum of a cone adapted to set over the ridge of u roof, as indi- cated in ihe elevation. PROBLEM 140. The Pattern of a Frustum of a Cone Intersected at its Lower End by a Cylinder, Their Axes Intersecting: * at Right Angles. Let S P K T in Fig. 491 be the elevation of the cylinder, and a G K H b the elevation of the frustum. Draw the axis of the cylinder, A B, which prolong, as shown by C D, on which construct a prolile of the cylinder, as shown by C E I) F. Produce the sides of the frustum, as shown in the elevation, until they meet in the point L, which is the apex of the cone. Draw the axis L K, which produce in the direction of O, and at any convenient point upon the same con- struct a plan of the frustum at its top, a b. In connection with the profile of the cylinder draw a correspond- ing elevation of the cone, as shown by K 1 a 1 b' K\ Produce the sides K 1 a 1 and K 3 b' until they intersect, thus obtaining the point L 1 , the apex corresponding to L of the elevation. Draw the axis L 1 E, as shown, which produce in the direc- tion of N 1 , and upon it draw a second plan of the frustum at a b, as shown by M' O 1 N 1 . Divide the plans M O N and M 1 O 1 N' into the same number of equal parts, com- mencing at corresponding points in each, as shown. With the T-square set parallel to the axis of the cones, and brought successively against the points in the plans, drop lines to the lines a b and a 1 b 1 , as shown. From L 1 draw lines through the points in o 1 b\ cut- ting the profile of the cylinder, as shown in K' E K 2 . and in like manner from the apex L draw lines indefi- nitely through the points in a b. ' Place the T-square parallel to the sides of the cylinder, and, bringing it against the points in the profile K 1 E K 2 just described, cut corresponding lines in the elevation, as shown at II K G. A line traced through these points of inter- section, as shown by H K G, will form the miter line between the two pieces as it appears in elevation. This miter line is not necessary in obtaining the pattern, but the method of obtaining it is here intro- duced merely to show how it may be done, should it be desired under similar circumstances in any other case. The development of the pattern in this case could be most easily accomplished by using L 1 as a center from which to strike arcs from the various points on the line ' K'. The same result is accom- N' M Fiy. 491. The Pattern of a Frustum of a Cone Intersected at Its Lower End by a Cylinder, Their Axes Intersecting at Rirjht Angles. plished, however, by continuing the lines drawn from K 1 E K 2 until they meet the side a G of the cone pro- longed, as shown from G to Z. Thus a Z becomes in all respects the same as a 1 K 1 . From L as center, and with radius L a, describe the arc // a*, upon which lay off a stretchout of the plan M N of the frustum. Through each of tin- points in this stretchout draw lines indefinitely, radiat- ing from L, as shown, Number the points in the Pattern Problems. 271 stretchout 2 b" corresponding to the numbers in the profile, commencing with the point occurring where it is desired to have the seam. Set the compasses to L Z as radius, and, with L as center, describe an arc cutting the corresponding lines drawn through the stretchout, as shown by 1, 3 and 1. In like manner reduce the radius to the second point in G Z, and de- scribe an arc cutting 2, 4, 4 and 2. Also bring the pencil to the third point and cut the lines correspond- ing to it in the same way. Then a line traced through the points thus obtained, as shown by H' K" G 1 , will be the pattern of the frustum. PROBLEM 141. The Pattern for a Conical Boss. The principles and conditions in this problem arc exactly the same as those in the one immediately pre- ceding (that is, the frustum of a cone raitering against Fig. 492. The Pattern fur a Conical Boss. a cylinder, their axis being at right angles), but its pro- portions are so different that it is here introduced as showing that the same application of principles often produces- results so widely differing in appearance as to be scarcely recognizable. Let A B C D of Fig. 492 represent the elevation of the boss that is required to fit against the cylindrical can, a portion of the plan of which is shown by the arc A B. The plan at the smaller end of the boss is represented by K K G II. Continue the lines A D and B until they intersect at K, which is the apex of the cone of which the boss is a frustum. An in- spection of the elevation will show that it is only necessary to describe one- fourth of the pattern, the remaining parts being duplicates. Divide one-quarter of the plan into any convenient number of parts, in the present instance four, as shown by the points in H E. . Drop lines from these points to the base D C, as shown. Draw lines from K through the points in the base until they intersect the arc at A B, which represents the body of the can. These points can be numbered to correspond with the points in the plan from which they are derived. At right angles to the line F K draw lines from the points on A B until they strike the line A K, where their true distances from K can lie measured. With K as a center, and K D as radius, strike the arc L M N, equal in length to the circumference of plan. If the whole pattern of boss is to be described from measurements derived from elevation it will be necessary to reverse the order of the numbers for each quarter, as shown. From K draw lines extending outwardly through these points, as indicated by the small figures. With K as center, draw an arc from the point 1' until it inter- sects radial lines 1 drawn from K, as shown at 0, Z and R. In the same manner draw an arc from 2' to lines '2, &c., as shown. A line traced through these points will produce the desired patterns, as shown by L S Z V R N. 272 The \> a- .\fctal Worker Pattern Book. PROBLEM 142. Pattern for the Lip of a Sheet Metal Pitcher. Let A B C D of Fig. 493 represent the side ele- vation of a pitcher top having the same flare all around, iind E F G H the plan at the base. By producing the lines A D and B C until they intersect in the point K, the apex of the cone of which the pitcher top is a sec- tion will be obtained. Divide one-half of the plan into any convenient number of equal spaces, as shown by the points in G H E. From these points drop lines lo tin 1 base D C, as indicated. Then draw radial lines from K, cutting the points in D C, and producing them until they intersect the curved line representing the top of the pitcher, otherwise an irregular cut through the cone, as shown by A B. For convenience in subsequent operations, number these points to cor- respond with the numbering of the points in the plan from which they are derived. Place the T-square at right angles to the axial line of the cone II K, and, bringing it against the several points in A B, cut the line DA, as shown in the diagram. By this means there will be obtained in the line K A the' length of radii which will describe area corresponding to points in the top line of the lip A B. With K as center and K D as radius, describe the arc L M N, which in length make equal to the circumfer- ence of the plan by stepping off on it spaces equal to the spaces originally established in the plan, all as in- dicated by the small figures. From K through each of the points in L M N thus established draw radial lines, extending outwardly indefinitely, as shown. Then with K as center and K", K,' K 3 , etc., as radii, strike arcs, which produce until they intersect radial lines of corresponding number just drawn, all as shown in the diagram. Then a line traced through the points thus obtained will be the required pattern, all as shown by L P R N M. The method above described is a strictly mathe- matical rule for obtaining such shapes when a design embodying the necessary curve at the top is at hand. As by the nature of the problem, this part of the pat- tern does not require to be fitted or joined to any other piece, it would be much easier to obtain, by the foregoing method, the principal points in the outside curve of the pattern and finish by drawing the re- mainder to suit the taste of the designer. In other words, after the arc L M N has been drawn and stepped off into spaces, draw radial lines from K through the points representing the highest and the lowest parts required in the top curve, as (." and 12, "upon which lines the required lengths can be set <>IT. Then these points can be connected by anv curve suitable for the purpose. The principle involved in the foregoing is exactlv the same as that of a hip or sitz bath given in the fol- \ Fig, 493, Pattern for the L

f the lv <>[ ;t hip bath having an equal amount of llaiv on all .sides, the plan of which is a circle. In describing At any convenient distance above D draw J K parallel to C E to be used as a regular base upon which to measure the circumference of the cone. Parallel to .1 K draw F II, and from a center obtained on F II by prolonging the axis A X draw a half -plan of the frustum, as shown by F G II. Divide this half-plan into any convenient number of equal parts, and from the points thus obtained carry lines parallel to the axis until they cut the line J K, and from there extend them in the direction of the apex A, thus cutting the curved line B D. Place the T-square parallel with J K, and bring- ing it against the several points in the curved line B D, cut the side E D, as shown. From A as center, with K 2 Fig. 494. Pattern for a Hip Bath of Regular Flare, the pattern for the body it will be considered as a sec- tion of a. right cone, the plane C E being at right angles to the axis and the base being represented hy the curved line B D, as shown. The sides E D and C B can be extended until they meet at A. Then A will be the apex of a cone of which C B D E is a frus turn having an irregular base B D. A K as radius, describe the arc K' K 2 , on which lay off a stretchout of either one-half or the whole of the plan, as may be desired. In this case a half is shown. From the extremities of this stretchout, as K' and K 2 , draw lines to the center, as K' A and K" A, and from the several points in the stretchout draw similar lines, as shown by 1, 2, etc. With one point of the dividers The New Metal Worker Pattern Book. set at A bring the pencil point to the point D in the side A l\. ami with that radius describe an arc, which produce until it cuts the corresponding line 1'2 in the stretchout, as shown at D 1 . In like manner, bringing the pencil point up to the several points between D and E in the elevation, describe arcs cutting lines of corresponding numbers in the stretchout. Then a line traced through these intersections will form the upper line of the pattern. From A as center, with A E as radius, describe the arc C' E 1 , cutting A K 1 and A K", as shown by C 1 and E 1 , forming the lower line of pattern. Then C 1 B l D 1 E 1 will be half the pattern for the side of the hip bath. As a feature <>!' design, the form produced in the pattern by a curved line B D drawn arbitrarily may not be entirely satisfactory. If, for instance, that part, of the pattern lying between lines 9 and 12 should not appear as desired, it can be moditied upon the pattern at will, as this edge of the pattern is not required to lit any other form. Such a modification is shown bv the dotted lines measured from F toward D or A unou the circumfer- ence of the cylinder. PROBLEM 146. Pattern of a Tapering; Article with Equal Flare Throughout, which Corresponds to the Frustum of a Cone Whose Base Is an Approximate Ellipse Struck from Centers, the Upper Plane of the Frustum Being- Oblique to the Axis. In Fig. 497, let H F G- A be the shape of -the article as seen in side elevation. The plan is shown by I L N O. In order to indicate the principle in- volved in the development of this shape, it will be necessary first to analyze the figure and ascertain the shape of the solid of which this frustum is a part. Since by the conditions of the problem the base is drawn from centers and the sides have equal flare, it follows that each arc used in the plan of the base is a part of the base of a complete cone -whose diameter can be found by completing the circle and whose altitude can be found by continuing the slant of its sides till they meet at the apex, all of which can be seen by an inspection of the engraving. Thus those parts of the figure shown in plan by K U T M and R U T P may be considered as segments cut from a right cone, the radius of whose base is either K or L R, and whose apex E is to be ascertained by continuing the alant of the side L J C 1 till it meets a vertical line erected from O 1 of the plan, which is the center of the arc of the base, all as shown in the end view. Also those parts of the plan shown by K U R and M T P are segments of a right cone whose radius is U I or T N" and whose altitude is found, as in the previous case, by continuing the slant of its side G.A (which is par- allel to C 1 I/) till it meets a vertical line erected from its center T, as shown in the side view. To complete the solid, then, of which FGA H is a frustum, it will only be necessary to take such parts of the complete cones just described as are included between the lines of the plan and place them together, each in its proper place upon the plan. The resulting figure would then have the appearance shown by H D C B A when seen from the side, and that of s C 1 L 2 when seen from the end. The lines of projection con- necting the various views together with the similarity of letters used will show the correspondence of parts. This figure is made use of in the second part of Chapter V, Principles of Pattern Cutting, to which the reader is referred for a further explanation of prin- ciples. Divide one-half of the plan into any convenient number of equal parts, as shown by the small figures, and from the points thus established carry lines verti- cally, cutting the base line H A, and thence carry them toward the apexes of the various cones from the bases of which they are derived. That is, from the points upon the base line H A derived from the arc K M draw lines toward the apex E, and from the points de- rived from the arc I K carry lines toward the apex D, and in like manner from the points derived from the arc M N carry lines in the direction of the apex B, all of which produce until they cut the top line F G of the article. From the points in F G thus established carry lines to the right, cutting the slant lines of the cones to which they correspond. Tims, from the points occurring between F and / draw lines cutting B A, being the slant of the small cone, as shown bv the points immediately below W. In like manner, from the points between g and G carry lines cutting the same line, as shown at G. The slant line of the large cone is shown only in end elevation, and there- fore the lines corresponding to the points between / and g must be carried across until they meet the line B 1 L 1 . Commence the pattern by taking any convenient point, as E 1 , for center, and E' L a as radius, and strike the arc L 3 S indefinitely. Upon this arc, commencing Pattern Problems. 277 at any convenient point, as K 4 , set off that part of the stretchout of tin- plan corresponding to the base of the, larger cone, as shown by the points o to 13 in the plan, ami as indicated by corresponding points from K* to M 2 in the arc. From the points thus established draw lines indefinitely in the direction of the center E 1 , as aa shown by/ 1 g\ Next take A B of the side eleva- lion as radius, and, setting one fool ( >f (lie compasses in the point K' of the arc, establish the point 1)' in the line K' K 1 , and in like manner, from M' J , with the saijie radius, establish- the point B 2 in the line M" K', which will be the centers from which to describe, those parts of the patterns derived from the smaller cones. From I) 1 and IV as centers, with radius B A, strike aroe from K' and M 2 , respectively, as shown by K' F and M' N 1 , upon which set off those parts of the stretchout cor- responding to the smaller cones, as shown b\- the arcs K I and M X of the plan. From the points thus estab- lished, being 5 to 1 and 13 to 17, inclusive, draw radial lines to the centers D' and B J , as shown. -Hif / IV / / \ \ /--/--I -J -l l j fiij. 497. Pattern of the Frustum of a Cone, the Base of which is an Approximate. Ellipse Struck from Centers, the Upper Plane of tlte frustum being Oblique to the Axis. shown. From E' as center, with radii corresponding to the distance from E 1 to points 5 to 13 inclusive, established in the line B 1 L 5 already described, cut cor- responding radial lines just drawn, and through the points of intersection thus established draw a line, all For that part of the pattern shown from F' to /', set the dividers to radii, measuring from B, corre- sponding to .the several points immediately below W of the side elevation, and from D' as center cut the cor- responding radial lines drawn from the arc. In like manner, for that part of the pattern shown from G 1 to g', set the dividers to radii measured from B, corre- sponding to the points in the line B A at G, with which, from B" as center, strike arcs cutting the correspond- ing measuring lines, as shown. Then F' G 1 N 1 F will be one-half of the pattern sought -in other words, the part corresponding to I K L M N of the plan. The whole pattern may be completed by adding to it a du- plicate of itself. 278 The New Metal Worker Pattern Book. PROBLEM 147. The Envelope of a Right Cone, Cut by a Plane Parallel to Its Axis. Let B A F in Fig. 498 be a right cone, from which a section is to be cut, as shown by the line C 1) in the elevation. Let G L H K be the plan of the cone in which the line of the cut is shown by D' D". For the pattern proceed as follows: Divide that portion of the plan corresponding to the section to be cut off, as shown by D 1 G D", into as many spaces as are nec- essary to give accuracy to the pattern, and divide the remainder of the plan into spaces convenient for laying off the stretchout. From A as center, with radius A B, describe an arc, as M N, which make equal to the stretchout of the plan G L H K, dividing it into the same spaces as employed in the plan, taking care that its middle portion, D" D', is divided to correspond with D'D'of the plan. From the points in M N correspond- ing to that portion of the plan indicated by D 1 G D' namely, 8 to 16 inclusive draw lines to the center A, as shown. From points of the same number in the plan carry lines vertically, cutting the base of the cone, as shown from B to D, and thence continue them toward the apex A, cutting C D, as shown. From the points in C D carry lines at right angles to the axis A E cutting the side of the cone, as shown by the points between C and B. From A as center, with radii corresponding to the distances from A to the several points between C and B, cut lines drawn from points of corresponding number in the stretchout, to A, and through the points of intersection thus obtained trace a line, as shown by D 3 (7 D 4 . Then the space indicated by D 3 C 2 D 4 is the shape to be cut from the envelope M A IS" of the cone to produce the shape to fit against the line C D in the elevation. To obtain the pattern of a piece necessary to fill the opening D 3 C" D* in the envelope, and represented by C D of the elevation, draw any vertical line, through which draw a number of horizontal lines corresponding in hight to the points in C D. The width of the piece upon each of these lines may be found by measuring 7 e K Fig. 498. -The Envelope of a Right Cone Cut by a Plane Parallel to its Axis. across the plan upon lines of corresponding number, as 11 13, 10 14, etc. Such a section is properly called a hyperbola (see Def. 113, Chap. I). PROBLEM 148. The Pattern for a Scale Scoop, Having Both Ends Alike. In Fig. 499, let A B C D represent the side ele- vation of a scale scoop, -of a style in quite general use, and E F II G a section of the same as it would appear cut upon the line B D, or, what is the same, Pattern Problems. so far as concerns the development of the patterns, an end elevation of the scoop. The curved line ABC, representing the top of the article, may be drawn at will, being, in this case, a free-hand curve. For the patterns proceed as follows : From the center K, by which the profile of the section or end elevation is Fiij. Vi'.i.~T.\e Pattern for a Scale Scoop. drawn, draw a horizontal line, which produce until it meets the center line of the scoop in the point 0. Produce the line of the side D C until it meets the line just drawn in the point X. Then X is the apex and X the axis of a cone, a portion of the envelope of which each half of the scoop may be supposed to be. Divide one -half of the profile, as shown in end elevation by E G, into any convenient number of spaces, and from the points thus obtained carry lines horizontally, cutting the line B D, as shown, and thence carry lines to the point X, cutting the top B C, as shown. With X D as radius, and from X as center, describe an arc, as shown by L N, upon which lay off the stretchout of the scoop, as shown in end elevation. From the points in L N thus obtained draw lines to the center X, as shown. From the points in B C drop lines at right angles to X, cutting the side D C, as shown. "With X as center, and radii corresponding to each of the several points between D and C, describe arcs, which produce until they cut radial lines of correspond- ing numbers drawn from points in the arc L N to the center X. Then a line traced through the points thus obtained, as shown by L M N, will be the profile of the pattern of one-half of the re- quired article. PROBLEM 149. The Patterns for a Scale Scoop, One End of Which Is Funnel Shaped. In Fig. 500 is shown a side view of a scale scoop by which it will be seen that the portion A B G II of the funnel-shaped end is a simple cylinder and, there- fore, need not be further noticed hen'. In Fig. 501 are shown a side and an end elevation of the tapering portions. It will also be seen that the part D E F of the side view is similar in all respects to the article treated in the preceding problem, and the pattern shown in connection with the same is obtained by exactly the same method as that there described and need not, therefore, be repeated. An inspection of the side elevation will show that the part G B C D F is a section of a cone of which I is the apex, H F the base and II' C" F' the plan of the base, and that this cone is cut by the lines B G and C D. To obtain the pattern of this part, first divide F' C" and C" H' of end elevation into any con- venient number of parts, and from the points thus ob- Fig. 500. Scale Scoop, One End of which is Funnel Shaped. tained carry lines cutting the miter line H F, as shown. From the points in H F carry lines to the 280 Tlie New Metal ll'//-/,-e/- Pattern Boole, apex I, cutting the curved line C D, as shown. From the points in C D drop perpendiculars cutting the sides the same as in Fig. ."><>1. With I of Pig. 502 as cen- ter, and IG and 1 F as radii, describe the arcs It 8 / END ELEVATION Fig. 501. Side and End View of Conical Portion of Scoop, with Pattern of Piece D E F. PATTERN 2 1 * Fig. BOS. -Pattern nf Piece B C D F G of Fig. 600. G F, as shown. For convenience in describing the pattern a duplicate of the side view of this part is shown in Fig. 502, in which similar parts are lettered and P Q. Upon P Q lay out twice the stretchout of H' C" F' of Fig. 501, if the pattern is desired in one piece. Thus the stretchout of II' ('" is represented l>v P (' and (^ \V, from the points in which draw lines to the center I. From I as center, and radii corresponding to the dis- tance from I to each of the various points in G F, descrilie ;nvs cutting lines of similar numbers. Trace lines through the points thus obtained, as shown by T U, V W. Tin- H T U W V S is the pattern for part of scoop shown in side elevation l>v BCD PROBLEM 150. The Pattern of a Conical Spire Mitering upon Four Gables. Let E I B in Fig. 503 be the elevation of a pin- nacle having four equal gables, down upon which a conical spire is required to be mitered, as shown. Produce the sides of the spire until they meet in the apex D. Also continue the side E F downward to anv convenient point below the junction between the spire Pattern Problems. 281 and the gables, as shown by II, which point may be considered the base of a cone of which the spire is a part. Let A" K L M be the plan of the gables^ The diagonal lines V L and M K represent the angles or valleys between the gables, while R S and T U repre- sent the ridges of the gables over which the spire is to be fitted. Through the point II in the elevation draw a line to the center of the cone and at right angles to its axis, as shown by II C. -This will represent the half diameter or radius of the cone at its base. With radius C H, and from center A" of the plan, describe a circle, as" shown, which will represent the plan of the cone at its base. At any convenient distance from the elevation, and to one side, project a diagonal section correspond- ing to the line M A" in the plan, as follows: From all the points in the side of the pinnacle draw horizon- tal lines indefinitely to the left, which will establish the hights of the corresponding points in the section. Fro in any vertical line, as D' A', as a center'line set off upon the horizontal lines the distances as measured upon the line M A" of the plan. Thus make B' A 1 5 433 F 2 1 Z 3 4 of the crossing of the two ridges of the gables, there- fore a line drawn from F 1 to B 1 will represent one of the valleys between the gables. Draw H' D 1 , the side of the cone. Its intersection with the line of the val- ley at G will then represent the hight of the lowest points of the spire between the gables, and a line pro- D 1 I. Fig. 504.-Pattern. M U Fig. 5na Plan, Elevation and Diagonal Section. The Pattern of a Conical Spire Mitering Upon Four Gables. equal to M A" and C' IP equal to A' 5, the radius of jected from this point back into the elevation, as shown, the cone at its base. The point F' represents the hight will locate those points in that view. Tli.e New Metal Worker Pattern Boole. To describe the pattern, first divide one-eighth of the plan of the cone, choosing the one which miters with the gable shown in the elevation, into any num- ber of equal spaces, as shown by the small figures. From these points carry lines vertically cutting the base of the cone H C, as shown, and thence toward the apex D, cutting the lineB J of the gable, against which this part of the cone is to miter. As the true distance of any one of the points just obtained upon the line B J from the apex D can only be measured on a drawing when that point is shown in profile, proceed to drop these points horizontally to the profile line D H, where they are marked I 1 , 2 1 , etc., and where their distances from D can be measured accurately. Next draw any straight line, as D' H" of Fig. 504, upon which set off all the distances upon the line D H of the elevation, all as shown, each point being lettered or numbered the same as in the elevation. With D" of Fig. 504 as a center, from each of these points draw arcs indefi- nitely to the left, as shown. Upon the arc drawn from IF set off spaces corresponding to those used in spac- ing the plan, beginning with IF, as shown by the small figures, and from each point draw a line toward the center D" cutting arcs of corresponding number drawn from the line F" IF. A line tra'ced through the points of intersection (g to F 2 ) will give the shape of the bottom of the cone to fit against the side of one of the gables, or one-eighth of the complete pattern. By repeating the space 1 5 upon the arc' drawn from H 1 seven times additional, as marked by the points lando, the point V will be reached, from which aline drawn to D 2 will complete the envelope of the cone. From the points marked 1 and 5 draw lines toward D 1 intersecting the arcs of corresponding number. This will locate all of the highest and lowest points of the pattern, after which the miter cut from g to F' can be transferred by any convenient means, as shown from [ the upper piece is iv quired to m,-ike with that f the lower, then bisect the angle J 1 II. as shown l>v the line I K. I>raw 1) K p;.ra!!"| to I \\ at the required hight of the lower piece, which will lie the miter line sought. Before completing the elevati >f 1 he 1 elbow it. will lie necessarv to notice a peculiarity of the oblique section of a eonc vi/.. that although the line A II bisects the cone and its base it does not bisect the oblique line 1) K, as by measurement the center of 1) K is found to be at x. Therefore, through the point/;, which is as far to the right of x as point " is to the left of it, draw auv line. 08 /> A', parallel to I .7 and make l> A' equal in length to a A, and draw A 1 I> and A 1 E. Xcxt draw (!' K' at right angles to A' />. representing the upper end of the elbow. Make' 1) F equal In K K 1 . and K (i equal to D G'. Then B F (I will be the elevation of a frustum of a cone, which, when cut in two upon the line D K, will, when the upper section is turned half-way around upon the lower part, form the elbow BDG 1 F 1 K ('. At an v convenient distance below the base of cone B draw half the plan, as shown by lj II M, which divide into any convenient number of equal spaces. From the points of division erect lines vertical! v, cut- ting the base of the cone 13 C, and thence carry them toward the point A, cutting the miter lino D E. Plac- ing the T-square parallel to the base line B C bring it successively against the points in D E, cutting the side of the cone, as shown below D. From A as center, with radii A 15 and A F, draw ares, as shown. Upon the are drawn from B, begin- ning at any convenient point, as N, step off a stretch- out of L II M, as shown by the small figures. From each of the points thus obtained draw measuring lines toward the. point A, and from the last point one cutting tli;- are drawn from F at <^. Placing one point of the compasses at the point A, bring the pencil point in turn to each of the points in the side of the cone below D and cut measuring lines of corresponding number. Then a line traced through the points of in- tersection, as shown from S to It, will be the miter cut between the two parts of the pattern of the frustum N I' Q necessary to form the patterns of the required elbow. As but half the plan of the cone was used in ob- taining a stretchout, the drawing shows but halves of the patterns. In duplicating the halves to form the co uplete patterns the upper piece can be doubled upon the line Q S and the lower part upon the line R N, thus bringing the joints on the short sides. If, according to the second condition stated at the beginning of this problem, the upper section of this elbow is required to have more or less flare than the lower section, thereby placing the apex A 1 nearer to or farther away from the line D E, a different course will have to be pursued in obtaining the pattern. If, for instance, the hight of the cone A II be reduced, the 1 base BC remaining the same, the proportions that is, the comparative length and width of the ellipse derived from the cut I) K would be different from thosederived from the same' cut were the proportions of the cone to remain unchanged. Therefore, since the shape of the lower piece at the line D E is a fixed factor, if the circle at G 1 F 1 be shifted up or down the axis, or, re- maining where it is, its diameter be changed, the piece D G' F 1 E becomes an irregular tapering article, in which case its pattern can most easily be obtained by triangulation. Patterns for pieces embodying those conditions can be found in Section 3 of this chapter, to which the reader is referred. PROBLEM 153. Patterns for a Three-Piece Elbow in a Tapering: Pipe. In Fig. 508 is shown a three-piece elbow occur- ring in taper pipe, in which the flare is uniform throughout the three sections. In solving this problem the simplest method will be to construct the elevation of the elbow and an elevation of an entire cone, from which several sections may be cut to form the re- quired elbow, at one and the same time. Therefore in the ele\ation of the cone E F G let L 1 M' be drawn at a distance from E F equal to the total length of the three pieces measured upon their center lines, and also let its length be equal to the diameter of the elbow at its smaller end ; then through E and L 1 and 286 Tlte Xew Metal Worker Pattern Book. through F and M' draw the sides of the cone, inter- secting in G. At any convenient point, as B, draw the line B A at the angle which the axis of the middle piece is required to make with that of the lower (in this case 45 degrees), and bisect the angle ABC, as shown, by the line B D. Parallel with B D draw P B at any re- upper piece is required to make with that of the middle piece (in this case also 45 degrees, or hori- zontal), and bisect the angle S U I, as shown, by U T. Then the miter line N can be drawn parallel with U T at any required distance from I, upon which locate the point i, making N i equal to O j. From i draw the axis of the upper piece of the elbow paral- Fiy. 508. A Three-Pica Elbow in a Tapering Pipe. quired -night, upon which locate the point I making P I equal to E J. (The reason for this is explained in the previous problem.) From the point I draw the axis of the second section of the elbow parallel with A B, making it (I H) equal to J G, and draw P II and R H. From any convenient point upon this axis, as U, draw U S at the required angle which the axis of the .lei with U S, making i K equal to j H. Next locate the line N upon the original cone, making N' P equal to R and 0' R equal to P N. Now make N L equal to N' L and M O equal to M' 0' and draw M L. It may be remarked -here that on account of the shifting of the positions of the axes of the several Pattern PrMems, 287 pieces upon the miter lines l>y turning them, as shown by L ,1 and i j, it will lie impossible to ascertain with extreme aeeiiraev the lengths of the various pieces upon their axes until the elevation E P N L M O R F is drawn, and therefore' to obtain the position which the line M L will occupy. This method of solving the problem is given upon the supposition that its simplicity will compensate for this slight inaccuracy, as usually ditl'erences of length can be made up in the parts with which the elbow may be connected. If the lines M h and K F are to be assumed at the outset as lixed factors between which a tapering elbow is to be constructed, it will be some- what difficult to ascertain the exact dimensions of a cone, K F G, which can be cut and its parts turned so as to constitute the required elbow. Hence, while two of the pieces (say the two lower ones) can easily be cut from an entire cone assumed at the outset, the third piece will have to be drawn arbitrarily to fit be- tween the last miter line N O and the small end M L, and will very likely be of different flare from that of the other two pieces. This will necessitate the last section being cut by the method of triangulation, problems in which are demonstrated in Section 3 of this chapter, to which the reader is referred. Having, as explained above, obtained the lines of cut through the cone, the patterns may be described as follows : Draw the plan V W Y, its center X fall- ing upon the axis of the cone produced, which divide in the usual manner into any convenient number of equal parts. Through the points thus obtained erect perpendiculars to the base E F, and thence carry lines toward the apex G, cutting the miter lines P R and N' O'. With the T-square at right angles to the axis G.C, and brought successively against the points in N' U' and P R. cut the side G F of the cone, as shown by the points above 0' and below R. From Gas center, with radius G F, describe the arc E 1 F 1 , upon which lay off the stretchout of the plan V W Y, as shown by the small figures 1, 2, 3, etc., and from these points draw measuring lines to the center G. From G as center describe arcs corresponding to the distance from G to the several points established in G F, which produce until they intersect lines of corresponding numbers drawn from the center G to the arc E 1 F 1 . Through these points of intersection trace lines, as shown by 0'' N 2 and P' R 1 . From G as center, with radius G M', describe the arc L' W. Then L' M' N 3 0' is the half pattern of the upper section, 0' N 3 R' P' that of the middle section, and P' R' F' E' that of the lower section. PROBLEM 154. The Patterns for a Regular Tapering Elbow in Five Pieces. Fig. 509. Diagram of Angles for a Five Piece Elbow. In this problem, as in the two immediately preced- ing, the various pieces necessary to form the elbow may be cut from one cone, whose dimensions must be determined from the dimensions of required elbow. The first essential will be to determine the angle of the cutting lines, which may be done the same as if the elbow were of the same diameter throughout. Such an elbow of five pieces would consist of three whole pieces and two halves; therefore, if it is to be a right angle elbow, divide any right angle, as A' B C' in Fig. 509, into four equal parts, as shown by the points 1, '2, 3. Bisect the part A' B 3 by the line A B and transfer the portion A' B A to the opposite side of the figure, as shown by C' B C. This gives the right angle ABC divided into the same number of pieces and half -pieces as would be em- ployed in constructing an ordinary five-piece elbow, 288 'fie Xcw Metal Worker l*uUern Book. M Fig. BIO. A. Fivt-Pieei Elbow in a Taptriny fife. Pattern Problems. The division lines in this diagram arc of the correct alible for the miter lines in the el how pattern, and there- fore can be used upon the diagram of the cone, out of which are to be obtained the pieces to compose the re- quired elbow. It is assumed that the amount of rise and projec- tion are not specified, therefore after having got the line of the angle or miter it becomes a matter of judg- ment upon the part of the pattern cutter what length Fig. all. Elevation of Five-Piece Tapering Elbow. shall be given to each of the pieces composing the elbow. In Fig. 510, let A B represent the diameter of the large end of the elbow. From the middle point in the line A B, as C, erect a perpendicular line, as indicated by C N, producing it indefinitely. On the line C N, proceeding upon judgment, as already mentioned, set off C X to represent the length of the first section of the elbow measured upon its center line. With X thus determined, draw through it the line D E, giving it the same angle with A B as exists between B C 1 of Fig. 509 and the horizontal B C. This, in all probability, can most readily be done by extending B A indefinitely beyond A and letting E D intersect with B A extended, producing at their intersection an angle equivalent to C B C 1 of Fig. 509. From the point X set off the dis- tance X Y, also established by judgment, thus deter- mining the position across the cone of the rniterlineof the next section. Through Y draw G F at the sam< angle as D E, already drawn, but inclined in the oppo site direction. In like manner locate the two other miter lines shown in the diagram, finally obtaining the point Z. From Z set off the width toward N of the last section of the pattern, and through the point N" tli us obtained draw the line M at right angles to ON, making it in length equal to the diameter of the small end of the elbow and placing its' central point at N. Through the points A M and B of the figure thus constructed draw lines, which produce indefinitely until they intersect the axis in the point P. Then P will be the apex of the required cone. Construct a plan of the base of the cone or large end of the elbow below and in line with the diagram, as shown in the drawing, which divide into any con- venient number of spaces, as indicated by the small figures, and from the points thus obtained carry lines vertically, cutting the base of the cone A B. From A B continue them toward the apex of the cone, cut- ting the several miter lines drawn. With the apex P of the cone for center, and with P B as radius, describe the arc T U, upon which eet off a stretchout of one- half the plan, all as indicated by the small figures. From the points thus established in T U carry lines to the center P. With the T-square placed at right angles to the axis N C of the cone, and brought against the points of intersection in the several miter lines made by the lines drawn from points in the base of the cone to the apex, cut the side O B of the cone, as shown. Then from P as center, with radii corresponding to the distance from P to the several points on B, as men- tioned, strike arcs cutting the lines of corresponding numbers in the pattern diagram, as shown. Then lines traced through the points thus obtained, as indicated bv D' E', F 1 G', etc., will cut the pattern W U T of the frustum in such a manner that the sections will consti- tute the half patterns of the pieces necessary to form the required elbow. In Fig. 511 is shown an elevation of the elbow resulting from the preceding operation. Til,' Xl'lf .]/,'/,(/ II'W.VV ['llllr PROBLEM 155. The Frustum of a Cone Intersecting a Cylinder of Greater Diameter than Itself at Other than Right Angles. la Fig. 512, K (i II F iv prose n Is an elevation of I obtained drop points parallel with the side (! II <>[ the the cylinder, and M N L K an elevation of the frustum of a cone intersecting it. V Z Q represents the profile or plan of the cylinder, to which it will l>e necessary to add a correctly drawn plan of the frustum before the miter lino in elevation can be obtained. At any con- cylinder and continue them indefinitely, cutting the line F 1 ()', which is drawn through the center of the plan of the cylinder at right angles to the elevation, all as shown in the ciurravinir. Make V \V V \\" of the lirst section c .instructed, hi like manner meas C E uis 12 n 10 a 'us 5^ 12 x x f-i-^f- iF W' Fig. 512. The frustum of a C/'une Iiitcrseetiiiy a, Cylinder uf Itiviter Diameter than Itself at. Other than Itirjkt venient point on the axial line T O of the cone construct the profile V Y X W, which represents a section through the cone on the line M N. Divide the section V Y X W into any convenient number of equal spaces in the usual manner, as shown by the small tignres 1. 2, 3, 4, etc. From each of the points thus established drop lines parallel with the axis of the cone cutting the line M N, From the intersections in M N thus distances from the center line V X of the iirst section to the points 2, 3, 4-, etc., and set oft corresponding spaces in the plan view, measuring from M' N', upon lines of corresponding numbers dropped from the in- tersections in M N, already described. Then a line traced through these points will represent a view of the upper end of the frustum as it would appear when looked at from a point directly above it. Produce the 21U sides <>!' the frustiua K. .\[ ;uid L N until they meet in tlie point < I. From (.) drop ;i line parallel to the side (r II of the cylindt.'r, cutting the line K 1 ()' in the point 0', thus establishing the positi< f the apex of the cone in the plan. From the point O' thus estaMislied draw lines through the several points in the section M' V N' W. which produce until they intersect the plan of the cylinder in points between Z and (.). as shown in the engraving. From 0, the apex of the cone in the elevation, draw lines? through the several points in M X alreadv determined, which produce until they cross G II, the side of the cylinder, and continue them inward indefinitely. Intersect these linos l>v lines drawn verticallv from the points of corresponding number be- tween Z and Q of the plan just determined. Then a line traced through these intersections, as indicated by K T L, will represent the miter between the frustum and cylinder, as seen in elevation. To lav off the pattern proceed as follows: From O as center, with O N as radius, describe the are P R, on which set oft a stretchout of the section Y V W X in the usual manner. From 0, through the several points in P It thus obtained, draw radial lines indefi- nitely. From the several points in the miter line K T L draw lines at right angles to the axis O T of the cone. producing them until they cut the sick: N L. From Oas center, with radii corresponding to the distance from to the several points in N L just obtained, describe ares, which produca until they intersect radial I in/s of corre- sponding number drawn through the stretchout P R. Then a line traced through these points of intersection, as indicated by S 17 U, will be the lower line of the pat- tern sought, and P S L' II R will be the complete pattern. The pattern for the cylinder and the opening in the same to lit the intersection of the cone is reallv a problem in parallel forms, with which problems (Section 1) it should properly be classed. F' Z Q is the pro- file of the cylinder, and L T K is the miter line. The stretchout R 1) is drawn at right angles to K F, the direction of the mold or cylinder. The points between Z' and Q" of the stretchout are duplicates of those be- tween Z and Q of the plan. Place the f-square at right, angles to the cylinder, and, bringing it successively against the points in the miter line K T L, cut lines of corresponding numbers. A line traced through the points of intersection thus formed, as shown by Z' K' Q' L', will be the shape of the required opening in the cylinder. PROBLEM 156. The Patterns of the Frustum of a Cone Joining: a Cylinder of Greater Diameter than Itself at Other than Right Angles, the Axis of the Frustum Passing to One Side of That of the Cylinder, Let E F II G in Fig. 513 be the elevation of a cylinder, which is to be intersected by a cone or frus- tum, 1) A .1 C, at the angle F D A in elevation, and which is to be set to one side of the center, all as shown by S PL M R of the plan. Opposite the end of the frustum, in- both elevation and plan, draw a sec- tion of 'it, as shown by T I' V W in the elevation and T 1 I*' V W in the plan. Divide both of these sec- tions into the same number of equal parts, commenc- ing at corresponding points in each, and number them as shown by the small figures in the diagram. From the points in T I" V \V carry lines parallel to the axis of the cone, cutting the line A .1, and thence drop them vertically across the plan. From the points in the section T 1 U' V \V draw lines parallel to the axis of the cone, as seen in plan, intersecting the lines of cor- responding number dropped from A J just described. Through these points of intersection 1 trace a line, as shown by L M. Then L M wdl show tin end of the frustum A J as it appears in plan. From X, the apex of the cone in elevation, drop a line vertically, cutting the axis of the cone in plan as shown at X'. From X draw lines through the points in A J and extend them through the side of the cylinder indefi- nitely. From X' through the points in L M draw lines cutting the plan of the cylinder, as shown from P to R, and from these points carry lines vertically, intersecting those of corresponding number in the elevation drawn from the apex X. Then a line traced through these points, as shown by D K C N, will be the miter line in elevation. For the pattern of the frustum, from X as center, wi.th radius X A, describe the arc A 1 J', upon which lay off a stretchout of the section T U V W, through the points in which, from X, draw radial lines indefi- nitely. From the points in 1) K C N carry lines at right angles to the axis of the cone, cutting the side A D extended, as shown from D to B. From X as 202 The New Metal Worker Pattern Book. center, with radii corresponding to the distance from X to the various points in the line D B, describe arcs cutting radial lines of corresponding number in tho pattern. Through the points of intersection in the tern of the frustum D A J C, inhering with the cylin- der at the angle described. The method of obtaining the pattern of the cyl- inder is analogous to that described in the preceding Fig. SIS. The Patterns of the Frustum of a Cone Joining a Cylinder of Greater Diameter than Itself at Other than Right Angles the Axis of the Frustum Passing to One Side of that of the Cylinder. pattern thus obtained trace a line, as shown by problem, and is clearly shown at the left in the C 1 K 1 N 1 D 1 . Then D' N 1 K 1 C' A 1 J 1 will be the pat- drawing. PROBLEM 157. The Patterns of a Cone Intersected by a Cylinder of Less Diameter than Itself, Their Axes Crossing at Right Angles. In Fig. 514, let B G E D F A C be the elevation of the required article. Draw the plan in line with the elevation, making like points correspond in the views, as shown by M S T U P N". Let D M E be a half Section of the cylinder in the elevation and D 1 M 1 E 1 a corresponding section in the plan. Pattern Problems. 293 Divide these sections into any convenient number of equal parts, commencing at the same point in each, as shown by the small figures, and draw the center line, of the cylinder in plan D'R. From each of the points in the section shown in elevation carry lines parallel to D H Fig. S14.A Cone Intersected by a Cylinder of Less Diameter than Itself at Right Angles to Its Axis. F cutting the side of the cone, and extend them some distance jnto the figure for further use. From the several points of intersection with the side of the cone, as shown by a, b, c, d and e, drop lines parallel to the axis of the cone cutting the line D' R of the plan, giving the points a,b',c\d l and e l , and through each of these points, from R as center, describe an arc, as indicated in the en- graving. From the points in the profile D' M 1 E 1 of the plan draw lines parallel to the sides of the cylinder, producing them until they meet the arcs drawn through corresponding points, giving the points indicated by I 1 , 2', 3', 4' and 5'. From these points carry lines vertically to the 'elevation, producing them until they meet the lines drawn from points of corresponding numbers in the profile of the cylinder in the elevation, giving the points T, 2% 3', 4 a and 5'. A line traced through these points, as shown from G to F, will be the miter line in elevation formed by the junction of the cvlinder and the cone. Fig. 515. Half Pattern of the Cone Shown in Fig. 514. To obtain the envelope of the cone with the open- ing to fit the intersecting cylinder proceed as follows: From any convenient point, as A 1 , Fig. 515, draw A 1 B 1 , in length equal to A B of the elevation. Set off points e', d', c", b' and a' in it, corresponding to e, d, c, b and a of A B, Fig. 514. From A 1 as center, with radius A' B 1 , describe the arc B' V, upon which lay off the stretchout of the plan of the cone, as indicated by the small figures outside of the pattern. (But one-half of the envelope of the cone is shown in the engraving.) From the same center- A 1 describe arcs from the points e", cT, c", V and a". From the center R of the plan draw lines to the circumference through the points 2', 3', 4', etc., giving the points in the circumference marked 2 s , 3 3 , 4 3 , etc. Set off by measurement corresponding points in the arc B' V, as shown by 3*, 2', 4*, 5', etc. From these points draw lines to the center A 1 , inter- secting the arcs of corresponding number drawn from 294 flic Ncio Metal \Vnrlvr Pattern % b\ e", etc. A line traced through these points of intersection, as shown l>y K 1 U' G' P', will be the shape of the opening to be cut in the side of the cone to fit the mitered end of the cylinder. The pattern for the cylindrical part is shown abov llii 1 elevation, and is obtained in accordance with 111. principles demonstrated in the first section of this chap- ter, which need not be here repeated. PROBLEM 158. The Patterns of a Cone Intersected by a Cylinder of Less Diameter than Itself at Rig-ht Angles to its Base, the Axis of the Cylinder Being: to one Side of that of the Cone. PATTERN OF CYLINDER In Fig. 51fi, let B A C represent the elevation of the cone, D E Gr H the elevation of the cylinder, which joins the cone at right angles to the base C C. J K L M N P Q is the plan of the articles, which is to be drawn in line and under the elevation, making like points correspond in the two views, as shown. Draw a section of the cylinder in line with the elevation, as shown by E F (> U. Divide the section of the cylinder into any convenient number of equal parts, as shown by the small figures. From the apex A drop a line through the plan, as shown by A M. Through the center of the section of the pipe, as shown in plan, draw a straight line to the center of plan of cone, as shown by J P. This line will also be at right angles to K M. From each of the points in the section of the pipe in elevation drop lines parallel to the sides of the pipe cutting the side of the cone, extending them to the line J P in plan, as shown by N a ft c, etc. Through each of these points, from P as center, describe circles, PLAN Fill. Slff.Plfin and Elevation of Cnne Tntersecteil l>i/ a Cylinder at RiijM Anijles to its Base. 205 MS shown, cutting the sides of the plan of cylinder. From each of the points of intersection with the side n|' the cone (A I!) draw lines parallel with the base, and extend them inward. Lf it is desired to show the miter line in elevation formed l>v the junction of pipe and cone, from the points f elevation, as shown by II r 1> a D. From the center A of pattern describe arcs cutting the points II cba D. It. is only necessary now to make each of these arcs ecpial in length to the one to which' it corresponds in the plan by any method most convenient. Thus make ad and n '/' equal to a d of the plan, I e and b e' equal to /> e of the plan and c/and cf equal to c/of the plan. A line traced through these points, as shown by H Q D O, will be the shape of the opening. Another method of making the measurements of the arcs is shown by FitJ. 517. Half Pattern of Cone Uluiuiti in Fig. SIS. fig. !il8. Perspective View of Cone and. Cylinder Shown in Fig. 510. The half pattern of the cone, with the opening to tit the cylinder, is shown in Fig. 517, to describe which proceed as follows: From any convenient point, as A in Fig. 517, with A B of Fig. 516 as radius, strike an arc indefinitely, as shown. From B of pattern set off each way the stretchout of .1 M and .] K of plan and connect K and M with A. Then K A M B is the half pattern of the cone, or as much as shown on plan by Iv .1 M. To obtain the shape of opening to be cut in cone to correspond with the shape of pipe, on A B, the center line of pattern, set off points corresponding to the radial dotted lines of the plan and pattern in the man- ner explained in the problem immediately preceding. The pattern for the cylinder is obtained in the manner usual with all parallel forms, its only pecul- iarity in this case being that its stretchout is taken from the irregular spaces upon the profile N O P Q of the plan, which are transferred to the line P' P", as shown. A pictorial representation of the finished article is shown in Fig. 51 S. upon which some of the lines of measurement shown in Fig. 510 have been traced. Tlie New Metal Worker Pattern BooJc. PROBLEM 159. Patterns of a Cylinder Joining: a Cone of Greater Diameter than Itself at Right Angles to the Side of the Cone. Let B A K in Fig. 519 be the elevation of a right cone, perpendicular to the side of which a cylinder, L S T M, is to be joined. The first operation will be to describe the miter line as it would appear in elevation. Draw the section U V of the cylinder, which divide into any convenient number of equal parts, as indicated by the small figures, and from these points drop lines parallel to L S, cutting the side A K of the cone in the points H, F and D, producing them until they cut the axis A X in the points G, E and C. In order to ascer- tain at what point each of these lines will cut the en- velope of the cone it will be necessary to construct sections of the cone as it would appear if cut on the lines G H, E F and C D. Draw a second elevation of the cone, as shown by B' A 1 K', representing the cone turned quarter way round ; the first may be regarded as a side elevation and this as an end elevation. Draw a plan under the side elevation of the cone, as shown by N R P 0, which divide into any convenient number of equal parts, and in like manner draw a corresponding plan or half plan under the end elevation, as shown by R 1 P 1 0'. Divide this second plan into the same spaces, numbering them to correspond with the other plan. From the points 1 to 5 in plan N R P O carry lines vertically to the base B K and thence toward the ;I]M'\ A, cutting the lines C D, E F and G II. In like man- ner, from the same points (1 to 5 inclusive) in the plan R' P 1 O 1 carry vertical lines to the base B' K 1 and thence toward the apex A'. Place the J-square at right, angles to the axes of the two cones, and, bringing it against the points of intersection of the lines from X to K with C D, cut corresponding lines in the second ele- vation, and through the points of intersection thus es- tablished trace a line, as shown by C 3 C 4 . Produce the axis X 1 A 1 to any convenient distance, upon whicli set off C 1 D 1 , in length equal to C D, in which set off the points corresponding to the points in C D, and through these points draw lines at right angles to C' D'. Place the T-square parallel to the v axis X 1 A', and, bringing it against the several points in C 3 C 4 , cut the lines of corresponding number drawn through C' D', as shown, and through the intersections thus established trace a line, as shown. Then C' D 1 is a section of the cone as it would appear if cut on the line C D. In like manner carry lines from the points upon E F across to the end elevation, intersecting them with lines of corresponding number, as shown from E 3 to E 4 , and thence carry them parallel to the axis, cutting lines drawn through E 1 F 1 , which with its points has been /'it/. '119. .1 Cylinder Joininy a Cone of Greater Diameter than Itxi-lf at Right Angles to the Side of the Cone. made equal to E F. The resulting profile E 1 F 1 is a section of the cone as it would appear if cut on the line K F. Also use the points in I,', which represents the cone as lurnc'il quarter \v:iy roiiml. Draw a corresponding plan under the end elevation, as shown bv S' <^' I" ()'. Divide this plan into the same number <>f equal parts, commencing t<> number them at the sa point as in the other plan that is. at the point Q. From the points 1 to 4 in- elnsive in S' K, and through these points draw lines at ri'jht angles to ])' K'. Place the J-square parallel to the axis Y' IV, and, bringing it against the several points in X' K' X' J , cut the lines drawn through D' E 1 , a < shown, and through these points of intersection trace u line, as shown by X 4 E' W. Then N' E 1 N 3 is a section of the cone as it would appear if cut on the line 1) K. Sections corresponding to F G and M A can be obtained in a similar manner. Having obtained sections of the cone correspond- ing to the several lines D E, F G and II A, it will next be necessary to project a plan at right angles to the axis of tie cvlindcr. in which each of these section.- shall lind its place. Therefore-, from all the points of the cylinder and of its intersections with the sides and axis of the cone project lines at right angles to N X indefinitely, through which at any convenient point, draw a line, as D" X', parallel to X X. I'poii this line, as a center of the plan about to be constructed, place the oblique sections just obtained so that each mav be in line with the line in the elevation which it repre- sents, and their center lines shall all coincide with D- X 1 , all as shown. Make T J 17 equal to T U and complete the plan of the cylinder, opposite the end of which draw a prolilc, as indicated bv V W X', commencing the divisions at the point V. From the several points in the profile V W X' drop lines paral- lel with the center line D 2 X 1 against the several proiiles ,/ Iv d',f G* f and /* A 2 /<', and thence drop the points back on the elevation, cutting correspond- ing lines in it. Thus, from the intersection of the line drawn from point W (3) with G 2 /' 1 of section cut the lino F G, which in the elevation corresponds to the point 3 in the profile V W X. From the intersec- tion of a line drawn from point '2 in V W with'A' J //' of section cut the line II A, and so on, as indicated by the dotted lines. A line traced through these points of intersection, as shown by the curved line M N, will be the miter line in elevation, from which the patterns can be obtained as follows: For the pattern of cylinder shown in elevation by M T U N, on T U extended lay off a stretchout of profile V W X, through the points in which draw the usual measuring lines. Place the J-square parallel with T U, and, bringing it against the points in the miter line M N, cut measuring lines of corresponding number. Trace a line through the points thus ob- tained, as shown from m to ?/*'. Then in / f ni is the pattern of the cylinder to lit against the cone, as shown in elevation by M T U N. To obtain the pattern of the frustum carrv lines from each of the points in the miter line M X liomini-* tallv across the elevation, cutting the side of the frus- tum c C, as shown by a 1 , I 1 and d' ; also through the same points draw lines from the apex 13. cutting the base line C L, and thence drop them on the plan, as sho.vn by 1, ~2, a and b. From any convenient point, as IV in Fig. ~>-l-2, as a center, with radii equal to B c and 1> C. describe arcs, as shown by O Q O' and c 1 /'. M ike O Q O 1 equal in length to the plan of cone S Q P and. upon it set off e-ich way from the point Q 300 Tlie New Metal Worker Pattern Book. spaces equal to those upon the plan between Q and S. From these points draw lines indefinitely toward the center B 5 . With B" as center describe arcs whoso radii correspond to B M, B a 1 , B b', B d 1 and B N", cutting lines of corresponding number or letter. Then a line traced through the intersections thus obtained will be the shape of opening to cut in the envelope of frustum where it joins the cylinder, and lines drawn from and 0' toward B 2 till they cut the arc c' I' in the points c' and I 1 will complete the pattern of the frustum. PROBLEM 161. The Patterns of Two Cones of Unequal Diameter Intersecting; at Right Angles to their Axes. Let U T V in Fig. 523 be the elevation of a cone, at right angles to the axis of which another cone or frustum of a cone, F G P, is to miter. Let L K N M be a section of the frustum on the line F G. Let U" W V s be a half plan of the larger cone at the base. The first step in describing the patterns is to obtain the miter line in the elevation, as shown by the curved line from to P. With this obtained the development of the pattern is a comparatively simple operation. To obtain the miter line O P proceed as follows : Divide the profile L K N M into any convenient num- ber of equal parts, as shown by the small figures. In- asmuch as the divisions of this profile are used in the construction of the sections or, in other words, since sections through the cone must be constructed to cor- respond to certain lines through this profile it is de- sirable that each half be divided into the same number of equal parts, as shown in the diagrams. Thus 2 and 2, 3 and 3, 4 and 4 of the opposite sides correspond, and sections, shown in the upper part of the diagram, are taken upon the planes which they represent. From the points in the profile L K N M draw lines parallel to B E cutting the end F G of the frustum. Produce the sides F and P G until they meet in E, which is the apex of the cone. Through the points in F G draw lines from E, producing them until they cut the axis of the cone, as shown at A, A 1 , A". Next construct sections of the cone as it would appear if cut through upon the lines A C, A 1 B, A' D. Divide the plan U" W V" into any convenient number of parts. From the points thus established carry lines vertically to the base line U V, and thence carry them toward the apex T, cutting the lines A C, A 1 B, A" D, all as shown. Through each of the several points of intersection in these lines draw horizontal lines from the axis of the cone to the side, all as shown. At right angles to the lines A C, A 1 B, A 2 D project lines to any convenient point at which to construct the re- quired sections. Upon the lines drawn from the points A, A 1 , A" locate at convenience the points A 3 , A 4 , A". Inasmuch as A 1 B is at right angles to the axis of the cone, the section corresponding to it will be a semicircle whose radius will be equal to A 1 B. There- fore, from A 3 as center, with radius A 1 B, describe the semicircle S B 1 R. For the section corresponding to A 2 D lay off from A 6 the distances A' S 1 and A' R", in a line drawn at right angles to A" D of the elevation, each in length equal to the horizontal line drawn through the point A" from the axis to the side of the cone. At right angles to S" R a draw A' D 1 , in length equal to A" D of the elevation. Set off in it points 5, 4 and 2, corresponding to similar points in A" D of the elevation. Through these points 5, 4: and 2, at right angles to A 6 D 1 , draw lines indefinitely. From A 5 as center, with radius equal to the length of horizontal line passed through point 5 in A" D of the elevation, describe an arc cutting line 5 of the section. From the same center, with a radius equal to the length of the horizontal line drawn through point 4 in the line A" 1 ) of the elevation, strike an arc cutting the line 4, etc. Then a line traced through these points, as shown by S 3 D' R", will be the section of the cone as it would appear if cut on the line A" D of the elevation. In like manner obtain the section S' C' R 1 , corresponding to A C of the elevation. These sections may, if preferred, be obtained in ths manner described in connection with Problems 159 and 160. As these sections are obtained solely for the pur- pose of determining at what point in their perimeters that is, at what distance from points C', B 1 and D' they will be intersected by the lines representing the points 2, 3 and 4 of the profile L K M N, it is not necessary that the complete half sections should be developed. In the engraving, the small intersecting cone has so little flare that the lines A C and A 3 D cross the large cone so nearly at right angles to its axis that sections 2 2 and 4 4 could be constructed with sufficient accu- Pattern Problems. 301 4-. P 2 . 523. The Patterns of Two Unequal Cones Intersecting at liirjht Angles tu their Axes. 302 77" New .!/ b, carry lines back to the elevation, intersecting the lines A C, A 1 B, A 2 D. Then a line traced through these several intersections, as shown from O to P, will be the miter line in elevation. Having thus obtained the miter line, proceed to describe the patterns, as follows : For the envelope of the small cone, from E as center, with radius E G, de- scribe the arc F 1 G 1 , upon which set off the stretchout of the section L K M X. Through the points in this arc, from E, draw radial lines indefinitely. From E as center, with radii corresponding to the several points in the miter line P, but obtained from the oblique sections above, cut corresponding radial lines. Thus with the radius E 11 / cut lines -i and -t, with the radius Iv ' cut lines i' and -2 and with radius E 1 b cut lines :! and 3. Then a line traced through these points of intersec- tion, as shown by P' ( )' P-', will be the shape of the pattern of the frustum 1" lit against the larger cone. For the pattern of the larger cone, from T us eenter,with radius T I", describe the are V U', in length equal to the circumference of the entire plan of the cone. From the points in the miter line O P earrv lines parallel to the base of the cone cutting its side T U, as shown between () 3 and P', also through the points in O P draw lines from the apex cutting the base and thence earrv them \crticallv to tin 1 plan. These points can be numbered upon the side of the cone to correspond with the plan, but entirely in- dependent of the system of numbers emplnved upon the smaller cone. Upon the arc V U' set off points corresponding to the points just obtained in the plan from the miter line, from which draw lines toward the center T. With one foot of the compasses set at the point T, bring the pencil point successively to the points between O 3 and P 3 and cut radial lines of corre- sponding number in the pattern. Then a line traced through these intersections, as shown by X V Z V. will be the shape of the opening to be cut in the en- velope of the larger cone, over which the smaller cone will fit, and T U' A*' will be the envelope of the entire cone. PROBLEM 162. The Patterns of the Frustums of Two Cones of Unequal Diameters Intersecting: Obliquely. In Fig. 524, let M X P O be the side elevation of the larger frustum and F' G' S R the side elevation of the smaller, the two joining upon a line between the points R and S, which line must be obtained be- fore the patterns can be developed. Produce the sides S G 1 and R F' until they meet in the point E. At any convenient place on the line of the axis of the smaller frustum draw the profile II F K G, corresponding to the end F 1 G 1 . Divide this profile into any convenient number of equal parts, as shown by the small figures 1, 2, 3, etc., and from these divisions, parallel to the axis of the cone, drop points on to F' G'. From the apex E, through these points in F 1 G 1 , carry lines cutting the side F P of the larger frustum, and pro- ducing them until they meet the center line, or the base O P, all as shown by 1? A. C A 1 and D A". The next step is to construct sections of the larger frustum as it would appear if cut on each of these lines, from which to obtain points of intersection with the lines of the smaller frustum for determining the miter line from R to S in the elevation. Draw the plan of the base of the larger frustum, as shown bv T U V \V. and divide one-half of it in the usual man- ner. From these points carry lines vertically to the base () P of the frustum. Produce the sides () M and P X until they meet in the point L. From the points Pattern Problems. 303 in the base line obtained from the plan carry lines project lines at right angles cutting A" B 1 , as shown, toward the apex L, cutting the section lines A B, in the points 4, 3 and 2. In like manner make A 4 C 1 . _ -ji- ^ O'ff 'TI Fig. 514' Patterns of the Frustums of Two Unequal Cones Intersecting Obliquely. A' C and A 3 D, as shown. Parallel to A B and of the same length, at any convenient point outside of the elevation, draw A 3 B 1 , and from the points in A B equal and parallel to A C, and from the points in A C project lines at right angles to it, cutting it as shown, giving the points 4, 3 and 2, Also make A' P' equal 304 The New Metal Worker PMern Book. to the section line A' D of the elevation, and cut it by lines from the points in A" D, obtaining the points 3 and 2, as shown. In order to complete these sev- eral sections, the width of the frustum through each of the points indicated is to be set off on correspond- ing lines drawn through A 3 B 1 , A 4 C 1 and A 6 D 1 . To obtain the width through these points first draw an end elevation of the larger frustum, as shown by M 1 N' P' 0'. Produce the sides, obtaining the apex L'. Draw a plan and divide it into the same number of spaces as that shown in T U V W, and commence numbering at a corresponding point, all as indicated by V IP T 1 W. From the points in the plan carry lines vertically to the base 0' P 1 , and thence toward the apex L 1 . Place the blade of the T-square at right angles to the axis of the cone, and, bringing it succes- sively against the points in the section line A B in the side elevation, draw lines cutting the axis of the end elevation, and cutting the lines corresponding in num- ber to the several points in A B, all as shown by a a, b b and c c. Make the length of the lines drawn through A 3 B' equal to the corresponding lines thus obtained, as shown by a 1 a 1 , b 1 b\ c 1 c 1 and d 1 d\ and through these extremities trace a line, as shown by d' B 1 d', which will be the section through the cone when cut on the line A B. In like manner complete the sections P C 1 I 1 and /' D' /'. As remarked in the previous problem, it is only necessary that these sections should be developed far enough from the points B', C 1 and D' to receive the lines representing the sections of the smaller frustum. Produce A 8 B 1 , making B' E 1 equal to B E of the ele- vation, and B' X 3 equal to B X' of the elevation. In like manner make C 1 E 3 equal to C E, and C' X 1 equal to G X. Make D 1 E 3 equal to D E, and D 1 X' equal to D X 3 . Through X 3 , at right angles to B' E', draw a line in length equal to the line 2 2 drawn across the profile F K G II, with which this section corresponds, as shown by 2' 2'. Through X 4 draw a line equal to II K, as shown by H 1 K', and through X s draw 4' 4', in length equal to the line 4 4 drawn through the profile F K G H. From E', through the extremities of 2' 2', draw lines cutting the section. In like man- ner draw lines from E 3 through the points H' K 1 , and from E 3 through the points 4' 4'. From the points at which these lines meet the sections, as a" a' in the first, o o in the second and m m in the third, carry lines back at right angles to and cutting the corresponding section lines in the elevation. A line traced through the points thus obtained, as shown by E S, is the miter line in elevation formed by the junction of the two frustums. Having thus obtained the miter line in elevation, proceed to develop the patterns as follows : From the points in B S, at right angles to A 1 E, which is the axis of the smaller cone, draw lines cutting the side K S, as shown by the small figures 1, 2, 3, 4 and o. These points are to be used in laying off the pattern of the smaller frustum. From E as center, with radius E G 1 , describe the arc F 3 G", upon which step off the stretchout of the profile F K H G, numbering Fig. 525. The Envelope of the Larger Frustum Shoivn in Fig. 524. the points in the usual manner. Through the points, from the center E, draw radial lines indefinitely. From the same center, E, with radius E 1 (of the . points in E S), cut the radial line numbered 1, and in like manner, with radii E2, E3, etc., cut the correspond- ing numbers of the radial lines. A line, E 1 S 1 , traced through the several points of intersection thus formed will be the larger end of the pattern for the small frus- tum, thus completing the shape of that piece, all as. shown by E 1 S 1 G' F 3 . Patient Problems. 305 To avoid confusion of lines, the manner of ob- taining the envelope of the large frustum is shown in Fig. ">-">, which is a duplicate of the side elevation and plan shown in Fig. r>i!-t, the miter line R 1 S' and the points in it being the same. Similar letters refer to corresponding parts in the several figures. From L* as center, with radius L" O' 1 , describe an arc, as shown by V Z, and from the same center, with radius I/' M", describe a second arc, as shown by y z. Draw V >/. and upon V Z lav off the stretchout of the plan V'WT, all as shown. Draw Z z. Then ZzyY will be the envelope of the large frustum. Through th? points in the miter line R' S' draw lines from the apex of the cone to the base, and from the base con- tinue them at right angles to it until they meet the circumference of the plan. Mark corresponding points in the stretchout Y Z, and insert any points which do not correspond with points already fixed therein. From eacli of the points thus designated draw a line across the envelope already described to the apex, as shown by 3 L a , x L" and 1 L'. Also, from the points in the miter line H' S 1 draw lines at right angles to the axis of the frustum cutting the side L a 0", as shown. From L" as center describe arcs corresponding to each of these points and cutting the radial lines drawn across the envelope of the cone. A line traced through the points of intersection between arcs and lines of the same number, as shown by h R 1 h' S 2 , will be the shape of the opening to fit the base of the smaller frustum. 306 The Xciv Metal \\'orkcr PiUttm Buuk. SECTION 3. (TRIANGULATION.) The class of subjects treated in this section will include all irregular forms which can be constructed from sheet metal by simple bending or forming, but whose patterns cannot be developed by the regular methods employed in the two previous sections of this chapter. These problems divide themselves naturally, in regard to the arrangement of the triangles used in the development of the patterns, into two classes, viz. : First, those in which the vertices of the triangles used in constructing the envelopes all terminate at a com- mon point or apex, and, second, those in which the relative position of the base and the vertex is reversed in each succeeding triangle, or, in other words, those in which the vertices of alternate triangles point in opposite directions. In the introduction to Section 2 (page 2-tO), at- tention is called to the difference between a scalene or oblique cone and a right cone with an oblique base. The scalene cone may be called the type or representa- tive of a large number of forms belonging to the first elass above mentioned, since many rounded surfaces entering into the construction of various irregular liar- ing articles are portions of the envelope of a scalene cone. The principles involved in this particular class of forms are explained in that part of Chapter V refer- ring to Fig. 271, page it-i. Inasmuch as triangulation is resorted to in all cases \vliere regular methods are not applicable, it is not sur- prising that the forms here treated, especially those included in the second elass above referred to, are more varied in character than those of anv other class to be met with in pattern cutting. An explanation of methods and principles governing these will be found in the third subdivision of Chapter Y, beginning on page 86. The last few problems of this class are devoted to the development of the horizontal surfaces of arches in cir- cular walls. The arrangement of problems in this section will be in accordance with the above classification although no headings will be introduced to distinguish the classes. PROBLEM 163. The Envelope of a Scalene Cone. In Figs. 526 and 527 are shown perspective rep- resentations of scalene or oblique cones. In Fig. .")2i> the inclination of the axis to the base is so great that a vertical line dropped from its apex would fall out- side the base, while in Fig. 527 a perpendicular from its apex would fall at a point between the center and the perimeter of its base. Supposing the circumference of the base in either case to be divided into a number of equal spaces, it is plain to be seen that lines drawn upon the surface of the cone from the points of division to the apex would be straight lines of unequal lengths, and that such lines would divide the surface of the cone into triangles whose vertices are at the apex of the cone and whose bases would be the divisions upon the base of the cone. It will be seen further that with the means at hand of determining the lengths of these lines forming the sides of the triangles, the pattern cutter possesses all that is necessary in developing their envelopes or patterns. Fig. 'CM. Scalene Cones of Different Fig. &!7. In Fig. 52S, D A II is an elevation of the cone shown in Fig. 526 and D G H is a half plan of the Pattern />/ /nce the base line of the elevation and the eenter line of the plan. Fig. .~>'2\l shows an elevation and plan of the eoiio shown in Fig. ->'1~ . drawn in the same manner. The principle involved in the development of the pat- terns of the two oblique cones is exactly the same and, as will be seen, letters referring to similar parts in tin- two drawings are the same; therefore the following demonstration will applv equally well in either ease. From the apex A drop a perpendicular to the base Fiij. MS. Pattern of Cone Shown in Fig. ~>M. line, locating the point N". Divide the base D / Cone Shown in J-'ii/. 527. points 1, 2, 3, etc., upon the line N D, describe arcs indefinitelv. From any point upon the arc drawn from point 1, as ->i , draw a line to A as one side of the pat- tern. Next take between the feet of the dividers a space equal to the spaces upon the circumference of the plan, and placing one foot of the dividers at the point , swing the other foot around till it cuts the arc drawn from point 2 ; then A ~2 will be the first triangle forming part of the envelope or pattern. \Vith the same space between the points of the dividers, and 2 of the pattern as center, swing the dividers ! around again, cutting the arc drawn from point 3. Repeat this operation from 3 as center, or, in other words, continue to step from one are to the next, until all the ares have been reached, as at '/. which in this 308 The New Metal Worker Pattern Book. case will constitute one-half the pattern ; after which, if desirable, the operation of stepping from arc to arc may be continued, as shown, finally reaching the point d. Draw d A and trace a line through the points ob- tained upon the arcs, as shown by ;i C carry lines toward the apex X, cutting o D, as shown. From X as center strike arcs from each of the points in p G indefinitely. Assume any point, as G 1 , upon the arc struck from point 1 as the first point in the pattern of the base, from which draw a line to X. Set the dividers to the space used in stepping off the plan, and, commencing at G 1 , step to the second arc, and from that point to the third arc, and so on, as shown in the engraving. A line traced through these points will be the boundary of a lower side of the semi- circular end. From each of these points just obtained draw a line toward the center X. Place one foot of i the dividers at X, and, bringing the pencil point suc- cessively to the points in o D, cut radial lines of cor- responding number just drawn. A line traced through these points of intersection, as shown by N 1 L 1 , will form the upper edge of the pattern of the end piece. From the point L 1 , which corresponds to L of the plan, as center, with L' F' as radius, describe the arc F' E 1 , and from F 1 as center, with radius equal to F B of the plan, intersect it at E 1 , as shown. Draw L 1 R 1 . Then L' E 1 F 1 is the pattern of one of the sides. To L' R 1 add a duplicate of the end piece already ob- tained, all as shown by L 1 E 1 S' N", and to S' N" add a duplicate of the side just obtained," as shown by S 1 N' G*, thus completing the pattern. Second Case. This case differs from the first onlv in the fact that the top of the article, being located near one end, is drawn concentric with the semicircle of the near end. As the result of this condition, that portion of its pattern shown by S E R L K N in the plan, Fig. 543, becomes one-half the envelope of the frustum of a right cone, the method of developing which is given in Problem 123 of the previous section of this chapter. In that portion of the article shown by E F H G S N the conditions are exactly the same as in the first case. In Fig. 543 corresponding parts have been let- tered the same as in Fig. 542 3 so that the demonstra- .314 Tlie New Metal Worker Pattern Book. tion given above is equally applicable to either figure. In the final make up of the various parts of which the a duplicate of L' F 1 G' N 1 , all as clearly shown in the engraving. The seam in this case has been located R f I'-- - Fig. 54S. Plan, Elevation and Pattern of Flaring Article with Round Top and Oblong Base, the Top being Located near One End. complete pattern is composed, the part R 1 S 1 N" L 1 is of course obtained as in Problem 123 instead of being upon the line N S of the plan instead of upon N G as before. PROBLEM 169. The Patterns of a Flaring Tub with Tapering Sides and Semicircular Head, the Head having: More Flare than the Sides. In Fig. 544, A B C D shows a side elevation of the tub, L M N P the plan at the top, and E F G II K the plan at the bottom, an inspection of which will show that the head, as shown by II O or C D, has more flare than the sides, whose flare is. shown by A J or A B, the flare of the sides and foot being the same. Inasmuch as the article is tapering in plan, the conical part of the pattern will include a little more than a semicircle, as shown. The points showing the junction between the straight sides and the conical part are to be determined by lines drawn from the cen- ters by which the top and bottom were struck, per- pendicular to the sides of the article. Therefore lay off in the plan T N and T P, drawn from the center T of the curved part of the plan of the top of the article, perpendicular to the sides M N and L P respectively. Pattern Problems. 31.5 And in like manner from S, the center by which the curved part of the bottom of the article is struck, draw S 44.l1tin, Elevation and J'attern of Flarituj Tub with. 'laperimj Sides and Semicircular Head. N G II K is part of the envelope of a scalene cone. To lind the apex of this cone, first drop lines from the points T and S vertically, cutting respectively the top and bottom lines of the elevation, as shown at T 1 and S 1 . A line connecting T 1 and S' will give the inclina- tion of the axis of the cone in that view, which con- tinue indefinitely in the direction of R 1 until it inter- sects the side C D continued, as shown at R'. Then R 1 is the apex of the cone. From R' draw R 1 R verti- cally, cutting the center line of the plan at R. Then R shows the position of the apex of the cone in the plan. As the pattern of the curved portion consists of two symmetrical halves when divided by the center lino of the plan, divide the curve N into any con- venient number of equal spaces, as shown by the small figures. Lines drawn from each of these points to R would represent the bases of a series of right angled triangles whose common altitude is V R 1 , and whose hypothenuses when drawn will represent the correct distances from the apex to the vari- ous points assumed in the base of the cone. The simplest method, however, of meas- uring these bases is to place one foot of the compasses at the point R, and, bringing the pencil point successively to the points in N O, draw arcs cutting the center line, as shown between T and 0. Now place the blade of the J-square parallel to R R 1 and drop lines from each of these points, cutting the line A D as shown. From the points obtained upon A D draw lines toward the apex R', cutting the bottom line of the tub B C. These lines drawn from the points in A D to R 1 will be the desired hypothenuses and may be used in connection with the spaces of the plan in developing the envelope of the scalene cone. Therefore from R 1 as center, and radii corresponding to the distance from R 1 to the several points in T 1 D, describe a set of arcs indefinitely, as shown. Assume any point upon the arc 0, as N', as a starting point, from which draw a line to R'. With the dividers set to the space used in divid- ing the plan N O, place one foot at the point N 1 and swing the other foot around, cutting the arc 1. Repeat this operation, cutting the arc 2, and so continue to step from arc to arc until all the arcs have been reached, which will complete the outline of one-half the pattern. The operation of stepping from arc to arc can be con- tinued, stepping back from arc 5 till arc is reached at P 1 , thus completing the top line of the pattern of the entire curved portion of the tub. From each of the points thus obtained draw lines toward the apex R 1 , as shown. Place one foot of the compasses at R', and, bringing the pencil point successively to the points in 316 The New Metal Worker Pattern Book. the line S' C previously obtained, cut radial lines of corresponding number in the pattern, as shown from G' to K 1 . Lines traced through the several points in the two outlines, as shown by G' H 1 K 1 and N 1 0' P', will complete the pattern of the conical part of the tub. The patterns of the sides and foot mav be obtained as described in Problem 74 and as indicated in the upper part of the engraving. PROBLEM 170. The Pattern of a Flaring; Article whici is Rectangular with Rounded Corners, Having: More Flare at the Ends than at the Sides. In Fig. 545, A B C D E F represents the plan at the top of a portion of the flaring article, whose general shape is rectangular with rounded corners. G II I J K L represents the plan of the bottom of the same, showing that the flare at the ends, represented by I G or J D, is greater than that of the sides, represented by A G. The arc E D of the plan of the top is struck from as center, while the arc J K of the bottom is struck from N. Since the top and bottom are parallel, as shown by P Q and P S of the side elevation, the corner J D E K is a portion of the envelope of the frustum of a scalene cone. To find the apex of the cone, drop lines from and N at right angles to P Q, cutting respectively the top and bottom lines in the side view, as shown at T and U, and draw TU and continue the same indefinitely in the direction of X. Also continue Q S until it in- tersects T U at the point X. Then X will be the apex of the cone. From X erect a line vertically, cutting the line D M of the plan at M ; then M will show the posi- tion of the apex of the cone in plan. Divide the arc E D into any convenient number of equal spaces, and from the points thus obtained draw arcs from M as center, cutting D, as shown. From the points in D drop lines vertically, cutting T Q, the top line of the side, otherwise the base of the cone. From the points thus obtained in T Q draw lines toward the apex, cut- ting U S. With one foot of the compasses set at X, bring the pencil point successively to the points in T Qand draw arcs indefinitely, as shown. From any convenient point upon the arc 0, as E', draw a line to X, forming one side of the pattern. Take between the points of the dividers a space equal to that used in stepping off the plan of cone E D, and, placing one foot at the point E 1 , swing the other foot around. Cutting the arc 1, from which intersection step to the next arc, and so continue until all the arcs have been reached at D', from which point draw a line to tin- 'ipc.\ X. Likewise from each of llie points between E' and 1)' draw line.- 1 toward the apex indefinitely. Finally, with one foot of the compasses at X, bring the pencil point to each of the points in L T S H/i PLAN N "T I JK M i j / l.'l -C Fig. 545. Plan, Elevation and Pattern uf Rounded Corner fur tin Article whose Sides and Ends Flare Unequally. and draw arcs, cutting radial lines of corresponding number in the pattern. Lines traced through the points between the points K' and J' and E' and D'will complete the pattern of the curved corner. Tin' pat- tern for the plain sides can easily be obtained after the manner described in Problem 74 and added to that of the corner as may be found practicable. Pattern Problems. 317 PROBLEM 171. t The Pattern of a Flaring Article which Corresponds to the Frustum of a Cone whose Base is a True Ellipse. In Fig. 540, let G H F E be the elevation of one side of the article, L M U R the elevation of an end, K' K' F 1 U' the plan of the article at the base, and T V S 1' the plan at the top. Produce E Cr and F II of the side elevation until they meet in the point I, the apex of the cone. Divide one-quarter of the plan E 1 R' into any con- Fig. 54B. Plan, Side and End Elevations of the Frustum of an Elliptical Cone. venient number of equal parts, as indicated by the small figures. From the points thus determined draw lines to the center C. These lines will form the bases of a series of right angled triangles whose common alti- tude is the hight of the cone, and whose hypothenuses when drawn will give the true distances from the apex p to tiie several points assumed in the base of the cone. Therefore at any convenient place draw the straight line D A of Fig. 54-7, in length equal to I E' ; . Make D B equal to 1 G'. From A and B of Fig. 547 draw perpendiculars to D A indefinitely, as shown by A O and B N. Take the distances C 5, C 4, C 3, etc., of the plan and set off corresponding distances from A on A 0, as shown by A 5, A 4, A 3, etc. From these points in A O draw lines to D, cutting B N. These lines are also shown in the elevation, but are not necessary in that view in obtaining the pattern. From ]) as center describe arcs whose radii are equal to the lengths of the several lines just drawn from D to the points in A 0. From any convenient point in the first arc draw a Fig. 547. Diagram of Triangles and Pattern of Frustum in Fig. 5j6. straight line to D, as shown by W D. This will form one side of the pattern. From W, as a starting point, lay off the stretchout of the plan E 1 R' F', etc., using the same length of spaces as employed in dividing it, stepping from one arc to the next each time, as shown. A line traced through these points will be the outline of the base of the pattern, one-half of the entire en- velope being shown in the pattern from W to Z. From the points in W Z draw lines to D, which intersect by arcs drawn with D as center and starting from points of corresponding number in B N. A line traced through the points of intersection will form the upper line of the pattern, as shown. Then \V X Y / will constitute the pattern of one-half of the envelope, to which add a duplicate of itself for the complete pattern. 318 T 'lie Xcw Jtfdal Worker Pattern Hook. PROBLEM 172. The Patterns for a Hip Bath. In Fig. 548, let II A L N bo the elevation of the bath, of whieh D 1 G E 1 IV is ;i plan on the line 1) E. Let the half section A' M C' B J represent the flare whieh the bath is required to have through its sides on a line indicated by A B in elevation. By inspec- tion of the elevation it will be seen that three patterns are required, which, for the sake of convenience, have been numbered in the various representations 1, 2 and 3. Since the plan of piece No. 1 on the line 1) B, M VF- Fig. 548. Plan, Elevation and Section of a Hip Hath. which is at right angles to its axis A F, is a semicircle, as shown by G D 1 B', and since its Hare at the side, as shown by C' B" A", is the same as at B D H, its pat- tern will be a portion of the envelope of a right cone. Patterns of this class have been treated in the previous section of this chapter to which the reader is re- ferred where, in Problem 143, an exactly similar subject has been treated. The operation of obtaining this pattern is fully shown in Fig. 549, and, therefore, need not be here described. Piece Xo. I', as shown in Fig. 548, is so drawn as to form one-half of the frustum of an elliptical cone. As its section at A B (shown at the right) must neces- sarily be the same as that of piece No. 1, against which it fits, the point F is assumed as the apex of the elliptical cone, and consequently the flare at the foot, E L, is determined by a continuation of the line drawn from the apex through E. Should it be decided to have more flare at the foot than that shown by E L, the point L may be located at pleasure, and the plan of the top, K L 1 A 1 , be drawn arbitrarilv ; after which its pattern may be developed by means of the alter- nating triangles alluded to in the introduction of this Fiij. 549. Pattern of Piece No. 1 of Hip Bath. section (page 306), examples of which will be found further on in this section. The plan G E' B', from which the dimensions of the pattern are to be determined, may be a true ellipse, or may be composed of arcs of circles, as shown, ac- cording to convenience. Divide one-half the plan G K 1 into any convenient number of equal spaces, as shown by the small figures, and from each point thus obtained draw lines to the center C. To avoid con- fusion of lines a separate diagram of triangles is con- structed in Fig. .550, in which M is the hight of tho Pattern Pmblems, 310 tiii) or frustum and M F the hight of the cone. Draw M L and (' K, each at right angles to M F. Upon (' K set off from C the lengths of the several lines C 1, 1584 s >L Fiy. 550. Pattern of Piece No. 2 of Hip Bath. (' -1, etc-., of the plan. Through each of the points in C E draw lines from F, cutting the line M L. From Fiy. 551. Diagram <>/ Radii for Pattern, of Foot. V as center draw ares indefinitely from each of the points in C E and also from the points in M L. Through any point upon arc 1 of the lower set, as G, draw a line from F and extend it till it cuts arc 1 of the upper set at K; then K G will be one side of the pattern. With the dividers set to the space used in dividing the plan G E 1 , place one foot at the point G of the pattern and step to arc 2, and so continue step- ping from one arc to the next till all are reached, as ;it E, and repeat the operation in the reverse order, finally reaching B and completing the lower line of the pattern. From each of the points in G E B draw lines radially from F, cutting arcs of corresponding V V- H- Fig. 552. Pattern of Foot of Hip Bath. number drawn from M L. Lines traced through these points of intersection will complete the upper line of the pattern. Then G E B A L K will be the required pattern of piece No. 2. As the plan D 1 G E 1 B' has been drawn entirely from centers (C, P, S and P'), the pattern of piece No. 3 is exactly similar to that described in Problem 134 of the previous section of this chapter, to which the reader is referred. In Fig. 551 is shown a diagram for obtaining the radii taken from dimensions given in Fig. 548, while Fig. 552 shows the pattern described by means of the radii given in Fig. 551. 320 Tlie New Metal Worker Pattern Book. PROBLEM 173. The Patterns for a Soapmaker's Float. In Fig. 553 is shown a perspective view of a soap- maker's float. In general characteristics it is very similar to piece No. 2 of the hip bath treated in the preceding problem. It also resembles the bathtub in that its bottom is bulged or raised with the hammer, and is therefore not included in the field of accurate pattern cutting. The sides are to be considered as parts of two cones having elliptical bases, the short diameters of which are alike, but the long diameters of which vary. In Fig. 554 is shown a plan and an inverted eleva- tion of the flaring sides, showing in dotted lines the completed cones of which the sides form a part. Thus L D 1 M represents one-half the base of an elliptical cone of which L M is the short diameter and D 1 K 1 one-half the long diameter. As all sections of a cir- fig. 553. Perspective View of Soapmaker's Float. cular cone taken parallel to its base are perfect circles, so all sections of an elliptical cone parallel to its base must be ellipses of like proportions with the base. Therefore the plan of the upper base of the frustum A' P must be so drawn that a straight line from A 1 to P will be parallel to a straight line joining D 1 and L. For the pattern of the portion shown by D A E F of the elevation, first produce the line F E of the ele- vation in the direction of K indefinitely. In like man- ner produce D A of the elevation until it reaches F E produced in the point K. Then D K F may be re- garded as the elevation of a half cone, of which that part of the vessel is a portion, and K F its perpendicular hight. Next, divide one-half the plan L D 1 M into any number of equal parts, as shown by the small figures 1, 2, 3, etc. Construct the diagram of trian- gles, shown in Fig. 555, by drawing the line D K 1 of indefinite length, and the line K' K at right angles to it, making K 1 K in length equal to F K, Fig. 554. Establish the point K 2 by making the distance K 1 K' J equal to E F of Fig. 554. Draw K 2 A parallel to K' 1). From each of the points 2, 3, 4, etc., of the plan draw lines to the center K 1 , and set off distances equal to these lines upon the line K 1 D of Fig. 555, measuring from K' toward D. From eacli of the points thus ob- tained draw lines to the point K, cutting A K'. With one foot of the compasses in the point K, and the other Fig. 554. Plan and Inverted Elevation of Soapmaker's Float. brought successively to the points 1, 2, 3, etc., in the line D K' and also to the points in the line A K 2 , de- scribe arcs indefinitely. Take in the dividers a space equal to the divisions in D 1 L of Fig. 554, and, commencing at the point a in arc 7 (Fig. 555), step to arc 6 and thence to arc 5, and thus continue stepping from one arc to the next until the entire stretchout of the half plan has been laid off, as shown in Fig. 555. From each of the points thus obtained in a d draw lines to K, cutting arcs of corresponding number drawn, from A K 2 . Then Pattern Problems. 321 a line traced through the several points of intersection thus obtained, as shown by l> < and a d, will be the boundary lines of tho pattern. The pattern for the other end of the article is to be, in the main, developed in the same manner as above described. One additional condition, however, exists in connection with this piece, vi/. : To determine the dimensions of the cone of which this piece (K 15 C F, Fig. ;">54) is a part, since the flare at B C is much greater than A D, while 1,1 10 flan; of both pieces at the side is the same as shown by P L of the plan. The quarter ellipse P B' of the plan being given, and also the point C 1 , it becomes necessary to draw from C 1 a quarter ellipse which shall be of like proportions with Q123J Mi? Fig. 555. Diagram of Triangles and Pattern of Piece L D> M of FiAj. 554. P B 1 , which, as remarked above, is a necessary condi- tion, both being horizontal sections of the same cone. To do this proceed as follows : Connect the points P and B 1 by means of a straight line. From the point C' draw a line parallel to P B 1 , and produce it until it cuts the line L G', which is a straight line drawn at right angles to L M. Then G 1 becomes a point in the lower base of the cone corresponding to the point P in the upper base. Draw the line G' P, and continue it until it intersects the long diameter in IT. Drop the point G 1 vertical from the plan on to the base line D C of the elevation, as indicated by the point G. Draw a line through the points G and E, which produce in- definitely in the direction of H. In like manner pro- duce the side C B of the cone until it intersects G E produced in tho point II. Then it will be found that the point II of the elevation and the point H 1 of the plan coincide, as indicated by the line H H 1 . H G of the elevation thus represents the axis and H one side of the cone of which the piece E B C G is a part, from which it will be seen that this cone is at once elliptical and scalene. . The operation of developing the pattern from this stage forward is the same as in the previous case, all as clearly shown in Fig. 556, save only in the addition of the triangular piece indicated by G E F of the elevation. After completing the other [8 ?Q ti 6 4 Fig. 556. Diagram of Triangles and Pattern of Piece L & M of Fig. 554. portions of the pattern, this triangular piece is added as follows : The distance H' L in Fig. 556 is to be set off on the line H' C in the same manner as the distances to the other points i.e., H 1 L is equal to H 1 L of Fig. 554. Then L is to be treated in the same manner as the other points, an arc being struck from it, as indi- cated in the engraving, by which to determine the corresponding point L" in the outline of the pattern. L" G 3 is made equal to L G 1 of the plan, Fig. 554. From L' draw a line to E. Then E L" G 2 will be the pattern of the triangular piece indicated in Fig. 554 by E F G. It is to be added upon the opposite end of the pattern in like manner, as indicated by E' G 1 L 1 . 322 Tlie New Metal Worker Pattern Book. PROBLEM 174. The Envelope of a Frustum of an Elliptical Cone Having an Irregular Base. The form E P K L J shown in Pig. 557, the lower line of which is an irregular section through an elliptical cone, is introduced here, not as representing any particular article or class of forms, but because it embodies a principle somewhat different from other sections of cones previously given, which may be use- ful to the pattern cutter. B A C is the side elevation of a cone having an elliptical base, one-half of which is shown by B 1 H C 1 . Divide one quarter of the plan, as H C', into any convenient number of equal parts, as shown by the small figures. From each of the points of division draw lines to the cen- ter D", and also erect lines cutting the base of the cone B C, from which carry them toward the apex, cutting the lines' E F and J L K. The lirst operation will be that of obtaining the envelope of the complete cone in the same manner as described in pre- vious problems. Construct a diagram of triangles, as shown at the right, in which A 1 D 1 is equal in hight to A D, and at right angles to G 1 W and D 1 V, ex- tensions respectively of E F and B C. Upon D' V, measur- ing from D 1 , set off the dis- tances from D" to the several points in H C 1 , as shown. From each of the points thus obtained draw lines toward A 1 , cutting G 1 W. Also from each of these points, with A' as center, describe arcs indefi- nitely. Take between the points of the dividers a space equal to that used in dividing the plan H C', and placing one foot upon the arc drawn from point 1 in the line D 1 V, step to arc 2, thence to arc 3 and so continue till one quarter of the stretchout is completed at 7, and, if desirable, continue the operation, taking the arcs in reverse order, thus completing the outline of one-half the envelope of the cone, as shown in the engraving. From each of the points in this outline or stretchout draw measuring lines toward the center A'. Place one point of the compasses at point A 1 , and, bringing the pencil point successively to the several points of intersection on the line G 1 W, cut measuring lines of corresponding number, as shown from G 1 to G a . Place the T-square parallel to the base B C, and, bringing it successively to the several points of inter- ' 123 4 6 87 DIAGRAM OF TRIANGLES -7 Fig. 557. Patterns for the Frustum of an Elliptical Cone Having an Irregular Base. section previously obtained in the curved line L K, cut lines of corresponding number drawn from the points in D 1 V to A 1 , as shown from X to Y. Finally, with one foot of the compasses at A 1 , bring the pencil point to each of the points of intersection last obtained and cut corresponding measuring lines in the pattern. Then lines traced through the points of intersection, as shown from L 1 to L 3 and from G' to G', will complete the pattern of one-half the frustum E F K L J. Should it be desirable to cut a pattern to fill the end J L K of the frustum, as for a bottom in the same, it will first be necessary to obtain a 'correct plan Pattern Problems. 323 of the line J K L. To accomplish this, set off on the lines D" 7, D'' 6, etc., of the plan the lengths of the several lines of corresponding number drawn from the lino G 1 D 1 to the intersections between X and Y, thus obtaining the desired line L 3 K 2 . Extend the center line B' C 1 of the plan, as shown at the right, upon which lay off a stretchout of the line L K, taking each of the spaces separately as they occur, all as shown by Z K 3 , through which draw measuring lines at right angles. Place the T-square parallel to B' C', and, bring- ing it to the several points in the line L 3 K', cut corre- sponding measuring lines. Then a line traced through the points of intersection, as shown by L* K 3 , will be the pattern of one-quarter of the desired piece, which may be duplicated as necessary for a half or for the entire pattern in one piece. PROBLEM 175. The Patterns of the Frustum of a Scalene Cone Intersected Obliquely by a Cylinder, their Axes Not Lying: in the Same Plane. In Fig. 558, let A B C D represent the frustum of an oblique cone, and T S K V U the cylinder that A D and B C are the outlines of the slanting sides. ' In Fig. 559 E F G H shows the plan of the frustum SECTIONS Fig. 558. Front Elevation of the Frustum of a Scalene Cone Intersected Obliquely by a Cylinder. joins the same at the angle indicated. The view here given of the frustum is that of its vertical side, so that Fig. 559. Elevation, Plan and Sections of the Frustum of a Scalene Cone Intersected Obliquely by a Cylinder. at its base and K I J G the plan of the top, from which the side elevation is projected at the left, D C 324 27;e Xew Metal Worker Pattern Book. being the base and A D the vertical side. The inter- secting cylinder is indicated by F L M G, and its pro- file by N P Q. The diameter of cylinder is the same as that of top of frustum. Divide the profile N O P Q into any convenient number of equal parts, and from the points thus ob- tained carry lines parallel with G M, cutting I J G and F G of plan, as shown. As the points in the profile of the cylinder lie in four vertical planes, indi- shown. With I" and d" as centers, strike the arcs G b' ;ui(l (! d\ thus forming sections of the cone in plan corresponding with a l> and c d of. elevation. The, four vertical sections above referred to are shown below the plan by I' F', e f, J' 7'>. The dotted lino K L will then be the distance from K in the base to E in the top. As it is necessary in obtaining the pattern for the entire envelope that the patterns of the parts shall suc- 9 1O 11 13 12 F Fig. 576. Diagrams of Triangles Obtained from fig. 575. ceed one another in the order in which they occur in the plan, the method of development here adopted is that of constructing separately each small triangle, as in the preceding problem, instead of by means of a number of arcs, as in Problem 177 and others preced- ing it. To begin, then, with the pattern of the part corresponding to F B G of the plan. The length F G of the pattern, in Fig. 577, is established by the length F G of the plan in Fig. 575. With F of pattern as center, and P 9 of Fig. 576 as radius, describe an arc, B, which intersect with one struck from G of the pat- tern as center, and N 9 of Fig. 576 as radius, thus es- tablishing the point B of the pattern. Then F B G is Pattern Problems. 333 tlic pattern for that part of the .article shown by F B G of tlic plan. From G of pattern as center, with radii corresponding to the hypothenuses of the tri- angles shown in N G of Fig. 576. strike the arcs shown. Thus G S of the pattern is equal to.N 8, G 7 of the pattern is equal to N 7, etc. With the dividers set to the same space used in stepping off the plan, with B or 9 of the pattern as center, strike a small arc intersecting arc S previously drawn, thus locating the point S. From S as center intersect arc 7, and so continue, locating the points <> and ">. Through the points thus obtained can be traced the line B D. Then With the dividers set to the same space as was used in stepping off the plan, and commencing at 5, intersect each succeeding arc from the point obtained in the one before it, as shown by the figures 4, 3, 2, 1. Trace a line through the points thus obtained, and connect E' H, as shown. Then E'D II is the pattern for that part of the article shown on plan by E D H. With II of pattern as center, and II K of plan as radius, describe a small arc, which intersect with one struck from E' of pattern as center, and L K of Fig. 576, or what is the same, E" II" of Fig. 574, as radius, thus establish- ing the point K' of the pattern. Connect H K' and / \ y / \/ ./ \/ F C Fig. 577. Pattern uf Transition Piece, Shown in Fig. 574. (i B 1) is the pattern for that part of the article shown on the plan by G B D. With G of pattern as center, and G H of plan as radius, strike a small arc, H, which intersect with one struck from D of pattern as center and L M of Fig. 576 as radius, thus establishing the point H of pattern. Connect G H and H D, as shown. Then G H D is the pattern for that part of the article shown in plan by G II D. With H of pattern as center, and the hypoth- enuses of triangles in M L H of Fig. 576 as radii, strike arcs, as shown, making H 4, II 3, H 2, H 1 of pattern equal to L 4, L 3. L 2, L 1 of the diagram of triangles. K' E', as shown, which gives the pattern for that part of the, article shown on plan by H K E. The radii for striking the arcs in A F B of the pattern are found in O P F of Fig. 576. The length F J of pattern is established by the length F J of the plan. The radii for striking the arcs in A J E of pat- tern are found in S li J of Fig. 576. J K of the pat- tern corresponds with J K of the plan, and E K of the pattern corresponds with L K of Fig. 576. Thus E A B D E' of the pattern is the stretchout of E A B D of the plan of the top, as K J F G H K' of the pattern is the stretchout of K J F G H of the plan of the base. PROBLEM 180. The Pattern for a Collar Round at the Top and Square at the Bottom, to Fit Around a Pipe Passing through an Inclined Roof. Let A B D C of Fig. 578 represent the side eleva- tion of the pipe and C D E F the side view of the col- lar, titting against the pitch of the roof shown by G H. Construct a plan below the elevation, as shown, making J K M L the plan view of tire pipe and N P R the plan view of the collar on a horizontal line, giving the 334 Tlie New Metal Worker Pattern Book. collar an equal projection at the bottom on the four sides, as shown. Through the center point X in plan draw a line parallel to N R, intersecting the circle at K and L ; likewise through the center X, and parallel to N, intersect the circle at J and M. From J and K draw lines to the corner N ; likewise from .1 and L, L and M, and M and K, draw lines to the cor- ners R, P and 0. It will be seen that by this opera- the bases of a series of right angled triangles, whose hypothenuses will give the correct distances across the pattern of the collar. To construct these triangles proceed as follows : Upon C Y extended assume any point, as S, at which erect the perpendicular ST, equal in hight to the cone C Y F, as shown bv the dotted line from F. From S on S C set off the lengths of the several lines in K N J of the plan, as shown by 1', 2', , 2 ' X S 3' 4' 5' ^G SIDE ELEVATION C Y' D 5' 6' 9' 8' 7' DIAGRAM OF TRIANGLES IN J K N Fig. 578. Plan and Side Elevation of a Collar to Fit Around a Pipe Pa siny throu h an Inclined Roof. tion the collar has been divided in such a manner that the four corner pieces are portions of oblique cones whose apices lay at the corners of the collar, while the side pieces between are simply flat triangular pieces of metal. The dotted lines connecting the plan with the elevation show corresponding points in the two views. Divide the quarter circles K J and J L into any convenient number of equal spaces, as shown by the small figures, and from points on each draw lines to the corners N and R. Then will these lines represent 3', etc., and from these points draw lines to T. In the same manner construct the diagram of triangles shown at the right. At U upon the line Y D extended erect the perpendicular U V, equal in hight.to the conn Y D E, as shown by the dotted line E V. From I' on U D set off the lengths of the lines in J R L and draw the hypothenuses, as shown. To develop the pattern, first draw any horizontal line, as A A 1 of Fig. 579, equal in length to P R in plan of Fig. 578. With A and A' as centers, and the Pattern Problems. 335 hypothenuse V 9' of Fig. 578 as radius, describe arcs intersecting each other at 9. Now, with 9 of the pat- tern us center, and 9 8 of the plan as radius, describe A A Fig. 579. Pattern of Collar Shown in Fig. 578. the arc S ; then with V 8' of Fig. 578 as radius, and A of the pattern as center, describe an arc, intersecting the arc previously drawn, thus establishing the point 8. Proceed in this manner, using alternately first the divisions on the quarter circle L J in plan, then the livpothenuses of the triangles whose bases are shown by the lines in J L K, until the point 5 in pattern has been obtained. Draw a line from 5 to A in Fig. 579. Then with A as center, and E F in side elevation, Fig. 578, as radius, describe an arc, shown at C of Fig. 579, and with 5 of the pat- tern as center, and the hypothe- nuse T 5' of Fig. 578 as radius, describe an arc intersecting the previous arc at C. Draw a line from 5 to C. Now proceed as above described, using alternately first the spaces on the quarter circle in J K in plan, then the hypothenuses of the triangles whose bases are shown in J K N in plan, until the point 1 in pat- tern has been obtained. Then with C of the pattern as center, and W N of the plan as radius, describe an arc, as shown at D, and with F C in side elevation as radius, and 1 of the pattern as center, describe an arc, inter- secting the arc previously de- scribed at D. Draw the lines 1 D, D C, C A, and through the intersections of the arcs trace a line, shown from 1 to 9 on pattern. This will complete one-half the pattern. The entire pattern may be completed by duplicating the part 1 5 9 A C D and adding the same to that already obtained in such a manner that the side 9 A will coincide with 9 A 1 , as shown by 9 5' 1' D 1 C' A 1 . PROBLEM 181. The Pattern for a Flaring Article Round at the Base and Square at the Top. The shape shown in Fig. 580 differs from that treated in Problem 176 principally in the fact that the round end is larger than the rectangular end instead of smaller as in Fig. 566 ; the conditions involved are, however, exactly the same as in the other problem Mini consequently the method of obtaining t\\e pattern must be similar. F G II J, in Fig. 580, represents the plan of the base, K L M N that of the top, and A B C E the elevation of a side of the article. Through 0, the center of the circle of the base, draw the diameters G J and F H parallel to the sides of the top. From the four points thus obtained in the cir- cumference of the base draw lines to the angles of the top, as shown by G M and H M, H N and J N, etc. It will be seen from this that the envelope of the ar- ticle consists of four flat triangles, of which L G M is a plan and B D C the elevation, and four rounded corners, either one of which, as J N H, is a portion of 336 The Xcw Metal /' shows the vertical rectangular end, and I d shows the angle at which the round end is placed, b I d being a half profile of the round end. As will be seen by an inspection of the front view, each quarter of the circu- lar opening is treated as the base of a portion of a scalene cone whose apex is in the adjacent angle of the rectangle, the intermediate surfaces being flat triangu- lar pieces. Thus B G and G D are the quarter bases of scalene cones whose apices are respectively at A and C ; A B E and CDF are triangles whose altitudes or profiles are shown respectively by a b and e d of the side view ; and A G C is a triangle whose profile appears at o u in the plan. Divide each quarter of the pro- file b I and I d of Fig. 583 into any number of equal spaces, as shown by the small figures ; also draw a duplicate of this half profile in proper relation to the plan, as shown by m g x, which divide as before, numbering the points in each to correspond, as shown. From the points in b I p drop lines at right angles to b d, cutting the same. From the points in m g x carry lines indefinitely to the left parallel to the center line gf, and intersect them by lines of corresponding number erected vertically from the points in I d. A line traced through the points of intersection will give a correct plan view of the opening in the round end. To avoid con- fusion of lines the intersections from the points between b and h or the upper half of the opening are shown only in the near or lower half of the plan from t to w, while the points belong- ing to the lower half (h to d) are shown only in the further half of the plan from p to q. From each of the points in p q of the plan draw lines to s, which is the projection of e of the side view or apex of the cone in the lower half, and from the points in t u draw lines to o, the apex of the cone of the upper half of the article. These lines represent only the horizontal distances from s and o to the points in the opening t u q p of the plan or B G D of the front 338 The X'-tr Metal Worker J\tttent Bouk. view. To ascertain the real distances between these points it will be necessary to first ascertain their ver- tical hights from an assumed horizontal plane and then to construct from these measurements a series of right angled triangles whose hypothenuses will give the de- sired distances. From the points in I h of the side view drop lines vertically, cutting a horizontal line drawn from a, as shown between v andj; and from the points in h d drop lines to w z drawn horizontally from e. To construct the triangles required in the top part, first draw the right angle R K, as shown in Fig. 584, and from on R set off the length of lines in b hj v of side view, as indicated by the small figures. From on O K set off the length of lines in o t u of plan of top, also as indicated by the small figures. Connect the points in R with those of similar number in K, as shown. To obtain the triangles required for the bottom part, proceed in a similar manner. Draw the right angle W S L, as shown in Fig. 585. From S on S W set off the length of lines in h d z w of side view, as indicated by the small figures. From S on S L set off the length of lines in s p q of plan of top, also as indicated by the small figures. Connect the points in S W with those of similar number in S L, as shown. For the pattern proceed as shown in Fig. 586. Draw the line 0', in length equal to A B of front view or o k of plan. Bisect 0' in C, and erect the D of pattern as center, and b 2 in b I of profile as radius, thus establishing point 2 of pattern. Proceed in this Fig. 584. Diagram of Triangles in Top Half. manner, using the length of lines in R K for dis- tances from of pattern, and the stretchout between points in b I of profile of side view for *he distance be- Fiij. 585. Diagram of Triangles in Bottom Half. tween points in D G of pattern; then draw D G ami G O. With point G of pattern as center, and 5 5' in Fig. 586. Pattern for Transition Piece Shovm in Fig. 583. perpendicular C D, in length equal to a b of side view, and draw O D, D O'. These lines are equal in length to R K of first diagram of triangles. With of pat- tern as center, and 2 2' in R K as radius, describe a small are, 2, which intersect with one struck from point W S L of triangles as radius, strike a small arc, E, which intersect with one struck from point of pat- tern as center and a e of side view as radius, thus estab- lishing point E of pattern. Draw G E and E O. With point E of pattern as center, and 6 6' of triangle in Pattern Problems. 339 W S L as radius, strike a small arc, 6, which intersect with one struck from point G of pattern as center, and I 6 of profile as radius, thus establishing point 6 of pat- tern. Proceed in this manner, using the length of lines in W S L as distance from E of pattern, and the stretchout between points in I d of profile of side view for the distance between points in G H of pattern, and draw G II and II E. With point H of pattern as cen- ter, and e d of side view as radius, strike a small arc, C, which intersect with one struck from point E of pattern as center, and o/of plan, or C of pattern, as radius, thus establishing point C of pattern ; then draw H C and C E. From E and C erect the perpendicu- lars E R and C F, in length equal to c e of side view, and draw F R. With of pattern as center, and a c of the side view as radius, strike a small arc, which intersect with one struck from E of pattern as center and e c of the side view as radius, thus establishing point K of pattern, and draw O K and K E. Then DGHFREKC represents the half pattern of article. The other half can be obtained in the same manner or by duplication, as may be found convenient. PROBLEM 183. The Pattern for a Flaring Article, Round at Top and Bottom, the Top Being: Placed to One Side of the Center, as Seen in Plan. In Fig. 587, the elevation of the article is shown liv A B D C, below which is drawn the plan of the Fiy. 687. Plan and Elevation of an Irregular Flaring Article, with Lines of Triangulation Shown in the Plan. same, corresponding parts in each being connected by the vertical dotted lines. There are two methods of triangulation available in the solution of this problem only one of which is given in this connection. Divide one-half of the circle representing the base of the article into any convenient number of spaces, as indicated by the small figures, 1, 2, 3, etc. In like manner divide the inner circle, which represents the top, into the same number of spaces, as indicated by 1', 2', 3', etc. Between the points of like numbers in these two circles, as for example between 2 and 2', 3 and 3', etc., draw lines, as shown; also connect the points in the inner circle with points in the outer circle of the next higher number, as indicated by the dotted lines. Thus, connect 1' with 2, and 2' with 3, and so on, as shown. These lines just drawn across the plan are the bases of a number of right angled triangles whose altitudes are equal to the hight of the article, and whose hypothenuses, when drawn, will give the cor- rect distances across the pattern, or envelope of the article, between the points in the top and those in the bottom in the direction indicated by the lines of the plan. The triangles having the solid lines of the plan as their bases are shown in Fig. 588, while those con- structed upon the dotted lines are shown in Fig. 589, and are obtained in the following manner : At any convenient point erect a perpendicular, E F, Fig. 588, which in length make equal to the straight hight of the article, as shown in the elevation. From F at right angles set off a base line of indefinite length. On this line, measuring from F, set off lengths equal to the several solid lines in the plan. For example, make the space F 10 equal to the length 10 3-iO Tlie New Metal Worker Pattern Book. 10' in the plan, and the space F 9 equal 9 9' in the plan, and so on, until F 1 is sot off equal to 1 1' in the plan. From the points thus established in the bast- line draw lines to the point E. The triangles thus constructed will represent, sections through the arti- cle on the solid lines in the plan. In other words, the several hypothenuses of the triangles shown in Fig. 588 are equal in length to lines measured at cor- responding points on the surface of the completed ar- ticle. In like manner construct the triangles shown in Fig. 589, representing measurements taken on the dotted lines shown in the plan. Draw the perpen- dicular K G, equal in length to the straight hight of the article. From K lay off a horizontal base line in- definite in length, drawing it at right angles to K G. From K set - off lengths equal to the dotted lines in the plan that is, making the distance K 10 equal to the Fig. 588. Diagram of Triangles Based upon the Solid Lines of the Plan in Fig. 587. Fig. 589. D'agram of Triangles Based upon the Dotted Lines of the Plan in Fig. 587. distance 9' 10 in the plan, and K 9 equal to the dis- tance 8' 9 in the plan, and so on until K 2 is made equal to the distance 1' 2 in the plan. From the points thus established in the base line draw lines to the point G. Then the hypothenuses of the triangles thus constructed will equal measurements along the surface of the completed article at points corresponding to the dotted lines in the plan. With distances thus established upon the surface of the ar- ticle, and with the stretchout of the required pattern determined at both top and bottom, it is easy to lay out the pattern upon the general plan of constructing a triangle when the three sides are given. The development of the pattern can be begun at either end according to convenience, and the operation is conducted as follows: Assume any line, as 1 1' of Fig. 590, which make equal in length to A B of the elevation, or, what is the same thing, equal to E 1 of Fig. 5S8, which is one side of the first triangle. The other two sides are respectively the distance 1 2 of the plan and the liypothenuse of the triangle shown in Fig. 589 corresponding to the line 2 1' of the plan. Accordingly, take the distance 1 2 of the plan in the dividers, and from 1 as center describe a short arc. Then, taking the distance G 2 of Fig. 589 in the di- viders, and with 1' as a center, intersect the arc already struck, thus establishing the point 2 of the pattern. Fig. 590. The Pattern of Article Shown in Fig. 587. The elements of the second triangle are the distance 1' 2' on the inner circle of the plan, the hypothennse of the triangle in Fig. 588 corresponding to the line 2 2' in the plan and the side just established in the pattern 2 1'. From 1' as center, with 1' 2' of 1he plan as radius, describe a short arc. Then from 2 as center, with E 2 of Fig. 588 as radius, describe a second arc, intersecting the one already made, thus establishing the point 2'. Proceed in this manner until triangles have been described adjacent to each Puttn'ii 341 other, corresponding to the divisions first established in the plan. Then linos traced through the points thus established, as shown from 1 to 10 and from 1' to In', will constitute the pattern of half the article. The other half may be obtained by any' convenient means of duplication and may be added on to either end of the half already obtained, according as it is desired to make the joint at the widest or narrowest part of the pattern. Fiij. r,91. Model of One-Half the Article Shown in Fig. 587, Illus- trating the Construction and Use of the Trinngles. In Fig. 591 is shown a model which may be con- structed of thin metal and wires, or of cardboard and threads, according to convenience, which will assist the student in forming a correct idea of the relation- ship existing between the various lines drawn upon the plan and the lines of which the pattern is con- structed. The top and bottom of the model are dupli- cates of the inner and outer circles of the plan. The piece forming the bottom should have the solid lines and the inner circle of the plan drawn upon it as a means of placing in position the several triangular pieces shown, which are duplicates of the several triangles shown in Fig. 588. These triangular pieces having been placed in position according to their numbers, and fastened at the top and bottom, their outer edges, or hypothenuses, will then represent the solid lines drawn across the pattern, and will bear the same relation or angle to the edges of the top and bottom pieces of the model that the solid lines of the pattern bear to the top and bottom outlines of the pat- tern. Finally, threads or wires having been attached, as shown, will represent the dotted lines drawn across the pattern, and will bear the same angle to the edges of the solid triangles, as measured upon the model, that the dotted lines of the pattern bear to the solid lines, as measured upon the pattern. .Since the top and the bottom of the shape here shown are both round and horizontal, the figure be- comes that of the frustum of a scalene cone; and, therefore, its sides, if continued upward, would ter- minate in an apex which can be made the common apex of a number of triangles whose bases are the spaces upon the outer line of the plan. This method of solving the problem as applied to a full scalene cone is given in Problem 163, which see. PROBLEM 184. The Pattern for a Flaring: Article, Round at Top and Bottom one Side Being Vertical. In Fig. 592, A B C D shows the elevation of the article, below which E F G H shows the plan at the bottom and E J K L the plan of its top, both circles being tangent at the point E. Divide the circle representing the plan of the top into any convenient number of equal spaces, as -repre- sented by the small figures between K L E in the dia- gram. In the illustration only one-half of the plan has been divided, which is sufficient for the purpose. Next divide a like portion of the plan of the base into the same number of equal parts, as shown by the figures between E H G. Connect these two sets of points, first by lines drawn between like numbers, as 1 and 1', 2 and 2', 3 and 3', etc. In a like manner connect 1 of the inner circle with 2' of the base, 2 with 3', 3 with 4', etc., all as shown by the dotted lines in the plan. These lines just drawn are the bases of a number of right-angled triangles, whose altitudes are equal to the vertical hight of the article, and whose hypothenuses, when obtained, will give the correct measurements across the pattern between the numbered points. For a diagram of triangles representing the solid 342 77te New Metal Worker Pattern Book. linos in plan erect the vertical line P S in Fig. 593, equal to A B of elevation. Then at right angles from S lay off. a base line, upon which set off distances, measuring from S, equal to the lengths of the several solid lines drawn across the plan in Fig. 592. Thus make S R equal to K Gr (1 1') and S 2 to 2 2', and so Fig. 592. Plan and Elevation of Flaring Article, Showing Method of Triangulation. on. From the points thus established in the base draw lines to the apex. Then the hypothenuses of the tri- angles will be equal to measurements on the surface of the finished article on lines drawn from the points in the base to corresponding points in the top. In the same way construct the diagram representing the triangles Imsed on the dotted lines in plan, as shown in Fig. ;V,4. Set off T V equal to the straight higlit of the article. From V draw the horizontal line V U, upon which, 7 54321 Fig. 593. Diagram of Triangles Based upon Solid Lines in the Plan in Fig. 592. measuring from V, set off distances equal to the length of the dotted lines across the plan. Thus make V 2 equal to 1 2', V 3 equal to 2 3', etc. From the points thus established in V U draw lines to the apex T. These lines will be equal to measurements upon the \Vx\x \ \\ X X\N \ \\\\ 10 118 S 7 6 5 4 S 8 Fig. 594. Diagram of Triangles Based Upon Dotted Lines in the Plan in Fig. 592. surface of the finished article between the points con- nected by the dotted lines in the plan. Having obtained the correct dimensions of all the Pattern Problems. 343 triangles assumed at the beginning of the work they may now be constructed consecutively, thus developing 8' Fig. 595. The Pattern of Flaring Article Shown in Fig. 592. the pattern in the following manner: Assume any straight line, as D C of Fig. 595, which make equal to D C of the elevation, or, what is the same thing, P R o* the diagram of triangles, Fig. 593. From C as a cen- ter, with a radius equal to 1' 2' of the plan, strike a small arc, which inter- sect with another small arc struck from D as center, with a radius equal to T 2 of Fig. 594, thus establish- ing the point 2' of the pattern. From D as center, with a radius equal to 1 2 of the plan of the top, Fig. 592, strike a small arc and inter- sect it with another struck from 2' of the pattern as center, and P 2 of Fig. 593 as a radius, thus establish- ing the point 2 in the top of the pattern. Proceed in this manner, using the hypothenuses of the tri- angles in Fig. 594 with the spaces in the Outer curve of the plan, Fig. 592, to establish the points in the bottom curve of the pattern; and the hypothenuses of the triangles in Fig. 593 with the spaces in the inner curve of the plan to establish the points in the top curve of the pat- tern. Lines traced through the points of intersection, as shown from C to B and from J) to A, will, with D C and A B, constitute the pattern for the half of the article shown by E H G K of the plan. The other half may be added, as shown, by any convenient means of duplica- tion. Since the top and the bottom of this figure are both round and horizontal, it becomes the frustum of a scalene cone, which permits of its being treated by a different, and perhaps simpler, method of triangu- lation, all of which is given in Problem 167, to which the reader is referred. PROBLEM 185. - The Pattern of an Article having: an Elliptical Base and a Round Top. Fig. 596 shows the plan and elevation of the ar- ticle for which the pattern is required. Divide one- quarter part of the plan of the base E G into any convenient number of equal spaces, and divide a cor- responding part of the plan of the top L K into the same number of spaces, numbering the points of divi- 344 The Neiv Metal Worker l>,itl>-ni Rook. sion the same in both, as indicated by the small figures l r 2, 3 and I 1 , 2', 3', etc. The article here shown pos- sesses some of the general features of the cone in that it is tapering in its sides, but inspection will show that the slant or taper of its sides varies in different pails of its circumference, or in other words, that different Fig. 596. Elevation and Plan of Flaring Article with Elliptical Base and Round Top. lines drawn through like numbers in the base and top would, if extended upward, meet the axis at different hights, hence some means must be devised for meas- uring the real distances between the points in the base and the points in the top, which may be accomplished in the following manner: First connect all points in the base in plan with points of the same number in c- C 2 / <7 o////// 23 i 507 Fig. 597. Diagram of Triangles Rased upon Solid Lines of the Plan in Fig. 596. Fig. 598. Diagram of Triangles Based upon Dotted Lines of the Plan in Fig. 596. the top by means of a solid line, as shown upon the plan by lines 1, I 1 , 2, 2', etc. Also draw the inter- mediate dotted lines connecting alternate points, as shown in the engraving by 2 1', 3 2' 4 3', etc., thus dividing the entire surface of the article into triangles. Construct a diagram, as shown by A 1 N 1 C 1 , Fig. 597, in which the actual distance between correspond- ing points in base and top shall be shown. Make C 1 N 1 equalto the straight hight of the article, C N of the elevation. At right angles to it set off N 1 A 1 , in length equal to the distance 1' 1 in plan. From N' set off also on N' A' spaces corresponding to 2' 2, 3' 3, 4 1 4, etc., of the plan, and from each of these points draw a line to C 1 , as shown. Then the lines converging at C 1 represent the distances which would be obtained by measurements made at corresponding points upon the article itself. Construct a like dia- gram of the distances represented in the dotted lines in the plan, as shown by C 3 N 2 O, Fig. 598. Make C' N" equal to C N of the elevation, and from N a set off at right angles the line N" 0. Upon this line make. the spaces N" 2, N" 3, N" 4, etc., equal to the length of the dotted lines 1' 2, 2' 3, 3' 4, etc., and from the Fig. 599. The Pattern of Article Shown in l-'iij. r,M. points thus obtained in W O draw lines to C". Then these converging lines represent the same distances as would be obtained if measurements were made between corresponding points upon the completed article. For the pattern, commence by drawing any line, as P X in Fig. 599, on which set off a distance equal t<> C' 1 of the first diagram, as shown by 1 I'. Then, with the distance from 1 to 2<>f the plan for radius, and I in pattern as center, describe an arc, which intersect by another arc struck from I 1 of the pattern as center, and C" 2 of the second diagram as radius, thus estab- lishing the point marked 2 in the pattern. Next, with I I 2' of the plan as radius, and from 1' of the pattern as center, describe an arc, which intersect bv another Pattern Problems. 345 arc drawn from 2 as center, and with C' 2 of the first diagram as radius, thus locating the point 2' of the pat- tern. Continue in this manner, locating eacli of the several points shown from X to Y and from P to R of the pattern, through the several intersections, tracing the lines of the pattern, as shown. Then X Y R P will lie one-quarter of the required pattern. Repeat this piece three times additional, as shown by W X P T, V WTU and Y Z S R, reversing each alternate piece, thus completing the pattern. PROBLEM 186. Pattern for an Irregular Flaring Article which is Elliptical at the Base and Round at the Top, the Top being so Situated as to be Tangent to One End of the Base when Viewed in Plan. In Fijr. ' il>() - l ct 1} <"> ^ K he the side elevation of the article and K N M one-half of the plan of the base. The half plan of the top is shown by K W L, the base and top being tangent in plan at the point K. The conditions and method of procedure in this problem do not differ materially from those of Problem 184. Next construct the diagrams of triangles, as shown in Fig. 601 at the right of the elevation, making A U in hight equal to D E of the elevation, and lay off U T at right angles to it. Let A represent all points in the circle representing the plan of the top of the article. Lay off from U upon U T the distance from each of Fig. 601. Diagrams of Triangles Obtained from Fiij. 600. Fig. 600. Elevation, and Plan of an Irregular Flaring Article with Elliptical Base and Round Top. The surface of this article may be divided into measurable triangles in the following manner : Divide the plans of ton and base into the same number of equal parts, as shown by 1, 2, 3, etc., in the base and 1', 2 1 , 3', etc., in the top, and connect similar points in the two bv solid lines, as shown by 6 6', 5 5', etc. Also connect each point in the plan of the top with the next lower number in the plan of the base, as shown by the dotted lines in the engraving, as 6 7', 5 6', etc. the several points in the circle to the corresponding point in the ellipse. Thus make U 7 equal to 7' 7 of the plan, U 6 equal to 6 1 6, etc. , and draw the radial lines A 2, A 3, etc. In like manner construct a correspond- ing diagram, as shown by C B V, using for the spaces in B V the lengths of the dotted lines between the circle and the ellipse in the plan, and draw C 2, C 3, C 4, etc. By means of these two sets of lines, converging at A and C respectively, and the stretchouts of the two 346 The New Metal Worker Pattern Book. curves of the plan, the actual dimensions of the tri- angles into which the surface of the article has been divided can be accurately measured. These are to be used in describing the pattern as follows : At any convenient place draw the straight line P R in Fig. 602, in length equal to G F of the elevation, or, what is the same, equal to A 7 of the first diagram. As but half of the plan of the article is shown, the pattern will also appear as one-half of the whole shape, and therefore P R will form its cen- tral line. From P as center, with radius C 6 of the second diagram, describe an arc, which intersect by a second arc struck from R as center, with radius 7 6 of plan, thus establishing the point 6 of the pattern. Then with radius A 6 of the first diagram, from 6 of the pattern as center, describe an arc, which cut with another arc struck from 7' of the pattern as center, and 7' 6' of the plan as radius, thus locating the point 6' of the pattern. Continue this process, locating in turn 5, 5', 4, 4 1 , etc., until points corresponding to all the points laid off in the plan are established. Lines traced Fig. 60S. Pattern for Article shown in Fig. 600, through the points 7, 6, 5, etc., and 7', 6', 5', etc., will, with P R and S, form the patter* 1 of one-half the article. PROBLEM 187. Pattern for a Flaring Article or Transition Piece Round at the Top and Oblong: at the Bottom, the Two Ends Being Concentric in Plan. ELEVATION C END ELEVATION Q, P PLAN Fig. 60S. Plan and Elevations of Flaring Article Bound at the Top and Oblong at the Bottom. In Fig. 603 are shown the side and end elevations and the plan of an article which might form a transi- tion between an oblong pipe below and a round pipe above. According to the conditions, as given in the en- graving, the problem is capable <>f two solutions. Since the upper and lower buses ;irc composed, either wholly or in part, of semicircles lying in parallel planes, those por- tions of the pattern of the article lying between the semicircles, as P N OP, must necessarilv form parts of the envelope of a scalene cone. Those portions of the pattern ma}- therefore be obtained, if desirable, by the method employed in Problems 168 and 169. The other solution, which is perhaps the nn in- simple, is given in Figs. ti<)4 to 606. An inspection of the plan will show that the article consists of l'< un- like quarters, therefore in Fig. ti<>4 is shown an en- larged plan and elevation of one-quarter of the article. Divide P T and N of the plan eaeli into the Pattern Problems. 347 same number of equal parts, and connect the points in P T with those in N. as indicated by the solid lines. Also connect points in the top with those in of dotted lines in plan. Thus make W 7 equal to T 7 and W 8 equal to 2 8, and so on. From the points thus established in the base draw lines to V. The ELEVATION K H 6 8910 |_ W 9 ,o 8 PLAN -UN Fiij. 604.- -Quarter Plan and Elevation Enlarged, Showing Method of Triangulation. the bottom, as shown by the dotted lines. These lines represent the bases of right angled triangles, the alti- tude of which will be equal to the straight hight of the article. For a diagram of the triangles representing the solid lines of the plan, draw any vertical line, as J K in Fig. 605, which make equal in hight to the hight of the article, as shown by the dottedjine D F. From K, at right angles to J K, draw K L, upon which set off dis- tances, measuring from K, equal to the lengths of the solid lines drawn across the plan. Thus make K 6 equal to T N and K 7 equal to 2 7 of plan, and so on. Also set off from K the distance II P of plan, as shown by K H. From the points thus established in K L draw lines to J. The hypothenuses thus obtained will give the distances across the finished article, as in- dicated by the solid lines of the plan. The next step will be to construct a diagram of triangles that will give the distances between points in tin 1 base and top, as indicated by the dotted lines in plan. This diagram is constructed in a similar man- ner, as shown at the right in Fig. 605. Draw the right angle V W X, making V W equal to tl.e straight hight of the article, and from W set off on W X the lengths A'lV/. G05. Diagram of Triangles. livpothenuses of the triangles thus obtained will give the distance from points in the base to points in the top, as indicated by the dotted lines in plan. To lay out the pattern first draw any line, as T N of pattern, in length equal to J 6 of first diagram of triangles, or, which is the same thing, D E of elevation. ' From N of pattern strike a short arc Firj. 606. Quarter Pattern of Article Shown in Fig. 60S. with a radius equal to N 7 of the plan, as shown. From T of pattern as center, with radius equal to V 7 of the second set of triangles, intersect this arc, thus establishing the point 7 of pattern. From T, with radius equal to T 2 of the plan, strike a small arc, as shown, and intersect it with another from point 7 of pattern as center, with J 7 of the diagram of triangles as radius, thus establishing the point 2 in pattern. 348 Tfie New Metal Worker Pattern Book. Proceed in this manner, using alternately the hypoth- enuses of the triangles in V W X of Fig. 605, the spaces in plan of base N, the hypothenuses of the triangles in J K L, Fig. 605, and the spaces in the plan of top, P T, in the order named, and as above ex- plained. The resulting points, as indicated by the small figures in the pattern, will be points through which the pattern line will pass. For the pattern of triangle P H O of pattern, with of pattern as center, and O II of plan as radius, strike a small arc in the direction of II. With P of pattern as center, and .1 IL of the diagram of triangles as radius, describe another small arc intersecting the one just struck. Draw () 11 and II P, thus completing the quarter pattern. PROBLEM 188. Pattern for a Transition Piece Round at the Top and Oblong at the Bottom. Two Cases. In Fig. 607 is shown the plan and elevations of a transition piece, constituting the first cw, such as is frequently required in furnace work when it is necessary to connect a round pipe with another pipe of equal area but flattened into an oblong shape. In Fig. 611 are shown the plan and elevations of a SIDE VIEW. 98 7 65432 101112 13 first case, shown in Fig. 6ii7. The principle involved in developing the patterns of the two shapes is exactly the same, consequently the following demonstration will apply equally well to either Fig. 607 or 611, in each of which corresponding points are lettered the same. (It will be noticed that separate diagrams of triangles and a separate pattern corresponding to Fig. Fig. 608. Diagrams of Triangles Based upon the Solid Lines of L the Plan, Fig. 607. Fig- 607. Plan and Elevation of Transition Piece First Case. transition piece of the second cose, answering the same general description as that given above, but differing only in the fact that the circle representing the top in the plan view does not touch the side of the line rep- resenting the bottom of the article. In other words, the side F H is slanting instead of vertical, as in the fill have not been given. While in reality they would differ somewhat from those shown in Figs. 608 to Uln. they would have the same general appearance, and in method of construction would be exactly the same, and therefore have not been considered necessary to the study of the problem.) 349 "N P Q represents a plan (if the shape described, above which A B 1) C shows an elevation of the front of the same, or as seen when looking toward Q. while to the left is shown a view obtained by looking to- ward the side X, in which K (1 corresponds to P X of the plan and F II to Q' g. Divide one-half of the plan of the top or round portion of the article into anv convenient number of equal spaces, in this case l:>. Since by the conditions of the problem one-half of the round end corresponds to the semicircular end of the oblong part, divide the semicircle J N L into the same number of equal parts. V 3 V' , 654 31 1 7 8 9 10 II 12 Fill. 009. Diar/rams of Triangles Based upon the Dotted Lines of the Plan, Fig. 607. Then connect points in the two lines of the plan of the same numerals. For example, 1 with 1, 2 with '2, :> with .'!, etc. In like manner connect the points in the end of the oblong portion with points of the next higher num- ber in the round end, as shown, as, for example, 1 with 2, 2 with 3, 3 with 4, etc. Upon all of these lines drawn in the plan it will be necessary to construct sections or triangles in which these lines form the bases and in which the vertical hight of the article \V V is the altitude. The various li ypothenuses thus obtained will then represent the true distances across the linished article upon the lines indicated in the plan. The triangles correspond- in^ to the solid lincsof tlu; plan are shown in Fig. f>OS, while those corresponding to the dotted lines of the plan are shown in Fig. <>u',. In order to avoid confusion each of these sets has been divided into two groups, as shown, and are constructed as follows: Lay off at any con- venient place the line A 15 (Fig. tlUS). equal to V W of Fig. t'.uT. From the point B, and at right angles to A B, draw the line B C and upon it set off the lengths of the several solid lines connecting the two outlines in the plan. Thus make B 1 equal to the distance 1 1 or J P of the plan. B 2 is equal to the distance 2 2 of the plan, and B 3 is equal to the distance 3 3 of the plan. etc. As already explained, D E is a duplicate of A B, and E F is drawn at right angles. On E F the spaces K 10, E 11, E 12 and E 13 are set off, being equal respectively to 10 10, 11 11, etc., of the plan. From the points thus established in the base lines B (J and E F draw lines to the apices A and D, thus completing the triangles. Then the hypothenuses A 1, A 2, A 3, etc., D 13, D 12, etc.,. correspond to the width of the pattern measured between points in- dicated by like figures in the plan. In the same general manner construct the triangles shown in Fig. 009, which correspond to sections on the dotted lines across the plan. G H and L M of Fig. 609 correspond to the hight V "W of the elevation. II K and M N are drawn at right angles to the perpen- diculars, and on these base lines spaces are set off, measuring from H and M respectively, corresponding to the length of the dotted lines across the plan. Thus H 1 corresponds to 1 2 of the plan, and M. 12 corres- ponds to 12 13 of the plan. From the points thus estab- lished in the base line lines are drawn to the apices G and L, thus completing the triangles. These hypothe- nuses are equal to the width of the pattern measured M I? 13 Fitj. 610. Pattern of Transition Piece Shown in Fig. 607. between points connected by the dotted lines in the plan. By the conditions of the problem, inasmuch as there are straight portions in the oblong end, there will bo portions of the pattern that will correspond to tri- angles the bases of which are equivalent to the length of the straight portion in the plan and the hights of which are equal respectively to the distances E G and F II of the side view. 350 Tlie New Metal Worker Pattern Book. SIDE VIEW Therefore, to describe the pattern proceed as fol- lows : At any convenient place, as shown by A B in Fig. 610, draw a line equal to the width of the pattern at a point corresponding to Q 1 Q in the plan. This would be the same as F H of Fig. 607 or 611. To com- plete the triangu- lar portion referred to set off from B the distance B C equivalent to Q L of Fig. 60Y, thus obtaining the point C. The dotted line A C in the pat- tern is drawn to show the portion obtained by this means. From C as center, with the space 13 12 of the plan of the oblong end as radius, describe a small arc, as shown to the left. Then from A as center, with radius L 12 of Fig. 609, corresponding to the width of the pat- tern measured on the dotted line 13 12 of the plan, de- rscribe another arc intersecting the one just drawn, thus establishing the point 12 in the lower edge of the pat- tern. From 12 as center, with D 12 of Fig. 608 as radius, being the width of the pattern on the line 12 12 of the plan, describe a short arc, as shown at 12 in the upper line in the pattern. Intersect this with another arc drawn from A as center, with 13 12 of the plan of the round end as radius, thus locating the point 12 in the upper line of the plan. Proceed in this manner, using in the order described the stretchout of the semi- circular end of the oblong section, the hypothenuses of the triangles corresponding to the dotted .lines in the plan, the hypothenuses of the triangles corresponding to the solid lines in the plan and the stretchout of the circular end, reaching finally the points D and E of the pattern, representing one side of the remaining triangu- lar section to be added. From E as center, with K J of the plan as radius, describe an arc. From D as center, with radius equal to D E of the pattern, strike a second arc intersecting the one just drawn, thus Fig. 611. Plan and Elevation of Transition Piece Second Case. locating the point F. Connect F and E and also F and D, thus completing the pattern of the part correspond- ing to K J P of the plan. The dotted line DG drawn across the pattern corresponds to the line X P of the plan, and D A B Gr will be one half of the finished pattern. PROBLEM 189. Pattern for an Offset Between Two Pipes, Oblong in Section, whose Long Diameters Lie at Right Angles to Each Other. In Fig. 612 are shown the plan and elevations of an offset or transition piece to form a connection be- tween two pipes of oblong profile which will be spoken of in the demonstration as the /14, and in Fig. til 8 are shown the diagrams of tri- angles derived from the solid and dotted lines in Q L K J of Fig. 613. X I hi 1 plan as Proceed in dicated by /'/(/. un.l>if a triangle whose base is E G. That portion of the circle of the top, therefore, between L and its corre- sponding point K in the other half of the article must be considered as the base of an oblique cone whose apex is at G. It is always advisable in the division of a surface into triangles that the solid and dotted lines crossing the plan should intersect the outlines of the top and the bottom as nearly at right angles as possible. Therefore, since the remainder of the top (Lto.J) and E H of the base are the bases of a surface which must be so divided as to best serve the purposes of triangulation, it is advisable to divide L .1 into inure spaces than E H, allowing the extra spares in the top nearest the point L to form the bases of a number <>I converging triangles, as shown. Thus lirst divide E II into any suitable number of spaces, as shown by the small I ig u res 5 to 0, then divide. L ,J into a greater number of equal spaces than E II, as shown bv the small figures ''> to l. Con- nect points of like number in the two outlines by solid lines, commencing at II and .1. as shown from i) 9 to 5 ~), drawing lines also from 4 and 3 of the top to 5 (E) of the bot- tom. Also draw the dotted lines 5 6, 6 7, etc., and the solid lines from points in L N to G. These solid and dotted lines will then form the bases of a series of right angled triangles whose hypothenuscs will give the real distances across the envelope of the finished article. These triangles are constructed, as shown in Fig. 621 at the right of the elevation, in the following man- ner. Extend A B and 1) C of the elevation, through which draw any vertical line, as ( j K. From Q on Q I* set off the lengths of all the solid lines of the plan. Thus make Q 9 equal to 9 9 or J II of the plan, Q 8 equal to S 8 of the plan, etc., and from the points thus established draw lines to E. In like manner draw the vertical line T V, and from T on T S set off the lengths of the dotted lines of the plan, as shown l>y the small figures, and from the points thus obtained draw lines to V, as shown. The small figures in S T correspond with the figures in L .1, the top line of the plan. In laying out the pattern shown in Fig. (>2 the joint is assumed upon the line J H of the plan. The pattern may be best begun by first laying out one of the large triangles forming a side of the article, as E L G or G K F of the plan, shown also by 1) N C of the elevation, Fig. 620. Draw any hori/ontal line, as E G of Fig. 622, equal in length to E G of the plan. Problems. 355 From E us center, with radius R ?, of Fig. 621, de- scribe :i small arc near I., which intersect with another arc drawn from G as ccntci 1 , with a radius equal to It 3' of Fig. 021. thus establishing the point L of the pattern. From (i of the pattern as center, with radii E G Fig. fcy. Pattern of Article Shmrn in Fig. 620. equal to It '2 and R lof Fig. 621, describe small ares, as shown between L and K of the pattern. Take between the points of the dividers a space equal that used in dividing the arc L K of the plan, and placing one foot of the dividers at L of the pattern step from arc to are, reaching K, as shown, and through the points thus obtained draw L K of the pattern; also draw K G. From E of the pattern as center , with radii equal to R 4 and It 5 of Fig. 021, describe small arcs to the left of L. With the dividers set to the space used in dividing the arc L J of the plan, place one foot at L (3) and step first to arc 4, then to arc 5, thus es- tablishing the points 4 and 5. With the last obtained point, 5, of the pattern as center, and a radius equal to the dotted line V 5 of Fig. 621, describe a small arc (6'), which intersect with another arc struck from point E of pattern as center, with a radius equal to 5 6 of the base line E H of the plan, thus establishing the point 6' of the pattern. With 6' of the pat- tern as center, and a radius equal to R 5 of Fig. 621, describe a small arc (6), which intersect with another arc struck from 5 of pattern as center, with a radius equal to 5 6 of the top line L J of the plan, thus estab- lishing the point 6 of the pattern. Proceed in this manner in the con- struction of the remaining triangles of the pattern, using alternately the lengths of the dotted and the solid hypothenuses in Fig. 621 corre- sponding to the dotted and the solid lines crossing the plan, in the order in which they occur, to determine the width of the pattern ; the spaces in E II of the plan to form the lower line E II of the pattern and the spaces in L J of the plan to form the upper line of the pattern, all as shown. The remaining parts of the pattern can % be obtained by any convenient means of duplication, K F G being a duplicate of LEG and K J 1 H' F being a duplicate of L J H E. PROBLEM 191. The Patterns for a Three-Piece Elbow, the Middle Piece of which Tapers. In Fig. 623, let A B D F II G E C be the side view of a three-piece elbow, the middle piece (C D F E) of which is made tapering. The piece C D F E may also be described as an offset between two round pipes of different diameters. A half profile of the upper and smaller of the two pipes is shown by a rfi b, while g n h shows that of the larger pipe. The straight por- tions A B D C and E F H G are in all respects similar to inanv pieces whose patterns have already been de- scribed in the first section of this chapter in Problems 38 to 45 inclusive. It will therefore be unnecessary to repeat the description in this connection. 356 Tfie New Metal Worker Pattern Boole. Since an oblique section through a cylinder is an ellipse, an inspection of the drawing will show that the sections C D and E F, the upper and lower bases of the middle piece, must be elliptical. The first opera- tion, therefore, will be to develop the ellipses, wliidi may be done in the following manner : Divide the fig 623. Elevation of a Three-Piece Elbow, the Middle Piece of which Tapers. profile ami into any convenient number of equal spaces, as shown by the small figures. From the points thus obtained drop lines vertically to a b and continue them till they cut the line C D. From the intersec- tions on C D carry lines at right angles to the same indefinitely, as shown. Through these lines draw any line, as c il, parallel to C D. Upon each of the lines drawn from C D, and measuring from c d, set off the lengths of lines of corresponding number in the profile a m l>, measuring from a I. A line traced through the points thus obtained, as shown by c k d, will be the required elliptical section. The section upon the line E F may be obtained in the same manner, all as shown byepf. In Fig. 62-4, C D F E is a duplicate of the middle piece of Fig. 623, below which is drawn its half plan made up of the elliptical sections just obtained, all as shown by corresponding letters. The piece thus be- 3 , ^v v ___fc__^ // ^ PLAN / P Fig. 624. The Middle Piece of Elbow in Fig. 623, with Plans of its Bases Arranged for Triangulation. comes an irregular flaring article or transition piece, the envelope or pattern of which may be obtained in exactly the same manner as described in Problems 184- or 186, to which the reader is referred. The eleva- tion in Fig. 623 is so drawn that C D and E F are parallel, and C E is at right angles to both. Should the elevation, however, be so drawn that C D is not parallel with E F the conditions will then become the same as in Problem 193 succeeding, which see; and should C E be drawn otherwise than vertically, the plan would then resemble that shown in Prob- lem 194. PROBLEM 192. The Patterns for a Raking: Bracket in a Curved Pediment. In Fig. 625, let C E F D be the front elevation of a portion of a curved pediment whose center is at K, and of which E K is the center line. C A B D of the same elevation represents the face view of a bracket having vertical sides, of which E G F is the normal profile. Since the bracket sides are vertical and are necessarily at different distances from the center line, it will be easily seen that they are of different lengths or hights ; that is, the side C D, being further from E K than the side A B, is longer. The patterns for Pattern 357 the two sides will therefore be different and the face piece will be really an irregular flaring piece. It will lirst be nn-cssary to obtain the pattern or profiles of the two sides. To facilitate this operation the normal profile of the bracket E G F has been so line E F, as shown. Thence carry lines around the arch from the center K, from which the same is struck, cutting the sides A 1> and C D. Conveniently near the lower side of the bracket draw any vertical line, as E 1 IV, as a base line upon which to construct the true '-ill PROFILE /12 13 RIGHT SIDE ELEVATION Fig. GiTi. -1 Raking Bracket in a Curved Pediment, Sluttving the Patterns for It* Face and Sides. placed that its vertical line or back coincides with the center line E K of the arch. Divide the face of this profile into any convenient number of parts, as shown by the small figures, and from these points carry lines at right angles to the back of the bracket, cutting the sides of the bracket. At any convenient position upon this line, above or below, as at G' E 1 F', draw a duplicate of the normal profile, so that its back or vertical line shall coincide with E 1 B 1 , and divide its face line into the same spaces as G F. Place the 358 T,,e ew T-squarc at right angles to the line E' B', and, bringing it successively against tin: points in the side D, draw lines cutting E 1 B 1 , continuing the same indefi- nitely to the left, as shown. At any convenient po- sition, as A 1 B', on the line E 1 B 1 transfer the spaces from A B, as shown, and from the points thus ob- tained draw lines indefinitely to the left also at right angles to E 1 B'. Place the T-square parallel to E' B', and, bringing it successively to the points in the normal profile G' F l , cut lines of corresponding number in the two sets of parallel lines just drawn. Lines traced through the points of intersection will give the re- quired patterns of the lower and upper sides, as shown respectively by C' M D 1 and A 1 N B 1 . Some of the lines of projection in the pattern of the upper side have been omitted to avoid confusion. At the ex- treme left of the engraving is shown a side view of the bracket as seen from the right, which is made up of the two sides just obtained and which have been placed in proper relation to each other, all as shown by the dotted lines projected to the left from the points A, B, C and D of the front elevation. Having now obtained all that is necessary, it re- mains to triangulate the face of the bracket preparatory to developing the pattern of the same. With this in view first connect all points of like number in the upper and lower sides of the front view by solid lines, as shown. Also connect them in-the'side elevation. Since points of like number in A B and C D have the same projection from the back of the bracket, it will be seen that the solid lines just drawn connecting them represent true distances across the face of the bracket. The four-sided figures produced by draw- ing these lines must now -be subdivided into triangles by means of dotted lines drawn diagonally through each. Therefore connect each point upon the profile of the lower side of the bracket with the point next higher in number upon the upper side, as shown in the side view. To determine the true length of these lines it will be necessary to construct a diagram of triangles, as shown by S V T in the upper part of the engraving. Draw S V and S T at right angles to each other. Make S V equal to the width of the bracket measured horizontally across the face, and upon S T, measuring from S, set off the lengths of the several dotted lines in the side view, as shown by the small figures. From each of points in S T draw lines to V. Then these lines will be the real distances between points of corresponding number on the lower side of the bracket and points of the next higher number upon the upper side. The figures in S T correspond with the figures upon the. lower side of the bracket, the point V representing, in the case of each line, the next higher number; thus -2 V is the distance from "2 to 3' across the face of the bracket, 3 V the distance 3 4', 4 V the distance 45', etc. The dotted lines in the side view representing the distances 1 2 and T 8 cannot, of course, be shown in that view, because tliey lie in surfaces which appear in profile; but since these surfaces are parallel with the pjane of the back of the bracket these distances for use in the pattern may be taken directly from the front view, as shown by the dotted lines 1 2' and 7 S' in that view. To lay out the pattern of the face piece first draw any line, as C 3 A 3 or 1 1' of the pattern, equal in length to 1 1' of the front view. From C 3 of the pattern as center, with a radius equal to 1 2' of the front view, de- scribe a small arc, which intersect with another arc drawn from A 3 as center, with a radius equal to 1' 2' of the front view, thus establishing the point 2' of the upper side of the pattern of the face. From 2' of the pattern as center, with a radius equal to 2 2' of the front view, strike a small arc, which intersect with another arc struck from 1 of the pattern as center, with a radius equal to 1 2 of the front view, thus establish- ing the position of the point 2 in the lower side of the pattern. From 2 of the pattern as center, with a radius equal to 2 V of the diagram of triangles, strike a small arc, which intersect with another arc struck from 2' of the pattern of center, with a radius equal to 2' 3' of the side view, thus establishing the point 3' of the pattern. From 3' of the pattern as center, with a radius equal to 3 3' of the front elevation, strike a small arc, which intersect with another arc .struck from 2 of the pattern as center, with a radius equal to 2 3 of the side view. So continue, using the distances acre >ss the face indicated by the dotted lines as found in the diagram of triangles in connection with the spaces in the profile A" W of the side view to form the upper side A 3 B 3 of the pattern, and the distances across the face as measured upon the solid lines of the front view in connection with the spaces upon the profile C 2 1>~ to form the lower side C :i D 3 of the pattern, -until the points 13 and 13' are reached. As the lines C A and D B of the front of the bracket must be cut to fit the curves of the moldings above and below, against which the bracket fits, the corresponding lines of the pattern can be drawn with radii respectively equal to K E and K K, as shown bv the curved lines C' A 3 and D 3 B 3 of the pattern. Prwums. PROBLEM 193. Pattern for a Transition Piece to Join Two Round Pipes of Unequal Diameter at an Angle. 359 In Fig. 626, DC Iv L shows u portion of the larger pipe, of which M P N O is the section ; II G B A ;i portion of the smaller pipe, of which E J F 1 is the section; and A B C I) tho elevation of the transition piece necessary to form a connection between the two pipes at the angle II A L. The drawing also shows that the ends of the two pipes to be joined are square, Fig. 626. Elevation of a Transition Piece Joining Two Round Pipes of Unequal Diameter at an Angle. or cut off at right angles, so that the lower base of A P> C D is a perfect circle whose diameter is D C (or M X ) and the upper base is a circle whose diameter. is A B (or K F). and also that the side A D is vertical. In the choice of a method of dividing the surface of the piece A BCD into triangles, either the elevation or the plan can be made use of for that purpose, ac- cording to convenience. In the demonstration here given the elevation has been used by way of variety, all as shown in Fig. 627. Proceed then to divide the plan of the upper base into any convenient number of equal spaces, as shown, and drop a line from each point at right angles to A B, cutting A B, and numbering each point to correspond with the number upon the plan. In like manner divide the plan of the lower base into the same number of equal spaces, and erect a perpendicular line from each, cutting the line D C, and numbering the points of intersection in the same order, or to correspond with the points in the upper base, all as shown. Connect the points in D C with points of similar number in A B by solid lines, also HALF PLAN OF LOWER BASE Fig. 627. Triangulation of Transition Piece Shown in Fig. 626. connect points in D C with points of the next higher number in A B by dotted lines, which will result in a triangulation suitable for the purpose. The next step will be to construct sections through the piece upon all of the lines upon the elevation (both solid and dotted), which operations are shown in Figs. 628 and 629, and which may be done in the following manner : Upon any horizontal line, as T S of Fig. 628, erect a perpendicular, as T U. Upon T S set off from T the several distances of the points in the lower base from the center line 1 7 of the plan, as measured upon the vertical lines (Fig. 627), all as indicated by the small figures. Upon T U set off the lengths of the 360 New Metal Worker Pattern Bool: solid lines of the elevation, numbering eacli point thus obtained to correspond with its line in the elevation. From each of the points upon T U draw horizontal 3 4 5 Figr. 628. Diagram of Sections Fig. 629. Diagram of Sections Taken on Solid Lines of Fig. Taken on Dotted Lines of Fig. 627. 627. lines to the right, making each in length equal to the distance of points of corresponding number in the plan of the upper base from the center line 1 7, as measured upon the lines at right angles to line A B, thus ob- taining the points 2', 3', 4', etc. Now connect these points with points of corresponding number in the base line T S by means of solid lines, as shown. In constructing the sections upon the dotted lines of the elevation, shown in Fig. 629, the same course- is to be pursued as that employed in Fig. 628. The base line W V is a duplicate of T S. Upon the per- pendicular line erected at W set off the lengths of the Fig. G31. Perspective View of Model. several dotted lines of the elevation, numbering each point thus obtained to correspond with the number at the top of its line in the elevation. From each point draw a horizontal line to the right as before, which make equal in length to the similar lines in Fig. 628, numbering each point as shown by the small figures 2', 3', 4', etc. Now connect each of these points with the point of next lower number in the base line V W by a dotted line. Having obtained all the necessary measurements, the pattern for one-half the envelope of A B C D may be developed in the following manner : Draw any line, as A D in Fig. 630, which make equal in length to A D of the elevation. With A as a center, and a radius equal to 1 2 of the plan of the upper base, Fig. 627, strike a small arc, which intersect with another struck from D as a center, with a radius equal to the dotted line 1 2' of the diagram, Fig. 629, thus establishing the position of the point 2 in the upper line of the pat- tern. From D as a center, with a radius equal to 1 2 of the plan of the lower base, Fig. 627, strike a small arc, which intersect with another struck from point 2, just obtained, with a radius equal to the solid line 2 2' 3 4 5 Fig. 630. Pattern of Transition Piece Shou-n in Fiy. G27. of the diagram, Fig. 628, thus fixing the position of point 2 in the lower line of the pattern. So continue, using the lengths of the dotted lines in the diagram, Fig. 629, in connection with the lengths of the spaces in the plan of the upper base, to develop upper line of the pattern, and the lengths of the solid lines in the diagram, Fig. 628, in connection with the lengths of the spaces in the plan of the lower base, to develop the lower line of the pattern, using each combination alter- nately until the pattern is complete. As each new point of the pattern is determined it should be num- bered, and the solid or dotted line used in obtaining the same may be drawn across the pattern if desired, merely as a means of noting progress, but these lines are not necessary, as each point is simply used as a center from which to find the next point beyond. Sometimes, in order to more thoroughly under- stand the method employed in such an operation as the foregoing, it is desirable to construct a small model, Pattern Problems. 361 which can be made from cardboard or thin metal, the details of which are clearly shown in Fiir. <>:>1. The pieees forming tho upper and lower bases of it should be duplicates of the half plans of the upper and lower bases shown in Fig. 627, having the lines there shown drawn upon them, and the piece forming the back is a duplicate of the plane figure A B C D. These three parts may be cut in one piece, after which a right angle bend on the lines A B and C I) will bring the two bases into correct relative position. Five quadrilateral figures corresponding to those shown in Fig. G28 may now be cut and fastened in position, according to their num- bers, between the two bases of the model. Threads or wires can be so plaeed as to correspond in position with the dotted lines shown in the elevation, Fig. 627, to complete the model. The model is only useful be- fore the pattern is developed to assist in showing the shapes and order or rotation of the various triangles; and one constructed to the dimensions of any problem which may occur to the student at the outset of his study of triangulation will serve to assist his imagina- tion in all subsequent operations. PROBLEM 194. The Pattern for a Flaring Collar the Top and Bottom of which are Round and Placed Obliquely to Each Other. B 3 Fig. 6SS. -Plan and Elevation of Flaring Collar. In Fig. 632 E F G H shows the side elevation of a flaring collar, the profile of the small end or top being shown at A B C D, and that of the bottom at K L M N of the plan. The- conditions embodied in this problem are in no respect different from those of the problem immediately preceding. A slight differ- ence in detail consists in the fact that in the former case the short side was at right angles to the larger end, while in the present case it is at right angles to the smaller end, but the pattern may be obtained by exactly the same method as that employed in the previous problem. However, as the elevation was there made use of to determine the triangulation, the plan will here be used upon which to determine the position of the triangles of which the pattern will sub- sequently be constructed. Divide A B C of the profile into any convenient number of parts, and, with the T-square at right angles to E F of the elevation, carry lines from the points in ABC, cutting E F, as shown. Extend the base line G II to the left indefinitely, and through the center of plan of base draw M, parallel with II G of elevation. Drop lines from the points in E F, extending them vertically through M, With the dividers take tin- distance across the profile A B C D on each of the sev- eral lines drawn through it, and set the same distance off on corresponding lines drawn through M. That is, taking A C as the base of measurement in the one case, and O M in the other, set off on the latter, on each side, the same length as the several lines measure on each side of A C. Through the points thus ob- tained trace a line, as shown by O P Q R, thus obtain- TV Xew Metal ll'o/7.r/' Pattern Rook. ing the shape of the upper outline as it would appear in the plan. As both halves of the plan when divided by the line M are exactly alike, it will only be necessary to use one-half in obtaining the pattern ; therefore divide K N M of plan into the same number of parts as was the half of profile, in the present instance six, as shown by the small figures. Number the points in K N M to correspond with the points in R Q, and connect corresponding points by solid lines, as shown by 1 1', 2 2', 3 3', etc. Also connect the points in O R Q with those of the next higher number in K N M, as shown by the dotted lines 1 2', 2 3', 3 4', etc. The solid and dotted lines thus drawn across the plan will represent the bases of a number of right angled triangles whose altitudes are equal to the vertical lines between E F and J Y of the elevation, and whose hy- V W X and on W X set off the lengths of dotted lines iu plan, and from W on W V the lengths of lines in E F Y J of elevation, excepting the line E J, or No. 7, which is not used. Connect the points in W V of diagram with those of tire next higher number in "W X, as shown by the dotted lines in diagram and by simi- lar lines in the plan. The resulting hypothenuses will give the correct distances from points in top of article to- points of next higher number in the plan of base. .y 2 3 , 4 5 6 \\ \ \ \ \ 43 21U W \ 2X Fig. 633. Diagram of Triangles Based upon the Solid Lines of the Plan in Fig. 63S. Fig. 634. Diagram of Triangles Based upon the Dotted Lines of the Plan in Fig. 6S2. pothenuses, when obtained, will give the real distances across the sides of the finished article in the direction indicated by the lines across the plan. To construct these triangles proceed as follows : Draw any right angle, as S T U in Fig. 633. On T U, measuring from T, set off the lengths of solid lines in plan, making T U of diagram equal to Q M of plan, T 2 of diagram equal to 2 2' of plan, T 3 of diagram equal to 3 3' of plan, etc. From T on T S set off the length of lines in E F Y J of elevation, making T S of diagram equal to F Y of elevation, T 2 of diagram equal to a 2 of elevation, T 3 of diagram equal to b 3 of elevation, etc. Connect points in T S with those of similar number in T U, as shown by the solid lines. The hypothenuses of the triangles thus obtained will give the distance from points in plan of base to points of similar number of top as if measured on the fin- ished article. The diagram of triangles in Fig. 634 is constructed in a similar manner. Draw the right angle i m Fig. .. Pattern of Flaring Collar. Having now obtained all the necessary measure- ments, the pattern may be developed as follows : Draw any line, as q m in Fig. 635, in length equal to S U of Fig. 633, or F G of elevation. With in of pattern as center, and M 2' of plan as radius, describe a small arc (2"), which intersect with one struck from point (j of pattern as center, and V X of diagram of triangles in Fig. 634 as radius, thus establishing the point 2" of pattern. With point 2" of pattern as center, and 2 2 of Fig. 633 as radius, describe a small arc (2), which Pattern Problems. 363 intersect with another struck from point q of pattern as center, and C ~2 of profile as radius, thus establish- ing the point 2 of pattern. With 2 of pattern as cen- ter, and '2 3 of Fig. 034 as radius, describe a small arc (3"), which intersect with another struck from 2" of pat- tern as center, and 2' 3' of plan as radius, thus estab- lishing point 3" of pattern. With 3" of pattern as center, and 3 3 of Fig. 033 as radius, describe a small arc (3), which intersect with one struck from point '2 of pattern as center, and the distance 2 3 of prolile as radius, thus establishing point 3 of the pat- tern. Proceed in this manner, using the hypothenuses of the triangles in Figs. 633 and 634 for the distances across the pattern ; the distances between the points in the plan of base for the stretchout of the bottom of pattern; and the distances between the points in the profile of top for the stretchout of the top of pattern. Lines drawn through the points of intersection, as shown by q o and m k, will, with q m and ok, constitute the pattern for half the article. The other half of the pattern q o' k' m can be obtained by any convenient means of duplication. PROBLEM 195. The Pattern for a Flaring Flange, Round at the Bottom, the Top to Fit a Round Pipe Passing through an Inclined Roof. In Fig. 630, let K L represent the pitch of the roof, A B C D the elevation of the (hiring flange, A J D the half plan of the base, and B F] C the half plan of round top through which the pipe passes. Fig. U3ti. Elevation of a Flaring Flange to Fit Against an Inclined Rouf. It will be seen by comparison that this problem embodies exactly the same principles as do the two immediately preceding, with the slight difference in detail that its short side is not at right angles to either upper or lower base. Also, in this case the bottom of the article appears inclined instead of the top. It will be seen at a glance that if the shape be considered as anything else than a flange against an inclined roof the drawing might be so turned upon the paper as to bring the line K L into a horizontal position, when it would present the same conditions as those of Problems 193 and 194 with the slight difference in detail above al- luded to. The method of triangulation employed in this case is exactly the same as in the problem immediately pre- \ s 6 5 3 2 Fig. 6S7. Diagram of Triangles Fig. 638. Diagram of Triangles Based upon the Solid Lines of Based upon the Dotted Lines of the Plan in Fig. 6S6. the Plan in Fig. 6S6. ceding, and the operation is so clearly indicated by the lines and figures upon the four drawings here given as scarcely to need explanation, if the previous problem has been read. The plans of both top and bottom are divided into the same number of equal parts, and a view of the top as it would appear when viewed at right angles to the base line K L, and as shown by F G II, is projected into the plan of base, as indicated by the lines drawn from B C at right angles to A D. 364 Tlie New Metal Worker Pattern Book. Points of like number in the two curves F H G and A J D are joined by solid lines, and the four-sided figures thus obtained are redivided diagonally by dotted lines. These solid and dotted lines become the bases of the several right angled triangles shown in Figs. 637 and 638, whose altitudes are equal to the hights given be- tween the lines B C and F G, and whose hypothenuses give correct distances across the pattern between points indicated by their numbers. The pattern is developed in the usual manner by assuming any straight line, as C D in Fig. 639, equal to C D of Fig. 636, as one end of the pattern, and then adding one triangle after an- other in their numerical order; using the stretchout of BEG, Fig. 636, to form the upper line of the pattern, the stretchout of A J L to form the lower side of the uattern and the various dotted and solid hypothenuses Fig. 639. Pattern of Flaring Flaitije Shown in Fiy. 636. in Figs. 637 and 638 alternately to measure the dis- tances across the pattern. PROBLEM 196. Pattern for an Irregular Flaring Article, Elliptical at the Base and Round at the Top, the Top and Bottom not Being Parallel. The conditions given in this problem are essen- tially the same as those of Problem 193, but the fol- lowing solution differs from that of the former problem in the method of finding the distances from points assumed in the base to those of the top, and is intro- duced as showing varieties of method: In Fig. 64-0, C G H D represents the side view of the article, of which G K H is a half profile of the top and C F D a half profile of the base. For convenience in ob- taining the pattern the half profiles are so drawn that their center lines coincide with the upper and lower lines of the elevation. Divide both of the half profiles into the same number of equal parts in the present instance eight. From the points obtained in the half profiles drop per- pendiculars cutting G H and C D. Connect the points secured in G H with those in C D, as a ?i, b m, etc. Also connect the points in G H with those in C D, as indicated by the dotted lines 1 n, a m, etc. Refer- ence to Problem 193 will show that in order to obtain the correct lengths represented by the several solid and dotted lines drawn across the elevation complete sections upon those lines were constructed, as shown in Figs. 628 and 629. In the present case these dis- tances will be derived from a series of triangles whose bases are the differences between the lengths of the lines drawn across the half profile of the top and those of the bottom. Therefore, to obtain the triangles giving the true dis- tances represented by the solid lines proceed as follows : G, Fig. 640. Elevation and Profiles of an Irregular r taring Article, Showing Method of Triangitlrttion. First set off from C D upon each line in the base C F D the length of the corresponding line in the top ; thus Pattern Problems. 365 make n o equal to a 2, mp equal to I 3, Z ? equal to c4, etc. For the bases of the triangles represented l>v the dotted lines set off from C D the length of correspond- W.. r' q' i}' u' R u' t' s' 1" 1 ll a" II II tl b" II II II c" II SI \ II \ 1 1 \ / c" \ \ \ \ III III /" 0' :\\ \ \ \ III . 1 \ ^ II 1 ' t \\ \ \ II 1 / u WX 3"2"X"Z" V c "7" 5 " 4 ' -u Representing Solid Lines of the Elevation. ing lines in G K H, as shown by the small figures in COD. Thus make m 2' equal to a 2, I 3' equal to I :i, / 4' equal to c 4, etc. The triangles represented by solid lines in the elevation, and shown in Fig. 641, arc obtained as follows: Draw the line P Q, and from 641. Diagram of Triangles Fig. 6J$. Diagram of Triangles Representing Dotted Lines of the Elevation. from C to are set off to the left of R, and the lengths from to D on R Q. Thus make R a' of diagram equal to n a of Fig. 640, R o of diagram equal to o 17 of half profile of base, and connect a' o'. Make R b' of diagram equal to m iof Fig. 640, R p of diagram equal to p 16 of the base, and connect p' b', etc. The triangles represented by dotted lines in C G H D are obtained in a similar manner. Draw the line T U in Fig. 642 and erect the perpendicular V W. From V, on V W, set off the lengths of dotted lines in C G II D of the elevation. Thus make V 1" equal to n 1 of Fig. 640, V a" equal to m. a, V b" equal to 1 6, etc. Upon V T or V U set off the lengths of the lines in C D F of Fig. 640, as indicated by the small figures. Thus make V X" equal to n 17 of the base in Fig. 640, and draw X" 1". Make V 2" equal to 2' 16 of the base, and draw 2" a", etc. To obtain the pattern first draw the line C G of Fig. 643, in length equal to C G of Fig. 640. From C, with radius equal to C 17 of the half profile of base, strike a small arc (o), which intersect with an- other arc struck from G as center, and 1" X'' of Fig. 642 as radius, thus establishing point o in the curve of the pattern. From G of pattern as center, and G 2 of the half profile of top as radius, strike a small arc, which intersect with another arc struck from \ r I \ 1 \ \ \ \ \ \ . \ \ \ \ \ \ \ \ \ Fig. 643. Pattern of Shape Shown, in Fig. a convenient point erect the perpendicular R S, From R on R S set off the lengths of solid lines in C G II D, and from R on P Q set off the lengths of correspond- in<> lines in C O D F, as indicated by the small letters o * ill COD. To avoid a confusion of lines the lengths o of pattern as center, and a o' of Fig. 641 as radius, thus establishing point a in the upper curve of the pattern. From o of pattern as center, and 17 16 in C F us radius, strike a small arc (p), which inter- sect with another arc struck from a of pattern as cen- 366 Tin .\\-n- Metal H'w/'/- I'utlir ter, and a" 2" of Fig. 64:i us radius, thus locating point p of the pattern. From a of the pattern as cen- ter, and 23 in G K as radius, strike a small arc, which intersect with anoth'er arc struck from /> of pat- tern as center, and p b' of Fig. 64-1 as radius, thus locating point /> of pattern. Proceed in this manner. using in the order named the spaces in the lower pro- lile, the hypotheuuses of triangles in Fig. 642, the spaces in the upper prolile. and the hypothenuses of triangles in Fig. 7 plan of the head piece, into the same number of equal parts, numbering each the same, as shown, and con- nect points of similar number by solid lines. Also connect each point in .J I, the line of the bottom, with tin- point of next higher number in G II, the line of the top, bv a dotted line, till as shown. The curves E F and L K, forming the plan of the foot piece, are also to be divided into spaces and the points connected by solid and dotted lilies in the same manner as those of the head. The solid and dotted lines thus drawn between the points in the two curves of the plan will form the bases of a series of right-angled triangles, whose hypoth- enuscs (after the altitudes are obtained) will give the real distance between the points whose number they bear upon the finished article. As, owing to the slant of the top line A B of the elevation, the triangles will Fig. 645. Half Pattern of Head Plice. Fig. 640. Half Pattern of Foot Piece. he of differing hights, the simplest way of constructing them will be as follows : Upon D C of the elevation extended, as a base, erect a perpendicular line, M N From N on the base line set off the various lengths of the solid lines in the plan of the head piece, as shown toward (.). From each of the points in the curve G H erect perpendicular lines, cutting A B of the eleva- tion; and from these points of intersection carry lines horizontally to the right, cutting the line M N, num- bering each point to correspond with, the points in G H, all as shown. Lines connecting points of similar num- ber at M and Q will be the hypothenuses required, or the real distances between points of similar number in the top and bottom of the finished article. In a sim- ilar manner erect another perpendicular, P, and set off from P on the base line the lengths of the several dotted lines in the plan of the head piece, as shown toward R. The hights of the points in the curve of the top can be determined upon the line P by con- tinuing the lines drawn from A B toward M till they intersect O P. Each point in the base P R is now to he connected with the point of the next higher number in () P by a dotted line. The various hypothenuses drawn between () and R will then be the correct dis- tances between the points connected by the dotted lines of the plan. (The numbers in P R correspond with those in I J of the plan and not with H G). The dia- grams from which the dimensions for the foot piece are obtained are shown at S T U and V W X at the left of the elevation and are obtained in a manner exactly similar to those just obtained for the head piece, all of which is clearly shown by the lines of the drawing. From an inspection of the drawing it will be seen that the line E F G H does not represent the true lengths or measurements taken on the top line of the tub, because A B, not being horizontal, is longer than the line E II, its equivalent in the plan, and therefore a true section on the line A B must be obtained, as shown above the elevation. This may be accomplished in the following manner : At any convenient distance above A B draw E' H 1 parallel to A B, and from all of the points previously obtained on A B carry lines at right angles to A B, cutting E 1 H', and extend them beyond indefinitely, numbering each line to correspond with the point in E F G H from which it is derived. On the line 2 set off from E' H' a distance equal to the distance of point 2 of the plan from the line E H as measured on the vertical line; on line 3 set off as before a distance equal to the distance of point 3 of the plan from E H, and so continue until the distances from E H of all the points in E F G H have been transferred in like manner to the new section. Then a line traced through these points, as shown by E 1 F 1 G 1 II 1 , will be a section or plan on the line A B, from which measurements can be taken in developing the upper edge of the pattern. Having obtained, by means of the various dia- grams constructed in connection with the elevation, the correct distances between all the points originally as- sumed in the plan, the pattern may now be developed by simply reproducing all of these distances or meas- urements in the order in which they occur upon the plan. For the pattern of the head piece assume any line, as I II of Fig. 345, which make equal in length to C B of the elevation, or what is the same thing, equal to 1 1 of the diagram M N Q. From H as a center, with a radius equal to 1 2 of the section of top, strike a small arc, which intersect with another arc struck 368 The Sew Mdal Worker Put/cm Book. from I as a center, with a radius equal to 1 2 of the diagram of dotted lines P R, thus establishing the point 2' of the pattern. From 2' as center, with a radius equal to 2 2 of the diagram of solid lines M N (}. strike a small arc, which intersect with another struck from point I as a center, with a radius equal to 1 2 of the line I J of the plan, thus establishing the point 2 of the pattern. Continue this operation, using in nu- merical order the distances taken from the top section, in connection with the distances obtained from the diagram of dotted lines O P R, to form the top line of the pattern, and the distances taken from the diagram of solid lines M N O, in connection with the distances measured upon the bottom line I J of the plan, to form the bottom line of the pattern, all as indicated by the solid and dotted lines drawn across a portion of the pattern. Then I H G J will be one-half the pattern of the head piece. The pattern for the foot piece is de- veloped in exactly the same manner by making E L equal to A D of the elevation, and using the diagram of dotted lines V W X to measure upon the pattern the distances indicated by the dotted lines upon the plan, and the diagram of solid lines S T U to measure upon the pattern the distances indicated by the solid lines across the plan, the distances forming the top line of the pattern being taken from E' F' of the top seetiou while the distanced forming the bottom line of the pattern are taken from the line L K of the plan. The pattern for the flat portion of the side F G J K can be obtained as follows: Parallel to K J of the plan draw any line, as K' J'. At right angles to K .1 of the plan project lines from points K. J, F and G, eutting K 1 J', as shown, establishing the points K 1 and I', and continuing the lines from points F and G in- definitely. From K 1 of the pattern as a eenter, with a radius equal to 8 8 of the diagram S T U, or of Fig. t>44, strike a small are, cutting the line projected from point F of the plan, as .shown at F" of the pattern. From J' of the pattern as a center, with a radius equal to 7 7 of the diagram M N Q, or of Fig. 644, strike a small arc, cutting the line projected from the point G of the plan, as shown at G 2 . Draw the lines K 1 F 2 , P G' J and G s J 1 ; then K 1 F 2 G'' J 1 will be the pattern of the flat portion of the side. The patterns of the several parts can be joined together, by any convenient method of duplication, in such a manner as to produce as much of the entire pattern in one piece as it is desired. PROBLEM 198. The Pattern for a Flaring: Flange to Fit a Round Pipe Passing through an Inclined Roof; the Flange to Have an Equal Projection from the Pipe on All Sides. In Fig. 647, let a b c d be the elevation of the pipe, E E' its plan, A B C D the elevation of the flange and C D the angle or pitch of the roof. Since the projection of the base of the flange is required to be equal on all sides, as shown by C 1 and D 1, the flange will appear in the plan as a perfect circle, F F'. To avoid confusion of lines another elevation of the flange G H K J is shown in Fig. 648, below which is drawn a half plan of its base, MEN, and above which is a half plan of its top, G L H, all of which will be made use of in dividing the surface of the flange into measurable triangles for the purpose of developing a correct pattern of the same. Divide the semicircle G L H into any convenient number of equal parts in the present instance 12 and from the points thus obtained drop perpendicular lines to G H. To obtain the shape of seetiou on roof line J K divide the half plan of base MEN into the same number of equal parts as was G L II, and from the points thus obtained carry lines at right angles to M N, cutting J K. From the points in J K draw lines at right angles to it, as shown by a 1, b 2, c 3, etc. On these lines, measuring from J K, set off the length of corresponding lines in M N B, thus making lines a 1, & 2, c 3, etc., in J K C equal to lines 1, &2, c 3, etc., in M N B. A line traced through these points, as shown by J C K, will give the shape of sec- tion on roof and furnish the stretchout of base for obtaining the pattern. In Fig. 649 is drawn a duplicate of the plan in Fig. 647, the spaces in its outer line 1> P being exact duplicates of the spaces in M B N of Fig. 648, Pattern 1'rublems. 369 and the spaces in its inner line O' D' P' being dupli- cates of those in G L II, all as shown liy the small liirures. Draw solid lines connecting similar points, as 1' 1, !' '1. .">' :!. etc. In like manner connect the ELEVATION PLAN Fig. 647. Plan and Elevation of Pipe and Flange. points in O' D' P' with those, of the next higher num- ber in D P, as with I', L with 2', 2 with 3', etc., with dotted lines. These solid and dotted lines will then form the bases of a series of right angled trian- gles, whose altitudes can be derived from the eleva- tion, and whose hypothenuses, when obtained, will be the correct measurements across the pattern between points of numbers corresponding with the lines across the plan. To construct the diagrams of triangles represented by solid and dotted lines in plan, extend G II of Fig. iJ.s indefinitely, as shown by H W. From the points in J K carry lines to the right indefinitely, parallel with G W, as shown by the lines between G W and K Y. At any convenient place, as R, and at right angles to GW, erect the lineR S, cutting the base line K Y. From R set off the distance R T, equal to the length of any of the solid lines in plan, Fig. 649, as P' P, which is the horizontal distance between the pipe and lower edge of the flange. Draw T U parallel with R S, and also draw lines from the points in R S to T. For convenience the points in R S can be num- bered to correspond with the points in J C K. Then the triangle T U S will correspond to a section through .2 10 Fig. 649. Plan of Flange, Showinij Trianyulation. the article on the line P' P in plan, the hypothenuse S T representing the distance between the pipe and lower edge of the flange. The diagram of triangles in VWYX is constructed in a similar manner ; draw W Y at right angles to G W, and set off the space W V equal to the length of one of the dotted lines in plan, Fig. 649, as 1', and draw lines from the points in W Y to V. In developing the pattern the stretchout of top of 370 The New Metal Worker Pattern, Book. llange where it joins the pipe can be obtained from the semicircle G L II. The stretchout of lower edge of flange where it joins the roof can be obtained from the section on the roof line J C K. The distance be- tween points in D P and 0' D' P' of plan, Fig. 649, length to T S of lirst diagram of triangles. With the dividers set to the distance K 1 in K C .1 of section strike a small arc (!') from the point <' of pattern. With the dividers set to the distance V 1 of second dia- gram of triangles strike a small arc from the point of W J SECOND DIAGRAM M ' 2 \ Of SE | 10 \ ) 2 \ e Sv ) 3 \ e ^-\^ ^xX 4 6 B My. 648. Elevation of Flanye, with Plan, Sections and Diagrams of Triangles. as indicated by the solid lines, is given in the diagram of triangles T R S. The distance between points as indicated by dotted lines in plan is given in the diagram V "W Y. For the pattern then proceed as follows : Draw any line, as IF K', Fig. 050, equal in pattern as center, cutting the first arc at 1' of pattern. From point 1' of pattern as center, and T 1 of first diagram of triangles as radius, describe a small arc (1), which intersect with one struck from of pattern as center, and 1 iu II L O as radius. Thus the points I'ntlrni Problems. 371 'Ml' and I 1 ' of pattern arc established. Proceed in this manner, using in the order described flic stretch- out obtained from the elliptical section K C J, the hy- pothenuse of triangle in second diagram corresponding to the dotted line in plan, the stretchout from the sec- tion (r L II, and the hypothenuse of triangle in the other hall of the pattern can be obtained by any means of duplication most convenient. Should it be required to construct such a flange to fit over the ridge of a roof, it is clear that that part of the flange shown in the plan, Fig. 649, by O O' D' D would be a duplicate of the part shown by D' P' P D, .' 12 J '*' II. fi.io. Half Pattern of Flange. Shown in Fig. (>47. first diagram corresponding to the solid line drawn across the plan, until all the measurements are used. Lines traced through the points thus obtained, as shown by H' G' and K' J', will be the half pattern. The and that, therefore, that portion of the pattern, Fig. 650, shown by 6 II' K' 6' would be one-quarter of the entire pattern, which could be duplicated so as to make either one-half or the whole pattern in one piece. PROBLEM 199. Pattern for the Hood of a Portable Forge. In Fig. 651, C A B D represents the front eleva- tion of a hood such as is frequently used upon a por- table forge, K L M N its plan and E F II J a its side view. The opening A B at the top of the hood is round, as shown by L P of the plan, while the base C D where it joins the forge is nearly semi-elliptical, as shown. by K L M of the plan. In the side elevation E shows the amount of flare and projection of the front of the hood, while the opening, shown in the front by C S D, appears as a simple straight line, a J. With these conditions given, the arch of the opening C S D of the front elevation can be determined in connection with the plan, by projection, as shown by the hori- zontal dotted lines, while if the arch of the front eleva- tion be assumed arbitrarily then its line (a J) in the side view must be obtained by projection, and will be either 372 Tim New \\\>rl-rr Potfrrn Book. straight or curved according to the nature of the curve employed in the front elevation. Assuming the straight line a J of the side view as the true profile of the arch, its curve in either the front elevation or the plan must he determined, as a means Fig. 651. Front and Side Elevations and Plan of Hwd, Showing System of Triangulation. of obtaining the pattern. As the flaring portion of the hood very much resembles a conical frustum having an oblique base, probably the simplest method of ar- riving at its true shape is to first determine the plan of this irregular frustum of which it is a part. Therefore produce the oblique line E a of the side elevation until it intersects the base line H J extended in the point G. Next set off from L on the center line of the plan a distance equal to H G of the side elevation, thus locating the point R. Through R, from a center to be determined upon the center line, draw the curve form- ing the front of the plan, with such length 52, as follows: Draw the vertical line A B, in length corresponding to the hight of the hood, as indicated by F II in the side elevation. At right angles to A B draw B C, in length corresponding to P R or 1 1 of the solid lines of the plan. From B set off also the spaces B 2, B 3, B 4, etc., corresponding in length to the lines 2 2, 3 3, 4 4, etc., of the plan. Con- nect the points in B C with the point A by solid lines. Then will these lines represent the true distances between points 1 and 1, 2 and 2, etc., of the plan. The second diagram of triangles is constructed in a similar manner. The vertical line E D is drawn, equal to F II of the side elevation. E F is- set off at right angles to it, in length equal to the dotted line 1 2 of the plan. From E are set off the distances E 3, E 4, etc., corresponding to the lines 2 3, 3 4, etc., of the plan. The points thus established in F E are then connected with D by means of dotted lines. Then will these lines represent the Fij, G-'j-i. Pattern of Hood. true distances between points 1 and 2, 2 and 3, etc.,- of the plan. To develop the pattern, first draw any vertical line, as L Z of Fig. 653, representing the center of the 374 The N&.O Metal Worker Pattern Boole. bark, which make equal to the liiglit of the hood I- 1 II. As the base of the hood is perfect ]y straight from L to the point 11, set off on a horizontal line from the point Z, in Fig. 653, a distance equal to L 11 of the plan, and draw 11 L of the pattern. With L as center, and 11 10 of the small circle in plan as radius, describe a short arc. Then, from 11 of the base in pattern as center, and 11 D of the second diagram of triangles as radius, describe a short arc intersecting the one lirst drawn, thus establishing the point 10 of the upper line of the pattern. Then from this point as center, with A 10 of the first set of triangles as radius, describe a short arc, and from 11 of the base of the triangular portion of the pattern, with 11 10 of the outer curve of the plan as radius, describe another arc intersecting it, thus establishing the point 10 in the lower line of the pattern. Proceed in this manner, using alternately the spaces in the inner line of the plan, the hypothe- nuses of the dotted triangles, the hypothenuses of the triangles indicated by solid lines, and the spaces in the outer line of the plan, obtaining the several points, as shown. Then lines traced through these points will be the pattern of the envelope of the shape indicated by F E Gr H of the side elevation, or in other words, of the frustum of which the hood forms a part. It now remains to cut away such a portum of this pattern a:: represents the part (r a J of the side elevation. To ac- complish this it is simply necessary to obtain the posi- tions of the points a, b, c, etc., of the plan and side elevation upon the lines 11,22, 3 3, etc., of the pat- tern. With the blade of the T-square set parallel to the base line G H of the side elevation bring it against the points of intersection made by the line a J with the radial lines, and cut the vertical line F H, as shown by the short dashes drawn through it. Transfer the points thus obtained in F II to the vertical line A B of the first set of triangles. Then with the blade of the T- square at right angles to A B, and brought successively against the points in it, cut the hypothenuses of the several triangles corresponding in number to the lines from which the points were derived in the side eleva- tion, all as indicated by the letters a, /;, e, 4. Elevations and 1'lnn of llond fur mi Oil Tank, Shnwinit System of Triangulation. side view will not be the correct bases of the triangles, therefore set off on A B, measuring each time from B, the correct lengths of the several solid lines of the plan, as indicate' 1 by the points near 1, 2, 3, etc., on the line C ols y Fiij. '>><. Diagram of Trianyles Based upon the Dotted Lines of the Plan. A B, from which points draw lines (shown dotted) to points of similar number in B C. Then the dotted lines 1 1, ^2, etc., of the side view will be the correct hvpothenuses of the triangles whose bases are indicated of the diagram equal to 2 1 of the plan, and in the same manner make B 2 and B 3 of the diagram equal to 3 2 and 4 3 of the plan respectively. From the points thus established in the base line of the diagram Fig. 656. Pattern for the Top nf Hood. draw lines to points of next higher number in the ver- tical line. These hypothenuses will then represent lengths of lines measured on the face of the hood corresponding to the diagonal dotted lines in the plan. To develop the pattern, first draw any line, as of Fig. 656, equal in length to A C of side, Fir. 4. From 0, at the right of the pattern, as center, with the distance between the points to 1 in the pro- file F K of the back as radius, describe a short arc. 376 The New Metal Worker Pattern Rook. Next take in the dividers the distance 1 of Fig. 655, and from the opposite end of the center lint- describe a short arc, intersecting the one already drawn at the point 1, thus establishing that point. From i as center, with dotted line 1 1 of the side view as radius, describe another short arc, which in turn intersect by an arc struck from of the left hand side of the pat- tern with 1 of the plan as radius. This will estab- lish the point 1 of the opposite side of the pattern. Continue in this way, intersecting the hypothenuse of the triangles whose bases are the dotted lines of the plan with the measurements taken from the back view, and the hypothenuse of the triangles which are shown l.iy the solid lines of the plan with the measurements taken from the circumference of the plan. In this manner all points in the profile of the pattern neces- sary to its delineation will be established. A free- hand line drawn through these points will give one half the required pattern, all as shown in Fig. fi/iti. The other half may be obtained by any convenient method of duplication. The shape of patterns forming the back and the vertical sides of the hood are clearly shown in the engraving and need no further explanation. PROBLEM 201. Pattern for an Irregular Flaring: Shape Forming: a Transition from a Round Horizontal Base to a Round Top Placed Vertically. In Fig. 657, let I D E F H represent the front elevation of the article, showing the circular opening D E F G forming its upper perim- eter or profile. The triangle A B shows the shape of the article as it appears when viewed from the side, below which is drawn the plan, showing its circular base, J K L M. The line N P shows the plan of the opening D E F G, which opening is shown in the side view by that por- tion of the line A B from A to Q. Opposite the front side of the plan N P is drawn a duplicate of the pro- file D E F G, as shown by E' F' G', so placed that its vertical center line E' G' shall coincide with the center line of the plan, as shown. As the article consists of two symmetrical halves it will only be necessary to develop one-half the complete pat- tern. Therefore divide one-half of both profiles E F G and E' F' G' into the same number of equal parts, numbering each in the same order, as shown by the small figures ; also divide the plan of the base into the same number of equal parts as the profile, numbering the points to correspond with the same. Drop lines from the points on the profile E' F' G' on to the line N P, at right an- gles to the same, as shown, and connect these points with those of similar number upon the plan of base by- solid lines, as shown. Also connect points upon the 4 M Fig. 6W. Elevations and Plan of an Irregular Shape Forming a Transit inn from a Round Horizontal Base to a Bound Top Placed Verticfi/h/. base with those of the next higher number upon the line N P by dotted lines. It will be noticed that the point J of the plan Pattern Problems. 377 represents at once the point 1 of tin.- l>asr and the points 1 and 7 of tin: profile, shown by B, Q ami A of the side elevation. The linos drawn across the plan represent the horizontal distances between the 3 I 2 B 3 I /'/;/. U38. Diagram of Triangles. points which they connect and will form the bases of a series of right angled triangles, whose altitudes can lie derived from the elevations, as will be shown, and whose hypothenuses, when drawn, will give the true distances between points of corresponding number across the finished article or its pattern. To obtain the altitudes of the triangles carry lines from the points in the half profile E F G horizontally across, cutting the line A Q, as shown; then the distances of the points in A Q from B will constitute the respective altitudes of the triangles. Therefore, to construct a diagram of all the triangles, draw any horizontal line, as D C, Fig. 658, near the center of which erect a perpendicular, B A. Upon B A set off from B the various distances from B to points in A Q of Fig. 657, numbering the same as shown by the small figures. From B set off on B C the lengths of the various solid lines drawn across the plan, Fig. 657, and connect points in B C with those of like number in B A. From B set off toward D the lengths of the various dotted lines drawn across the plan and connect them by dotted lines with points of the next higher number in the line B A, all as shown ; then these various hypothenuses will constitute the true distances across the finished article between points of corresponding number indicated on the plan and elevations. The distances between points in the base line forming the larger or outer curve of the pat- tern can be measured from the base line in plan, while spaces forming the upper or shorter side of pattern can be measured from either of the profiles. To develop the pattern it is simply necessary to construct the various triangles whose dimensions have been obtained in the previous operations, beginning at either end most convenient and using the dimensions in the order in which they occur until all have been used and the pattern is complete. Therefore, upon any straight line, as A C of Fig. 659, set off a distance equal to A C of Fig. 657 or the solid line 7 7 of Fig. 658. From C as a center, with a radius equal to 7 6 of the plan, Fig. 657, describe a small arc to the left, which intersect with another small arc struck from A as a center, and with a radius equal to the dotted line 7 6 of the diagram of triangles, Fig. 658, thus establishing the position of the point 6 of the pattern. From A of Fig. 659 as center, with a radius equal to 7 6 of the profile, Fig. 657, describe a small arc, which intersect with another struck from point 6 of pattern as center, with a radius equal to the solid line 6 6 of Fig. 2, thus locating the position of the point 6' of the pattern. fig. 6Z9. Pattei'ii of &it, and connect the points thus obtained \vith Y and Z, as shown, thus completing the side view. Commence by dividing one-quarter of the plan of the base K H into any convenient number of spaces. as shown by points 1, 2, 3, etc. For greater accuracy these spaces may be made shorter as they approach the ends of the base, where the line has more curve than near the middle. Having established the points 0, 1, 2, 3, etc., in K H, draw a line from each of them to the center of the plan M. By this means the quarter of the circle representing the top of the article, measurements, as shown at the right of C E, the hypothenuses of which will represent the real distances between the required points. Therefore from the points established in II K drop lines vertically cutting the section line A .1) 1>, as indicated, then carry lines from the points on A D horizontally till they cut the Hue C K and continue them indefinitely to the right. The points at which these lines cross the center line E C will represent the hights of the several triangles. On these horizontal lines, measuring fnua the center line C E, which is assumed as the common perpendicular for all the tri- angles, set off the bases of the several triangles, trans- ferring the distances from the plan. From the points 1 OK Y E Z Fuj. 662. Top and Side View of Boss, Showing System of Trianyulation. and shown in the diagram by P R, will be divided in the same manner or proportionately to the plan of the base, all as shown by points I 1 , 2 1 , 3', etc. It will be seen that these lines divide the surface of the boss into a number of four-sided figures, each of which must now be redivided diagonally so as to form trian- gles. Therefore connect with I 1 , 1 with 2', etc., by means of dotted lines, as shown. These solid and dotted lines drawn across the top view represent the horizontal distances between the points given, while the vertical distances between the same can be measured on lines parallel to C E ; hence it will be necessary to construct a series of triangles from these thus established draw lines to E. which will give the hypothenuses of the several triangles. For example, on the line drawn from the point 5", in A D B, meas- uring from C, set off a distance equal to 5' 5 and also a distance equal to 6' 5 In the top view. The difference between these two is so small as to be im- perceptible in a drawing to so small a scale as this. In like manner, on the line drawn from 4" set oil' a distance equal to the length of the diagonal lines 4 5' and 4 4' in the top view, and in the same manner on the line drawn through 3 s set off the distance equal to 3 4 1 in the top view and also 3 ?>'. Then, as liefore remarked, lines drawn from the points thus established Pattern PrMvnits. 381 in the horizontal lines toward K will be the hypoth- cnuses of tin; several triangles corresponding to sections represented by the diagonal lines in the to]> view. In view of the fact that the base of the boss is mi-veil as shown bv A 1) it will be notieeil that the measurements from K to II in the top view do not represent the real distances, because the distance 11 M is less than the distance A D. In ease extreme ac- curacy is required it will therefore be necessary to develop an extended section on the base line A D, \vliich mav !" done as follows: Kxtend the line M TT required width of the pattern on one end, as shown by Y 6" in Fig. 662. With these two points established proceed to obtain other points in both lines of the pattern by striking arcs with radii equal to the spaces established in the plan of both base and top of the article and to the hypothenuses of the triangles already described. Thus, from S as center, with radius equal to the distance 6' 5 4 of the stretchout of the base, de- scribe a short arc, as shown at 5 in the pattern. Then from T as center, with radius equal to E 6' of the tri- angles, intersect it by a second arc, as shown. From T as center, with radius equal to 6' 5' of the plan of T V U Fig. 663. Pattern for Bos*. of the top view, as shown at the left, upon which place a correct stretchout of A D ; that is, make D 1 1* equal to I) 1% I 3 -2* equal to 1* 2", etc., and through each of the pofnts thus obtained draw measuring lines at right angles to D' M. Place the T-square parallel to H M, and, bringing it successively against the points in the line K II, drop lines into the measuring lines of cor- responding number, as shown by 0*, 1', 2 4 , etc. Then will the distances 0* I 4 , 1* 2 4 , etc., be the correct dis- tances to be used in developing the pattern instead of the distances 01, 1 2, etc. The pattern may now be developed as shown in Fig. 663. 1/iv oil' the line S T, in length equal to the the top of the article, describe a small arc, as shown, and from 5 of pattern as center, and radius equal to E 5' of the triangles, intersect it by another arc, thus deter- mining the second point in the top. Proceed in this manner, adding one triangle after another in the order in which they occur in the top view, using the spaces of the plan of top and of the stretchout of the bottom and the hypotheneuses of the triangles as above described. Lines traced through the points thus obtained, as shown from S to N and from T to M, will give the pat- tern of one-quarter. This can be duplicated as often as is necessary to make the entire pattern in one piece, or to produce it in halves, as shown. PROBLEM 204. Patterns for a Ship Ventilator Having a Round Base and an Elliptical Mouth. In Fig. 66-1 are presented the front and side ele- vations of a ship ventilator of a style in common use. A 1 B shows the section or plan of its lower piece A E F B, as well as of the pipe to which it is joined, while K S P is the shape of its mouth, or a section upon the line C D. The curves E C and F D connecting the two ends of the ventilator and forming the general outlines of the same may be drawn at the discretion of the designer. As the ventilator is constructed after the manner of an elbow, it mav be divided into as 382 T/iu New Metut Worker Putter n Bwk. many sections or pieces as desired. Therefore divide the curved lines E C and F D into the same number of spaces, and connect opposite points by straight lines, as shown by G II, K L and M N. These lines should be so drawn as to produce a general equality in the appearance of the different pieces without refer- ence to equality in the spaces in either outline. The next step is to establish a profile or section upon each one of these lines. These profiles can be drawn arbitrarily, but each should be so proportioned that the series will form a gradual transition from the circle A 1 B' to the ellipse R O S P. All the profiles will, therefore, be elliptical, those nearer the mouth being more elon- gated than those nearer the base or neck. Since the lower piece is cylindrical and is cut obliquely by E F, the section at E F must neces- sarily be a true ellipse and can be developed by a method frequently explained in connection with various problems in the first section of this chapter, and as also explained in Geometrical Problem 68 on page 61. Of the remaining sections, their major axes are, of course, equal to the lengths of the lines G H, K L and M N, and their minor axes may be determined by any method most convenient, or in the following man- ner : Draw R U and S V, repre- senting a front view of the curved lines passing through the points n, m, k, g and e of the side view. From the points g, k and m project lines horizontally across to the front view, cutting the lines R U and S V and the center line T. Then / I, d o and b c will be respectively one-half the minor axes of the sections above referred to. With the major and minor axes of the several sections given, they may be drawn by any method producing a true ellipse, or in case the mouth has been drawn by means of arcs of circles the other sections may be drawn in the same manner. Each of the several pieces of which the ventilator is composed (except the lower piece) becomes, as will be seen, a transition piece between two elliptical curves not lying in the same plane, and in that respect is the same as the form shown in Problem 191. The pat- tern for each piece must, therefore, be obtained at a separate operation, that for the piece M N D C only being given. To avoid confusion of lines a duplicate of it is transferred to the opposite side of the front elevation, as shown by W Y Z X. Drop points from Y and Z perpendicular to the center line T of the elevation, thus locating the points M 3 and N 2 . Make the distance tf c equal to I c. Then draw the ellipse M" I? N 2 , which will be a front view of the sec- tion M N of the side elevation. On a line parallel with Y Z construct the section M' V N 1 , as follows: Let M 1 N' be equal to and opposite Y Z. Let the distance c l>' be equal to the distance c b of the sec- w K i , ._LYd-_ fffi- Fig. 664. Elevations and Sections of a Ship Ventilator. tion. With these points determined, draw through them the semi-ellipse M 1 V N'. Divide the sec- tions M 1 b' N 1 and S P into the same number of equal parts, as indicated by the small figures in the engraving. Drop the points 1, 2, 3, 4, etc., on to and perpendicular to the line Y Z ; thence carry them per- pendicular to the center line O P of the front eleva- tion, cutting the section M 2 b'' N" in the points l' J , 2 2 , 3*, etc., thus dividing it into the same number of spaces as were given to the original section M 1 b 1 N 1 . Next connect the points of like numbers in the two sections of the front elevation by solid lines, thus : Connect 2' with 2", 3' with 3", 4' with 4 2 , etc. ; also connect the points 2' with 1', 3' with 2', 4' with 3' J , Pattern Problems. etc., by dotted lines, all as shown in the engraving. These lines represent the bases of right angled tri- angles, whose altitudes may be measured on the hori- zontal lines cutting the lines W X and V Z. The next step, therefore, is to construct diagrams of these triangles, as shown at A and B of Fig. 665. l>ra\v any two horizontal lines as bases of the triangles, and erect the perpendiculars E C and F I). On both K C and F 1) set: off the various 1 lights of the tri- angles, measured as above stated and as indicated by the points 1, 2, 3, 4, etc. Next set off the length of the bases of the triangles as follows: In diagram A, let C 1 equal the distance 1' T of Fig. 664; make C 2 equal to 2' 2" and G 3 equal to 3' 3", etc. Connect the points in the vertical line with the points in the horizontal line of the same number, thus obtaining ' ;;* D Fitj, 665. Diagrams of Triangles. hypothenuses of the triangles, or the true distance between the points I 1 1*, 2' 2", etc., of the elevation. In diagram B, let the distances D 2, D 3, D 4, etc., represent the distances 1" 2', 2" 3', etc., of the eleva- tion. Having located these points, connect 1 in the vertical line with 2 in the base; also 2 in the vertical line with 3 in the base, and proceed in this manner for the other points. This will give the hypothenuses of the triangles, whose bases are 1' 2', 2" 3 1 , etc., in the elevation. Having thus obtained the dimensions of the vari- ous triangles composing the envelope of the first section of the ventilator, proceed to develop the pattern for it, as shown in Fig. 666. On any straight line, asC M, set off a distance equal to 1 1 in diagram A. From C as center, with radius equal to 1' 2' of the elevation, Fig. 664, draw an arc, which cut by another arc drawn from M as center, with radius equal to 1 2 of diagram B, thus establishing the point 2. From 2 as center, with radius equal to 2 2, diagram A, draw an arc, which cut with another arc drawn from 1' as center, with radius equal to 1 2 of the elevation, thus estab- lishing the point 2'. Proceed in this manner, next locating the point 3, then the point 3' ; next the point 4, and then 4', etc. It will be noticed that, after C M - Fig. 666. Pattern of First Section of Ship Ventilator. passing points 6 and 6', 7' is obtained before 7. This is for the sake of accuracy, as it will be seen by in- spection of the elevation that the distance 7' 6' is less, and therefore more accurately measured in the eleva- tion, than the distance from 6 3 to 7'. Having thus located the points 1, 2, 3, etc., I 1 , 2', 3', etc., trace the lines C D and N M, and connect D with N, as in- dicated in Fig. 666. Then D N M C will be the pattern for one-half the section M N D C of the eleva- tion. The pattern of the section E A B F will be the same as that for the corresponding piece in an ordinary elbow, and, therefore, need not be specially explained here. 384: Tku Suv Mdal Wurkvr Pattern Uvok. PROBLEM 205. Patterns lor the Junction of a Large Pipe with the Elbows of Two Smaller Pipes of the Same Diameter. The elbows of the smaller pipes in the problem here presented are such as would, if each were com- pleted independently of the other, form six-piece elbows. The junction between the two elbows occurs between the fifth pieces, which pieces unite to form the transition from the smaller diameters of the elbows to the diameter of the larger pipe, or sixth piece. A pictorial representation of the finished work is shown in Fig. 667, in which, however, the upper section, or larger pipe, is omitted to more fully show the shape and junction of the transition pieces. A front view or elevation of the various parts is shown in Fig. 608. The side view given in Fig. 669 shows more fully the amount of lateral flare of the transition piece necessary to form a union between the varying diameters of the larger and smaller pipes. The drawing of that portion of the elbows in the smaller pipes from the horizontal parts up to the line a h t in Fig. 668 is exactly the same as that employed in drawing a six-piece elbow. The piece A G It , occupying the place of what would otherwise be the fifth piece of the elbow, becomes in this case an irregular shape, the lower end or opening, A, of which is nearly circular while its upper end, A G, is a perfect semicircle. This piece unites with its mate G D t h on the line G //, thus form- ing the complete circle at A D, a plan of which is shown immediately below the ele- vation. The relative proportion between the diameters of the larger and smaller pipes is such that the junction between the elbows is carried somewhat below the fifth pieces, mitering the fourth pieces for a short distance, as shown from k to L. The method of cutting the lower parts of the elbow, how- ever, is the same as that employed in all elbow patterns where the pipe is of a uniform diameter throughout, numerous examples of which are given in Section 1 of this chapter, to which the reader is referred. As the section or profile of all the parts forming the elbow is a perfect circle when taken at right angles to the sides of the pipe, as at Q F or M N, it will be seen that a section on the line a It will be somewhat elliptical; it will therefore be necessary to obtain a correct drawing of this section from which to obtain the stretchout of the lower end of the piece A G h a. with which it joins, and also a drawing of it as it will appear in plan. Therefore between two parallel lines drawn from M and N at right angles to M X construct a profile or section, as shown below at the left, which divide 'into any convenient number of equal spaces, as shown by the small letters a, I, c, etc. From each of these points carry lines back to M X at right angles to the same, and continue them in either direction till they cut the miter line a H of the elevation, as shown by the small letters, and the center line n a of the section. From the points in a n of the elevation draw lines at right angles to the same indefinitely, as shown above the elevation, across which at any convenient point .draw a line, as B' C', at right angles to them. From B' C 1 set off on the lines last drawn distances equal to the distances from the circumference to the diameter on corresponding lines in the section below, all as shown by a", i% c' J , etc. A line traced through fig. 667. Perspective View of the Junction of a Large Pipe ivith the Elbows of Two Smaller Pipes. these points will be the correct section on the miter line a M. It will be noticed that the section has not- been carried further than the point It', the balance of the curve not being required by reason of its intersec- tion with the corresponding piece in the other elbow. Below the elevation and in line with the same, as shown by the center line G T, is drawn the plan of the larger pipe A B C ]). It will be necessary to add to this the plan of the curve on the line // of the eleva- tion, in order that the horizontal distances between the points assumed in the two curves may be accurately measured. Therefore from the points on the miter line a h drop lines vertically through the plan, cutting the transverse center line X Y. From X Y set off distances on these several lines equal to the distances Pattern Probkms. 385 of corresponding points from the line a n of the original section, as shown by ', //, etc., from X to S. A line traced through these points will give' the correct posi- tion of the intersection of the smaller pipe- as seen from above. This entire, line is shown in the plan, although the part from S to X will not be recpiired, for the reason given above. An inspection <>f the plan will show that the side of the plan from V T to the right would be an exact duplicate of the left side if it were completed, and that therefore the plan consists of four symmetrical quarters, one of which, X R T, is completely shown in the plan. Hence the pattern for this quarter will snllice by duplication for the entire transition piece. Divide the quarter of the plan of the larger pipe P T, adjacent to the curve X S, into the same number of equal spaces as are found in the inner curve from X to S, as shown by the small figures 1, 2, 3, etc. Connect corre- sponding points in the two lines us shown by the solid lines It,' 8, ,j' 7, ./" li, etc. Next sub- divide the four-sided figures thus obtained by their shortest diagonal, as shown by the dot- ted lines ,j S,/' 7, etc. These solid and dotted lines across the plan represent the bases of a series of right angled triangles whose altitudes can easily be obtained from the elevation, and whose hypothenuses when obtained will give correct dis- tances across the finished piece between points connected on the plan. These lines have also been drawn across the ele- vation from corresponding points in the same for illustrative purposes, but such an operation is not neces- sary to obtain the pattern. Neither is the side view shown in Fig. 669 necessary to the work, but is here introduced merely to assist the student in forming a more perfect conception of the operations described. From the points a, b, c, etc. on the miter line a h of the elevation carry lines horizontally across, cutting the vertical line G L, as shown by the points from .s- to //. The distances of these points from G will then represent the vertical distances of corresponding points in X S of the plan from the plane of upper base of the transition piece shown by A D of the elevation and V P T of the plan. To obtain the hypothenuses of the various tri- angles above alluded to, or in other words, the true Fig. 668. Front Elevation, Plan and Sections, Shoiviny Method of Triangutotion. lengths of the lines dividing the surface, as shown in the two elevations and plan, it will be necessary to construct a series of diagrams, as shown in Fig. 670. Therefore draw any two lines, as A S and h h', at right 386 The Xt-iv Metal Worker Pattern Bwk. angles to each other; make h 8 equal to h' 8 of the ])lan, Fig. 668, and h h' equal to h 8 of the elevation, and draw h' 8. Next draw any two lines, as rj 8 and . E S is drawn at right angles to E T and upon it are set off the bights of the points in A F D the same as in M N of Fig. 674. Upon E T set off from B the lengths of the several dotted lines drawn across the 11 . . ; : 10,12 O I 8,13 . " - M a b c d e Fig. 674. Diagram of Sections Upon Solid Lines of Fig. 673. elevation, as shown by corresponding letters, and from the points thus obtained erect perpendiculars also as in Fig. 674. Finally connect by dotted lines such points as correspond with those connected by dotted lines in the elevation. Thus from 9 in E S draw a line to point 1 in the base line, corresponding to the line / 1 of the elevation, Fig. 673. Lines from 10 to 2 and from 11 to 3 of the diagram will correspond respectively to g a and h b of Fig. 673. To develop the pattern from the dimensions now obtained proceed as follows : At any convenient place from TT of pattern as center, describe an arc. whidi cut with another arc struck from point 9 of pattern as center, and 9 ~2 of Fig. IJ74 as radius, thus establish- ing point > of pattern. With point 2 of pattern as center, and 2 10 of Fig. 675 as radius, describe an arc, which intersect with another arc struck from point 9 of pattern as center, and 9 10 of profile as radius, 11 ^-^^ ' 10,12 C: 3 .-- 5 ^-'*" 1 * : > 9,13 X ^ ."*" N * """ _N fff . ' R lab c d e Fig. r,75. Diagram of- Sections Upon Dotted Lines of Fig. 673. thus establishing point 10 of pattern. With point 10 of pattern as center, and 10 3 of Fig. 674 as radius, describe a small arc, which intersect with one struck from point 2 of pattern as center, and 2 3 of G L II as radius, thus establishing point 3 of pattern. Continue this process, locating in turn the remaining points in pattern, as shown. Lines drawn through the points thus obtained, as indicated by G II and D A, will be one-half of the required pattern. The other half of the pattern can be obtained in a similar manner, or by tracing and transferring. The pattern for the Fig. 676. Pattern for Lower Piece of Elboiv. draw the straight line H D in Fig. 676, in length equal to H D of Fig. 673. From II of pattern as center, with radius 1 9 of Fig. 675, describe an arc, which intersect by a second arc struck from D as center, with radius D 9 of profile, thus establishing the point 9 of pattern. Then with radius II 2 of G L H, Fig. 673, other part of elbow, as shown in Fig. 672 by G C B H. can be obtained by the same method. The shape G L H of Fig. 673 is to be drawn to the left of the miter line G H, and the operation continued, usiiiL: the same process, as shown in Figs. 674, 675 and 676. Pattern Problems. 389 PROBLEM 207. The Pattern for a Y Consisting of Two Tapering Pipes Joining a Larger Pipe at an Angle. In Fig. 677, BCD K represents ilie elevation of a portion of the larger pipe and C' K D' L its prolile. This pipe is cut. oil: square at its lower end, witliwliieh the liranches of the Y are to be joined. A 15 O II J and G II E F are the elevations of the two similar tranches joining each other from II to 0, and the larger l-'iii. >!77. Elevation find Profiles nf Y with Tapering Branches. pipe on the line B E. A' N J' M is the profile of one of the tapering branches at its smaller end. Since the article consists of two symmetrical halves when divided from end to end on the lines A' J' or C' D' of the profiles, and since the two branches are alike, the pattern for one-half of one of the branches, as A B () II J, is all that is necessary. The dividing surface A B II J, lying as it were at the back of the half of the branch shown in eleva- tion by the same letters, will then form a plane or base from which the hights or projection of all points in the surface of the branch piece can be measured. As the branch piece A B H J is an irregular tapering form, its surface must be divided into a series of measurable triangles before its pattern can be ob- tained. Therefore divide the half profile C' L D' into any convenient number of equal parts in the present instance six, as shown by the small letters fghj k and from these points drop lines parallel with C B, cutting the line B K, as shown. In a similar manner divide the half profile A' N" J' into the same number of equal Fig. 678. Elevation of One Branch of Y, Shouting Method of Triangvlation. parts as was C' L D', as shown by the small letters a i bed e. From the points' thus obtained carry lines par- allel with J' J, cutting A J. Connect the points in A J with those in B E, as shown. To avoid a confusion of lines the subsequent oper- ations are shown in Fig. 678, in which A B II .1 is a duplicate of the piece bearing the same letters in Fig. 677. The profiles B L K and A X .1 are also dupli- cates of those shown in Fig. 677 and arc for conven- ience here placed adjacent to the lines which they rep- resent. B L of the upper profile then represents a section on the line B 0. and A N J that upon the line A J, but the section on the line H, the miter be- tween the two branches, is as yet unknown. To obtain this it will be necessary to first obtain sections upon 390 New Metal Worker Pattern Book. the various lines drawn across the elevation from B V, to A J in Fig. 678, or in other words, diagrams upon which the true lengths of those lines can be measured. In the diagram of sections shown in Fig. 679 fig. 680. Diagrams of Sections Upon Solid Lines of the Elevation. S T represents the dividing surface or base plane al- luded to above and is made equal in length to 2 2' of Fig. 678. At either extremity of this line erect the perpendiculars S b and T/, as shown. Make T/ equal in hight to 2' /of profile B L E, and upon S b set off from S the hight S a, equal to 2 a of the profile A N J, and draw the line a /. On S T, measuring from T, set off the distance 2' 2", and erect the perpendicular 2"/", cutting af at f". Then will a f represent the true distance between the points 2 and 2' in Fig. 678, and af" will represent the true distance from 2 to 2",- '/ be equal to the true distance between 3 and 3' of Fig. 678, b g" will be the true distance from 3 to 3" of Fig. 678 and 3" g" will be the hight of the point 3". To construct the section on the line II, at points 3" and 2 "draw 3" y' and 2"/' at right angles to II, making them respectively equal ?," y" and 2"/" of Fig. 679. As the profile B L E is a semicircle the hight of point 4' that is, 4' h is equal to E ; therefore through the points E, /',/' and II draw the curve sliown, which will be the true section on line II, from which the stretchout can be taken for that portion of the pattern. The sections on the remaining lines (4 4', 5 5' and 6 6') of the elevation are shown in Fig. 680 and are constructed in exactly the same manner as those shown in Fig. 679, giving c Ji, dj and e k as the true lengths O 1 iii. (183. Pattern of Tapering Branch. of those lines. Before the pattern can be developed the four-sided figures into which the surface of the branch pipe has been divided by the solid lines must be subdivided into triangular spaces, as sliown by the dotted lines in the elevation. Sections upon these lines must also be constructed, in order that their true lengths can be obtained. These are shown in two groups in Figs. 681 and 682 and are constructed in a manner exactly similar to that described in connection with Fig. 679. They may be easily identified by cor- respondence between the figures on the base lines V V and W X and those of the elevation. To describe the pattern proceed as follows : Draw any line, as J H in Fig. 683, in length equal to J H of Fig. 678. With J of pattern as center, and J' a of smaller profile as radius, describe a small arc (a), which 391 Cut with one struck from IF of pattern as center, and 1" a of Ki^. lixl as ,-ailius. thus establishing the point a of pattern. With u of pattern as center, and n /"'of Fig. 679 as radius, describe another small are (/'), which intersect with one struck from II of pattern as center, and II /' of profile II K as radius, thus estab- lishing the point /'of pattern. In a similar manner, a l> of pattern is struck with n l> of profile as radius; /' b of pattern with/" l> of Fig. >sl as radius; // "!> as ratlins, and y" \ Gr F in the plan and shown in eleva- tion by J K L M N. In Fig. 686 are shown the sec- tions having for their bases the solid lines of the elevation, which are con- structed in the following manner : I 'pon any horizontal- line, as P Q, set off from P the lengths of the several solid lines of the elevation, as indicated by the small figures corresponding with those in J N M. At P, which corre- sponds with all the points in K L of the elevation, erect a perpendicular, P H, upon which set off the hights of the points in K R L, as 2' 2, 3' 3, etc., shown by P 2, P 3, etc. At each of the points near Q erect a perpendicu- lar, which make equal in hight to the length of line drawn from the point of corresponding number in G C II of the plan to the line G II. Thus make. 9' 9, 10' 10, etc., equal to 9" a, 10" b, etc., of the plan. From the points 9, 10, etc., draw solid lines to the points in H P, connect- ing points correspondingly connected by the solid lines of the elevation. The sections having for their bases the dotted lines of the elevation are shown in Fig. 687, and are constructed in exactly the same manner. Upon Y Z, set off from Y the lengths of the dotted lines of the elevation, numbering the points near / to correspond with those in J N M of the elevation. The perpendiculars erected from these points are tin- same as those similarly loraled in Fig. (ISti. and the perpendicular X Y is a duplicate of II P ( Fig. li.st;. From the points 9, 10, 12, etc.. draw dotted lines to - points in X Y, connecting points correspondim'-lv con- * nected by dotted lines of the elevation. To describe the pattern shown in Fig. 088 pro- 1 PROFILE ELEVATION />(/. >;ft!),Plan and Elevation of Three-Pronged Fork. ceed as follows: Draw any line, as .1 K, in length equal to J K of elevation. Fig. 685. With K of pat- tern as center, and K 2 of profile as radius, describe a small arc (2), which cut with one struck from J of pattern as center, and 8' 2 of Fig. 687 as radius, thus establishing point 2 of pattern. With point 2 of pat- tern as center, and 9 2 of Fig. 686 as radius, describe another small arc (9), which intersect with one struck Pattern Problems. 393 from J of pattern as center, ant] .1 !' of elevation ;i radius, thus establishing the point 9 of pattern. Pro- Fig. 686. Diagram nf Sections Upon Solid Lines in J K L M N of Fig. 6Sr>. ceecl in this manner until the remaining points are located, all as clearly indicated by the solid and dotted lines in Fig. 688. By drawing lines through the ""LIT = ~"~r 4 n =-= 3 5 _ _==. - " H 3 6 -=. ^--=^--=~=~ """ I Ficj. 687. Diagram of Sections Upon Dotted Lines in J K L M N of Firj. 686. points thus obtained the half pattern shown by K Q L M N J is the result. The other half, as shown by Fiy. 688. Pattern of Taperinr/ Branch. K Q' L' M' N' J, can be obtained in a similar manner, or by duplication. PROBLEM 209. The Pattern for an Offset to Join an Oblon? Pipe With a Round One. In Fig. 689, B C F G represents the side elevation of the offset, A B G H a portion of a round pipe join- ing it below, and C D E F a portion of the oblong pipe joining it above. In the plan immediately below, T K L M shows the plan of the round pipe and N P Q II S that of the oblong pipe, while the distance. L T shows the amount of the offset. The piece forming the offset is similar in shape to that shown in Problem 189, the difference being that its bases B G and C F are neither horizontal nor parallel to each other and that sections on the lines of the bases are not given. Since the article required consists of symmetrical halves when divided on the line J T of the plan, the plane surface ABCDEFGH lying as it were back of the half shown by the eleva- tion may, as in Problem 207, be regarded as a base from which to measure all hights, or projections, in obtaining the required profiles and sections necessary in developing the pattern. The first steps necessary will be to obtain true sections upon the lines C F and B G of the elevation. In Fig. 690, C D E F rep- resents a duplicate of the part bearing the same letters in the elevation. Upon D E as a base line construct a duplicate of the half section of oblong pipe N P T of Fig. 689, as shown by D N P E. Divide the semicircle N P into any convenient number of equal parts, as shown by the small figures. With the blade of the T-square placed at right angles The Xi-ir Mi-Utl \VorL-iT I'ntU-riL to 1) E, drop lines cutting C F. "With the T- s 1 uarc placed at right angles to C F, mid brought Against the points in C F, draw lines, extending them indefinitely, as shown. Measuring in each instance from C F, set off on the lines just drawn the same length as similar lines in D N O P E, and through the points thus ob- M Fit/. 689. Plan and Elrrntinn nf Offset. tained trace a line, as shown by n p. Then C F p O n is the half shape of cut on line C F. In Fig. 691, A B G II is a duplicate of the elevation of the round pipe, below which is drawn a half profile of same, A M II. To obtain the shape of cut on line B G, divide the half profile A M H into the same number of parts as was N P, and, with the T-square placed parallel with A 1>, and brought successively against the points in A M II, carry lines cutting ]>'._', upon C V and B G of which, as base lines, are drawn duplicates of the sections obtained in Figs.. <>!0 and 1191, all as shown. From points in u p drop lines at right angle's to C !'. cutting the same, and from points in B in (1 drop lines at right angles to B G, cutting it. Connect points in these lines iu consecutive order by solid lines, as shown, and subdivide the four sided figures thus obtained by dotted lines representing their shorter diagonals. The surface of the offset or transi- tion piece is thus divided into a series of very tapering triangles, the lengths of whose bases or shortest sides Fiij. .'. Middle Piece of Offset, Showiny Method of Triangulation. are given in the two sections C n p F and B m G. In order to obtain the correct lengths of their longer sides two diagrams or series of sections must be con- structed for that purpose, which arc shown in Figs. 093 and W . the opposite lialf of the article by ;i large plain trian- gular surface, S J N, on the upper side, and by another, Q L P, on its lower side, which must be added to the pattern of the curved portion after it has been de- veloped. It will also lie seen that V W 1' of Fig. 093 is one-half of J N S. Therefore' to develop the pattern, first draw any line, as j x in Fig. (595, in length equal to B C of Fig. 689, or V W of Fig. 693. With / as center, and 1' V of Fig. 693 as radius, describe a small arc (near ,v), which intersect with another small intersect with another .small arc struck from point 1 of pattern as center, and a radius equal to 1 2 of the pro- file (' n () ji F of Fig. 692, thus establishing the position of point 2 of pattern. Proceed in this manner, using the dotted oblique lines in Fig. t>;4, the lengths of the spaces in B in G in I^ig. 692, the lengths of the solid oblique lines in Fig. 693 and the lengths of the spaces in n O p F of Fig. 692 in the order named until the line 7 14 is reached. Lines traced through the points of intersection fromj to I and from s to t will give the Fig. 69't. Pattern of Offset. arc struck from x as center, and with a radius equal to W 1' of Fig. 693, thus duplicating the triangle V W 1'. From s, or point 1, as center, with a radius equal to 1 9' of Fig. 694, describe a small arc (near 9), which intersect with another small arc struck from j or 8 of pattern as center, with a radius equal to 8 9 of the pro- file B m G, Fig. 692, thus establishing the position of point 9 of pattern. From 9 as a center, with a radius equal to 9 2' of Fig. 693, describe a small arc, which shape of the curved portion, of the pattern. From 14 of pattern as center, with a radius equal to G F of Fig. I'.'.H. or V 7 of Fig. 693, describe a small are, which intersect with another small arc struck from point 7 of pattern as center, and a radius equal to 7 7' of Fig. 'It::, or T P of the plan. Draw 7 t and t /; then will /_/ .': .s rj t be one-half the pattern required. The other half can be obtained by any means of duplication most convenient. PROBLEM 210. Pattern for an Offset to Join a Round Pipe with one of Elliptical Profile. This problem differs from the preceding one only in the shape of the pipe having the elongated profile, which profde in the preceding problem consists of two semicircles joined by a straight part, whereas in this ease its curve is continuous throughout; its pattern therefore will consist throughout of a series of triangles Pattern Problems. having short buses instead of having a large llat trian- gular surface uniting its curved portion as in the pre- vious ease. Ill Fig. li'.Mi. 1) (' 1! A represents the elevation of the otl'set. (' F K 15 that of a portion <>f the round pipe with which it is required to connect at its upper end ?'; r 7 !? -& i I ' i i ; i i i | ELEVATIOM \ \ PROF Fiij. G96. Elevation and Sections of Offset, Shonriny Method of Triangulation. and II 1) A. U that <>f the elliptical pipe joining it be- low. M I' X is the half profile of the round pipe and K J L that of the elliptical pipe. The plan, or top view is not shown,' and is not necessary to the work of obtaining the pattern. Since the profiles given neces- sarily represent sections on lines at right angles to the respective pipes, as at I 1 ' K and II 0, it will .first be necessary to derive from them sections on the joint or miter lines C 1> and 1> A, from which to obtain correct stretchouts of the two ends of the pattern of the offset piece. As the pattern required consists of svinmet- rical halves, one-half only will be given, and one-half of the profiles only need be used. Therefore divide the half profile M 1' X into any convenient number of equal spaces, as shown by the small figures, and from the points thus obtained draw lines at right angles to F E, cutting M N and C 13. To avoid confusion of lines a duplicate of C B is shown at the left by C 1 B 1 . From the points on C' B' draw lines at right angles to it indefinitely, and upon each of these lines, measuring from C' B', set off the lengths of lines of corresponding number in the profile M P N measured from M N. Thus make the distance of point 2' from C' B' equal to the distance of point 2 from line'M N, the length of line 3' equal to that of line 3, measuring from the same base lines as before, etc. A line traced through the points of intersection, as shown by C 1 B 1 , will be the correct section on the line C B of the elevation. The method of obtaining the section on the line D A, shown at D' I A', is exactly the same as that just described in con- nection with the round pipe, all as clearly shown in the lower part of the engraving. The next operation will consist of dividing the surface of the transition or offset piece into measurable triangles, making use of the spaces used in the profiles; therefore connect points in C B with those of similar number in D A by solid lines, as 1 with 1', 2 with 2', .etc., and connect points in C B with those of the next higher number in D A by dotted lines, as 1 with 2', 2 with 3', 3 with 4', etc. The surface of the transition piece is thus divided into a series of triangles the lengths of whose bases or short sides are found in the two sections C' B 1 and D 1 I A 1 . As the bights of corresponding points in the two sections, measuring from their center or base lines, differ very materially, it will be necessary to construct two diagrams of sections from which the lengths of the various solid and dotted lines can be obtained. In Fig. 697 is shown a diagram of sections through A B C D taken on the solid lines drawn across the eleva- tion, in which the base line P Q represents the sur- face of a plane dividing the offset into symmetrical halves. At P erect a perpendicular, P E, upon which set off the hight of the points in the profile K J L or the section I)' I A 1 , measuring upon the straight lines joining them with the base line K L, as shown by the small figures. From P, upon P (j, set off the lengths 398 Xvw Metal \YUI-JM- Pattern Bwk. of the various solid lines drawn across the elevation, also shown by small figures, and at each of the points thus obtained erect a perpendicular, which make equal in hight to the distance of point of corresponding num- ber in profile M P N from M N, measuring on the per- pendicular line. Thus, make line 2 of Fig. 697 equal IB f'iy. .' '2 "I Fig. 697, strike a small are, which intersect with an- Fig. 690. Half Pattern of Offset Piece. other small are struck from of pattern as center, and a radius equal to 1 2 of section C 1 B', thus establish- ing the position of point 2 of pattern. So continue. using alternately the dotted and the solid oblique lines in Figs. 698 and 697 to measure the distances across the pattern, the spaces from the section 1)' 1 A 1 to form the stretchout of the lower end (I) A) of pattern, and the spaces from the section C' O B 1 to form tin- stretchout of the upper end (C B) of the pattern. Lines traced through the points of intersection, as from C .to B and from I) to A, will complete one-half the required pattern. Pattern Problems. 399 PROBLEM 211. The Patterns for a Funnel Coal Hod. In Fig. Too HIV shown the drawings for a funnel IK-IT: constructed in two pieces, the front being in one coal hod of a style in general use. In preparing such piece joined together on the line B C of the elevation a set of drawings it is necessary that care should lie or B 3 C 3 of the plan, and joined to the hack piece on taken to have a correspondence of all the principal the line II D. As will be seen by an inspection of c' Fill. 7'iJOriiin, KIrration inul Xfrtions of a Funnel Coal Hod, Shoiriiifi Method of Trian (filiation. parts in the two views, as shown by the dotted lines, leaving the final drawing of the curves to be more accurately performed as circumstances may require in subsequent parts of the' work. The design is capable of any degree of modification so far as the proportions of its parts are concerned without in the least affect- ing the method of obtaining its patterns. Thus, hights, lengths, diameters or curves may be changed at the discretion of the designer. The coal hod is the elevation, the front piece consists of a flat tri- angular piece. II .1 D, joined to two irregular flaring pieces, A .1 II (i and 15 J D C. On account of the taper or slant of the flat portion of the front piece, as shown by J 2 I) 2 of the plan, the line D 2 H' has been drawn somewhat obliquely from X-, the center of the bottom, instead of at right angles to A 2 E". The section at A B is assumed to be a perfect circle and should be drawn exactly opposite, as 400 The New Metal Worker i'attcru Book. shown, its vertical center line A 1 B 1 being placed parallel to A B. Divide each quarter of this, as A 1 J' and J 1 B', into any number of equal spaces, as shown by the small figures, and through the points thus ob- Q 87' fl' 5 Fig. 701. Diagram of Sections on Solid Lines in A J H G of Fig. 700. tained draw lines cutting A 1 B 1 and A B. From J, the middle point on A B, draw lines to D and to H. Also divide H' G 1 of the plan into the same number of equal spaces as A 1 J 1 , numbering the points to cor- respond. From the points thus obtained erect lines perpendicularly, cutting G H of the elevation. Con- nect points ot like number on A J and G H, as 5 with 5, 6 with 6, etc., by solid lines, as shown; also, con- nect each point on A J with that of next higher number on G H by a dotted line, as 5 with 6, 6 with 7, etc. These solid and dotted lines just drawn are the lines upon which measurements are to be taken in obtaining the pattern, and upon which sections must ftl - ______ _ ~" ^_"~r"~- - --^1- J ,U T 98' 7' 6 Fig. 702. Diagram of Sections on Dotted Lines in AJHGof Fig. 700. be constructed before their true lengths can be ob- tained. In Fig. 701 are shown the sections having the solid lines in A J H G as their bases, which are con- structed in the following manner: Draw any right angle, as P Q K. Upon P Q set off the hights of the several points in the section A 1 J' from the line A 1 B 1 , as measured upon the straight lines joining them with A' B 1 ; thus make Q 5 and Q 6 equal to the distance of points 5 and 6 from the line A 1 B 1 . From Q on Q B, measuring from Q, set off the lengths of the several solid lines in A J H P, as indicated by the small figures, and from the points thus obtained erect perpendiculars equal in hight to the length of lines drawn from points of corresponding number in G' 11' of the plan to the line G 1 X; thus make the perpen- diculars at points 5', 6', etc., equal to the length of the lines drawn from points 5 and 6 in G' II' to G 1 X. Connect the points thus obtained with points of cor- responding number in P Q. The oblique lines thus obtained will be the true distances represented bv lines of corresponding number in the elevation. The diagram in Fig. 7(>2 shows the sections upon the dotted lines in A J II G and is constructed in the same man- ner. Upon T U, .measuring from T, are set off the lengths of the several dotted lines. S T is the same as P Q of Fig. 701, and the perpendiculars at U are equal to those of corresponding number in Fig. 701. Points in S T are then connected with the perpen- diculars of next higher number by dotted lines, which L 3'3' 4' 5 Fig. 70S. Diagram of Sections on Solid Lines in J B C D of Fiij. 7. give the true lengths represented by the dotted lines of the elevation. That portion of the front piece shown by .1 B (' 1> of the elevation must be triangulated in exactly the same manner as the portion just described, and sec- tions constructed upon the several solid and dotted lines there drawn, as shown in Figs. 703 and 704. However, as no outline is given in either the plan or the elevation from which a correct stretchout of C 1> can be obtained, a section must be constructed for that purpose, which can be done in the following manner : First draw C 1 M 1 as the vertical center line of a rear ;r -i.i. I.- -,W !" 3 4 -.w Fig. 704. Diagram of Sections on Dotted Lines in J B C D of Fig. 700. elevation. From points C and D project lines hori- zontally to the ri^'lit, cutting' C' M 1 at C' and M. l'| D M, measuring from M, set off half the widtli of the front piece at D" of the plan; that is, make M D 1 Pattern Problems. 401 equal to M" T>\ Any desirable curve may then be drawn from 1)' to C', representing the rear elevation of curve represented by D of the side elevation. As the distance from C to D is much greater than C' M, an extended profile, as measured upon C D, must n<>\v lie developed from which to obtain a correct stretch- out of that portion of the pattern. Therefore divide the curve C' D 1 into the same number of parts as the quarter circle B' J', and from the points thus obtained carry lines horizontally to the left, cutting C D. Upon C 1 M extended, as C* M 1 , set off spaces equal to those in C D, as shown, and through Fig. 705. Half Pattern of Front Piece of Funnel Coal Hod. the points thus obtained draw lines to the left indef- initely. From the points in C 1 D 1 drop lines verti- cally, cutting those just drawn, all as shown. A line traced through the points of intersection, as shown from D" to C 2 , will give the desired stretchout. In dividing the curve C' D 1 into spaces it is advisable to make those nearest to D 1 less than those near the top of the curve in order to compensate for the increase in the spaces in C" M' as they approach the bottom; thus obtaining a set of nearly equal spaces upon the final profile D 2 C 2 , all of which will appear clear by an inspection of the drawing, As above stated, the diagrams of sections in Figs. 7<>3 and 704 are constructed in the same manner as those of Figs. 701 and 702. The bights in K L and V W are taken from J 1 B' of Fig. 700 and are the same as those in P Q and S T of Figs. 701 and 702. The distances upon L N and W Y are those of the solid and dotted lines in J B C D of Fig. 700, and the hights of the perpendiculars near N and Y are equal to the lengths of the lines drawn from points of cor- responding number in the profile D 2 C" of Fig. 700 to the lines C' M 1 . To develop the pattern of the front piece, first draw any line, as A G in Fig. 705, equal in length to A G of Fig. 700. From G as a center, with a radius equal to the dotted lines 9 8 of Fig. 702, describe a short arc (near 8), which intersect with another arc drawn from A as center, with a radius equal to 9 8 of the section A 1 J' B 1 of Fig. 700, thus establishing the position of point 8 in the upper line of the pattern. From 8 of the pattern as center, with a radius equal to 8 8 of Fig. 701, describe a short arc (near 8'), which intersect with another arc drawn from G of the pat- tern as center, with a radius equal to 9 8 of the plan, Fig. 700, thus establishing the point 8' in the lower line of the pattern. Continue in this manner, using the lengths of the oblique dotted lines in Fig. 702 in connection with the spaces in the section A 1 J 1 B 1 of Fig. 700 as radii to determine the points in the upper line of the pattern, or the side forming the mouth, and the lengths of the oblique solid lines of Fig. 701 in connection with the spaces in the plan of the bottom (G 1 H') as radii with which to determine the points in the lower line of the pattern or the side to fit against the bottom. Having reached the points 5 and 5', next add to the pattern the flat triangular surface shown by J H D of the elevation. From H (5') of the pattern as center, with a radius equal to 5 5 of Fig. 706, the side of the last triangle in the pattern of the back piece, describe a short arc (near D), and intersect the same with another are struck from J (5) of the pattern as center, with a radius equal to the oblique line 5 5 of Fig. 703, and draw H D and B J. Using D J of the pattern as one side of the next triangle, take as radii the dis- tances 5 4 of Fig. 704 and 5 4 of the section D 2 C 1 of Fig. 700 to locate the position of point 4' of the pat- tern, as shown in Fig. 705. With 4 4 of Fig. 703, and 5 4 of the section B' J' of Fig. 700 as radii locate the point 4 of the pattern, as shown, and so continue until C D is reached. Lines traced through the points of intersection from B to A, C to D and H to G will complete the pattern of one-half the front piece. 402 Tlie New Metal \Vorkei- 1'uttcni The method of triangulating the piece forming tlie back of the coal hod and the development of the pattern of the same are so clearly shown in Figs. 706, 707 and 70S, in addition to the plan and elevation, Fig. 700, as to need only a brief description. Divide H' F' and D a E 2 of the plan, Fig. 700, into the same Fig. 706. Diagram of Sections on Solid Lines in D E F H of Fig. 700. number of equal parts, and from the points thus ob- tained erect lines vertically cutting the corresponding- lines II F and D E of the elevation, as shown by the dotted lines. Connect points of like number in that view by solid lines and points in D E with those of next lower number in II F by dotted lines. Since D E, being inclined, is longer than M 2 E 3 , its equivalent in the plan, it will be necessary to develop an ex- tended section upon the line D E of the elevation, as shown by D 5 E 3 of the plan, which may be done in the same manner as the section on the line C D above explained. Upon M a E a extended, as E" E', set off the Fig. 707. Diagram of Sections on Dotted Lines in D EF H of Fig. 700. spaces in D E, and through the points thus obtained draw lines at right angles, as shown, which intersect with lines drawn parallel with M* E" from points of corresponding number in D" E", thus establishing the curve D 6 E 3 , from which a correct stretchout of the top of the back piece may be obtained. In Figs. 7o(i and 7<>7, the hights of the various points upon the perpendiculars from X and Z are equal to the lengths of the straight lines drawn from points of corresponding numbrr in II' F 1 of the plan, Fig. 700, to the line M J F 1 . The distances set off to tin- right upon the horizontal lines from X and Z are equal to the lengths of the several solid and dotted lines ir> D E F II of the elevation, and the hights of the per- pendiculars at the right ends of the bases are equal to the straight lines drawn from points in the section D s E 3 to the line E" E 3 . The several oblique solid and dotted lines are, therefore, the true distances repre- sented by the solid and dotted lines of corresponding number in the elevation. Fig. 708. Half Pattern of Back Piece of Funnel Coal Hud. In Fig. 708, E F is equal to E F of -Fig. 700 and is made the base of the first triangle, from which base the several triangles constituting the complete pattern may be developed in numerical order and in the usual manner from the dimensions obtained in Figs. 706 and 707 and in the plan and section in Fig. 700, all as clearly indicated. The pattern for the piece forming the foot of the coal hod is a simple frustum of an elliptical cone, the method of obtaining which is fully explained in Prob- lem 171. In Fig. 546 of that problem the lines E F ami Gr H are drawn much further apart than the proportions of the foot in the present case would justify, but the operation of obtaining its pattern is exactly the same. Pattern Problems. PROBLEM 212. 4u:; Patterns for a Three-Piece Elbow to Join a Round Pipe with an Elliptical Pipe. In Fig. 709, let A B C D represent the profile of the round pipe and EFGHIJKL the elevation of the elbow. In the plan the profile of round pipe is represented l>y A 1 B 1 C" D 1 , the elbow by P l M' G 1 M' PROFILE C ELEVATION Fig. 709. Plan and Elevation of Three-Piece Elbow, Round at One End and Elliptical at the Other. P", and the shape of elliptical end of elbow by J 1 M 1 G 1 M". The section J G II I of elevation is without Hare, and sections K F G J and E F K L are flared, as shown in plan. Through the plan draw E 1 G 1 and M 1 M", and carry M' M' 1 through the center of elevation, as shown by M N () P. Perpendiculars dropped from the points K O F of elevation, cutting P' M', E' G 1 and P' M 2 , as shown by K' O' F 1 O 2 , will give the shape of iiiik'r line K K in plan. As J G II I of elevation is without flare the pat- tern for this part is procured in the ordinary manner as for a pieced elbow. Since J' M' G' M" is the profile of an elliptical cylinder, the section upon the oblique line J G of the elevation, cut- ting the same, must necessarily be an ellipse whose major axis is equal to M' M 3 and whose minor axis is equal to J G. In like manner, the section at K F of the elevation may be assumed as an ellipse whose minor axis is K F and whose major axis is equal to O' 0'. In Fig. 710, a duplicate of K F G J of elevation is shown by T R U W. Bisect T E in d and erect the perpendicular d S, and make d S equal to Q' 0' of plan. Through the points T, S and R trace the half ellipse, as shown. In a similar manner bisect W U in I, and erect the perpendicular I V, in length equal to N 1 M 1 of plan. Through the points thus obtained trace the half ellipse W V U. Divide T S R into any convenient num- ber of equal parts, and from the points thus obtained drop per- pendiculars cutting T R, as shown. Also divide "W" V U into the same number of equal parts as was T S R, and from the points thus obtained drop perpendiculars cutting W U. Connect points in T R with those opposite in W U, as shown by the solid lines. Thus connect a with A, b withy, c with &, etc. Also connect the points in T R with those in W 0, as indi- cated by the dotted lines. Thus connect 1 with h, a with y, b with &, etc. The next step will be to construct a series of sec- tions upon the several solid and dotted lines just drawn for the purpose of obtaining the true distances which they represent. The sections represented by solid lines 404 Tlie New Metal Worker Pattern Book. are shown in Fig. 711. To construct these sections proceed as follows : For section a, draw the line a A, 13 Fig. 710. Elevation of Middle Section of Elbow, Showing Method of Triangulation. 2 1 II 1 17 1 12' 16 1 15' d Fig. 711. Sections on Solid Lines of Fig. 710. in length equal to a A of Fig. 710. From a erect a by a 2', and from h erect a perpendicular, in length equal to h 11 in W V U, as shown by h II 1 , and connect 2' 11', as shown. For section b, drawft/, in length equal to bj of Fig. 710. From b erect the perpendicular b 3", in length equal to b 3 in T R S, and from/ erect a per- pendicular, in length equal to j 12 in W V U, and con- nect 3 1 12', etc. For the sections representing the dotted lines, as shown in Fig. '712, proceed in a similar manner. For section A, draw 1 A, in length equal to 1 h of Fig. 710. From h erect a perpendicular, in length h 12" g is 13 2 5" 15 3 i d Fig. 712. Sections on Dotted Lines of Fig. 710. equal to A 11 in W V U, and connect 1 with 11'. For section j, draw aj, in length equal to a j of Fig. 710. From a erect the perpendicular a 2 3 , in length equal to a 2 in T S R, and from / erect a perpendicular, equal in length toy 12 in W V U, and connect 2 2 12', etc. To describe the pattern for part of article repre- sented in Fig. 710 by T R U W, as shown in Fig. 71:;, proceed as follows : Draw any line, as R V of pattern, perpendicular, in length equal to a 2 in T S R, as shown ; in length equal to R U of Fig. 710. With R of pat- I'ntti-rn Problems. 405 tern as center, and 1 II 2 of the diagram of sections, Fig. 712, as radius, describe a small are, which intersect with one struck from point V of pattern as center, and 1" 1 1 of Fig. Tin as radius, thus establishing the point 11 of pattern. With point 11 of pattern as center, and 2' 11' of diagram of sections, Fig. 711, as radius, describe a tances between points represented by numbers in Fig. 711 for the length of solid lines in pattern, and the spaces in E S T for the stretchout of R T" of pattern. Lines drawn through the points thus obtained, as indi- cated by J{ T' \V V, will be one-half of the required pattern. The other half, as shown by R T W V, can Fig. 713. Pattern of Middle Section of Elbow. small arc, which intersect with one struck from R of pattern as center, and R 2 of Fig. 710 as radius, thus establishing point 2 of pattern. Proceed in this man- ner, using the dotted lines in Fig. 712 for the distances in pattern represented by dotted lines; the spaces in \\ V U for the stretchout of V W of pattern; the dis- be obtained by a repetition of the same process or by duplication. The pattern for E F K L of elevation can be ob- tained in a similar manner, A B C of profile being one- half the shape on E L, and R S T of Fig. 710 being the half section on F K. PROBLEM 213. Patterns for a Right Angle Piece Elbow to Connect a Round with a Rectangular Pipe. In Fig. 714 is shown the design of a right angle elbow of which one end is rectangular, as shown by N O P Q, and the other round, as shown by A B C D. Such an elbow may be constructed in any number of pieces, the elevation for which may be drawn in the manner described in the case of an ordinary piece elbow. In the present instance the elbow consists of seven pieces. In adjusting the transition from the rectangle to the circle, it is evident that the flat sides of each of the five intermediate pieces must become shorter in each piece as the round end of the elbow is approached ; and that the quarter circles forming the corners of each of the intermediate pieces must be of shorter radius as the corners of the rectangular end are approached. This may be accomplished upon the elevation in the following manner : Through the center of the eleva- tion draw the lines c to m, and divide L T into the number of parts there are pieces in the elbow subjected to the change in shape, in the present instance five. From k' set off each way four spaces, as shown by k k' and k' k". Set off from f three spaces, as shown by / /' and//'. Continue this operation and connect the points L k j h y f and g" h" f k" I/, thus showing in side elevation the change from the rectangle N O P Q to the circle A B C D. To show a similar shape on the outer curve of elbow, draw any line, as E M, in Fig. 715. From E on E M set off the spaces E F, F G, etc., to L M of Fig. 714. As it is only necessary to show the half 406 Tlte New Metal Worker Pattern Boole. shapes, from M and L erect perpendiculars, in length equal to N n of profile, and connect same, as shown by M'' L''. Connect L 2 F, and from the points in E M I will indicate the method to be .sections. followed in the other ELEVATION K PROFILE rf. 714. Elevation and Profiles of an Elbow to Connect a Hound With, a Rectangular Pipe. erect perpendiculars cutting F L*. The shapes on in- ner curve of elbow, as shown in Fig. 716, are obtained in the same manner as described for Fig. 715. For the hights of section on c m of the elevation, on any line, as E M, in Fig. 717, starting from E, set off the distances cf,fy', g' h', etc., and from the points thus obtained erect perpendiculars, as shown. From E and F set off the distance C' B of profile of circular end and draw E' F" From M and L set off the dis- tance N n of profile of rectangular end and draw M s L*. Connect L 3 with F 3 , thus completing the section. The method for obtaining the patterns for sections E F F' E' and L M M' L' is the same as for an ordinary pieced elbow. The method for obtaining the pattern for one of the remaining sections will be shown, which In Fig. 718, F G G' F' is a duplicate of the sec- tion having similar letters in elevation. The shape F II F' is the half of an ellipse, because F' F of Fig. 71-t is an oblique section of a cylinder of .which A B C D is the plan, and can be de- scribed in any convenient manner. From G erect the perpendicular G S, equal to G G' of Fig. 715, and from G' erect another perpendicular, equal to G G' of Fig. 716. From points g and y' erect perpendiculars equal to G 3 G of Fig] 717, and connect T T', as shown. p Connect S T and T' S' by a quarter of an ellipse. Divide S T, T' S', F U and U F' into any convenient number of equal parts, and from the points thus obtained drop perpendiculars cut- G Fig. 715. Shape of Flat Part of Fig. 716. Shape of Flat Part of Outer Curve of Elbow. Inner Curve of Elbow. ting G G' and F F'. Connect these points, as shown by the solid and dotted lines in G G' F' F. The next operation will consist in constructing sections upon these solid and dotted lines for the pur- Pattern Problems. 407 pose of ascertaining the correct distances which they represent. These sections are shown in Figs. Tl'J and 720. To construct the sections represented by solid Fig. 718. Elevation of One of the Pieces, Showing Method of Trianyulation. lines in Fig. 718, proceed as follows: Draw the line F G of Fig. 719, in length equal to F G of Fig. 718, and from point G erect a perpendicular, in lengtlrequal to G S of Fig. 718. Connect F S, which gives the distance between points F and S. For the second section draw v o, in length equal to v o of Fig. 718, H 3 f Fig. 717. Extended Section on Line c m of Fig. 714. and make the perpendiculars o 6 and v 7 of the section equal to lines having similar letters in Fig. 718. The other sections are obtained in a similar manner. To construct the sections represented by dotted lines in Fig. 718, proceed as follows: Draw the line F o of Fig. 7i'i, ia length equal to F o of Fig. 718, and from o erect the perpendicular o 6, equal to o (5 of Fig. 718, and connect F 6. For the second section draw v p, in length equal to vp of Fig. 718, and from points v and p erect perpendiculars equal to v 7 and p 5 of Fig. 718. Connect points 7 5, which give the distance between corresponding points in Fig. 718. In Fig. 721, G G' F' F is the pattern of part of 7 1 G V F" Fiii. 719. Sections on Solid Lines of Fig. 718. article shown by similar letters in Figs. 714 or 718. The distances represented by solid lines in pattern are obtained from the sections in Fig. 719, as indicated by corresponding figures, and the distances represented by dotted lines in pattern are obtained from the sections in Fig. 720. The stretchout of G G' of the pattern is obtained from G T T' G' of Fig. 718, as the stretchout of F U F' of the pattern is obtained from F U F' of Fig. 718. 408 The New Metal Wwker Pattern Book. To develop the pattern from tlie sections above constructed, first draw G F of'Fig. 721, in length equal to G F of Fig. 718 or 719, upon which duplicate the 10 1 - p Fig. 720. Sections on Dotted Lines of Fig. 718. triangle G F S of Fig. 719, as shown. From S of pat- tern as center, with S 6 of Fig. 718 as radius, de- scribe an arc, 6, which intersect with one struck from F of pattern as center, with radius F 6 of Fig. 720, thus establishing point 6 of pattern. Then with radius F 7 of Fig. 718, from F of pattern as center, describe an arc, which intersect with a second arc struck from point 6 of pattern as center, and 7 of Fig. 719 as radius, thus establishing point 7 of pattern. Continue this process until the various points indicated in pat- tern are located. Lines drawn through the points S'G' F 7 Fig. 721. Half Pattern of F G G' F' of Fig. 714. thus obtained, as indicated by G S T T' S' G' and F' U F, will complete one-half of the required pattern. The other half can be obtained by duplication or by a repetition of the above process. In obtaining the pattern for anv one of the re- maining pieces first draw a duplicate of its elevation as taken 'from Fig. 714, upon either side of which con- struct the proper section, obtaining the points in the same from Figs. 715, 71 <> and 717 as was done in Fig. 718, after which the subsequent operation is analogous to that above described. PROBLEM 214. Pattern for the Soffit of a Semicircular Arch in a Circular Wall, the Soffit Being Level at the Top and the Jambs of the Opening Being at Right Angles to the Walls in Plan. Two Cases. First Case. In Fig. 722, let A B C represent the outer curve of an arch in a circular wall corresponding to A' H C' of plan, and let E B D represent the inner opening in the wall, as shown by E' F' D' in plan. Then A E B C D will represent the soffit of the arch in elevation and A' H C' D' F' E' the same in plan. In the engraving the outer curve of the arch is a per- fect semicircle, and the inner curve is stilted, as shown, so as to make the soffit level at B. Instead of the stilted arch, the inner curve may, if desired, be drawn as a semi- ellipse of which E D is the minor axis and F B one-half of the major axis. Divide A B of elevation into any convenient num- ber of equal parts, shown by the small figures. With the T-square parallel with the center line B B', drop lines from the points in A B, cutting A' II of plan. as shown. Since that portion of the inner arch from E to 12 is drawn vortical, as above explained, divide 12 B into the same number of parts as was A B, and, with the T-square parallel with the center line B B', drop lines to E' F', as shown. Connect opposite points in A' II with those in K !". as shown l>v the solid lines in plan. Also divide the four-sided figures thus produced by means of the diagonal dotted lines 6 8, 5 9, etc., as shown. The several triangles thus produced will represent in plan the triangles into which the soffit, or under side, of the arch is divided for the purpose of obtaining its pattern. In order to ascer- 1'attern Problems. 409 tain the real Jistances across tlie surface of the arch which the solid and dotted lines represent, it will be necessary to construct a series of sections of which these lines are the bases, as shown in Figs. 723 and 734. In constructing the diagram shown Fig. 723, the several solid lines of the plan, though not exactly equal in length (because tliev an! not, drawn radiallv from the center of the curve A' II C'), may be con- sidered as of the same length. Draw the right angle P Q R as in Fig. 723, and from Q set off horizontally the distance II F' of plan, as shown by Q R. Draw R S parallel with Q P, and, measuring from Q, set off on Q P the length of lines dropped from points in A ELEVATION B fig. 722. Plan and Elevation of Arch in a Circular Wall. first Case. B to A F, as shown by corresponding figures 2 to 6. Likewise set off from R on R S the length of lines dropped from points in E B to E F, as shown by the figures 12 to 7, and connect the points in P Q with those in S R, as indicated by the solid lines in plan. Thus connect 1 with 12, 2 with 11, 3 with 10, etc. To construct the diagram based upon the dotted lines of the plan, draw the right angle M N in Fig. 724, and, measuring in each instance from N, set off on N M the same distances as in Q P of Fig. 723. Starting from N, set off on N the lengths of dotted lines in plan, as shown by the small figures in N 0. With the T-square parallel with M N, draw lines from the points in N 0, and, in each instance measuring from N 0, make these lines of the same length as lines of similar number dropped from points in E B of eleva- tion to E F. Connect the points in these lines with points in M N, as indicated in plan by the dotted I] 10 1211-," to Fig. 723. Diagram of Sections an Fig. 724. Diagram of Sections on Solid Lines of Plan, fig. 722. Dotted Lines of Plan, fig. 722. lines. Thus connect 6 with 8, 5 with 9, 4 with 10, etc. The next step is to obtain the distances between points in A B of elevation as if measured on the outer opening in the curved wall. To do this, on F A ex- tended set off a stretchout of A' H of plan, as shown by the small figures 5', 4', etc., and with the T-square at right angles to the stretchout line J F, draw the usual measuring lines. With the "[-square parallel with J F, carry lines from the points in A B to lines of similar number drawn from the stretchout line. A line traced through these points, as shown by J B, will give the true distances desired between the points in the outer curve of the arch. The distances between points in E B, the inner curve, are obtained in a similar manner. To avoid a confusion of lines, the stretchout of F' E' of plan is 13 12 11 Fig. 125. One-Half Pattern of Soffit of Arch Shown in fig. 7%2. set off on F C of elevation, as shown by the small figures, 7', 8', 9', etc. B D is also divided into the same parts as was E B, and from the points thus ob- tained lines are drawn to the right parallel with F C. With the T-square parallel with B F, carry lines 410 J7ie New Metal Worker Pattern Book, from the stretchout points in F K, cutting lines of similar number drawn from the points in B D. A line traced through the points thus obtained, as shown by B K, will give the distance between points as if measured on the inner curved line of the wall. From the several sections now obtained the pat- tern may be developed in the following manner : At any convenient place draw the line a e in Fig. 725, making it in length equal to Q K of Fig. 723, or A' E' thus establishing the point 11 of pattern. Continue in this way, using the tops of the sections in Figs. 7 _'."> ami 724 for measurements across the pattern, the spaces in J B for the distances along the edge a h of pattern, and the spaces in B K for the distances along the inner edge e f, establishing the several points, as shown. Through the points in a h and e f lines are to be traced, while f h is to be connected by a straight line, thus completing one-half the pattern. The other 123 4 5 6|< DEVELOPMENT OF OUTER CURVE PLAN Fig. 726. Plan and Elevation of Arch in a Circular Wall. Second Case. of plan. At right angles to a e draw e 12, in length equal to R 12 of Fig. 723, and connect a with 12 if it is desired to show the triangle. From a as center, and J 2 of elevation as radius, describe a small arc, 2, which intersect with one struck from point 12 of pat- tern as center, and 12 2 of Fig. 724 as radius, thus es- tablishing the point 2 of pattern. With 2 of pattern as center, and 2 11 of Fig. 723 as radius, describe another arc, 11, which intersect with one struck from 12 of pattern as center, and 12 11 of B K as radius, half of pattern can be obtained by the same method or by any convenient means of duplication. If the arch were semi-elliptical instead of semi- circular, the method of procedure would be the same as above described. Second Case. In Fig. 726, ABC represents the outer curve of an arch in circular wall, as shown by A' H C' in plan. E B D represents the inner curve in elevation, as does E' D' the same in plan. Thru A E B D C represents the soffit of the arch in eleva- Pattern Proilt in*. 411 tion and A' II 0' D' E' the same in plan. The con- ditions given in this ease differ from those of the first case only in the fact that the inner curve of the arch in this ease is straight in plan, as shown by E' D', instead of curved to the radius of the wall as in Fig. 722. The method of procedure in this case is exactly the same as before, but one less operation will lie. nec- essary, since measurements upon the inner curve may lie taken directly from E B D of the elevation. To avoid a confusion of lines, a duplicate ef 4' 3' 2' " 1 2 345" 1* """ 3 ---,2' l| 12 3 456 fiij. 7-'7.I>i(iyraui of Sections on Fig. 728. Din gram of Sections on Solid Lines of Plan, Fiij. 796. Dotted Lines of Plan, Fig. 7K. of E B I) of elevation has been drawn in plan, as shown bv E' B' D'. To obtain the divisions on plan divide A B into any convenient number of equal parts, and from the points thus obtained drop lines parallel with the center line B B' to A' II of plan, as shown. Divide E' B' in a similar manner, and from the points thus obtained drop lines to E' F' of the plan, as shown. Connect points in A' II' with those of similar number in E' F' by solid lines. Also connect points in A' H with those of next lower number in E' F' by dotted lines. These solid and dotted lines just drawn will .form the bases of a series of sections, shown in Figs. 727 and 728, whose upper lines will give correct dis- tances across the pattern of the soffit. To construct the sections based upon the solid lines of the plan, first draw the right angle P Q R in Fig. 727, and set off on Q P, measuring from Q, the length of the vertical lines in A B F of elevation. Starting from Q, set off on Q R the length of solid lines in A' H F' E' of plan, as shown by the small figures in Q R. With the T-square parallel with P Q, draw lilies from the points in Q R, and, measuring from Q R, set off on these lines the length of lines of corresponding number in E' F' B' of plan, and connect the points with points of similar number in P Q. The diagram of sections based upon the dotted lines of the plan, shown in Fig. 728, is constructed in the same manner, using the length of dotted lines in plan for the distances in N 0, the length of lines in E' F' B' of plan for the length of lines set off at right angles to N 0, and the length of lines in A B F of elevation for the distances in M N. Connect the points as in- dicated by the dotted lines of the plan, all as shown. The next step is to obtain the correct distances between points in A B of elevation, or A' II of plan. To do this lay off horizontally J K, on which set off a stretchout of A' H of plan, and, with the T-square at right .ingles with J K, draw the usual measuring lines. With the T-square parallel with J K, carry lines from the points in A B to lines of similar num- ber. A line can be traced through these points, as shown by J L, from which the correct stretchout of the outer side of the pattern can be obtained. To describe the pattern first draw any line, as a e of Fig. 729, equal to A' E' of plan. With e of pattern as center, and E' 1 of the inner curve of the arch as radius, strike a small arc, 1', which intersect with one struck from a of pattern as center, and Q 1' of Fig. 7'27 as radius, thus establishing the point 1' of pat- tern. With a of pattern as center, and J 2 of Fig. 72n A' J5 of plan, as the lines dropped from F E to F G give the hight of sections on F' E' of plan. To construct the sections shown in P n ig. 7-!l, rep- resented in plan by the solid lines, proceed as follows: Draw the right angle a a 1, making a ' equal to E' B' of plan, and a' 1 equal to G B (n 1) of elevation. Dra\\ a 1 -' parallel to of elevation. Con- nect 11 with -2 of section, which will give the -ictnal distance between points 11 and 2 of elevation or plan. The remaining sections are constructed in a similar manner, each of the sections representing a vertical section through the head on the lines of corresponding numbers in the plan. The sections based upon the .lotted lines of the plan, shown in Fig. V32, are con- structed in exactly the same manner. Draw It a, in length equal to 1 1 1 of plan, and erect the two perpen- diculars, as shown. Make a 1 of section equal to // I of elevation, and also make I 11 of section equal to I Pattern Problems. 413 11 of elevation, and connect- 11 with 1. The remain- ing sections are constructed in the same manner. Before ilie pattern can be obtained it will be nec- essary k> develop extended sections of the inner and b 4 e j u k Fiij. "',!. Sections Baaed I'poti the Solid Lines of the Plan, Fig. 730. outer curves, as shown to the left and right of the ele- vation. This is done for the purpose of obtaining the actual distance between points shown in elevation. For convenience, on (I A extended, as H K, lay off a stretchout, of A' B' of plan, and from the points therein contained erect the perpendiculars, as shown. From the points in A B of elevation draw lines parallel with H G, cutting perpendiculars of similar number erected from H K. A line drawn through these points of in- a d 3 e f k Fig. 732. Sections Based Upon the Dotted Lines of the Plan, Fig. 730. tersection, as shown by II J, will show the shape of A B of elevation as laid out on a flat surface. The development of the inner curve is shown to the right of the elevation. On L N is laid out the stretchout of F' E' of plan, and on the perpendiculars erected from the points in the line are set off the same distances as on lines of similar number in F E G of elevation. A line traced through these points, as shown by L M, also shows the shape of F E, as laid out on a flat surface, and gives the distance between points as if measured on the finished article. To obtain the pattern, using the distances between points in II J and L M of Fig. 730, and the diagrams in Figs. 731 and 732, proceed as follows: Draw any line, as B E in Fig. 733, in length equal to 1 12 of the first section in Fig. 731. With E as center, and Mil of inner curve as radius, describe a small arc, 11, which intersect with one struck from B as center, and 1 11 of Fig. 732 as radius, thus establishing the point 11 of pattern. With 11 of pattern as center, and 11 2 of the second section in Fig. 731 as radius, describe an- Fig. 733. Pattern of Splayed Arch in Fig. 730. other small arc, 2, which intersect with one struck from point B of pattern as center, and J 2 in J H of outer curve as radius, thus establishing the point 2 of pattern. Continue in this way, using the tops of the sections in Figs. 731 and 732 for the measurements across the pattern, the spaces in the inner curve L M and in the outer curve H J for the distances about the edges of the pattern, establishing the several points, through which draw the lines shown. Then B A F E is the half pattern of splayed head, shown in elevation by A B E F. The other half can be obtained by any convenient means of duplication. A semicircular splayed arch can be developed in the same manner as above described. The pattern for a blank for a curved molding, either semicircular or semi-elliptical, for an arch in a ' circular wall comprises really the same relations of parts as are shown in Fig. 730, and could be obtained as above described. 414 77(e New Metal Worker Pattern Book. PROBLEM 216. Pattern for a Splayed Arch in a Circular Wall, the Larger Opening Being on the Inside of the Wall. In Fig. 734 is shown the plan and elevation of an arch in a curved wall, such as might be used as the head of a window or door, the jambs and head to have the same splay. In the plan, C E D represents the inner curve of wall and A F B the outer. J G M in elevation represents the inner curve and K II L iho outer. In order to arrive at a system of triangles by means of which to measure the splayed surface, lirst the inner and outer lines of plan, as shown. The solid and dotted lines of the plan will form the bases of a series .if right angled triangles whose altitudes can be found in the elevation, and whose h vpothcmises will give correct measurements across the face of the splayed surface. To determine the bights of these triangles extend to the left the horizontal lines drawn through Iv II of the ELEVATION G -H-l -I <4 ~~ "L J i pflj ill 1 1 III DEVELOPMENT OF INNER CURVE DEVELOPMENT OF OUTER CURVE B PLAN Fig. 734. Plan, Elevation and Extended Sections of Splayed An-h. divide J G of the elevation into any convenient num- ber of parts, in this case five, as indicated by the small figures. From the points in the curve thus established drop lines parallel to the center line G F, cutting the inner curve of the plan C E, as shown. Next carry lines from points in J G in the direction of the center X, intersecting the outer curve and establishing the points 7, 8, 9, etc., in it. From these points drop lines to the outer curve in plan A F, establishing the points 7, 8, 9, etc. Connect opposite points in J G with those in K H, as 1 and 7, 2 and 8, 3 and 9, 4 and 10, 5 and 11, and 6 and 12. Likewise connect 1 and 8, 2 and 9, 3 and 10, etc., as shown by the dotted lines. In the same manner connect corresponding points in elevation until they intersect lines dropped from J G, as shown by the points b, <1. /and //. Then // -2. .">: For the lirst triangle, G II 7, draw the right angle G II 7, making the altitude equal to G II of elevation and the base equal to E F, 1 7, of plan, and draw the hypothennse. which represents the actual distance between the points 1 and 7 of the elevation or plan. In like manner the hypothcimso of the second triangle 258 shows the actual distance between the points 2 and S of the elevation; '2 // in Fig. 73/5 is made equal to 2 It of the elevation, while I S equals 2 8 of the plan. Pattern Problems. 415 The remaining triangles in Fig. 735 arc con- structed in a similar manner, each of the triangles rep- resenting a Vertical section tlirongh the head on the lines of corresponding numbers in the plan. The triangles shown in Fig. T.'iii correspond to similar sections taken on the dotted lines in plan. The base a S of the first triangle is ei|iiul to 1 S of the plan, and the hight 1 a to 1 a of elevation, the point n of the elevation being on a level with s, as shown bv the dotted line s a. The hypothennso 1 S is then drawn, which gives the distance between the points 1 Fig. "In. Diagram of Triangles Rased Upon the Solid Lines of the Plan. and 8 in plan or elevation. The bases of the remain- ing triangles are derived from the dotted lines in plan, and the hights from the distances 2 c, 3 e, -i y and 5 of elevation. Before the correct measurements or stretchouts for the inner and outer lines of the pat- tern can be obtained it will be neces- sary to develop extended sections of the inner and outer curves of the arch in elevation, as shown at the left and right of that view. For the develop- ment of the outer curve, as shown to the right by A' H' F'. proceed as fol- lows : On X M extended lay off a 8tretch0ti1 equal to A- F of the plan, transferring it point by point. From the points thus established in A' F' carry lines vertically, extending them indefinitely, as shown, and then from the points in the outer curve K II of the elevation cany lines horizon- tally to the right, intersecting the corresponding lines just drawn from A' F', and through the points thus ' established trace the curved line A' II'. The de- velopment of inner curve, as shown to the left, is accomplished in a similar manner. On X J ex- tended lay off a stretchout of C E of plan, and from the points thus established carry lines vertically. From the points in the inner curve J G carry lines horizontally to the left, intersecting lines of similar number, and through the points thus established trace the curve C' G'. To describe the pattern shown in Fig. 737, first a 3 X \ Fig. 736. Diagram of Triangles Based Upon the Dotted Lines of the Plan. draw the line g h, or 1 7, in length equal to the hypoth- enuse 1 7 of the first triangle in Fig. 735. From 7 as center, with 7 8 of the development of the outer curve as radius, strike a small arc, as shown at 8 in the pattern. From 1 as center, with 1 8 of the first triangle in Fig. PATTERN Fig. 737. Pattern for Splayed Arch. 736 as radius, intersect the arc last struck, thus estab- lishing the point 8. From 1 as center, with radius equal to 1 2 of the development of inner curve, strike a small arc, as shown a-t 2. Then from 8 as center, with 8 2 of the second triangle in Fig. 735 as radius, 410 Tlie New Metal Worker Pattern Book. intersect the arc at 2 already drawn, thus definitely establishing the point 2 in the upper line of the pattern. Continue in this way, using the hypothenuses of the several triangles, as shown, for measurements across the pattern, and the spaces of the inner and outer curves as developed in Fig. 734 for the distances along the edges of the pattern, establishing the several points, as shown. Lines traced through the points from c to , and drop lines from these points to E' F' of plan, as shown. Connect opposite points in E' F' of plan with those in A' II. as indicated b\- the solid lines. Also ELEVATION B M 1211 10 9 8 ?N SHAPE ON E F'oF PLAN Fig. 7i8. Plan, Elevation and Extended Sections of an Arch in u Circular Wall. ing line of the inner curve, as E 12, being straight and vertical. The pattern of the soffit could have been obtained in exactly the same manner had the elevation of this outer curve been a semi-ellipse. Divide A B of the elevation line into any con- venient number of equal parts, and with the "T-square parallel with center line B H drop lines from the points in A B, cutting A' H of plan, as shown by the small figures 1 to 6. As the semicircle representing connect the points of the plan obliquely, as shown l>v the dotted lines, thus dividing the plan of the soffit of the arch into triangles. In order to ascertain the true distances which these lines drawn across the plan repre- sent it will be necessary to construct a series of sections^ of which they are the bases, as shown in Figs. I'.W a.nd 740. In Fig. 739 is shown a diagram of sections corre- sponding to the solid lines in plan, to construct which Pattern Problems. 417 proc-ivd ;is follows: Draw the right angle P Q K, and, measuring from 0, set off on of elevation to A K. as shown by the figures 2 to t>. Likewise set off 1'r 1,1 "ii (^ I { the length of solid lines in plan, as shown l>y the small figures 7 to 12. With the "["-square parallel with P Q, ereet lines from the points in 1!. and, measuring from () It. set oil' on these lines the length of lilies of corresponding number in E 1? V of elevation, /'/;/. ;.'.'. lii'ii/i-tnii with 8, 4 with 9, etc. To construct a diagram of sections corresponding to the dotted lines of the plan draw the right angle ST I'. Fig. 740, and set off on S T the length of lines dropped from points in A B of elevation to A F. or tranl'er the distances in P Q, Fig. 73!. On T U set off the length of dotted lines in plan, and from the points thus obtained draw lines parallel with S T, making these lines of the same length as those dropped from points in E 13 of elevation to E F. Connect the upper ends of lines 8 to 12 with points in S T, as indi- cated by the dotted lines in plan. Thus connect 2 with 12, 3 with 11, -t with 10, etc. The next step is to obtain the distance between points in A I> and K B of elevation as if 'measured upon the curved surfaces of the wall. It is therefore neces- sary to develop extended sections of the two curves of the arch as shown at the left in the engraving. The development of the curve of the inner side of the arch is projected directly from the elevation in the follow- ing manner: ( )u A 1'' extended lav off J K, equal to the curve 1 line A' II of plan, miking the straight line equal to the stretchout of half of the curve of the plan, as indicated by the small figures. At right angles to J O CO J K, and from the points in the same, erect lines, making them the same length as lines of similar num- ber dropped from points in A B, or, with the "["-square parallel with J K and A F. earrv lines from the points iu A B intersecting the vertical lines of similar num- ber. A line traced through the points of intersection, as shown by .1 L. will be the desired shape. The shape on F' K' of the plan, corresponding to E 15 F of eleva- tion, is obtained in a similar manner. On M X in Fig. 7:Js set off the stretchout of E' F' of plan, as indicated by the small ligures, from the points of which erect vertical lines. On these lines set off the same length as the lines of similar number in E B F of elevation. A line traced through the points thus obtained will give the desired outer curve of the arch. To develop the pattern from tne several sections obtained proceed in the following manner: Draw the line a e of Fig. 7-H, in length equal to Q K of Fig. 739, and with e as center, and M 12 of the curve M O as radius, strike a small arc, 12, which intersect with one struck from to the win- dow cap shown in Fig. 742 it is necessary to lirst draw tin- lines / ./ of the profile in elevation, ami pr of the plan, establishing the flare and necessary width or stretchout at those points, after which the points M 8 76 5 t 3 1 N a fig. 742. Elevation and Plan of an Arched Wind**"- <'HI< '" " Circular Wall. ji and r may be dropped upon the line M N of the elevation, and the points k and // earned hori/.ontalk to the center line K X, as shown. The points thus obtained in M N and K L must then be connected bv the necessary ares struck from X as center, thus completing an elevation of the llaring piece. Likewise the projection of the points // and /; from K L must l.e sel off from C on C G of the plan, as shown at >/" and &", and the points thus obtained connected with Pattern Problems. 419 r and p by arcs struck from the center used in describ- ing the plan of tlie wall. Should the width and llaiv of the blank, as deter- mined by the raked jirolile Y at the foot of the aivh. vary so much from those of the normal jiroiile at the top that parallel curved lines could not be drawn to connect the points at the foot with those at the top in either or both views, then centers must bo found upon 73(>; after which the demonstration given in that problem may be followed to obtain the required pattern. It will thus be seen from the foregoing that no matter what form of arch be used or in what manner it be placed upon the wall, the method of obtaining the pattern of its flaring surfaces remains the same. It might under some circumstances be desirable Fig. 743. Method of Obtaining a Section through Molding. the center lines of the plan and of the elevation from which arcs can be drawn connecting the required points. Having thus completed a plan and elevation of the flaring piece or blank, it will-be seen by reference to Problem 215 that the lines p' r and k' g' of the elevation, Fig. 742, correspond with A and B E of Fig. 730; and that p r of the plan. Fig. 742. and the ares drawn from it correspond with IV A'. V K' of Fig. to construct stays at several intervals between the top and the foot of an arch, similar to that shown at Y in Fig. 742, for the purpose of more accurately determin- ing the flare in all parts of its sweep,or for the purpose of constructing a form of templet to assist in the operation of raising the mold. The profiles of such stays can be obtained by the usual operation of rak- ing, which is fully described in numerous problems in 420 Tlie Kew Metal Worker Pattern Book. the first section of this chapter. In Fig. 743 the operation of obtaining a profile upon the line X' X" is graving must be obtained by developing extended sections of those lines on E C of the plan, as de- Fig. 744- Perspective View nf Templet for Use in Raising the Curved Mold. carried out in all its detail, resulting in the form shown at Q, and needs no further comment. In constructing a form or templet, as shown in Fig. 74-4, the outlines of the arch shown in the en- scribed in Problem 214 and those following, after which the stays can be placed upon lines drawn to correspond with those from which the respective sec- tions were taken. Note. An alphabetical list of Terms, arranged as an index, will be found on page 15. The Geometrical Problems giveu in Chapter IV are not indexed, but as each problem is illustrated by a special diagram, its nature may be readily determined. Problems. Page. ADJACENT GABLES. The miter between the mold- ings of adjacent gables of different pitches upon an octagon pinnacle 98 208 ADJACENT GABLES. The miter between the mold- ings of adjacent gables of different pitches upon a pinnacle with rectangular shaft 97 207 ARCH. Pattern for the soffit of an arch in a cir- cular wall, the soffit being level at the top and the jambs of the opening being splayed on the inside 217 416 ARCH. Pattern for a splayed elliptical arch in a circular wall, the opening being larger on the outside of the wall than on the inside 215 412 ARCH Pattern for a splayed arch in a circular wall, the larger opening being on the inside of the wall 216 414 ANVIL. The patterns for an anvil 63 162 ARTICLE. The pattern for an article forming a transition from a rectangular base to an ellip- tical top 178 330 ARTICLE. The pattern for an article forming a transition from a rectangular base to a round top, the top not being centrally placed over the base 179 331 ARTICLE The pattern for an article rectangular at one end and round at the other, the plane of the round end being parallel to that of the rec- tangular end 182 337 ARTICLE. The pattern for an article with rectan- gular base and round top 177 328 ARTICLE. Pattern of an article having elliptical base and round top 185 343 BALL. The pattern for the miter between the moldings of adjacent gables upon a square shaft formed by means of a ball 73 176 BALL. To construct a ball in any number of pieces of the shape of gores 30 ]24 BALL. To construct a ball in any number of pieces of the shape of zones 124 351 BATH. Pattern for a hip bath of regular flare 143 373 BATH The patterns for a hip bath 172 313 BATHTUB The patterns for a bathtub 197 366 Problems. Page. BATHTUB. Pattern for the lining of the head of a bathtub 303 378 BIFURCATED PIPE. The patterns for a bifurcated pipe, the two arms being the same diameter as the main pipe and leaving it at the same angle 45 137 BLANK for a curved molding 126 254 BLANK. Pattern for the blank for a curved mold- ing in an arch in a circular wall 218 418 BLOWER FOR A GRATE. The pattern for a blower for a grate 73 180 BOILER COVER Pattern for a raised boiler cover with rounded corners 165 309 Boss. The pattern for a conical boss 141 271 Boss. The pattern for a boss to fit around a faucet 203 379 Boss. The pattern of a boss fitting over a miter in "a molding 35 120 BRACKET. The patterns for a raking bracket 88 193 BROKEN PEDIMENT. To obtain the profile and the patterns of the returns at the top and foot of a segmental broken pediment 94 204 BROKEN PEDIMENT. To obtain the profile of a horizontal return at the top of a broken pedi- ment necessary to miter with a given inclined molding and the patterns of both 93 202 BUTT MITER against a curved surface 5 99 BUTT MITER against an irregular or molded surface 8 103 BUTT MITER against a plain surface oblique in ele- vation i 97 BUTT MITER against a plain surface oblique in plan 2 97 BUTT MITER of a molding inclined in elevation against a plain surface oblique in plan 70 173 CAN Boss. Pattern for a can boss to fit around a faucet 79 131 CAPITAL. The construction of a volute for a capital 49 142 CHIMNEY TOP. The pattern for a chimney top 176 327 CIRCULAR WALL. Pattern for a splayed arch in a circular wall, the larger opening being on the inside of the wall 216 414 CIRCULAR WALL. Pattern for a splayed elliptical arch in a circular wall, the opening being larger on the outside of the wall than on the inside 215 412 422 Index of Probknls. Problems. Page. CIRCULAR WALL. Pattern of the blank for a curved molding in an arch in a circular wall 218 418 CIRCULAR WALL. Pattern for the soffit of an arch in a circular wall, the soffit being level at the top and the jambs of the opening being splayed on the insi !e 217 416 CIRCULAR WALL. Pattern for the soffit of a semi- circular arch in a circular wall, the soffit being level at the top and the jambs of the opening being at right angles with the walls in plan. Two cases 214 408 COAL HOD. The patterns for a funnel coal hod .... 21 1 399 COLD AIR Box. The patterns for a cold air box in which the inclined portion joins the level por- tion obliquely in plan. Two solutions 99 210 COLD AIR Box. The patterns for the inclined por- tion of a cold air box to meet the horizontal portion obliquely in plan ... 100 214 COLLAR. The pattern for a collar, round at the top and square at the bottom, to fit around a pipe passing through an inclined roof 180 333 COLLAR. The pattern for a flaring collar, the top and bottom of which are round and placed obliquely to each other 194 361 CONE. Pattern for the frustum of a cone fitting against a surface of two inclinations 139 269 CONK. Patterns of a cylinder joining a cone of greater diameter than itself at right angles to the side of the cone 159 296 CONES. The patterns of the frustums of two cones of unequal diameters intersecting obliquely 162 302 CONES. The patterns of two cones of unequal di ameter intersecting at right angles of their axes 161 300 CONE. The envelope of a frustum of an elliptical cone having an irregular base 174 322 CONE. The envelope of a frustum of a right cone 123 250 CONE. The envelope of a frustum of a right cone contained between planes oblique to its axis 144 274 CONE. The envelope of an elliptical cone 164 308 CONE. The envelope of a right cone 122 249 CONE. The envelope of a right cone cut by a plane parallel to its ax's 147 278 CONE. The envelope of a right cone whose base is oblique to its axis 136 265 CONE. The envelope of a scalene cone 163 306 CONE. The envelope of the frustum of a cone the base of which is an elliptical figure 132 260 CONE. The envelope of the frustum of a right cone the upper plane of which is oblique to its axis 135 265 CONE. The envelope of the frustum of a scalene 167 311 CONE. The frustum of a cone intersecting a cyl- inder of greater diameter than itself at other than right angles 155 290 Problems. Page. CONE. The frustum of a cone whose base is a true ellipse ...171 317 CONE. The pattern of a cone intersected by a cyl- inder at its upper end, their axes crossing at right angles 145 275 CONE. The pattern of a frustum of a cone inter- sected at its lower end by a cylinder, their axes intersecting at right angles 140 270 CONE. The pattern of a tapering article with equal flare throughout which corresponds to the frus- tum of a cone whose base is an approximate ellipse struck from centers, the upper plane of the frustum being oblique to the axis 1-16 276 CONE. The patterns of a cone intersected by a cyl- inder of less diameter than itself at right angles to its base, the axis of the cylinder being to one side of that of the cone 158 294 COSE. The patterns of a cone intersected by a cyl- inder of less diameter than itself, their axes crossing at right angles 157 29.3 CONE. The patterns of a cylinder joining the frus- tum of a cone in which the axis of the cylinder is neither at right angles to the axis nor to the side of the cone 160 298 CONE. The patterns of the frustum of a cone join- ing a cylinder of greater diameter than itself at other than right angles, the axis of the frus- tum passing to one side of that of the cylinder 156 291 CONE. The patterns of the frustum of a scalene cone intersected obliquely by a cylinder, their axes not lying in the same plane 175 CONICAL Boss. The pattern for a conical boss 141 CONICAL SPIRE. The pattern of a conical spire mitering upon eight gables 151 CONICAL SPIRE. The pattern of a conical spire mitering upon four gables 150 CORNER PIECE. Patterns for the corner piece of a mansard roof embodying the principles upon which all Mansard finishes are developed 11 COVER. The pattern for a raised boiler cover with rounded corners 165 COVER. The pattern of an oblong raised cover with semicircular ends 129 CRACKER BOAT. The pattern for a cracker boat 138 CURVED MOLDINGS The patterns for simple curved moldings in a window cap 127 CURVED MOLDING. Pattern of the blank for ;\ curved molding in an arch in a circular wall 218 CURVED MOLDING. The blank for a curved mold- ing 126 CURVED MOLDING. The pattern for the curved molding in an elliptical window cap 128 CURVES. To obtain the curves for a molding cov- ering the hip of a curved mansard roof Ill CYLINDER. Patterns of a cylinder joining a cone of greater diameter than itself at right angles to the side of the cone 159 323 271 282 280 10.V 309 253 267 265 418 25! 256 238 296 Index of Problems, 423 Problems. Page. CYLINDER. The pattern of a cone intersected by a cylinder at its upper end, their axes crossing at right angles 145 275 CYLINDER. The patterns of a cone intersected by a cylinder of less diameter than itself at right angles to its base, the axis of the cylinder being to one side of that of the cone 158 294 CYLINDER. The patterns of a cylinder joining the frustum of a cone in which the axis of the cyl- inder is neither at right angles to the axis nor to the side of the cone 160 2P8 CYLINDER. The patterns of a cylinder mitering with the peak of a gable coping having a double wash 69 171 CYLINDER. The patterns of the frustum of a scalene cone intersected obliquely by a cylinder, their axes not lying in the same plane 175 323 DECAGON. The patterns for a newel post, the plan of which is a decagon 22 117 DIAGONAL BRACKET The patterns for a diagonal bracket under the cornice of a hipped roof 90 198 DODECAGON. The patterns for an urn, the plan of which is a dodecagon 23 118 DORMER. Pattern for the molding on the side of a dormer, mitering against the octagonal side of a tower roof 65 165 DOUBLE ELBOW. The pattern for the intermediate piece of a double elbow joining two other pieces not lying in the same plane 53 147 DROP. The pattern for a drop upon the face of a bracket .24 119 ELBOW. Patterns for a right angle piece elbow to connect a round with a rectangular pipe 213 405 ELBOW. Patterns for a three-piece elbow in a tapering pipe 153 285 ELBOW. Patterns for a three-piece elbow to join a round pipe with an elliptical pipe 212 403 ELBOW. Patterns for a two-piece elbow in a taper- ing pipe 152 284 ELBOW. The pattern for the intermediate piece of a double elbow, joining two other pieces not lay- ing in the same plane 53 147 ELBOW. The patterns for a five-piece elbow 42 133 ELBOW. The patterns for a four-piece elbow 41 133 ELBOW. The patterns for an elbow at any angle 44 136 ELBOW. The patterns for a regular tapering elbow in five pieces 154 287 ELBOW. The patterns for a right an'gle two piece elbow, one end of which is round and the other elliptical 206 387 ELBOW. The patterns for a three-piece elbow 40 131 ELBOW. The patterns for a three-piece elbow, the middle piece of which tapers 191 355 ELBOW. The pattern for a two-piece elbow 38 130 Problems. Page. ELBOW. The patterns for a two-piece elbow in an elliptical pipe. Two cases 39 130 ELLIPTICAL CONE. The envelope of a frustum of an elliptical cone hiving an irregular base 174 822 ELLIPTICAL CONE. The envelope of an elliptical cone 164 308 ELLIPTICAL PIPE. Joint between an elliptical pipe and a round pipe of larger diameter at other than right angles. Two cases 58 155 ELLIPTICAL PIPE. The pattern of an elliptical pipe to fit against a roof of one inclination 32 125 ELLIPTICAL VASE. The patterns for an elliptical vase constructed in twelve pieces 81 183 EQUILATERAL TRIANGLE. A pattern for a pedestal of which the plan is an equilateral triangle 16 111 FACE AND SIDE. Patterns of the face and side of a plain tapering keystone 10 104 FACE MITER at right angles, as in the molding around a panel 12 106 FINIAL. The patterns for a finial the plan of which is an irregular polygon 82 185 FINIAL. The patterns of a finial the plan of which is octagon with alternate long and short sides 85 189 FIVE-PIECE ELBOW. The patterns for a five-piece elbow 43 133 FLANGE. A conical flange to fit around a pipe and against a roof of one inclination 137 266 FLANGE. The pattern for a flaring flange round at the bottom, the top to fit a round pipe passing through an inclined roof 195 363 FLANGE. The pattern for a flaring flange to fit a round pipe passing through an inclined roof . the flange to have an equal projection from the pipe on all sides 193 363 FLANGE. The pattern for a pyramidal flange to fit against the sides of a round pipe which passes through its apex 50 144 FLANGE. The pattern of a flange to fit around a pipe and against a roof of one inclination 37 129 FLANGE. The pattern of a flange to fit around a pipe and over the ridge of a roof 36 128 FLARING ARTICLE. Pattern for an irregular flaring article which is elliptical at the base and round at the top, the top being so situated as to be tan- gent to one end of the base when viewed in plan . . . 186 345 FLARING ARTICLE. Pattern for an irregular flaring article whose top is a circle and whose base is a quadrant 190 354 FLARING ARTICLE. Pattern of an irregular flaring article both top and bottom of which are round and parallel but placed eccentrically in plan. Otherwise the envelope of the frustum of a sca- lene cone 167 311 FLARING ARTICLE. The pattern for an irregular flaring article, elliptical at the- base and round 424 Index of Problems. Problems. Page. at the top, the top and bottom not being paral- lel 196 364 FLARING ARTICLE. The pattern for a flaring ar- ticle or transition piece, round at the top and oblong at the bottom, the two ends being con- centric in plan 187 346 FLARING ARTICLE. The pattern for a flaring article, round at the top and bottom, the top being placed to one side of the center as seen in plan 183 339 FLARING ARTICLE. The pattern for a flaring ar- ticle round at the base and square at the top 181 335 FLARING ARTICLE. The pattern for a flaring ar- ticle, round at top and bottom, one side being vertical 184 341 FLARING ARTICLE. The pattern of a flaring article of which the base is an oblong and top square 74 177 FLARING ARTICLE. The pattern of a rectangular flaring article 9 104 FLARING ARTICLE. The pattern of a regular flar- ing article which is oblong with semicircular ends 130 258 FLARING ARTICLE. The pattern of a flaring article, the top of which is round and the bottom ob- long, with semicircular end. Two cases 168 312 FLARING ARTICLE. The pattern of a flaring article which corresponds to the frustum of a cone whose base is a true ellipse 171 317 FLARING ARTICLE. The pattern of a flaring article which is rectangular, with rounded corners, having more flare at the ends than at the side? 170 316 FLARING COLLAR The pattern for a flaring collar, the top and bottom of which are round and placed obliquely to each other 194 361 FLARING END. The pattern of the flaring end of an oblong tub 76 178 FLARING FLANGE. The pattern for a flaring flange, round at the bottom, the top to fit a round pipe passing through an inclined roof 195 363 FLARING FLANGE. The pattern for a flaring flange to fit a round pipe passing through an inclined roof, the flange to have an equal projection from the pipe on all sides 198 368 FLARING OBLONG ARTICLE. The pattern of a reg- ular flaring oblong article with round corners 131 259 FLARING PAN. The pattern of an oval or egg shaped flaring pan 134 263 FLARING SECTION. The pattern of the flaring sec- tion of a locomotive boiler 77 179 FLARING SHAPE. Pattern for an irregular flaring shape forming a transition from a round hori- zontal base to a round top placed vertically 201 376 FLARING TRAY. The pattern of a heart-shaped flaring tray 133 261 FLARING TUB. The patterns of a flaring tub with tapering sides and semicircular head, the head having more flare than the sides 169 314 Problems. Page. FLOAT. The patterns for a soap makers' float 173 320 FORGE. Pattern for the hood of a portable forge 199 371 FORK. Pattern for a three-prong fork with taper- ing branches 2' 8 391 FOUR-PIECE ELBOW. The patterns for a four-piece elbow 41 132 FOUR- PIECE ELBOW.- The patterns for a pipe inter- secting a four-piece elbow through one of the miters 61 159 FOUR-SIDED FIGURE. The patterns of a molding mitering around an irregular four-sided figure 14 109 FRUSTUM OF A CONE intersecting a cylinder of greater diameter than itself at other than right angles 155 290 FRUSTUM OP A CONE. The envelope of the frustum of a cone the base of which is an elliptical figure. . .132 260 FRUSTUM OF A CONE whose base is a true ellipse 171 317 FRUSTUM OF A PYRAMID. The envelope of the frus- tum of a pyramid which is diamond shape in plan 75 178 FRUSTUM OF A RIGHT CONE. The envelope of th<> frustum of a right cone the upper plane of which is oblique to its axis 135 265 FRUSTUM. Pattern of a tapering article with equal flare throughout which corresponds to the frus- tum of a cone whose base is an approximate el- lipse struck from centers the upper plane of the frustum being oblique to the axis 146 276 FRUSTUMS. The patterns of the frustums of two cones of unequal diameters intersecting obliquely. . . 162 302 FRUSTUM. The envelope of a frustum of an ellip- tical cone having an irregular base 174 322 FRUSTUM. The envelope of a frustum of a right cone 123 250 FRUSTUM. The envelope of a frustum of a right cone contained between planes oblique to its axis 144 274 FRUSTUM. The envelope of the frustum of an octag- onal pyramid 116 243 FRUSTUM. The envelope of the frustum of a scalene cone , 167 311 FRUSTUM. The envelope of the frustum of a square pyramid 1 15 242 FRUSTUM. The pattern for the frustum of a cone fitting against a surface of two inclinations 139 269 FRUSTUM. The pattern of a frustum of a cone in- tersected at its lower end by a cylinder, their axes intersecting at right angles 140 270 FRUSTUM. The patterns of the frustum of a cone joining a cylinder of greater diameter than itself at other than right angles the axis of the frus- tum passing to one side of that of the cylinder 156 291 FRUSTUM. The patterns of the frustum of a scalene cone intersected obliquely by a cylinder, their axes not lying in the same plane 175 323 Index of Problems. 425 Problems. FUNNEL COAL HOD. The patterns for a funnel coal hod on GABLE COPING. The patterns for a square shaft of curved profile, mitering over the peak of a gable coping, having double wash 68 GABLE COPING. The patterns of a cylinder miter- ing with the peak of a gable coping having a double wash 69 GABLE CORNICE.. The pattern for a gable cornice mitering upon an inclined roof 64 GABLE MITERS. The patterns of simple gable miters 15 GABLE MOLDING. The pattern for a gable molding mitering against a molded pilaster 62 GORE PIECE. The pattern for a gore piece forming- a transition from an octagon to a square, as at the end of a chamfer 86 191 Page. 399 169 171 164 110 161 GORE PIECE. The pattern for a gore piece in a molded article forming the transition from a square to an octagon 87 192 GORES. To construct a ball in any number of pieces of the shape of gores 30 134 GORE. The pattern for a three piece elbow, the middle piece being a gore S3 1^7 GRATE. The pattern for a blower for a grate 78 ISO HEXAGONAL PYRAMID. The envelope of a hexag- onal pyramid H4 341 HEXAGON. The pattern for a pedestal the plan of which is a hexagon 19 114 HEPTAGON. The pattern for a vase the plan of which is a heptagon. ... 20 115 HIP BAR. Patterns for the top and bottom of the hip bar in a skylight ]03 219 HIP BATH. Pattern for a hip bath of regular flare. . . . 143 273 HIP BATH. The patterns for a hip bath 172 313 HIP FINISH. The pattern for a hip finish in a curved mansard roof, the plan of the hip being a right angle fi IQI HIP MOLDING. Pattern for a hip molding mitering against the planceer of a deck cornice on a man- sard roof which is square at the eaves and octa- gon at the top 107 227 HIP MOLDING. The pattern for a hip molding upon a right angle in a mansard roof mitering against a bd molding at the top 102 217 HIP MOLDING. Patterns for the fasdas of a hip molding finishing a curved mansard roof which is square at the base and octagonal at the top no 236 HIP MOLDING. Patterns for a hip molding miteriug against the bed molding of a deck cornice on a mansard roof which is square at the base and octagonal at the top 108 HIP MOLDING. The pattern of a hip molding upon an octagon angle of a mansard roof mitering upon an inclined wash at the bottom 106 230 225 Problems. Page. HIP MOLDING. The pattern of a hip molding upon a right angle in a mansard roof mitering against, the planceer of a deck cornice 101 215 HIP MOLDING. The patterns for the miter at the bottom of a hip molding on a mansard roof which is octagon at the top and square at the bottom 109 233 HIP MOLD. The pattern of a hip mold upon an oc- tagon angle in a mansard roof mitering against a bed molding of corresponding profile 105 223 HOOD. Pattern for the hood of a portable forge 199 371 HOOD. The patterns for the hood of an oil tank 200 374 HORIZONTAL MOLDING. From the profile of a given horizontal molding to obtain the profile of an in- clined molding necessary to miter with it at an octagon angle in plan, and the patterns for both arms of the miter 95 205 HORIZONTAL MOLDING. From the profile of a given inclined molding to establish the profile of a horizontal molding to miter with it at an octa- gon angle in plan, and the patterns for both arms. . ; .96 206 HORIZONTAL RETURN. To obtain the profile of a horizontal return at the foot of a gable necessary to initer at right angles in plan witli an inclined molding of normal profile, and the miter patterns of both 91 200 HORIZONTAL RETURN. To obtain the profile of a horizontal return at the top of a broken pedi- ment necessary to miter with a given inclined molding, and the patterns of both 93 202 INCLINED MOLDING. From the profile of a given horizontal molding to obtain the profile of an in- clined molding necessary to miter with it at an octagon angle in plan, and the patterns for both arms of the miter 95 305 , INCLINED MOLDING. The pattern for an inclined molding mitering upon a wash including a re- turn ............................................. 66 166 INCLINED MOLDLNG.--TO obtain the profile of an in- clined molding necessary to miter at right an gles in plan with a given horizontal return, and the miter patterns of both ........................ 93 OQI INCLINED ROOF. The pattern for a gable cornice mitering upon an inclined roof .................... 64 164 JACK BAR. Pattern for the top of a jack bar in a skylight ......................................... 104 221 JOINT at other than right angles between two pipes of different diameters, the axis of the smaller pipe being placed to one side of that of the larger one ......................................... 60 158 JOINT between an elliptical pipe and a round pipe of larger diameter at other than right angles. Two cases ........................................ 53 155 JOINT between two pipes of different diameters in- tersecting at other than right angles ............... 57 154 JOINT between two pipes of the same diameter at other than right angles ............................ 54 IQI 426 Index of Prollefns. Problems. Page. JOINT. The patterns for a T- joint between pipes of the same diameter 47 139 JUNCTION. Patterns for the junction of a large pipe with the elbows of two smaller pipes of the same diameter 205 384 KEYSTONE. The patterns for a keystone having a molded face with sink 26 120 LEVEL MOLDING. The pattern for a level molding mitering obliquely against another level mold- ing of different profile 67 167 LIP. Pattern for the lip of a sheet metal pitcher 142 272 LOCOMOTIVE BOILER. The pattern for the flaring section of a locomotive boiler 77 179 MANSARD ROOF. Pattern for a hip molding miter- ing against the planceer of a deck cornice on a mansard roof which is square at eaves and oc- tagon at the top 107 227 MANSARD ROOF. Patterns for a hip molding miter- ing against the bed molding of a deck cornice on a mansard roof which is square at the base and octagonal at the top 108 230 MANSARD ROOF. Patterns for the corner piece of a inaneard roof embodying the principles upon which all mansard finishes are developed 11 105 MANSARD ROOF. The pattern for a hip finish in a curved mansard roof the plan of the hip being aright angle 6 101 MANSARD ROOF. The pattern for a hip molding upon a right angle in a mansard roof mitering against a bed molding at the top ;102 217 MANSARD ROOF. The pattern of a hip molding upon an octagon angle of a mansard roof mitering upon an inclined wash at the bottom 106 225 MANSARD ROOF. The pattern of a hip molding upon a right angle in. a mansard roof mitering against the planceer of a deck cornice 101 215 MANSARD ROOF. The pattern of a hip mold upon an octagon angle in a mansard roof mitering against a bed molding of corresponding profile 105 223 MANSARD ROOF. The patterns for the fascias of a hip molding finishing a curved mansard roof which is square at the base and octagonal at the top. MANSARD ROOF. The patterns for the miter at the bottom of a hip molding on a mansard roof which is octagon at the top and square at the bottom 109 MANSARD ROOF. To obtain the curves for a mold- ing covering the hips of a curved mansard roof. ... Ill MITER. A butt miter against a curved surface 5 MITER. A butt miter against an irregular or molded surface .110 236 MITER. A butt miter against a plain surface oblique in elevation 8 .1 MITER. A butt miter against a plain surface oblique in plan 233 238 99 103 97 97 Problems. Page. MITER. A face miter at right angles, as in the mold- ing around a panel 12 106 MITER. A return miter at other than a right angle, as in a cornice at the corner of a building 4 98 MITER. A square return miter or a miter at right angles, as in a cornice at the corner of a build- ing 3 98 MITER between two moldings of different profiles 7 102 MITER LINE. To obtain the miter line and pattern for a straight molding meeting a curved mold- ing of the same profile 55 153 MITERS. Patterns of simple gable miters 15 110 r.IiTER. The pattern for the miter between the moldings of adjacent gables upon a square shaft formed by means of a ball 73 176 MOLDED BASE. The patterns for a molded base in which the projection of the sides is different from that of the ends 80 182 MOLDED FACE. The patterns for a keystone having a molded face 26 120 MOLDED PILASTER. The pattern f or a gable molding mitering against a molded pilaster fi2 161 MOLDING. A butt miter of a molding inclined in el- evation against a plain surface oblique in plan 70 173 MOLDINGS. Miter between two moldings of differ- ent profiles 7 102 MOLDING. Pattern for the molding on the side of a dormer mitering against the octagonal side of a tower roof 65 165 MOLDINGS. Pattern for the moldings and roof pieces in the gables of an octagon pinnacle' 72 174 MOLDINGS. Patterns for the moldings and roof pieces in the gables of a square pinnacle ... 71 173 MOLDING. The patterns of a molding mitering around an irregular four-sided figure 14 109 MOLDINGS. The patterns of the moldings bound- ing a panel triangular in shape 13 108 MOLDING. To obtain the niiter line and pattern for a straight molding meeting a curved molding of the same profile 55 15'3 NEWEL POST. The patterns for a newel post the plan of which is a decagon 22 117 OBLONG TUB. The pattern of the flaring end of an oblong tub 76 178 OBLONG VESSEL. Pattern for the end of an oblong vessel which is semicircular at the top and rec- tangular at the bottom 166 310 OCTAGONAL PEDESTAL. The patterns for an octag- onal pedestal 21 115 OCTAGONAL PYRAMID. Envelope of the frustum of an octagonal pyramid having alternate long and short sides 117 244 OCTAGONAL PYRAMID. The mvelopeof the frustum of an octagonal pyramid 116 243 Index of Problems, 427 Problems. OCTAGONAL SHAFT. The patterns of an octagonal shaft the profile of which is curved fitting over the ridge of a roof 29 OCTAGON PINNACLE. Pattern for the moldings and roof pieces in the gables of an octagon pinnacle 72 OCTAGON PINNACLE. The miter between the mold- ings of adjacent gables upon an octagon pinna- cle 98 OCTAGON SHAFT mitering upon the ridge and hips of a roof 35 OCTAGON SHAFT. The pattern of an octagon shaft fitting over the ridge of a roof 33 OCTAGON SHAFT. To describe the pattern of an oc- tagon shaft tofit against a ball 28 OCTAGON SPIRE. The pattern for an octagon spire mitering upon a roof at the junction of the ridge and hips 121 OCTAGON SPIRE. The pattern of an octagon spire mitering upon eight gables 119 OCTAGON SPIRE. The pattern of an octagon spire mitering upon four gables 120 OCTAGON.. The patterns of a finial the plan of which is octagon with alternate long and short sides 85 OFFSET. Pattern for an offset to join a round pipe with one of elliptical profile 210 OFFSET. The pattern for an offset between two pipes, oblong in section, whose long diameters meet at right angles to each other 1 89 OFFSET. The pattern for an offset to join an oblong pipe with a round one 209 PAX. The pattern of an oval or egg-shaped flaring pan 134 PANEL. Patterns of the moldings bounding a panel triangular in shape 13 PEDESTAL. The pattern for a pedestal of which the plan is an equilateral triangle 16 PEDESTAL. The pattern for a pedestal square in plan 17 PEDESTAL. The pattern for a pedestal the plan of which is a hexagon 19 PEDESTAL. The patterns for an octagonal pedestal 21 PENTAGON. The patterns for a vase the plan of which is a pentagon 18 PINNACLE. -^ The miter between the moldings of ad- jacent gables of different pitches upon a pin- nacle wHli rectangular shaft 97 p IPE s._ A joint between two pipes of the same di- ameter at other than right angles 54 PIPES. A T-joint between pipes of different diam- eters i 56 PIPES. A T-joint between pipes of different diam- eters, the axis of the smaller pipe passing to one side of that of the larger 59 Page. 128 174 208 127 126 122 248 245 246 189 396 350 393 263 108 111 113 114 115 112 207 151 153 156 Problems. PIPES. Joint at other than right angles between two pipes of different diameters, the axis of the smaller pipe being placed to one side of that of the larger one 60 PIPES. The joint between two pipes of different diameters intersecting at other than right an- gles 57 PIPE. The pattern of an elliptical pipe to fit against a roof of one inclination 32 PIPE. The pattern of a flange to fit around a pipe and against a roof of one inclination 37 PIPE. The pattern of a flange to fit around a pipe and over the ridge of a roof 36 PIPE. The pattern of a round pipe to fit against a roof of one inclination 31 158 154 125 129 128 125 127 137 134 159 146 253 140 272 185 PIPE. The pattern of a round pipe to fit over the ridge of a roof 34 PIPE. The patterns for a bifurcated pipe, the two arms being of the same diameter as the main pipe and leaving at same angle 45 PIPE. The patterns for a pipe carried arountt a semicircle by means of cross joints 43 PIPE. The patterns for a pipe intersecting a four- piece elbow through one of the miters 61 PIPE. The patterns for a rectangular pipe intersect- ing a cylinder obliquely 52 PIPE. The patterns for a semicircular pipe with longitudinal seams 125 PIPE. The patterns for a square pipe describing a twist or compound curve 48 PITCHER. Pattern for the lip of a sheet metal pitcher ... 1 42 POLYGON. The patterns for a finial the plan of which is an irregular polygon 82 PROFILE. From the profile of a given horizontal molding to obtain the profile of an inclined mold- ing necessary to miter with it at an octagonal angle in plan and the patterns for both arms of the miter 95 205 PROFILE. From the profile of a given inclined mold- ing to establish the profile of a horizontal mold- ing to miter with it at an octagon angle in plan and the patterns for both arms 96 206 PROFILE. To obtain the profile and patterns of the returns at the top and foot of a segmental broken pediment 94 204 PROFILE. To obtain the profile of a horizontal re- turn at the foot of a gable necessary to miter at right angles in plan with an inclined molding of normal profile, and the miter patterns of both 91 200 PROFILE. To obtain the profile of a horizontal re- turn at the top of a broken pediment necessary to miter with a given inclined molding, and the patterns of both . . ." 93 202 PROFILE. To obtain the profile of an inclined mold- ing necessary to miter at right angles in plan 428 ProbUms. with a given horizontal return, and the miter patterns of both 93 PYRAMIDAL FLANGE. The pattern for a pyramidal flange to fit against the sides of a round pipe which passes through its apex r>"i PYRAMID. The envelope of the frustum of a square pyramid 1 15 PYRAMID. The envelope of a hexagonal pyramid 1 14 PYRAMID The envelope of a square pyramid 1 1 3 PYRAMID. The envelope of a triangular pyramid 1:2 PYRAMID. The envelope of the frustum of an octag- onal pyramid 1 K; PYRAMID. The envelope of the frustum of an oc- tagonal pyramid having alternate long and short sides 117 PYRAMID. The evelope of the frustum of a pyra- mid which is diamond shape in plan 7.3 PYRAMID. The patterns for a square pyramid to fit against the sides of an elliptical pipe which passes through its center 51 RAKING BRACKET. The patterns for a raking bracket 88 RAKING BRACKET. The patterns for a raking bracket in a curved pediment 192 RAISED PANEL The pattern for a raised panel on the face of a raking brack-t 89 RECTANGULAR FLARINO ARTICLE. -The pattern of a rectangular flaring article. ... 9 RECTANGULAR PIPE. -The patterns for a rectan- gular pipe intersecting a cylinder obliquely 53 RETURN MITER at other than a right angle, as i i a cornice at the corner of a building I RETURNS. To obtain the profile and patterns of the returns at the top and foot of a segmental broken pediment. 9j RIGHT CONE. The envelope of a right co:io 123 RIGHT CONE. The envelope of a right cone cut by a plane parallel to its axis 147 RIGHT CONE. The envelope of a frustum of a right cone .03 RIGHT CONE. The envelope of a frustum of a right cone contained between planes oblique to its 144 of Problems. Page. 201 141 242 241 241 240 24:! 241 178 RIGHT CONE. The envelope of a right cone whose base is oblique to its axis 136 RIGHT CONE. The envelope of the frustum of a right cone the upper- plane of which is oblique to its axis 135 ROUND PIPE. The pattern of a round pipe to Ft against a roof of one inclination 31 ROUND PIPE. The pattern of a round pipe to fit over the ridge of a roof 34 SCALENE CONE. The envelope of a scalene cone 163 274 265 125 127 306 lit:! 358 190 104 140 98 204 219 Problems. Page. SCALENE CONE The envelope of the frustum of a scalene cone 1(57 311 SCALENE CONE. The patterns of the frustum of a scalene cone intersected obliquely by a cylinder, their axes not lying in the same plane 175 323 SCALE SCOOP. The pattern for a scale scoop having both ends alike 143 373 SCALE SCOOP. The patterns for a scale sci, () p, one end of which is funnel shaped 149 379 SEGME.NTAL BROKEN PEDIMENT. To obtain the pro- file and patterns of the returns at the top and foot of a segmental broken pediment 94 304 SHIP VENTILATOR. Patterns for a ship ventilator having a round base and an elliptical mouth 20 1 381 SIDE Patterns of the face and side of a plain taper- ing keystone 10 104 SKYLIGH r BAR. The patterns for the top and bot- tom of a " common " skylight bar 46 138 SKYLIGHT. -Pattern for the top of a jack bar in a skylight 11,4 2 21 SKYLIGHT. The patterns for the top and bottom of the hip bar in a skylight 103 219 SOAP MAKKRS' FLOAT. The patlerns for a soap makers' float 173 330 SOFFIT. Pattern for the soffit of an arch in a circu- lar wall, the soffit being level at the top and the jambs of the opening being splayed on the inside. . .217 416 SOFFIT. Pattern for the soffit of a semicircular arch in a circular wall, the soffit being level at the top and the jambs of the opening being at right angles to the walls in plan. T\\>> cases SPLAYED ARCH. Pattern for a splayed arch in a circular wall, the larger opening being on the inside of the wall 315 SPLAYED ELLIPTICAL ARCH. P. it tern for a splayed elliptical arch in a circular wall, the opening be- ing larger on the outside of the wall tlian on the inside 315 SPIRE. The pattern for an octagon spire initering upon a roof at the junction of the ridge and hips 121 HPIRE. The pattern of a conical spire miterin.,' upon four gables 150 SPIRE. The pattern of a conical spire initering upon eight gables ir,i SPIRE. The pattein of an octagon spire initering upon eight gables 119 SPIRE. The pattern of an octagon spire initering upon four gables 130 SPIRE The pattern of a 1-qiia.re spire initering upon four gables .118 SQUARE PINNACLE. Patterns Tor the moldings and roof pieces in the gables of a square pinnacle 71 SQUARE PIPE. The patterns for a square pipe de- scribing a twist or compound curve 49 214 408 414 412 946 280 282 245 246 244 173 140 In'!'. i- i if 429 Problems. Page. SQUARE PYRAMID. The envelope of a square pyra- mid 113 241 SQUAKE PYRAMID. The envelope of the frustum of a square pyramid 115 242 SQUARE PYRAMID. The patterns for a square pyra- mid to fit against the sides of an elliptical pipe which passes through its center 51 145 SQUARE RETURN MITER, or miter at right angles, as in a cornice at the corner of a building 3 98 SQUARE SHAFT. The patterns for a square shaft of curved profile mitering over the peak of a gable coping having a double wash 68 169 SQUARE SHAFT. The pattern of a square shaft to fit against a sphere 27 122 SQUARE SPIRE. The pattern of a square spin' miter- ing upon four gables 118 244 TASK. The patterns for the hood of an oil tank 200 374 TAPERING ARTICLE. The patterns of a tapering arti- cle which is square at the base and octagonal at the top 84 188 TAPERING ARTICLE. The pattern of a tapering arti- cle with equal flare throughout which corre- sponds to the frustum of a cone whose base is an approximate ellipse struck from centers, the upper plane of the frustum being oblique to the axis 146 276 TAPERING ELBOW. The patterns for a regular ta- pering elbow in five pieces 154 287 THREE-PIECE ELBOW. Patterns for a three piece elbow in a tapering pipe 153 285 THREE-PIECE ELBOW. --Patterns for a three-piece elbow to join a round pii>e with an elliptical pipe 212 403 THREE PIECE ELBOW. The pattern fora three-piece elbow, the middle piece being a gore 83 187 THREE-PIECE ELBOW. The patterns for a three- piece elbow 40 131 THREE-PIECE ELBOW. The patterns for a three- piece elbow the middle piece of which tapers 191 355 THREE-PRONG FORK. Pattern for a three-prong fork with tapering branches 208 391 T-Joint between pipes of different diamaters 56 153 T-Joiut between pipes of different diameters, the axis of the smaller pipe passing to one side of that of the larger 59 156 T- Joint. The patterns for a T-joint between pipes of the same diameters 47 139 TRANSITION from a round horizontal base to a round top placed vertically 201 376 TRANSITION PIECE. Pattern for a transition piece round at the top and oblong at the bottom. Two cases 188 348 Problems. Page. TRANSITION PIECE. Pattern for a transition piece to join two rou:i 1 pipes of unequal dianr-terat an angle 193 359 TRANSITION PIECE. The pattern for a flaring article or transition piece round at the top and oblong at the bottom, the two ends being concentric in plan 187 346 TRANSITION. The pattern for a gore piece forming a transition from an octagon to a square, as at the end of a chamfer 86 191 TRANSITION. The pattern for a gore piece in a molded article forming a transition from a square to an octagon 87 192 TRANSITION. The pattern for an article forming a transition from a rectangular base to an ellipti-' cal top 178 330 TRANSITION The pattern for an article forming a transition from a rectangular base to a round top. the top not being centrally placed over the base 179 331 TRAY. The pattern of a heart-shaped flaring tray 138 261 TRIANGULAR PYRAMID. The envelope of a trian- gular pyramid 112 240 TWO-PIECE ELBOW. Patterns for a two-piece el- bow in a tapering pipe 152 384 Two PIECE ELBOW. The pattern for a two piece elbow 38 130 TWO-PIECE ELBOW. The patterns for a two-piece elbow in an elliptical pipe. Two cases 39 130 TWO-PIECE ELBOW. The patterns for a right angle two piece elbow one end of which is round and the other elliptical 206 387 URN. The patterns for an urn the plan of which is a dodecagon 23 118 VASE. The pattern for a vase the plan of which is a heptagon 20 115 VASE. The pattern for a vase the plan of which is a pentagon 18 112 VENTILATOR. Patterns for a ship ventilator having a round base and an elliptical mouth 204 881 VOLUTE. The construction of a volute for a capital 49 142 WASH. The pattern for an inclined molding miter- ing upon a wash, including a return 66 166 WINDOW CAP. The pattern for the curved molding in an elliptical window cap 128 256 WINDOW CAP. The patterns for simple curved moldings in a window cap 127 255 Y. The patterns for a Y consisting of two tapering pipes joining a larger pipe at an angle 207 389 ZONES. To construct a ball in any number of pieces of the shape of zones 124 251 YH ! 8 ! 5 I*t5*1 mat I I ^^^^^^^^f*\ ?* ~ * , ,