\\:^X' -yy. ^"A Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/descriptivegeomeOOanthrich TECHNICAL DRAWING SERIES DESCRIPTIVE GEOMETRY BY GARDNER C. ANTHONY, Sc.D. II GEORGE F. ASHLEY OF T-HE UNIVERSITY sEdi-IFORNil BOSTON, U.S.A. D. C. HEATH & CO^ PUBLISHERS 1910 GENERAL Copyright, 1909, By D. C. Heath & Co. PREFACE An extended experience in engineering practice and teaching has emphasized the importance of certain methods for the presen- tation of principles and problems in graphics which the authors have gradually' developed into their present form. Most of the subject- matter which we now publish has been pre- sented in the form of notes and used during the past eight years, and nearly all of the problems have received the critical test of the classroom. Previous to the preparation of the notes, which were made the foundation of this treatise, the third angle of projection had been adopted for problems, and almost exclusively used, as conforming more nearly to engineering practice. It has been the aim of the authors to make a clear and concise statement of the princi- ples involved, together with a brief analysis and enumeration of the steps to be taken, so that the essentials of each problem shall be clearly set forth. The illustrations have been chosen with care and ari-anged so as to appear opposite the descriptive text. Too much stress cannot be laid on the im- portance of preparing problems with such care that they may clearly illustrate the principles involved. In general it should not be neces- sary for the student to prescribe the condi- tions governing the relations of points, lines, and surfaces because of the time consumed and the probable failure to bring out the salient features of the problem. The graphic presentation of the problems to be solved should facilitate the lay-out by the student, and enable. the instructor to judge quickly of their character and adaptability to the special cases under consideration. Two sets of prob- lems have been prepared to illustrate most cases, and the number may be further in- 196407 IV PREFACE creased by reversing, or inverting, the illus- trations given. Tlie unit of measurement may also be changed to adapt the problems to any chosen size of plate, and if the propor- tion 1)3 maintained it will be possible to solve the problems within the given space. If it is desired to change the assignment annuall}-, the even numbers may be used for one year and the odd numbers for the following year. It is hoped that the number of problems is sufficiently great to admit of considerable va- riety in the course. Tufts College, January, 190*.). Although the authors believe that an ele- mentary course in tlie orthographic and iso- metric projections of simple objects should precede the more analytical consideration of line and plane, as herein presented, yet this treatise has been proven adequate to meet the demand of those who have not received such preparation. By thus making it available for students of science and mathematics, as well as for engineers, it is hoped to promote a more general knowledge of the grammatical con- struction of Graphic Language. GARDNER C. ANTHONY GEORGE F. ASHLEY. TABLE OF CONTENTS ART. 1. 2. 3. 4. 9. 10. CHAPTER I Definitions and First Principles Descriptive Greometry Projection Coordiuate Planes . Quadrants or Angles Orthographic Projectioi Notation . Points Lines Line parallel to a coordinate plane Line perpendicular to a coordinate plane 1 n. Line lying in a coordinate plane 1 , 12. Line parallel to a coordinate plane 2 13. Lines parallel in space 3 14. Lines intersecting in space 3 1.5. Lines intersecting the ground line 4 16. Traces of a line 6 17. To define the position of a line 7 18. Planes 8 19. GL the trace of V and H 8 8 8 8 10 10 10 10 12 12 20. 21. 22. ii! 24. 2.5. 26. CHAPTER n Points, Lines, and Planes Operations required for the solution of prob- lems To determine three projections of a line Projection of point in 2 Q Revolving of P To determine the traces of a line . To determine the traces of a line parallel to P To determine the projections of a line when its traces are given 14 14 15 15 16 18 18 27. Conditions governing lines lying in a plane . 19 28. Conditions governing lines lying in a plane and parallel to //or F 29. An infinite number of planes may be passed through any line . . . ' . 30. To pa.ss a plane through two intersecting or parallel lines. Case 1 20 31. To pass a plane through two intersecting or parallel lines, Case 2 20 19 20 VI TABLE OF CONTENTS ART. 32. To pass a plane through two intersecting or parallel lines. Case 3 33. To pass a plane through a line and a point . 34. To pass a plane through three points not in the same straight line 35. Given one projection of a line lying on a plane, to determine the other projection 36. Given one projection of a point lying on a plane, to determine the other projection 37. To locate a point on a given plane at a given distance from the coordinate planes . ^38. To revolve a point into either coordinate plane 39. To determine the true length of a line. Case 1 40. To determine the true length of a line, Case 2 41. Relation of the revolved position of a line to its trace 42. The revolved position of a point lying in a plane 43. The revolved position of a line lying in a plane 44. To determine the angle between two inter- secting lines ....... 45. To draw the projections of any polygon . 46. Counter-revolution, Construction 1 . 47. Counter-revolution, Construction 2 . 48. Counter-revolution, Construction 3 .• 49. To determine the projections of the line of in- tersection between two planes. Principle . 50. To determine the projections of the line of in- tersection between two planes. Case 1 . 51. To determine the projections of the line of in- tersection between two planes. Case 2 •52. To determine the projections of the line of intersection between two planes when two traces are parallel \(.K AUT. 53. 21 22 54. 22 55. 22 24 .56. 24 57. 25 27 58. 28 59. 28 30 60. 31 61. 31 32 62. 32 33 34 63. 64. 35 65. 35 66. 36 67. 68. 37 To determine the projections of the line of in- tersection between two planes when all traces meet in (jL 37 To determine the projections of the line of in- tersection between two planes. Case 3 . 38 To determine the projections of the line of in- tersection between two planes when one plane contains (iL 38 Revolution, Quadrants, and Counter-revolu- tion 40 To determine the point in which a line pierces a plane. General Method .... 42 To determine the point in which a line pierces a plane. Case 1 .42 To determine the point in wliicii a line pierces a plane. Case 2 42 To determine the point in which a line piel-ces a plane. Case 3 . . . . . .42 To determine the point in which a line pierces a plane. Case 4 44 If a right line is perpendicular to a plane, the projections of that line will be peipendicu- lar to the traces of tlie plane ... 44 To project a point on to an oblique plane . 44 To project a given line on to a given oblique plane 45 To determine the shortest distance from a point to a plane . . . . . .45 Shades and Shadows 46 To determine the shadow of a point on a given surface 47 To determine the shadow of a line upon a given surface 47 TABLE OF CONTEXTS Vll ART. PAGE 69. To determine the shadow of a solid upon a given surface '48 ' 70. Through a point or line to pass a plane hav- ing a defined relation to a given line or plane .50 ' 71. To pass a plane through a given point parallel to a given plane 50 ^ 72. To pass a plane through a given point perpen- dicular to a given line 50 73. To pass a plane through a given point parallel to two given lines 51 74. To pass a plane through a given line parallel to, another given line 52 -475. To pass a plane through a given line pei-pen- dicular to a given plane .... 76. Special conditions and methods of Art. 70 77. To pass a plane through a given point perpen- dicular to a given line. Special case . 78. To pass a plane through a given line perpen- dicular to a given plane. Special case 79. To determine the projections and true length of the line measuring the shortest distance between two right lines not in the same plane . . . . . . . . .54 .—480. To determine the angle between a line and a plane 54 52 52 5.3 81 . To determine the angle between a line and the coordinate planes 82. To determine the projections of a line of defi- nite length passing through a given point and making given angles with the coordi- nate planes 56 83. To determine the angle between two planes. Principle 57 84. To determine the angle between two oblique planes 85. To determine the angle between two oblique planes by perpendiculars .... 86. To determine the angle between an inclined plane and either coordinate plane 87. To determine the bevels for the correct cuts, the lengths of hip and jack rafters, and the bevels for the purlins for a hip roof . 88. Given one trace of a plane, and the angle be- tween the plane and the coordinate plane, to determine the other trace. Case 1 . 89. Given one trace of a plane, and the angle be- tween the plane and the coordinate plane, to determine the other trace, Case 2 . 90. To determine the traces of a plane, knowing the angles which the plane makes with both coordinate planes 91. Fourth construction for counter-revolution 55^ -„^ 57 58 58 60 62 62 CHAPTER ni Generation and Classification of Surfaces 92. Method of generating surfaces 93. Classification of Surfaces 65 65 94. Ruled Surfaces 95. Plane Surfaces 65 66 Vlll TABLE OF CONTENTS 96. Single-curved Surfaces 97. Conical Surfaces . 98. Cylindrical Surfaces 99. Convolute Surfaces 66 100. A Warped Surface 66 101. Types of Warped Surfaces 67 102. A Surface of Revolution 67 103. Double-curved Surfaces . PAGE 68 68 70 70 CHAPTER IV Tangent Planes 104. Plane tangent to a single-curved surface . 72 105. One projection of a point on a single-curved surface being given, it is required to paas a plane tangent to the surface at the ele- ment containing the given point . . 73 106. To pass a plane tangent to a cone and through a given point outside its surface . . 74 107. To pass a plane tangent to a cone and parallel to a given line . . . . . .75 108. To pass a plane tangent to a cylinder and through a given point outside its surface . 75 109. To pass a plane tangent to a cylinder and parallel to a given line .... 76 110. Plane tangent to a double-curved surface . 77 111. One projection of a point on the surface of a double-curved surface of revolution being given, it is required to pass a plane tangent to the surface at that point ... 78 112. Through a point in space to pass a plane tangent to a given parallel of a double- curved surface of revolution ... 79 113. To pass a plane tangent to a sphere at a given point on its surface 80 114. Through a given line to pass planes tangent to a sphere 80 CHAPTER V Intersection of Planes with Surfaces, and the Development of Surfaces 115. To determine the intersection of any surface with any secant plane .... 82 116. A tangent to a curve of intersection . . 83 1 17. The true size of the cut section . , .83 118. A right section 83 119. The development of a surface . . .83 120. To determine the intersection of a plane with a pyramid .83 121. To develop the pyramid .... 84 122. To determine the curve of intersection be- tween a plane and any cone ... 86 TABLE OF CONTENTS IX 123. To determine the development of any oblique 128. cone ' . .86 129. 12-1. To determine the curve of intersection be- 130. tweeu a plane and any cylinder ... 88 125. To develop the cylinder . . . . 88 131. 126. The development of a cylinder when the axis 132. is parallel to a coordinate plane . . 91 127. To determine the curve of intersection be- tween a plane and a prism ... 92 To develop the prism The Helical Convolute To draw elements of the surface of the heli- cal convolute To develop the helical convolute . To determine the curve of iutersection be- tween a plane and a surface of revolu- tion . 93 93 94 95 96 CHAPTER VI IXTERSECTIOX OF SURFACES 133. General principles 99 134. Character of Auxiliary Cutting Surfaces . 99 135. To determine the curse of intersection be- tween a cone and cylinder with axes oblique to the coordinate plane .... 100 136. Order and Choice of Cutting Planes . . 103 137. To determine if there be one or two curves of iutersection 103 138. To determine the visible portions of the curve 104 139. To determine the curve of intei-section be- tween two cylinders, the axes of w hich are oblique to the coordinate planes . . 104 140. To determine the curve of intersection be- tween two cones, the axes of which are oblique to the coordinate planes 141. To determine the curve of intersection be- tween an ellipsoid and an oblique cylinder 142. To determine the curve of intersection be- tween a torus and a cylinder, the axes of which are pei-pendicular to the horizontal coordinate plane 143. To determine the curve of intersection be- tween an ellipsoid and a paraboloid, the axes of which intersect and are parallel to the vertical coordinate planes . CHAPTER Vn Warped Surfaces 144. "Warped Surface. Classification . . . 109 145. Having given three curvilinear directrices and a point on one of them, it is required to determine the two projections of the element of the warped surface passing through the given point .... 104 105 106 108 110 TABLE OF CONTENTS 146. Having given two curvilinear directrices and a plane director, to draw an element of the warped surface, Case 1 . . . 147. Having given two curvilinear directrices and a plane director, to draw an element of the warped surface. Case 2 . . . 148. Modifications of the types in Arts. 145 and 146 149. The Hyperbolic Paraboloid .... 150. Through a given point on a directrix to draw an element of the hyperbolic paraboloid . 151. Having given one projection of a point on an hyperbolic paraboloid, to determine the other projection, and to pass an element through the point »AOE ART. 152. 153. 111 154. 155. 156, 112 157. 112 114 158. 116 159. 160. 116 PAGB Warped Helicoids 118 Right helicoid 118 General type of warjied helicoids . . .118 Hyperboloid of Revolution of one Nappe . 118 Through any point of the surface to draw an element 120 The Generatrix may be governed by Three Rectilinear Directrices .... 121 The Generatrix may be governed by Two Curvilinear Directrices and a cone Director 121 Tangent plane to an hyperboloid of revolu- tion . . . ' 121 Through a right line to pass a plane tangent to any double-curved surface of revolution. General solution 122 161. Directions for solving the problems CHAPTER Vni Problems . 124 162. Problems 125 DESCRIPTIVE GEOMETRY OF THE UNIVERSl^ ' OF DESCRIPTIVE GEOMETRY CHAPTER I DEFINITIONS AND FIRST PRINCIPLES I. Descriptive Geometry is the art of graph- ically solviug problems involving three dimen- sions. By its use the form of an object may be graphically defined, and the character, relation, and dimensions of its lines and sur- faces determined. To the student it presents the most admir- able training in the use of the imagination, such as the engineer or architect is called upon to exercise for the development of new forms in structure and mechanism, and which must be mentally seen before being graphically ex- pressed. To the engineer and architect it supplies the principles for the solution of all problems relating to the practical representation of forms. as illustrated by the various types of working drawiugs which are used by artisans to exe- cute designs. It is the foundation for the understanding of the different systems of projection such as orthographic, oblique, and perspective. 2. Projection. The representation of an object is made on one or more planes by a process known as projection, the picture, draw- ing, or projection of the object being deter- mined by the intersection of a system of lines with the plane. The lines are known as pro- jectors and are drawn from the ojaect to the plane of projection, or picture plane^ If these lines, or projectors, are perpendicular to the plane of projection, the system is known as DESCRIPTIVE GEOMETRY Orthographic Projection, arid at least two pro- jections, or views, are required to fully repre- sent the object, Fig. 1 represents two views of a box by orthographic projection. This system is the one commonly employed for working drawings, and for the solution of problems in Descriptive Geometry. If the projectors are parallel to each other, and oblique to the plane of projection, the system is known as Oblique Projection.* Fig. 2 represents the application of this method to the illustration of the object shown in Fig. 1. This system is used for the purpose of producing a more pictorial effect, but one which is easily executed and susceptible of measurement. If the projectors converge to a point on the opposite side of the plane of projection, the system is known as Perspective. Fig. 3 is an application of this method to the representa- tion of the object previously illustrated. This system is used to produce the pictorial effect *For a treatise on Oblique Projection, and that branch of Orthographic Projection known as Isometric Projection, see " Elements of Mechanical Drawing" of this series. obtained by the camera, and is chiefly em- ployed by architects for the representation of buildings as they will appear to the eye of an observer. 3. Coordinate Planes. The planes upon wliich the representations are made are called coordinate planes, or planes of projection, and are usually conceived to be perpendicular to each other and indefinite in extent. Fig. 4 shows their relative positions, but for conven- ience of representation they are limited in extent. The plane designated by 5" is called the horizontal coordinate plane, and the repre- sentation made thereon is known as the hori- zontal projection, plan, or top vieiv. The plane designated by F'is called the vertical coordinate plane, and the representation made thereon is known as the vertical projection, elevatioti, or front view. The plane designated by P is called the profile coordinate plane, and the representation made thereon is known as the profile projection, side elevation, end or side view. The line of intersection between the V and ^planes is known as the ground line. ORTHOGRAPHIC PROJECTION 4. Quadrants or Angles. The poi-tion of space lying in each of the four diedral angles formed by the vertical and horizontal coordi- nate planes is designated as follows : 1st quadrant, above H and before V. 2nd quadrant, above R and behind V. 3rd quadrant, below ^and behind V. 4th quadrant, below H and before V. 5. Orthographic Projection. The projection* of a point on any coordinate plane is obtained by letting fall a perpendicular from the point in space to the coordinate plane, its intersec- tion with that plane being the projection of the point. The perpendicular is called the projector or projecting line. In Fig. 4, a is the point in space and its projections are designated by the same letter with r, A, or p written above and to the right, as a% signifying the vertical projection, a^, the horizontal projec- tion, and flP, the profile projection. For convenience of representation, the verti- cal coordinate plane is conceived as revolved * " Projection " used without a qualifying adjective always means orthographic projection. HORIZONTAL PROJECTION VERTICAL PROJECTION Fig. I. OBLIQUE PROJECTION Fig. 2. DESCRIPTIVE GEOMETRY about the ground line to coincide with the hori- zontal coordinate plane in such a way that the first and third quadrants will be opened to 180°, and the second and fourth quadrants will be closed to 0°. In comparing Figs. 5, 6, and 7 it will be observed that any point in the fourth quadrant, as point a, has, after the planes have been folded together, both pro- jections below the ground line; any point in the third quadrant, as point 6, has the vertical projection below, and the horizontal projection above, the ground line ; any point in the second quadrant, as point (?, has both projections above the ground line; and any point in the first quadrant, as point d, has the vertical projection above, and the horizontal projection below, the ground line. Thus in P'ig. 7 it is evident that the portion of the paper above the ground line represents not only that part of the horizontal codrdinate plane which lies behind the vertical coordinate plane, but also that part of the ver- tical coordinate plane which lies above the horizontal coordinate plane. Likewise the paper below the ground line represents that portion of the horizontal coordinate plane ly- ing in front of the vertical coordinate plane, and also that portion of the vertical coordinate plane lyiug below the horizontal coordinate plane. The profile coordinate plane is commonly revolved about Gr^L nsan axis until it coincides with the vertical coordinate plane, as in Figs. 8 and 9; but it may be revolved about Gr^ L as an axis until it coincides with the horizontal coordinate plane, as in Figs. 10 and 11. It makes no difference as to the correct solution of the problem whether the profile coordinate plane be revolved to the right or to the left, but care should be used to revolve it into such a position as will cause the least confusion with other lines of the problem. 6. Notation. Points in space are designated by small letters, as a, b, c, etc., and their vertical, horizontal, and profile projections by the same letters with v, A, and p placed above and to the right, as a*, a'', a^. When revolved into one of the coordinate planes, the points will be designated by a', i', or a", 6", etc. CXX>RDINATE PLANES ORTHOGHAPHIC PROJECTION Fig. 7. Fig. I I , 6 DESCRIPTIVE GEOMETRY A line in space is designated by two of its points, as ah, or by a capital letter, usually one of the first ten of the alphabet, as A, and its projections are designated by a^'J", a'^U', aPlP, or A\ A\ A^. When revolved into one of the coordinate planes, the lines will be desig- nated by a'h\ a"h", or A', A", etc. The trace of a line, i.e. the point in which it pierces a cocirdinate plane, may be designated by V-tr, Htr, or P-tr, according as the line pierces the vertical, horizontal, or profile coordinate plane. A plane in space is determined by three points not in the same straight line ; by a line and a point ; or by two parallel or intersecting lines. In projection a plane is usually desig- nated by its traces, i.e. the lines in which it pierces the coordinate planes. These traces are designated by the last letters of the alpha- bet beginning with M, as VM, HM, PM, accord- ing as the plane M intersects the vertical, horizontal, or profile coordinate plane. Abbreviations. The following abbreviations are used in connection with the figures and problems : V, the vertical coordinate plane. If, the horizontal coordinate plane. P, the profile coordinate plane. CrL, ground line, the line of intersection between F'and IT. VP, the intersection between T'^and P. HP, the intersection between ^and P. IQ, 2Q, 3Q, 4Q, the first quadrant, second quadrant, etc. In general, an object in space is definitely located by two projections only, usually the vertical and the horizontal. Hereafter, only these two projections Avill be considered, unless otherwise indicated. The character of the lines employed is as follows : Given and required lines. Invisible and projection lines. . All lines not designated above. 7. Points. Two projections are necessary to locate a point with reference to V and H. From Figs. 12 and 13 it is obvious that these two projections must always lie on a perpen- dicular to the ground line after V has been POINTS revolved to coincide with H. The distance from the point in space to H is equal to the distance from its vertical projection to the ground line, and the distance from the point in space to T"is equal to the distance from its horizontal projection, to the ground line. A point on either coordinate plane is its own projection on that plane, and its other projec- tion is in the ground line, as points c, c?, and e. Figs. 12 and 13. The points represented in Figs. 12 and 13 are described as follows: Point rt, in 5^, 5 units from T''and 8 units from H. Point 5, in 5^, 4 units from V and 7 units from H. Point /, in 4Q, 2 units from V and. 6 units from H. Point c, in JT, between IQ and 4Q^ A: units from X'. Point rf, in T'i between IQ and 2Q, 1 units from ff. Point e, in GL. 8. Lines. Since a line in space is deter- mined by any two of its points, the projections of these two points determine the projections of the line. Figs. 14 and 15. A line may also be projected by passing planes through it Fig. 12. 'a* I A f Fig. 13. aCAL£ lllllllilllllllll 8 DESCRIPTIVE GEOMETRY perpendicular to the coordinate planes. The intersections of these planes with V, JI, and P will determine respectively the vertical, hori- zontal, and profile projections of the line. Such auxiliary planes are known us plane pro- jectors or projecting planes of the lines. • In -Pig. 14 the plane ahh^a^ is the vertical project- ing plane of the line ah; the plane abb''a^ is the horizontal projecting plane of the line a5, and the plane abb^a^ is the profile projecting plane of the line ab. g. A line parallel to a coordinate plane will have its projection on that plane parallel to the line in space, and its other projection will be parallel to the ground line. Li^he A, Figs. 16 and 17, is a line parallel to V. A" is parallel to A in space, and A'' is parallel to GrL. Line B is parallel to ff, line C is parallel to both F'and IT, and line D is parallel to P. 10. A line perpendicular to a coordinate plane will have for its projection on that plane, a point, and its other projection will be per- pendicular to the ground line. Line JE, Figs. 18 and 19, is a line perpendicular to V. U" is a point and U'' is perpendicular to GrL. Line Fis perpendicular to ^, and line ^is perpen- dicular to P, K" and K'' being perpendicular to VP and IfP, respectively, and JC^ being a point. 11. Aline lying in either coordinate plane is its own projection on that plane, and its )ther projection is in the ground line. Jn Figs. 20 and 21, line A lies in ff and line B lies in V. 12. A line parallel to one coordinate plane and oblique to the other has its projection on the plane to which it is parallel equal to the true length of the line in space, and the angle which this projection makes with the ground line is the true size of the angle which the line makes with the plane to which it is oblique. Line A, Figs. 16 and 17, is seen in its true length in its vertical projection, and it makes an angle, of 30° with II. 13. If two lines are parallel in space, their projections will be parallel, Figs. 22 and 23. Lines (7 and D are ]iarallel; therefore C and D' are parallel, and C^ and D'' are parallel. LINES Fig. 17 Fig. 18. J^\ A' 1 > : «* ! Fig. 20. Fig. 22. Fig. 19 A' ' «* I Fig. 21. Fig. 23. 10 DESCRIPTIVE GEOMETRY t^. If two lines intersect in space, they have one point in common; hence, the projec- tions of the lines will intersect each other in the projections of the point, as at a* and a\ Figs. 24 and 25. If the projections of the lines do not intersect on a common perpendicu- lar to the ground line, the lines in space do not intersect. Lines A and B, Fig. 26, do not intersect. 15. If a line intersects the ground line, as in Fig. 27, its projections will intersect the ground line in the same point. 16. The traces of a line are the points in which the line pierces the coordinate planes. In Figs. 28 and 29, a is the horizontal trace, and b the vertical trace, of line C. The ver- tical projection of the horizontal trace is in the ground line, as is likewise the horizontal projection of the vertical trace. 17. For convenience of expression, it is cus- tomary to define the position of a line with respect to the coordinate planes by giving the quadrant in which it lies, together with its inclination with, and distance from, the coordi- nate planes. Line (7, Figs. 28 and 29, lies in the first quadrant, and if read from a toward b, inclines upward, backward, and toward the right. The vertical projection indicates that the inclination is upward; the horizontal pro- jection indicates that the inclination is, at the same time, backward; while either projection indicates that the inclination is to the right. The angle of inclination will be considered later. If the line be read from b toward a, the inclination would be downward, forward, and to the left. Either is correct. Fig. 30 illus- trates four other lines as follows : Line ab, in 3Q, inclined upward, forward, and to tlie right. Line cd, in 2Q, inclined downward, back- ward, and to the right. Line ef, in 4Q, parallel to IT, inclined for- ward and to tlie right. Line gk, in 3Q, parallel to P, and inclined upward and forward. LINES 11 Fig. 26. 12 DESCRIPTIVE GEOMETRY i8. Planes. The position of planes may be represented in projection as follows: 1. By the projections of two intersecting or parallel lines. 2. By the projections of a line and a point. 3. By the projections of three points not in the same straight line. 4. By the lines of intersection with the coordinate planes. (Traces.) All planes being indefinite in extent must intersect one or botli of the coordinate planes. Such lines of intersection are called the traces of the planes. Fig. 31 illustrates the inter- sections of the planes N with V and ff, the vertical and horizontal traces being lettered FTVand JIN^, respectively. The orthographic representation is shown in Fig. 32, save that it is not always customary to continue the traces beyond the ground line, the horizontal trace being drawn on one side and the vertical trace on the other, as in Fig. 40. Since the vertical trace of a plane is a line lying on V, it may also be lettered as the verti- cal projection of a line. Thus, in Figs. 31 and 32, VN may be lettered A^ and, according to Art. 11, Page 8, A'* must coincide with CrL. Likewise, JIN is a line lying in II and may be lettered B'', while B^ must lie in CrL. The following positions of planes are illus- trated by Figs. 33 to 48 inclusive: Perpendicular to JSTand parallel to V, Figs. 33 and 34. Perpendicular to P^and parallel to ff, Figs. 35 and 36. Inclined to F'and H, but parallel to GrL^ Figs. 37 and 38. Inclined to ff and perpendicular to FJ Figs. 39 and 40. Inclined to F'and perpendicular to ff, Figs. 41 and 42. Perpendicular to V and H, Figs. 43 and 44. Inclined to F'and IT, but containing CrL, Figs. 45 and 46. Inclined to V, H, and GrL, Figs. 47 and 48. The traces of parallel planes are parallel. 19. From the foregoing illustrations it will be observed that the ground line is the hori- zontal projection of the vertical coordinate PLANES 13 plane, and that any point, line, or plane lying in V will have its entire horizontal projection in the ground line. Likewise, it will be ob- served that the ground line is the vertical projection of the horizontal coordinate plane, and that any point, line, or plane lying in ff will have its entire vertical projection in the ground line. Fig. 33. Fig. 37. HM Fig. 34. HM VN Fig. 38. Fig. 41. Fig. 42. YN Fig. 35. Fig. 36. % X ^\ ^ / y Fig. 39. Fig. 40. Fig. 43. Fig. 44. Fig. 32. Fig. 47. Fig. 48. CHAPTER II POINTS, LINES, AND PLANES 20. Three distinct operations are required for the solution of problems in Descriptive Geometry. First, a statement of the Principles in- volved. Second, an outline of the Method to be observed, by the enumeration of the steps necessitated. Third, the graphic Construction of the problem. The first two operations are purely mental, and the last is the mechanical opera- tion of executing the drawing. 21. To determine three projections of a line. Principle. The projections of two points of a line determine the projections of the line. Method. 1. Determine the vertical, hori- zontal, and profile projections of two points of the line. 2. Connect the vertical projections of the points to obtain the vertical projection of the line ; connect the horizontal projections of the points to obtain the horizontal projec- tion of the line ; and connect the profile pro- jections of the points to obtain the profile projection of the line. Construction. Figs. 49 and 50. Let it be required to determine the projections of a line passing through point a, in IQ, 4: units from F", 7 units from If, and through point b, in 4Q, 16 units from V, 9 units from IT; point a to be 12 units to the left of point b. On any perpendicular to GL lay off J", 9 units below GrL and b\ 16 units below GL (Art. 7, page 6). On a second perpendicular, 12 units to the left of b" and b'^, lay off a", 7 units above GL and a\ 4 units below GL. Connect a" and b" to obtain the vertical projection of the line, and connect a'' and b'^ to obtain the hori- zontal projection of the line. U PROJECTIONS OF A LINE 15 To obtain a^ and J^, assume the position of the profile plane, shown in Fig. 49 by its inter- section with V and ff as VP and ffP. Re- volve P about VP as an axis to coincide with V. Then a^ will lie on a line drawn through a" parallel to CrL, and at a distance from VP equal to that of a'' from GL. Tliis is obtained bj projecting a'' to HP and revolving HP about X as a center to coincide with GL, and projecting perpendicularl}^ to meet the parallel to GL through a" at a^. Obtain b^ in like manner. Connect a'' and b^ to obtain the pro- file projection of the line. 22. If point e in 2Q be one of the given points, ^ and e" having been determined, obtain e^ as follows: Project e'' to HP, re- volve ffP to GL, and project perpendicularly to meet a line drawn parallel to GL through e" at «**; but after e* has been projected to HP it is imperative that HP be revolved in the same direction that it was revolved when the profile projections of the other points were de- termined. If the horizontal projection of one point be revolved, then the horizontal projec- tions of all points must be revolved, and ver- tical projections must not be revolved. 23. P may be revolved in either of the directions shown by Figs. 49 and 51. Fig. 51 represents the line when P has been revolved about HP as an axis to coincide with H. Here it vdll be observed that the vertical pro- jections, and only these, have been revolved. Fig. 49 Fig. 51. 16 DESCRIPTIVE GEOMETRY 24. To determine the traces of a line. Principle. The traces of a line are the points in which the line pierces the coordinate planes. The projections of these traces must, therefore, lie in the projections of the line, and one projection of each trace will lie in the ground line (Art. 7, page 6). Method. 1. To obtain the vertical trace of the line, continue the horizontal projection of the line until it intersects the ground line ; this will be the horizontal projection of the vertical trace, and its vertical projection will be perpendicularly above or below the ground line in the vertical projection of the given line. 2. To obtain the horizontal trace of the line, continue the vertical projection of the line until it intersects the ground line ; this will be the vertical projection of the horizon- tal trace, and its horizontal projection will be vertically above or below the ground line in the horizontal projection of the given line. Case 1. When the line is inclined to V, H, and P. Construction. Figs. 52 and 53. Let it be required to determine the vertical trace of line A. Continue A^ until it intersects GL in c*, which is the horizontal projection of the vertical trace. Next project this point to yl", as at c", wliich is the vertical projection of the vertical trace. The horizontal trace is similarly determined thus : Continue A^ until it intersects GL in d", which is the vertical projection of the hori- zontal trace. Next project this point to A^,, as at c?'', which is the horizontal projection of the horizontal trace. To determine the profile trace, consider P, Fig. 52, to be represented by its intersections with F'and H^ as VP and HP^ and to be re- volved to the right about VP as an axis until it coincides with V. Continue A^ until it intersects VP in/", which is the vertical pro- jection of the profile trace. Continue A^ until it intersects HP in/'', which is the horizontal projection of the profile trace, revolve to 6ri, using the intersection of HP with CrL as a center, and project to f^ by a line drawn TRACES OF A LINE 17 through /" parallel to CrL. This is the profile projection of the profile trace. Fig. 53 is the oblique projectioa of the line, and clearly shows that the line passes from one quadrant to another at its vertical and horizontal traces, i.e. line A passes from IQ to 2Q at its vertical trace, c, and from i^ to 4Q at its horizontal trace, d. XoTE. The vertical projection of the verti- cal trace of a line is often called the vertical trace of the line, since this trace and its ver- tical projection are coincident. Likewise the horizontal projection of the horizontal trace is called the horizontal trace of the line. Fig. 521 18 DESCRIPTIVE GEOMETRY 25. Case 2. When the line is inclined to V and H and is parallel to P. Construction. Fig. 54. If the given line is parallel to P, its vertical and horizontal traces cannot be determined by the above method, since the projections of the line are perpendicular to the ground line; hence, a profile projection of the line is necessary. T*?" Fig. 54. Let it be required to determine the vertical and horizontal traces of the line ab. P is assumed at will and is indicated by HP and VP. Determine a^b^, the profile projection of the line (Art. 21, page 14), by revolving P about VP as an axis until it coincides with V. That portion of P which before revolution was in 5^ now falls below GrL and to the right of VP ; that portion which was in 4Q falls below GrL and to the left of VP; that portion which was in i ^ falls above CrL and to the left of HP. Tlius the profile projec- tion of the line, when continued, indicates that the line passes through 3Q into 2Q lit point df, d^ being the profile projection of the horizontal trace. Likewise the line passes from 5^ into 4Q at point c, c^ being the profile projection of the vertical trace. Counter- revolve P to its original position and obtain dl" and d", which are the horizontal and vertical projections of the horizontal trace, and c" and c'', which are the vertical and horizontal pro- jections of the vertical trace. Since line ab is parallel to P, it has no profile trace. 26. To determine the projections of a line when its traces are given (Art. 21, page 14). Method. 1. Determine the horizontal projection of tha vertical trace, and the ver- tical projection of the horizontal trace. 2. Connect the horizontal trace with the hori- LINES LYING IN PLANES 19 zontal projection of the vertical trace to obtain the horizontal projection of the line. Connect the vertical trace with the vertical projection of the horizontal trace to obtain the vertical projection of the line. 27. Conditions governing lines lying in a plane. Since tiie traces of a plane are lines of the plane, they must intersect all other lines of the plane, and conversely, all lines of a plane must have their traces in the traces of the plane. See lines \A and B^ Figs. 55 and 56. 28. If a line is parallel to H^ jt-^ntersects H at infinity; therefore, its horizontal trace is at infinity and its horizontal projection is parallel to the horizontal trace of the plane in which it lies, its vertical projection being parallel to the ground line. See line C, Figs. 57 and 58. Likewise if a line is parallel to V^ its vertical projection is parallel to the vertical trace of the plajie in which it lies, and its horizontal projection is parallel to the ground line. Fig. 56 Fig 57. Fig. 58. 20 DESCRIPTIVE GEOMETRY 29. If the traces of any plane be drawn through the traces of a line, the plane must contain the line ; therefore, an infinite number of planes may be passed through any line. In Fig. 59, planes iV, R, T, and S all contain line E. 30'. To pass a plane through two intersect- ing or parallel lines. Pkinciple. The traces of the plane must contain the traces of the lines (Art. 27, page 19). Case 1. Method. 1. Determine the traces of the given lines. 2. Connect the two liori- zontal traces of the lines to obtain the horizon- tal trace of the plane, and connect the two vertical traces of the lines ' to obtain the vertical trace of the plane. Construction. Fig. 60. A and B are the given intersecting lines. Determine their horizontal traces <^ and dl^^ and their vertical traces e" and /" (Art. 24, page 16). Con- nect the horizontal traces to determine HT and the vertical traces to determine VT. Since the traces of the plane must meet in GiL., only three traces of the lines are necessary. Check. Both traces of the plane must intersect the ground line in the same point. Note. The vertical projection of the ver- tical trace of a plane will always be spoken of as the vertical trace of the plane, but it must constantly be borne in mind that the vertical trace of a plane is a line lying on V and that its horizontal projection is in the ground line. Likewise the horizontal projection of the hori- zontal trace of a plane will be spoken of as the horizontal trace of that plane, but, as before, the horizontal trace is a line lying in H and its. vertical projection is in the ground line. 31. Case 2. Method. If the traces of the given lines cannot readily be found, new lines intersecting the given lines may be assumed, which, passing through two points of the plane, lie in it, and their traces, there- fore, are points in the traces of the required plane. Construction. Fig. 61. The two given intersecting lines are A and B, the traces of which cannot be found within the limits of THE PLANE OF: LINES 21 the drawing. Line C is an assumed line join- ing point e of line A and point/ of line jB, the traces of which are easily located at g and k. Line P is a second similar line. ST^ connect- ing the horizontal traces of lines G and D, is the required horizontal trace of the plane of lines A and B. Likewise YT^ connecting the vertical traces of lines C and i), is the required vertical trace of the plane of lines A and B. 32. Case 3. Method. If one of the given lines is parallel to the ground line, the required plane will be parallel to the ground line, and, therefore (Figs. 37 and 38, page 13), the traces of the plane will be parallel to the ground line. This problem may be solved by Case 2, or by the following method: Determine the profile trace of the required plane by ob- taining the profile traces of the given lines. Having found the profile trace of the plane, de- termine the horizontal and vertical traces. Fig. 59 22 DESCRIPTIVE GEOMETRY Construction. Fig. 62. Let J. and J5 be the given lines parallel to the ground line. Assume P and draw H^ and VP. Continue the horizontal and vertical projections of lines A and B to intersect HP and FP, respectively. The horizontal and vertical projections of the profile trace of line A will lie at c* and e", and the profile trace at d\ Similarly determine d^^ the profile trace of line B. Through cf and dP draw PiV, the profile trace of the required plane. E^ will be tlie profile projection of the horizontal trace of tlie required plane, and Kp the profile projection of the vertical trace. By counter-revolution obtain ^iVand VN. 33. To pass a plane through a line and a ' point. Method. Connect the given point with any assumed point of the line and proceed as in Art. 30, page 20. "^ 34. To pass a plane through three points not in the same straight line. Method. Connect the three points by aux- iliary lines and proceed as in Art. 30, page 20. In Fig. 63, a, J, and c are the given points. 35. Given one projection of a line lying on a plane, to determine the other projection. Principle. The traces of the line must lie in the traces of the plane (Art. 27, page 19). Method. Determine the traces of the line and from them determine the unknown pro- jection of the line. Construction. Fig. 64. LetiTTVand FiV be the traces of the given plane, and A'' one projection of line A lying in iV^. Continue A'^ to nieet HN in a\ the horizontal trace of line A; a" is in GL (Art. 24, page 16). Con- tinue A!' to meet GiL in 5\ the horizontal pro- jection of the vertical trace; 5", the vertical trace, is in VN. Connect a" and 5'' to obtain A^\ the required vertical projection of line A. If ^" had been the given projection. A'' would have been similarly determined. If the given projection, J?", Fig. 65, be par- allel to GL, B" will be parallel to HS, for, if a line lying in an inclined plane has one 'pro- jection parallel to the ground line., the other pro- jection is paralleVto the trace of the plane; and PROJECTION OF LINES IN PLANES 23 conversely, if a line lying in an inclined plane has one projection parallel to the trace of the plane^ the other projection is parallel to the ground line (Art. 28, page 19). Therefore, to determine B*" continue B" to meet VS in c% the vertical trace of the line. Its horizon- tal projection, c^, will be in GL, and B^ will pass through c'' parallel to ^*S'. E" HN X* * c" Xc" B" 1/* 1 Xrf" 1 iXx" A" c" B' d' ^^-^ VN Fig. 64 Fig. 62. 24 DESCRIPTIVE GEOMETRY 36. Given one projection of a point lying on a plane, to determine the other projection. Principle. The recjuired projection of the point will lie on the projection of any line of the plane passed through the point. Method. 1. Through the given projec- tion of the point draw the projection of any line lying o;i the plane. 2. Determine the other projection of the line (Art. 35, page 22). 3. The required projection of the point will lie on this projection of the line. Construction. Figs. 66 and 67. Let a^ be the given projection of point a on plane N. Through a!^ draw B^^ the horizontal projection of any line of plane N passing through point a. Determine B" (Art. 35, page 22). Then a" lies at the intersection of B^ and a per- pendicular to GL through a^. The same result is obtained by using line G lying in plane N and parallel to F^ or using line J) lying in plane iVand parallel to H. If the vertical projection of the point be given, the horizontal projection is determined in a similar manner. In general, solve the problem hy the use of an auxiliary line parallel to V or H. 37. To locate a point on a given plane at a given distance from the coordinate planes. Principle. The required point lies at the intersection of two lines of the given plane; one line is parallel to, and at the given dis- tance from, F", and the other line is parallel to, and at the given distance from, H. Method. 1. Draw a line of the plane parallel to, and at the required distance from, one of the coordinate planes (Fig. 65, page 23). 2. Determine a point on this line at the required distance from the other coordi- nate plane. Construction. Fig. 68. Let it be re- quired to locate point h on plane R^ at x distance from H and y distance from V. Draw line A in plane R^ parallel to, and at y distance from, V. A^ will be parallel to, and at y distance from, (ri, and A" will be parallel to VR (Art. 35, page 22). On A^ determine J" at x distance from 6ri, and project to A!" to determine h^. Or, point TO REVOLVE A POINT 25 Fig. 68 h may be located by passing line C in plane R^ parallel to, and at x distance from, H. If the plane be parallel to GL, use the fol- lowing method: 1. Draw on the given plane any line oblique to V and H. 2. Locate the required point thereon at the required distance from F'and H. 38. To revolve a point into either coordi- nate plane. Principle. The axis about which the point revolves must lie in the plane into which the point is to be revolved. The revolving point will describe a circle wliose plane is perpen- dicular to the axis, and whose center is in the axis. The intersection of this circle and the coordinate plane is the required revolved posi- tion of the point. Method. 1. Through that projection of the given point on the plane in which the axis lies draw a line perpendicular to the axis. 2. On this perpendicular lay off a point hav- ing its distance from the axis equal to the hypotenuse of a rigiit triangle, one leg of which is the distance from the projection of the point 26 DESCRIPTIVE GEOMETRY to tlie axis, and the other leg of Avhich is equal to the distance from the second projection of the point to the ground line. Let a, Fig. 69, be the point in space to be revolved into H about I) D^ lying in H as an . axis. If a point be revolved about an axis, its locus will be in a plane perpendicular to the axis. X is this plane, and BX, its hori- zontal trace, is perpendicular to D*. Point a, in revolving, will describe a circle with ah as a radius. The points a' and a", in which this circle pierces S, are the required revolved positions of the point a. Since angle aa''b is a right angle, ha is equal to the hypotenuse of a right triangle, one leg of which is a''b, the distance from a'^ to the axis, and the other leg Fig. 71. Fi^. 72. Tffrfc LENGTH OF A LINE o? is a*«, or the distance from a* to (rZ. C<»NSTRUCTiON. Fig. 70. The point a is represented by its two projections, a*" and a*. Through (^- draw ^X perpendicular to 7>*. The revolved position, a' or a", will lie in HX at a distance from D'' equal to the hypotenuse of a right triangle, one leg of which is the dis- tance from a* to the axis D*, and the other leg is the distance from a* to GL. If the axis lies in T', the revolved position of the point will lie in a line passing through the vertical projection of the point perpen- dicular to the axis and at a distance from this axis equal to the hypotenuse of a right tri- angle, one leg of which is the distance from the vertical projection of the point to the axis, while the other leg is the distance from the •horizontal projection of the point to the ground line. 39. To determine the true length of a line. Principle. A line is seen in its true length on that coiirdinate plane to which it is parallel, or in which it lies (Figs. 16 to 21, page 9). Method. Revolve the line parallel to, or into either coordinate plane, at which time one of its projections will be parallel to the ground line and the other projection will measure the true length of the line in space. Case 1. When the line is revolved paral- lel to a coordinate plane. Construction. Fig. 71. Let ah he a. line ill the first quadrant inclined to both T'^and H. Neither projection will equal the true length of the line in space. Revolve ah about the projecting line aa*. as an axis until it is parallel to V. Point a will not move ; hence, its projections, a" and a*, will remain station- ary. Point h will revolve in a plane parallel to H^ to ftj ; hence, J* will describe the arc 6*ftj*, and a^Jj* will be parallel to CrL. Then 6'' will move parallel to GL to its new position 5*, and a'ftJ^ the new vertical projection, will equal the true length of line ah. Fig. 72 is the orthographic projection of the problem. It is of interest to note that the angle which this true length makes with GL is the true size of the angle which the line in space makes with H. 28 DESCRIPTIVE GEOMETRY Figs. 73 and 74 represent the line ah re- volved parallel to JjT, thus obtaining the same result as to the length, of line. Here the angle which the true length of the line makes with QL is the true size of the angle which the line in space makes with V. 40. Case 2. When tlie line is revolved into a coordinate plane. Construction. Figs. 75 and 76 represent the line ah of the previous figures revolved into S about its horizontal projection a^h^ as an axis. Point a revolves in a plane perpen- dicular to the axis a*6* (Art. 38, page 25); therefore, its revolved position, a\ will lie in a line perpendicular to a*J* and at a distance from a^ equal to aa"*^ or the distance from a" to aL* Similarly h' is located. 41. From Fig. 76 it will be seen that if the revolved position of the line, a'h\ be continued, it will pass through the point where a6, con- tinued, intersects its own projection, which is * This does not contradict Art. 38, page 26, in that one leg of the triangle is equal to zero. the horizontal trace of the line, and since it is in the axis, it does not move in the revolution. Thus, in every case where a line is revolved into one of the coordinate planes, its revolved position will pass through the trace of the line. Figs. 77 and 78 illustrate a line cd piercing V in the point e. In order to determine its true length it has been revolved into K about d^d'' as an axis. Since the point e lies in the axis and does not move, point e revolves in one direction while point d must revolve in the opposite direction, thus causing the revolved position to pass through the vertical trace of the line. Again it is of interest to note that when a line is revolved into a coordinate plane to de- termine its true length, the angle whicli the revolved position makes with the axis is the true size of the angle which the line in space makes with the coordinate plane into which it has been revolved. Thus in Figs. 77 and 78, angle c'e'^&' is the true size of the angle which line ctT makes with T''. TRUE LENGTH OF A LINE 29 Fig. 76. Fig. 77. Fig. 78. 30 DESCRIPTIVE GEOMETRY 42. Given a point lying on a plane, to determine its position when the plane shall have been revolved about either of its traces as an axis to coincide with a coordinate plane. Fkinciple. . This is identical with the prin- ciple of Art. 38, page 25, since eitlier trace of the plane is an axis lying in a coordinate plane, one projection of which is the line itself, and the other projection of which is in the ground line. Method. See Art. 38, page 25. Construction. Fig. 79. ^iVand FiV^are the traces of the given plane and 6* and 6", the projections of the point. If the plane with point h thereon be revolved into H about HN as an axis, the point will move in a plane per- pendicular to fl!ZV, and will lie somewhere in V*h'. Its position in this line must be deter- mined by finding the true distance of the point from HN. This distance will equal the hy- potenuse of a right triangle of which V*d^ the horizontal projection of the hypotenuse, is one leg, and 5''c, the distance of the point from H^ is the other leg. By laying off V'e / 1 1 1 \ J^ \ e "^ Fig. 79 W __- ANGLE BETWEEN LINES 31 eqyal to cJ^", and perpendicular to 6*t7, the lengtli required, de, is determined, and when laid off on db' from rf, will locate the revolved position of point b. If it were required to revolve the point into F" about or as an axis, it would lie at b" in a perpendicular to T'jV through 6^ The dis- tance kb" will equal kf, the hypotenuse of a right triangle, one leg of which is kb", and the other leg of wliich is e(|nal to ci*. 43. The revolved position of a line lying in a plane is determined b}"^ finding the re- volved position of two points of the line. If the line of the plane is parallel to the trace which is used as an axis, its revolved position will also be parallel to the trace. If a line has its trace in the axis, this trace will not move during the revolution ; there- fore, it will be necessary to determine but one other point in the revolved position. Fig. 80 represents the lines C and D of the plane M revolved into JS" about JIR as an axis. The revolved position of point a is determined at a\ through which C is drawn parallel to IIR (since line C of the plane is parallel to HR), and jy through «*, the horizontal trace of line D. 44. To determine the angle between two intersecting lines. Pkinciple. The angle between intersect- ing lines maybe measured when the plane of the lines is revolved to coincide with one of the coordinate planes. Method. 1. Pass a plane through the two intersecting lines and determine its traces (Art. 30, page 20). 2. Revolve the plane with the lines thereon, about either of its traces as an axis, until it coincides with a co- ordinate plane (Art. 43, page 31). As it is necessary to determine the revolved position of but one point in each line, let the point be one common to both lines, their point of inter- section, and, therefore, the vertex of the angle between tliem. 3. The angle between the re- volved position of the lines is the required angle. 32 DESCRIPTIVE GEOMETRY' 45. To draw the projections of any polygon having a definite shape and size and occupying a definite position upon a given plane. Principle. The polygon will appear in its true size and shape, and in its true position on the plane, when the plane has been revolved about one of its traces as an axis to coincide with a coordinate plane. Method. 1. Revolve the given plane into one of the coordinate planes about its trace as an axis. 2. Construct the revolved position of the polygon in its true size and shape, and occupying its correct position on the plane. •S. Counter-revolve the plane, with the poly- gon thereon, to its original position, thus obtaining the projections of the polygon. Construction. Fig. 81. Let it be re- quired to determine the projections of a regu- lar pentagon on an oblique plane iV, the center of the pentagon to be at a distance x from H^ and y from V, and one side of the pentagon to be parallel to F", and of a length equal to z. Determine the projections of the center as at 0" and 0'' (Art. 37, page 24). Revolve plane N about one of its traces as an axis until it coincides with one of the coordinate planes (in this case about VN until it coin- cides with F'), 0' being the revolved position of the center (Art. 42, page 30). About 0' as a center draw the pentagon in its true size and shape, having one side parallel to VN, and of a length equal to z. Counter-revolve the plane to obtain the projections of the pentagon. 46. Counter-revolution. The counter- revolution may be accomplished in several ways, of which three are here shown. Construction 1. Through 0', Fig. 81, draw any line C to intersect VN in w", and connect if with 0" to obtain C". Through d' draw B' parallel to C to intersect VN in wi", from which point draw D" parallel to C. From d' project perpendicularly to VN to in- tersect D" in c?", the vertical projection of one point in the required vertical projection of the pentagon. Similarly determine the vertical projections of the other points by drawing lines parallel to O. PROJECTIONS OF A POLYGON 33 47. Construction 2. A sometimes shorter method of counter-revolution, to obtain the vertical projection of the pentagon, is as follows. Fig. 82. Assume the revolved position of the pentagon to have been drawn as described above. From any point of tlie pentagon, as c', draw a line through the center o\ and continue it to meet the axis of revolu- tion in m'. Then m^o' is the vertical projec- tion of this line after counter-revolution (Art. 43, page 31). and c' will lie on nfo' at its inter- section with a perpendicular to \1^ from c' (Art. 42, page 30). Produce c'h' to VN, thus determining the line c'6^ and, therefore, J". Since h'a' is parallel to VN^ a' may be found by drawing a parallel to VN from h^ to meet the perpendicular from a! (Art. 85, page 22). Likewise d^ is determined, since c'd' is parallel to VN. Next continue e'd' to the axis at n"; draw n'd' to meet o'o' produced at e". The horizontal projection of the pentagon is best determined by Art. 35, page 22. Fig. 81, V Fig. 82. 34 DESCRIPTIVE GEOMETRY 48. Construction 3. Fig. 83. Let it be assumed that the plane iV" has been revolved into V about VN as an axis, and that the revolved position of ITN^ has been determined, as UN', by revolving any point g of the hori- zontal trace of the plane about FTVas an axis (Art. 43, page 31). The angle between UN' and FiV is the true size of tlie angle between the traces of the plane ; also the area between UN ' and VN is the true size of that portion of plane iV lying between its traces. Next let it be assumed that the polygon has been drawn in its revolved position, and occupying its correct location with respect to VJV and UN, e' and d' being two of its vertices.. To coiinter-revolve, continue e'd' to intersect UN' in k', and VJV in n". Since A: is a point in the horizontal trace of the plane, its vertical pro- jection will lie in GrL. Then k^ will lie in CrL at the intersection of a perpendicular to VNirom k', and k'^ is in SN. Also the inter- section of e'd' with the axis VWm w", and its horizontal projection Avill lie in GL at w*. Draw w"^", producing it to intersect perpen- diculars to in^ from d' and e', in d" and e". Draw n''k'', producing it to intersect perpen- diculars to GrL from c?" and e", at <7^ and e*. Similarly determine the jDrojections of the other points of the pentagon. A fourth construction for counter-revolu- tion is explained in Art. 91, page 63. UNIVERSIT OF ^*^^^^£SS%NTERSECTION BETWEEN PLANES 35 49. To determine the projections of the line of intersection between two planes. Principle. The line of iutersection be- tween two planes is common to each plane; therefore, the traces of this line must lie in the traces of each plane. Hence, the point of intersection of the vertical traces of the planes is tbe vertical trace of the required line of intersection, and the point of intersection of the horizontal traces of the planes is the hori- zontal trace of the required line of intersec- tion between the planes. There may be three cases as follows : Case 1. When no auxiliary plane is re- quired. Case 2. When an auxiliary plane parallel to F'or J3'is used. Case 3. When an auxiliary plane parallel to P is required. 50. Case 1. When like traces .of the given planes may be made to intersect, no auxiliary plane is required for the solution of the problem. Method. 1. Determine the points of in- tersection of like traces of the planes, which points are the traces of the line of intersection between the planes. 2. Draw the projectfons of the line (Art. 26, page 18). Construction. Figs. 84 and 85 illustrate the principle and need no explanation, line ch being the line of intersection between the given planes N and S. Fig. 84. Fig. 85. 36 DESCRIPTIVE GEOMETRY Figs. 86 and 87 illustratie a special condi- tion of Case 1, in which one of the phines is parallel to a coordinate ^plane. Plane iV is inclined to both Fand H, and plane R is par- allel to H. Since plane R is perpendicular to F", its vertical trace will be the vertical pro- jection, e^", of the required line of intersec- tion between planes N and R. As e^ is the vertical trace of line ef, the horizontal projec- tion of this line will be g'/*, which is parallel to HN (Art. 35, page 22). 51. Case 2. When one or both pairs of like traces of the given planes do not intersect within the limits of the drawing, and are not parallel, or when all the traces meet the ground line in the same point, auxiliary cut- ting planes parallel to either V or IT may be used for the solution of the problem. Method. 1. Pass an auxiliary cutting plane parallel to V or JT, and determine the lines of intersection between the auxiliary plane and each of the given planes. Their point of intersection will be common to the given planes, and, therefore, a point in the re- % quired line. 2. Determine a second point by passing another auxiliary plane parallel to V or H, and the required line will be determined. Construction. Fig. 88. represents two planes, T and M, with their horizontal traces intersecting, but their vertical traces inter- secting beyond the limits of the drawing. By Case 1, point d is one point in the line of intersection between the planes. To obtain a second point an auxiliary plane X has been passed parallel to V. Then JIX is parallel to GrL, and C"^, the horizontal projection of the line of intersection between planes X and M, lies in IIX, while (7" is parallel to VM (Case 1). Likewise J5*, the horizontal projection of the line of intersection between planes X and T, lies in ffX, while B" is parallel to VT. Then point a, the point of intersection between lines B and (7, is a point in the line of inter- section between planes i^f and T^ since point a lies in line of plane iHf, and in line B of plane T. Line ad is, therefore, the required line of intersection between the two given planes. INTERSECTION BETWEEN PLANES 37 Fig. 86. Fig. 87. Fig. 89. If neither the vertical nor horizontal traces of the given planes intersect within the limits of the drawing, it will be necessary to use two auxiliary cutting planes, both of which may be parallel to F", both parallel to IT, or one parallel to F'and one parallel to H. "52. If two traces of the given planes are parallel, the line of intersection between them will be parallel to a coordinate plane, and will have one projection parallel to the ground line and the other projection parallel to the parallel traces of the given planes. 53. Fig. 89 represents a condition when all four traces of the given planes intersect GL in the same point, neither plane containing GrL. This point of intersection of the traces, i, is one point in the required line of inter- section between the planes (Case 1). A second point, a, is obtained by passing the auxiliary cutting plane X parallel to H^ inter- secting plane N in line and plane aS' in line D. Point a, the point of intersection of these lines, is a second point in the required line of intersection, ah, between the given planes N and S (Case 2). * 38 DESCRIPTIVE GEOMETRY 54. Case 3. When both intersecting planes are parallel to the ground line, or when one of the intersecting planes contains the ground line. Method. 1. Pass an auxiliary cutting plane parallel to P. 2. Determine its line of inter- section with each of the given planes. 3. The point of intersection of these two lines is one point in the required line of intersection between the two given planes. It is not nec- essary to determine a second point, for, when both given planes are parallel to the ground line, their line of intersection will be parallel to the ground line. When one plane contains the ground line, one point in the line of intersection is the point of intersection of all the traces. (Figs. 91, 92.) Construction. In Fig. 90 the two planes iVand /S'are parallel to CrL. Pass the auxil- iary profile plane P intersecting plane N in the line whose profile projection is i>^, and the plane S in the line whose profile projection is B^. A^, the intersection of these two lines, is the profile projection of one point in the line of intersection between planes iV and S ', in fact A^ is the profile projection of the re- quired line, and A* and A'^ are the required projections. 55. In Fig. 91 the plane S contains GL and, therefore, it cannot be definitely located with- out its profile trace, or its angle with a coordi- nate plane, and the quadrants through which it passes. Plane iVis inclined to V,II, and P, and plane S contains (ri, passing through 1 Q and 3Q at an angle 6 with V. As in the pre- vious example the profile auxiliary plane is required and PiV, the profile trace of N, is determined as before. PS, the profile trace of S, is next drawn through 1 Q and 3 Q and making an angle 6 with VP. Then d^, the intersection of PiV and PS, is the profile projec- tion of one point in the required line of inter- section between planes N and S, and d^ and d'^ are the required^ projections of this point. The horizontal and vertical traces of this line of intersection are at 5, for it is here that VN and VS intersect, and also where ^iVand ^aS' intersect. The projections of two points of INTERSECTION BETWEEN PLANES 39 the line having now been determined, the pro- jections of the line, dh, may be drawn. Case 3 is apj^licable to all forms of this prob- lem, and Fig. 92 represents the planes N and aS' of Fig. 89 with their line of intersection, ah, determined by this method. HS .^- 1^5 / k / Fig. 90. [Fig. 92. 40 DESCRIPTIVE GEOMETRY 56. Revolution, Quadrants, and Counter rev- olution. Fig. 93. Let it^ be required to pass a plane >S' through the iirst and third quadrants, making an angle with F'and intersecting the oblique plane iVin the line A. The problem is solved by Art. 5-1, page 38, but this question may arise: In what direction shall Pas' be drawn and with what line shall it make the given angle 6? Conceive the auxiliary profile plane P to be revolved about VP as an axis until it coincides with F, that portion of P which was in front of V to be revolved to the right. Then the line VP HP represents the line of intersection between VnndP; (ri represents the line of intersection between H and P; and PN, the line of intersection between iVand P. That portion of the paper lying above GrL and to the right of VP will represent that portion of P which, before revolution, was in IQ; above (rX and to the left of VP, in 2Q; below GL and to the left of VP, in 3Q; and below GL and to the right of VP, in 4Q. Since jS is to pass through IQ and 3Q, making an angle 6 with v., the line PS, in which S intersects P, should be drawn on that portion of P which is in IQ and 3Q, and should pass through the point of intersection of VP and GL, making the angle 6 with VP. P/S iind PiV^ intersect in d^, the profile projection of one point in the required line of intersection. In counter- revolution, that is, revolving P back into its former position perpendicular to both Fand H, rotation will take place in a direction opposite to that of the first revolution, and d" and d'^ will be as indicated. Fig. 94 represents the same problem when the auxiliary profile plane P has been revolved in the opposite direction to coincide with F. Figs. 95 and 96 are examples of the same problem when the auxiliary profile plane P has been revolved about IIP as an axis until it coincides witli II. Then GL will represent the revolved position of the line of intersection between F and P, and the line VP HP will represent the revolved position of the line of intersection between ^and P. PS will then make its angle B with GL, and its direction will be governed by the rotation assumed. OOU NTER - REVOLUTION 41 Fig. 95 2 Q V k ^^/^' y \ 6*6^^ ""i 1 N 1 y^ vs HS \ -^4^ ^ rf* A^ 1Q ft. % \^ \ 4Q Fig. 96. 42 DESCRIPTIVE GEOMETRY 57. To determine the point in which a line pierces a plane. Method. 1. Pass a^ii auxiliary plane through the line to intersect the given plane. 2. Determine the line of intersection between the given and auxiliary planes. 3. The re- quired point will lie at the intersection of the given line and the line of intersection between the given and auxiliary planes. There may be four cases as follows: Case 1. When any auxiliary plane contain- ing the line is used. Case 2. When the horizontal or vertical projecting plane of the line is used. Case 3. When the given line is parallel to P, thus necessitating the use of an auxiliary profile plane containing the line. Case 4. When the given plane is defined by two lines which are not the traces of the plane. 58. Case 1. When any auxiliary plane con- taining the line is used. Construction. Fig. 97. Let A be the given line and iV the given plane. Through line A pass any auxiliary plane Z ( Art. 29, page 20) intersecting plane iV in line C (Art. 50, page 35). Since lines A and die in plane Z, d is their point of intersection, and since C is a line of plane iV, point, ci is common to both line A and plane N; hence, their intersection. 59. Case 2. When the horizontal or verti- cal projecting plane of the line is used. Construction. Fig. 98 represents the same line A and plane iVof the previous figure. Pass the horizontal projecting plane X of line A (Art. 8, page 7), intersecting plane iV in line C (Art. 50, page 35). Lines A and C in- tersect in point d, the required point of pierc- ing of line A and plane N". Fig. 99 is the solution of the same problem by the use of plane Y, the vertical projecting plane of line A. 60. Case 3. When the given line is parallel to P, thus necessitating the use of an auxiliary profile plane containing the line. Construction. Fig. 100. Let N be the given plane and line ab, parallel to P, the given line. Pass an auxiliary profile plane P through the given line ah, intersecting plane iVin the PIERCING OF THE LINE AND PLANE 43 line (7, liaving C" for its profile projection profile projection of the required point of (Art. 54, page 38). Also determine a^J^, the piercing of line oA and plane N. The vertical profile projection of the given line (Art. 21, and horizontal projections of this point are page 14). C-P and a^hp intersect in dp^ the £?" and rf*. Fig. 99 44 DESCRIPTIVE GEOMETRY 6i. Case 4. When the given plane is de- fined by two lines which are not its traces. . Construction. Figs. 101 and 102. Let the given plane be defined by lines B and 2), and let A be the given line intersecting this plane at some point to be determined. This case should be solved without the use of the traces of any plane. Pass the horizontal pro- jecting plane of line A intersecting line B in point e, and line D in point/, and consequently intersecting the plane of lines B and D in line ef. Because this horizontal projecting plane of line A is perpendicular to H all lines lying in it will have their horizontal projections coinciding, and e^f^ will be the horizontal pro- jection of the line of intersection between the auxiliary plane and the plane of lines B and D. Draw the vertical projection of this line through the vertical projections of points e and /; its intersection with vl*', at c?", will be the vertical projection of the required point of piercing of line A with the plane of lines B and B. The same result may be obtained by the use of the vertical projecting plane of line A. 62. If a right line is perpendicular to a plane, the projections of that line will be perpendicular .to the traces of the plane. In Figs. 103 and 104 HN and VN are the traces of a plane to which line A is perpendicular. The horizontal projecting plane of line A is perpendicular to jET. by construction; it is also perpendicular to plane N because it contains a line, A, perpen- dicular to iV; therefore, being perpendicular to two planes it is perpendicular to their line of intersection, HN. But HN'\& perpendicular to every line in the horizontal projecting plane of line A which intersects it, and, therefore, perpendicular to A''. Q.E.D. In like manner VN may be proved to be perpendicular to A . The converse of this proposition is true. 63. To project a point on to an oblique plane. Principle. The projection of a point on any plane is the intersection with that plane of a perpendicular let fall from the point to the plane. Method. 1. From the given point draw A PERPENDICULAR TO A PLANE 45 Fig. i03. Fig. 104. a perpendicular to the given plane (Art. 62, page 44). 2. Determine the point of piercing of this perpendicular and the plane (Art. 57, page 42). This point of piercing is the re- quired projection of the given point upon the given oblique plane. 64. To project a given line on to a given oblique plane. Method. If the line is a right line, project two of its points (Art. 63, page 44). If the line is curved, project a sufficient number of its points to describe the curve. 65. To determine the shortest distance from a point to a plane. Pkinciple. The shortest distance from a point to a plane is the perpendicular distance from the point to the plane. Method. 1. From the given point draw a perpendicular to the given plane (Art. 62, page 44). 2. Determine the point of piercing of the perpendicular and the plane (Art. 57, page 42). 3. Determine the true length of the perpendicular between the point and the plane (Art. 39, page 27). 46 DESCRIPTIVE GEOMETRY 66. Shades and Shadows. The graphic representation of objects, especially those of an architectural character, may be made more effective and more easily understood by draw- ing the shadow cast by the object. When a body is subjected to rays of light, that portion which is turned away from the source of light, and which, therefore, does not receive any of its rays, is said to be in shade. See Fig. 105. When a surface is in light and an object is placed between it and the source of light, intercepting thereby some of the rays, that portion of the surface from which light is thus excluded is said to be in shadow. That portion of space from which light is excluded is called the umbra or invisible shadow. (a) The umbra of a point in space is evi- dently a line. (6) The umbra of a line is in general a plane. (c?) The umbra of a plane is in general a solid. (c?) It is also evident from Fig. 105 that the shadow of an object upon another object is the intersection of the umbra ofHhe first object with the surface of the second object. The line of separation between the portion of an object in light and the portion in shade is called the shade line. It is evident from Fisr. 105 that the shade line is the boundary of the shade. It is also evident that the shadow of the object is the space inclosed by the shadow of its shade line. The source of light is supposed to be at an infinite distance; therefore, the rays of light will be parallel and will be represented by straight lines. The assumed direction of the conventional ray of light is that of the diag- onal of a cube, sloping downward, backward, and to the right, the cube being placed so that its faces are either parallel or perpendicular to F; H, and P (Fig. 106). This ray of light makes an angle of 35° 15' 52" with the coordi- nate planes of projection, but from Figs. 106 and 107 it will be observed that the projections of the ray are diagonals of squares, and hence, they make angles of 45° with GL. Since an object is represented by its projec- / SHADES AND SHADOWS 47 tions, the ray of light must be represented by its projections. An object must be situated in tlie first quadrant to cast a shadow upon both F'and H. 67. To determine the shadow of a point on a given surface pass the umbra of the point, or as is generally termed, pass a ray of light through the point and determine its intersec- tion with the given surface by Art. 24, page 16, or by Art. 57, page 42, according as the shadow is required on a coordinate or on an oblique plane. 68. To determine the shadow of a line upon a given surface it is necessary to determine the intersection of its umbra with that surface. If the line be a right line, this is generally best accomplished by finding the shadows of each end of the line and joining them. If the line be curved, then the shadows of several points of the line must be obtained. In Figs. 108 and 109 cb is an oblique line having one extremity, c, in H. By passing the ray of light through h and locating its horizontal trace, the shadow of h is found to fall upon H at J**. Since c lies in J5i it is its own shadow upon jGT; therefore, the shadow of line cb upon H is c*i**. 48 DESCRIPTIVE GEOMETRY 69. To determine the shadow of a solid upon a given surface it is necessary to determine the shadows of its shade lines. When it is diffi- cult to recognize which lines of the object are its shade lines, it is well to cast the shadow of every line of the object. The outline of these shadows will be the required result. Fig. 110 represents an hexagonal prism located in the first quadrant with its axis per- pendicular to H. Its shadow is represented as falling wholly upon FJ as it would appear if H were removed. Fig. Ill represents the prism when the shadow falls wholly upon H^ as it would appear if F'were removed. Fig. 112 represents the same prism when both V and H are in position, a portion of the shadow falling upon F", and a portion falling upon H. It is now readily observed that the shade lines are he, ed, de, ep, ps, sg, gk, and kh, and that the shadow of the prism is the polygon in- closed by the shadows of these shade lines. When an object is so located that its shadow falls partly upon V and partly upon H^ it is generally best to determine first, its complete shadow upon both V and 5", and to retain only such portions of the shadow as fall upon F" above H, and upon ^before V. Fig. 113 represents a right hexagonal pyra- mid resting upon an oblique plane N and having its axis perpendicular to that plane. Its shadow has been cast upon iV, and the construction necessary to determine the shadow of one point, a, has been shown (Art. 59, page 42). From the foregoing figures these facts will be observed : If a point lies on a plane, it is its own shadow upon that plane. See point C, Figs. 108, 109, and the apex of the pyramid, Fig. 113. If a line is parallel to a plane, its shadow upon that plane will be parallel, and equal in length, to the line in space. Lines hk and ep. Fig. 110, are parallel to V', lines he and 8jo, cd and QB, de and gk, Fig. Ill, are parallel to H, and the sides of the hexagon. Fig. 113, are parallel to N. If two lines are parallel, their shadows are parallel. Observe that the shadows of the SHADES AND SHADOWS 49 opposite sides of the liexagons, and of the lat- eral edges of the prisms, are parallel. If a line is perpendicular to a coordinate plane, its shadow on that plane will fall on the projections of the ra^s of light passing through it. Lines hk and ep. Fig. Ill, fulfill this condition. a'd'/' Ce'd' a' b'f c'e' d Fig. 113. a" b' r c'e' d' 50 DESCRIPTIVE GEUMETRY 70. Through a point or line to pass a plane having a defined relation to a given line or plane. There may be five cases, as follows : Case 1. To pass a plane through a given point parallel to a given plane. Case 2. To pass a plane through a given point perpendicular to a given line. Case 3. To pass a plane through a given point parallel to two given lines. Case 4. To pass a plane through a given line parallel to another given line. Case 5. To pass a plane through a given line perpendicular to a given plane. In cases 1 and 2 the directions of the re- quired traces are known. In cases 3, 4, and 5 two lines of the required plane are known. 71. Casev I. To pass a plane through a given point parallel to a given plane. Principle Since the required plane is to be parallel to the given plane, their traces will be parallel, and a line through the given point parallel to either trace will determine the plane. Method. 1. Through the given point pass a line parallel to one of the coordinate planes and lyiJig in the required plane (Art. 9, page 8). 2. Determine the trace of this line, thus determining one point in the re- quired trace of the plane. 3. Draw the traces parallel to those of the given plane. Construction. Fig. 114. Let N be the given plane and b the given point. Through b pass line A parallel to V and in such a di- rection that it will lie in the required plane, A'' being parallel to GL and A" parallel to VN (Art. 35, page 22). Determine d, the hori- zontal trace of line A, and through d'^ draw ffS, the horizontal trace of the required plane, parallel to HK VS will be parallel to A\ 72. Case 2. To pass a plane through a given point perpendicular to a given line. Principle. Since the required plane is to, be perpendicular to the given line, the traces of the plane will be perpendicular to the projec- tions of the line (Art. 62, page 44). Hence, a line drawn through the given point parallel to either coordinate plane, and lying in the required plane, will determine this plane. RELATION BETWEEN LINES AND PLANES 51 Fig. 116. Method. 1. Through the given point pass a line parallel to one of the coordinate planes and lying in the required plane. 2. Deter- mine the trace of this line, thus determining one point in the required trace of the plane. 3. Draw the traces perpendicular to the pro- jections of the given line. Construction. Fig. 115. Through point h draw line C parallel to T^and lying in the required plane S. O' will be perpendicular to A' (Art. 62, page 44). RS and VS are the required traces. 73. Case 3. To pass a plane through a given point parallel to two given lines. Principle. The required plane will con- tain lines drawn through the given point parallel to the given lines. Method. 1. Through the given point pass two lines parallel to the two given lines. 2. Determine the plane of these lines (Art. 30, page 20). Construction. Fig. 116. Given lines A and B and point b. Through point b pass lines C and D parallel respectively to Hues A and B. Si the plane of these lines, is the required plane. 52 DESCRIPTIVE GEOMETRY 74. Case 4. To pass a plane through a given line parallel to another given line. Principle. The required plane will con- tain one of the given lines and a line inter- secting it and parallel to the second. Method. 1. Through any ]wint of the given line pass a line parallel to the second given line. 2. Determine the plane of these intersecting lines (Art. 30, page 20). 75. Case 5. To pass a plane through a given line perpendicular to a given plane. Principle. The required plane will con- tain the given line and a line intersecting this line and |)erpendicular to the given plane. Method. 1. Through any point of the given line pass a line perpendicular to the given plane. 2. Determine the plane of these lines. CoNSTRUcrTnifr-^fiig. 117. Given line A and plane N. Through any point of line A pass line O perpendicular to plane iV (Art. 62, page 44). Determine S, the plane of lines A and C (Art. 80, page 20). 76. Special conditions and methods of Art. 70. Case 1.* To pass a plane through a given point parallel to a given plane. Fig. 118 illustrates a condition in which the given plane iVis parallel to (ri/, which neces- sitates the use of an auxiliary profile plane. Through the given point b pass the profile plane P. Determine b^ and PiV, and tliroug^i bP draw PS^ the profile trace of the required plane, parallel to PiV", whence VS and ^*S', the traces of the required plane, are determined. This condition may also be solved, Fig. 119, by passing a line A through the given point i, parallel to ani/ line C of the given plane iV. Through the traces of line A the traces of the required plane *S' are drawn parallel to ^iVand VN, respectively. This method is applicable to all conditions of Case 1. 77. Case 2.* To pass a plane through a given point perpendicular to a given line. Fig. 120 illustrates a condition in which the projections of the given line are perpendicular to GrL ; hence, the traces of the required plane will be parallel to GrL. Let ac be the given * The numbers of the cases are the same as those in Art. 70, page 50. RELATION BETWEEN LINES AND PLANES 53 vs y VN > 1 ! j/\ ! ^/ \/ HN ^' HS b — i' y * ,-^' HS Fig. 122. line and h the given point. Pass an auxiliary profile plane P, and determine a^e^, the pro- file projection of the given line, and &^, the profile projection of the given point. Through h^ draw PS^ the profile trace of the required plane, perpendicular to a^e^, whence VS and ^*S' are determined. 78. Case 5.* To pass a plane through a given line perpendicular to a given plane. Fig. 121 illustrates a condition in which the given line ah is parallel to P. This may be solved by the use of an auxiliary profile plane, or by the following method: Through two points of the given line ab pass auxiliary lines C and D perpendicular to the given plane iV, and determine *S, the plane of these parallel lines. Then S is the required plane contain- ing line ah and perpendicular to plane N. Fig. 122 illustrates a condition in which the given plane N is parallel to GL. This solu- tion is identical with that of Art. 75, page 52, save that to determine the traces of the auxil- iary line B, an auxiliary profile plane P has been used. Since ^ is to be perpendicular to N^ Bp must be perpendicular to PIf. 54 DESCRIPTIVE GEOMETRY 79. To determine the projections and true length of the line measuring the shortest distance between two right lines not in the same plane. Principle. The shot-test distance between two right lines not in the same plane is the perpendicular distance between them, and only one perpendicular can be drawn termi- nating in these two lines. Method. 1. Through one of the given lines pass a plane parallel to the second given line. 2. Project the second line on to the plane passed through the first. 3. At the point of intersection of the first line and tlie projection of the second erect a perpendicular to the plane. This perpendicular will inter- sect the second line. 4. Determine the true length of the perpendicular, thus obtaining the required result. Construction. Figs. 123 and 124. Given lines A and B. Pass the plane S through line A parallel to line B (Art. 74, page 52). Project line B on to plane S at 5j, using the auxiliary line I) perpendicular to plane S and intersecting it at point k (Art. 64, page 45).' Since line B is parallel to plane S, B^ will be parallel to B ; hence, their projections are parallel (Art. 13, page 8). B^ intersects A at w, at which point erect the required line E perpendicular to plane S (Art. 62, page 44), intersecting line B at 0. Determine the true length of line E (not shown in the figure) by Art. 39, page 27. 80. To determine the angle between a line and a plane. Principle. The angle which a line makes with a plane is the angle which the line makes with its projection on the plane, or the com- plement of the angle which the line makes with a perpendicular which may project any point of the line on to the plane. The line in space, its projection on the plane, and the projector, form a right-angled triangle. Method. 1. Through any point of the given line drop a perpendicular to the given plane. 2. Determine the angle between this perpendicular and the given line. The angle thus determined is the complement of the required angle. ANGLE BETWEEN LINE AND PLANE 55 Construction. Fig. 125. Given line A and plane N. Through any point 133. ^a 60 DESCRIPTIVE GEOMETRY 87. To determine the bevels for the correct cuts, the lengths of hip and jack rafters, and the bevels for the purlins for a hip roof. Fig. 135 represents ' an elevation and plan of a common type of hip roof having a pitch equal to and a width of c'J". ah is the center line of the hip rafter ; E vg, a cross sec- tion of the ridge ; F, K^ X, M^ are jack rafters. Hip. To find the true length of the hip rafter gJ, and the following angles: Down cut, I: The intersection of the hip with the ridge. Heel cut, 2: The intersection of the hip with the plate. Side cut, 3: Intersection of the hip with the ridge. Top bevel of hip, 4. The true length of the hip eh may be ob- tained by revolving it parallel to F^ as at g^6^ or parallel to JjT, as at e^^. The down cut bevel, 5, is obtained at the same time. The bevel of the top edge of hip is found by pass- ing a plane perpendicular to ab intersecting the planes of the side and end roofs. HZ is the horizontal trace of this plane, and the bevel, 4, is obtained as in Art. 84, page 57. Jack Rafters. The down cut bevel, 5, and the heel cut bevel, ^, of the jack rafters are shown in their true values in the elevation ; and the side cut, 7, is shown in the plan. The true lengths of the jack, rafters are ob- tained by extending the planes of their edges to intersect the revolved position of the hip rafter, as at WjO*. Purlin. It is required to determine the down cut, 5, side cut, 9, and angle between side and end face of purlin, 10. To obtain the down cut revolve side face parallel to ^and the true angle, 5, will be obtained. Similarly, the side cut made on the top or bottom face is obtained by revolving that face parallel to ^, the true angle being indicated by bevel, 9. In order to obtain the angle between the side and end faces, the planes of which are indi- cated in the figure by ^S* and ^, find the inter- section between these planes and determine the angle as in Art. 84, page 57. ANGLE BETWEEN PLANES 61 Fig. 135. 62 DESCRIPTIVE GEOMETRY 88. Given one trace of a plane, and the angle between the plane and the coordinate plane, to determine the other trace. There must be two 6ases, as follows : Case 1. Given the trace on one coordinate plane and the angle which the plane makes with the same coordinate plane. Construction. Fig. 136. Let HT be the given trace and a the given angle betAveen T and S. Draw A* perpendicular to HT, it being the horizontal projection of the line of maxi- mum inclination with H (Art. 86, page 58). If this line be revolved into H, it will make the angle a with A'^, and d' will be the resolved position of the vertical trace of the line of maximum inclination with H. The vertical projection of this trace must lie on the vertical trace of the horizontal projecting plane of A and at a 'distance from CiL equal to d'^d'. Therefore, d" will be a point in VT, the re- quired trace. There may be two solutions, as VT may be above or below GL. 89. Case 2. Given the trace on one coor- dinate plane and the angle which the plane makes with the other coordinate plane. Construction. Fig. 137. Let ^y be the given trace and ^ the given angle between T and V. Consider B, the line of maximum inclina- tion with V, as revolved about the horizontal trace of its vertical projecting plane and mak- ing an angle, /8, with GiL. d' will be the revolved position of the vertical trace of B. From e, with radius ed', describe arc d'd'^. VT, the required trace, will be tangent to this arc. There may be two solutions, as VT may be above or below CrL. 90. To determine the traces of a plane, knowing the angles which the plane makes with both coordinate planes. Construction. Fig. 138. Let it be re- quired to construct the traces of plane T, mak- ing an angle /3 with iTand angle a with H. The sum of a and y8 must not be less than 90° nor more than 180°. Conceive the re- quired plane T as being tangent to a sphere, the center of which, c, lies in CrL. From TO DETERMINE TRACES OF PLANES 63 Fig. 138 the point of tangency of the sphere and plane T, conceive to be drawn the lines of maximum inclination with ^and V. On revolving the line of maximum inclination with V, into JJ", about the horizontal axis of the sphere, as an axis, it will continue tangent to the sphere, as at A\ making an angle with GL equal to the required angle (between V and T'), and its horizontal trace will lie in the axis of revolution at a*, its vertical trace lying in the circle described from c as a center, with a radius cm'. Similarly revolve the line of maximum inclination with 5^ into F^ as at B\ making an angle with GL equal to the re- quired angle (between J^and T}, and its ver- tical trace will lie in the axis of revolution at i*, its horizontal trace lying in the circle de- scribed from c as a center, with a radius ce'. Then ffT will contain a* and be tangent to are 0*^^, and VT will contain Jif and be tan- gent to arc k'tn'. The traces must intersect GL in the same point. Both HT and VT may be either above or below GL. 91. In Art. 48, page 34, reference was made 64 DESCRIPTIVE GEOMETRY to a fourth construction for counter-revolution. This construction involves the angle of maxi- mum inclination between the oblique and coordinate planes. Let it be required to draw the projections of a regular pentagon lying in plane JSf, Fig. 139, when its revolved position is known. Pass plane X perpendicular to both JV and F", intersecting plane i\r in the line of maximum inclination with V, shown in revolved position as gf'. Since this line shows in its true length, all distances on plane N perpendicular to FiY may be laid off on it. Then ge" represents the distance from point e to FTV", and e* will lie on a line through e" parallel to FTV. Likewise other points of the pentagon are determined. Also e"k is the distance from point e in space to V; hence, e'' will lie at a distance from GrL equal to e"k. Fig. 139. CHAPTER III GENERATION AND CLASSIFICATION OF SURFACES 92. Every surface may be regarded as hav- ing been generated by the motion of a line, which was governed by some definite law. The moving line is called the generatrix, and its dififerent positions are called elements of the surface. Any two successive positions of the generatrix, having no assignable distance between them, are called consecutive elements. The line which may direct or govern the gen- eratrix is called the directrix. 93. Surfaces are classified according to the form of the generatrices, viz. : Ruled Surfaces, or such as may be gen- erated by a rectilinear generatrix. Double-curved Surfaces, or such as must be generated by a curvilinear generatrix. These have no rectilinear elements. Tiie ruled surfaces are reclassified as dei^el- opable, and nondevelopable or warped surfaces. \Rvded SUBFACES Plane Single-curved (developable). Cylinder Cone Convolute Warped (nondevelopable) [Double-curved (nondevelopable) 94. Ruled Surfaces. A right line may move so that all of its positions will lie in the same 6& 66 DESCRIPTIVE GEOMETRY plane ; it may move so that any two consecu- tive elements will lie in the same plane ; or it may move so that any two consecutive ele- ments will not lie inHhe same plane. Thus, ruled surfaces are subdivided into three classes, as follows : Plane Surfaces : All the rectilinear ele- ments lie in the same plane. Single-curved Surfaces: Any two con- secutive rectilinear elements lie in the same plane, i.e. they intersect or are parallel. Warped Surfaces: No two consecutive rectilinear elements lie in the same plane, i.e. they are neither intersecting nor parallel. 95. Plane Surfaces are all alike. The rec- tilinear generatrix may move so as to touch one rectilinear directrix, remaining always parallel to its first position ; so as to touch two rectilinear directrices which are parallel to each other, or which intersect ; or it may re- volve about another right line to which it is perpendicular. 96. Single-curved Surfaces may be divided into three classes, as follows : Cones: In which all the rectilinear ele- ments intersect in a point, called the apex. Cylinders: In which all the rectilinear elements are parallel to each other. The cylinder may be regarded as being a cone with its apex infinitely removed. CoNVOLUTES : In which the successive rec- tilinear elements intersect two and two, no three having one common point. 97. Conical Surfaces are generated by the rectilinear generatrix moving so as always to pass through a fixed point, called the apex, and also to touch a given curve, called the directrix. Since the generatrix is indefinite in length, the surface is divided at the apex into two parts, called nappes. The portion of a conical surface usually considered is included between the apex and a plane which cuts all the elements. This plane is called the base of the cone and the form of its curve of inter- section with the conical surface gives a dis- tinguishing name to the cone, as, circular, elliptical, parabolic, etc. If the base of the cone has a center, the right line passing SINGLE CURVED SURFACES 67 through this center and the apex is called the axis of the cone (Fig. I-IO). A Right Cone is one having its base per- pendicular to its axis. Fig. 140. Fig. 141. 98. Cylindrical Surfaces. The cylinder is that limiting form of the cone in which the apex is removed to infinity. It may be gen- erated by a rectilinear generatrix which moves so as always to touch a given curved directrix, having all of its positions parallel. A plane cutting all the elements of a cylindrical sur- face is called its Ja«e, and the form of its curve of intersection with the surface gives a dis- tinguishing name to the cylinder, as in the case of the cone. If the base has a center, the right line through this center parallel to the elements is called the axu (Fig. 141). A cylinder may also be generated by a cur- vilinear generatrix, all points of which move in the same direction and with the same velocity. A Right Cylinder is one having its base perpendicular to its axis. 99. Convolute Surfaces may be generated by a rectilinear generatrix which moves so as always to be tangent to a line of double cur- vature.* Any two consecutive elements, but no three, will lie in the same plane. Since there is an infinite number of lines of double curvature, a great variety of convolutes may * A line of double curvature is one of which no four consecutive points lie in the same plane. 68 DESCRIPTIVE GEOMETRY exist. One such form which may readily be generated is the helical convolute (Fig. 142). It is the surface generated by the hypotenuse of a right triangle under the following condi- tions: Suppose a right triangle of paper, or some other thin, flexible material, to be wrapped about a right cylinder, one leg of the triangle coinciding with an element of the cylinder. If the triangle be unwrapped, its vertex will describe the involute of the base of the cylin- der, and the locus of the points of tangency of its hypotenuse and the cylinder will be a helix, the hypotenuse generating the helical convolute. The convolute may also be re- garded as being generated by a rectilinear generatrix moving always in contact with the involute and helix as directrices, and making a definite angle with the plane of the involute. 1 00. A Warped Surface is generated by a rectilinear generatrix moving in such a way that its consecutive positions do not lie in the same plane. Evidently there may be as many warped surfaces as there are distinct laws re- stricting the motion of the generatrix. Any warped surface may be generated by a rectilinear generatrix moving so as to touch two linear directrices, and having its consecu- tive positions parallel either to a given plane, called a plane director, or to the consecutive elements of a conical surface, called a cone director. loi. The following types, illustrated by Figs. 142 to 148, indicate the characteristic features of warped surfaces : Hyperbolic Paraboloid, Fig. 143. Two rectilinear directrices and a plane director, or three rectilinear directrices. Conoid, Fig. 144. One rectilinear and one curvilinear directrix and a plane director. Cylindroid, Fig. 145. Two curvilinear directrices and a plane director. Right Helicoid, Fig. 146. Two curvi- linear directrices and a plane director. Oblique Helicoid, Fig. 147. Two curvi- linear directrices and a cone director. Hyperboloid of Revolution, Fig. 148. Two curvilinear directrices and a cone direc- tor, or three rectilinear directices. WARPED SURFACES 69 Fig. 144 Fig. 147. 7a DESCRIPTIVE GEOMETRY 102. A Surface of Revolution, Fig. 149, is the locus of any line, or generatrix, the posi- tion of which remains unaltered with reference to a fixed right line about which it revolves. This fixed right line is called the axis of revo- lution. A circle of the surface generated by any point of the generatrix is called a parallel, and planes perpendicular to the axis will cut the surface in parallels. Any plane contain- ing the axis of revolution is called a merid- ian plane, and the line cut from the surface by this plane is called a meridian line. All meridian lines of the same surface are obvi- ously identical, and any one of them may be considered as a generatrix. That meridian plane which is parallel to a coordinate plane is called the principal meridian. If the generatrix be a right line lying in the same plane as the axis, it will either be parallel with it or intersect it ; in the former case the surface generated will be a cylinder, in the latter, a cone, and these are the only single-curved surfaces of revolution. If the generatrix does not lie in the same plane with the axis, the consecutive positions are neither parallel nor intersecting. The surface must then be warped and its meridian line will be an hyperbola. This is the only warped surface of revolution. It may also be generated by revolving an hyperbola about its conjugate axis, and is known as the hyperbo- loid of revolution of one nappe (Fig. 148). 103. Double-curved Surfaces. With the ex- ception of the cylinder, cone, and hyperboloid of revolution all surfaces of revolution are of double curvature. They are infinite in num- ber and variety. Representative types are : The Sphere: Generated by revolving a circle about its diameter. The Prolate Spheroid: Generated by revolving an ellipse about its major axis. The Oblate Spheroid : Generated by revolving an ellipse about its minor axis. The Paraboloid : Generated by revolv- ing a parabola about its axis (Fig. 149). The Hyperboloid of Two Nappes: Generated by revolving an hyperbola about its transverse axis. DOUBLE-CURVED SURFACES 71 The Torus (annular or not): Generated by revolving a circle about a line of its plane other than its diameter. Fig. 150 illustrates an annular torus. The Double-curved Surface of Traxs- POsiTiox — Serpentine : Generated by a sphere the center of which moves along an helix (Fig. 151). PARABOLOID Fig. 149. TORUS (AMMULAfi'} Fig. 150. SERPENTINE Fig. 151, *^BR-A^ OF UNIVERSITY CHAPTER IV TANGENT PLANES 104. A plane is tangent to a single-curved surface when it contains one, and only one element of that surface. The two lines com- monly determining a tangent plane are the tan- gent element and a tangent to the surface at some point in this element. If this second line lies in one of the coordinate planes it will be a trace of the tangent plane. In Fig. 152 point d is on the surface of a cone to which a tangent plane is to be drawn. The line drawn through point d and the apex of the cone will be the element at which the plane is to be tangent and, therefore, one line of the tangent plane. If a second line, ck, be drawn through point d, and tangent to any section of the cone containing this point, it will be a second line of the tangent plane. The traces of these lines will determine JjTaS' and VS, the traces of the required tangent plane. If the base of the single-curved sur- face coincides with one of the coordinate planes, as in Fig. 152, bk can be used for the tangent line, thus determining the hori- zontal trace directly. 72 TANGENT PLANES 73 Fig. 152. 105. One projection of a point on a single- curved surface being given, it is required to pass a plane tangent to the surface at the ele- ment containing the given point. Principle. The tangent plane will be de- termined by two intersecting lines, one of which is the element of the surface on which the given point lies, and the second is a line intersecting this element and tangent to the single-curved surface, preferably in the plane of the base. Method. 1. Draw the projections of the element containing the given point. 2. In the plane of the base draw a second line tangent to the base at the tangent element. 3. Determine the plane of these lines. If the plane of the base coincides with one of the coordinate planes, the tangent line will be one of the traces of the required tangent plane. Note. In this and the following problems the base of the single-curved surface is consid- ered as lying on one of the coordinate planes. 74 DESCRIPTIVE GEOMETRY Construction. Fig. 153. Let the single- curved surface be a cone which is defined by its projections, and tlje base of which lies in one of the coordinate planes; in this case in H. Let S' and HN of the tangent planes S and iV, will be tangent to the base at 1(f^ and ArJ. The vertical traces of the tangent elements, l" and o", determine the necessary points in VS and FiV, the required vertical traces of the tangent planes. 1 06. To pass a plane tangent to a cone and through a given point outside its surface. Pkinciple. Since all tangent planes con- tain the apex of the cone, the required plane Fig. 153. PLANE TANGENT TO GONE 75 must contain a line drawn through the apex of the cone and the given point. It must also have one of its traces tangent to the base of the cone, since this base is supposed to lie in a coordinate plane. Method. 1. Draw the projections of a line passing througli the apex of the cone and the given point. 2. Determine its traces. 3. Through that trace of the auxiliary line which lies in the plane of the base draw the trace of the required plane tangent to the base of the cone. 4. Draw the other trace of the plane through the other trace of the auxiliary line. There are two possible tan- gent planes. 107. To pass a plane tangent to a cone and parallel to a given line. Pkixceple. The tangent plane must con- tain the apex of the cone and a line through the apex parallel to the given line. Method. 1. Through the apex of the cone draw a line parallel to the given line. 2. Obtain the traces of this auxiliary line. 3. Draw the traces of the required tangent plane through the traces of the auxiliary line, making one of them tangent to the base of the cone. There are two possible tangent planes. 108. To pass a plane tangent to a cylinder and through a given point outside its surface. Princii'LE. The tangent plane will be de- termined by two intersecting lines, one of which is drawn through the given point parallel to the elements of the cylinder, and the second is drawn tangent to the cylinder from some point in the first line, preferably in the plane of the base. Method. 1. Through the given point draw the projections of a line parallel to the elements of the cylinder. 2. Determine the traces of this auxiliary line. 3. Through that trace of the auxiliary line which lies in the plane of the base of the cylinder draw one trace of the required plane tangent to the base. 4. Through the other trace of the auxiliary line draw the second trace of the tangent plane. There are two possible tangent planes. 76 DESCRIPTIVE GEOMETRY Construction. Fig. 154. Let a" and a^ be the projections of a given point through which the plane is to be passed tangent to the cylinder. The base of the cylinder rests on H. B" and B'' are the projections of the auxiliary line drawn through point a parallel to the ele- ments of the cylinder. Through d}^ the hori- zontal trace of this line, HS and HN^ the horizontal traces of the two possible tangent planes, may be drawn tangent to the base of the cylinder. The vertical traces of these planes, VS and VN^ must contain c", the verti- cal trace of the line B. The elements at which planes iVand 8 are tangent are lines F and K^ respectively. 109. To pass a plane tangent to a cylinder and parallel to a given line. Principle. The tangent plane must be parallel to a plane determined by the given line and a line intersecting it, which is parallel to the elements of the cylinder. Method. 1. Through any point of the given line draw a line parallel to the elements PLANE TANGENT TO CYLINDER 77 of the cylinder. 2. Determine the traces of the plane of these two lines. 3. Draw the traces of the tangent plane parallel to the traces of the auxiliary plane (Art. 18, page 12), one of the traces being tangent to the base of the cylinder. There may be two tan- gent planes. CoNSTKUCTiox. Fig. 155. Through any point, c, of the given line A, draw line B paral- lel to the elements of the cylinder. The tan- gent planes will be parallel to X, the plane of these lines. N and *S' will be the required tangent planes. no. A plane is tangent to a double-curved surface when it contains one, and only one point of that surface. The two lines com- monly determining the tangent plane are the lines tangent to the meridian and parallel at the point of tangency (Art. 102, page 70). A Normal is the line perpendicular to the tangent plane at the point of tangency. A Normal Plane is any plane containing the normal. 78 DESCRIPTIVE GEOMETRY III. One projection of a point on the surface of a double -curved surface of revolution being given, it is required to pass a plane tangent to the surface at that point. Principle. Planes tangent to double- curved surfaces of revolution must contain the tangents to the meridian and parallel at the point of tangency. There may be two methods. 1st Method. 1. Through the given point draw a meridian and a parallel (Art. 102, page 70). 2. Draw tangents to these curves at the given point. 3. Determine the plane of these tangents. Construction. Fig. 156. Given the ellip- soid with its axis perpendicular to H. A plane is required to be drawn tangent to the surface at the point having e* for its horizontal pro- jection. With f'e'^ as a radius, and /'' as a center, describe the circle which is the hori- zontal projection of the parallel through e. One of the possible vertical projections of the parallel is tliat portion of B" lying within tlie ellipse. Project e^ on to this line to obtain e". the vertical projection of the given point e. Through point e draw line B tangent to the parallel, and, therefore, in the plane of the parallel. That portion of A'' lying within the circle, which is the horizontal projection of the ellipsoid, will be the horizontal projec- tion of the meridian drawn through point e. Revolve this meridian line about fk as an axis until it coincides with the principal meridian (Art. 102, page 70), the vertical projection of which will be shown by the ellipse. The re- volved position of the vertical projection of point e will now be at ej, and a line may be drawn tangent to the meridian at this point, its projections being A\^ A'[. Counter-revolve this meridian plane to determine the true posi- tion of the tangent line, shown by its projec- tions at A'\ A''. The traces of the tangent lines A and B will determine the traces of the required tangent plane. There are two possi- ble tangent planes. 2nd Method. 1. Draw the projections of a cone tangent to the double-curved surface of revolution at the parallel passing through PLANE TANGENT TO ELLIPSOID 79 Fig. 156. the given point. 2. Pass a plane tangent to the auxiliary cone at the element drawn through the given point (Art. 105, page 73). Construction. Fig. 156. J.* and A*" are the projections of the element of a cone tan- gent to the ellipsoid and containing point e. A plane tangent to the cone at line A will be the required plane tangent to the ellipsoid at point e. If the vertical trace of line A lies beyond the limits of the paper, the direc- tion of FiV may be determined by observing that the trace of the required plane must be perpendicular to the normal D at the point e (Art. 110, page 77). There are two possible tangent planes. 112. Through a point in space to pass a plane tangent to a double-curved surface of revolution at a given parallel. ^Iethod. 1. Draw a cone tangent to the double-curved surface, of revolution at the given parallel. 2. Pass a plane through the given point tangent to the cone (Art. 106, page 74). There are two possible tangent planes. 80 DESCRIPTIVE GEOMETRY 113. To pass a plane tangent to a sphere at a given point on its surface. Method. This may be solved as in Art. Ill, or by the following method. 1. Through the given point draw a radius of the sphere. 2. Pass a plane through the given point per- pendicular to the radius (Art. 72, page 50), and this will be the required plane. 114. Through a given line to pass planes tangent to a sphere.* Principle. Conceive a plane as passed through the center of the sphere and perpen- dicular to the given line. It will cut a great circle from the sphere and lines from the re- quired tangent planes. These lines will be tangent to the great circle of the sphere and intersect the given line at the point in which it intersects the auxiliary plane. The planes determined by these tangent lines and the given line will be the required tangent planes. * For a general solution of problems requiring the draw- ing of tangent planes to double-curved surfaces of revolu- tion, and through a given line, see Art. 160, page 122. Method. 1. Through the center of the sphere pass a plane perpendicular to the given line (Art. 72, page 50) and determine its traces. 2. Determine the point in which this plane is pierced by the given line (Art. 57, page 42). 3. Into one of the coordinate planes revolve the auxiliary plane containing the center of the sphere, the great circle cut from the sphere, and the point of intersection with the given line. 4. From the latter point draw lines tangent to the revolved position* of the great circle of the sphere. These lines will be lines of the required tan- gent planes. 5. Counter-revolve the auxil- iary plane containing the lines of the tangent planes. 6. Determine the planes defined by the given line and each of the tangent lines obtained by 4. These will be the required tangent planes. Construction. Fig. 157. The given line is A, and the center of the sphere is e. HX and VX are the traces of the auxiliary plane perpendiculjlr to A and passing through e PLANE TANGENT TO SPHERE 81 (Art. 72, page 50). The point/ is that of the intersection of line A and plane X. Revolve plane X into the vertical coordinate plane and e' will be the revolved position of the center of the sphere, and f the revolved position of the point /. Draw the great circle of the sphere and the tangents C and B'. In counter-revolution these lines will be at C and B, intersecting J. at /. C and A will be two intersecting lines of one of the required tangent planes, S, and the second plane, JV, will be determined by lines A and B. Fig. 157. CHAPTER V INTERSECTION OF PLANES WITH SURFACES, AND THE DEVELOPMENT OF SURFACES 115. To determine the intersection of any surface with any secant plane. General Method. 1. Pass a series of auxiliary cutting planes which will cut lines, straight or curved, from the surface, and right lines from the secant plane. 2. The inter- sections of these lines are points in the re- quired curve of intersection. This method is applicable alike to prisms, pyramids, cylinders, cones, or double-curved surfaces of revolution. The auxiliary cutting planes may be used in •dny position, but for convenience they should be chosen so as to cut the simplest curves from the surface, that is, straight lines or circles. With solids such as prisms, pyramids, single-curved, or other ruled surfaces, the above method consists in finding the intersec- tion of each element with the oblique plane by Art. 61, page 44. 82 INTERSECTION OF PLANE WITH PYRAMID 83 ii6. A tangent to the curve of intersection of a jDlane with a single-curved surface may be drawn by passing a plane tangent to the sur- face at the point assumed (Art. Ill, page 78). The line of intersection of the tangent plane with the secant plane will be the required tangent. 117. The true size of the cut section may al- ways be found by revolving it into one of the coordinate planes, about a trace of the secant plane as an axis. 118. A right section is the section cut from the surface by a plane perpendicular to the axis. 119. The development of a surface is its true size and shape when spread open upon a plane. Only surfaces having two consecutive elements in the same plane can be developed, as only such surfaces can be made to coincide with a plane. Therefore, only single-curved surfaces, and solids bounded by planes, can be developed. Solids bounded by planes are developed by finding the true size and shape of each successive face. Single-curved sur- faces are developed by placing one element in contact with the plane and rolling the surface until every element has touched the plane. That portion of the plane covered by the sur- face in its revolution is the development of the surface, 120. To determine the intersection of a plane with a pyramid. Principle. Since a pyramid is a solid bounded by planes, the problem resolves itself into determining the line of intersection be- tween two planes. Again, since the pyramid is represented by its edges, the problem still further resolves itself into determining the points of piercing of these edges with the plane. Method. 1. Determine the points in which the edges of the given pyramid pierce the given plane. 2. Connect the points thus obtained in their order, thereby determining the required intersection between the plane and pyramid. 84 DESCRIPTIVE GEOMETRY Construction. Fig. 158. Given the pyra- mid of which lines A, B^ . (7, D, and E are the lateral edges, or elemenjts, intersected by plane iV^. The points of piercing, a, 6, e, d, and e, of the elements with plane iV have been de- termined by the use of the horizontal project- ing planes of the elements (Art. 59, page 42), and the lines ab, be, cd, de, and ea, joining these points of piercing in their order, are the lines of intersection of plane and pyramid. Since all the auxiliary planes used contain point I, the apex of the pyramid, and are per- pendicular to ff, they must contain a line through I perpendicular to JI; hence, o, the point of piercing of this line and plane JV, is a point common to all the lines of intersection between plane iV and the auxiliary planes. By observing this fact the work of construc- tion can be slightly shortened. The true size of the cut section is obtained by revolving each of its points into V about FiVas an axis (Art 43, page 31). 121. To develop the pyramid. Principle. If the pyramid be laid on a plane and be made to turn on its edges until each of its faces in succession has come into contact with the plane, that portion of the plane which has been covered by the pyramid in its revolution will be the development of the pyramid. From the above it is evident that every line and surface of the pyramid will appear in its true size in development. Method. 1. Determine the true length of each line of the pyramid. 2. Construct each face in its true size and in contact with adjacent faces of the pyramid. Construction. Figs. 158 and 159. The true lengths of the elements and their seg- ments have been determined by revolving them parallel to F", as at ^j, B^, etc., a^, 5j, etc., Fig. 158. The edges F, Gr, J, J", and K are already shown in their true lengths in their horizontal projections, since the plane of these lines is parallel to H. Hence, through any point I, Fig. 159, representing the apex, draw a line -B, equal in length to B^. With Z as a center and with a radius equal to Cj, draw an arc of indefinite length. With the end of B as a center and with a radius equal to (T^ draw an arc intersecting the first in point w, DEVELOPMENT OF PYRAMID 85 thus definitely locating lines (7 and Cr. The other faces are determined in like manner. The development of the cut is obtained by ^^ laying off from Z, on its corresponding ele- ^\ ment, the true lengths of the elements from \^ \^ the apex of the pyramid to the cut section, and joining the points thus obtained. The base and cut section may be added to complete the development of the pyramid. Fig. 159 86 DESCRIPTIVE GEOMETRY 122. To determine the curve of intersection between a plane and any cone. Principle. The problem is identical with that of the intersection of a plane with a pyra- mid, for a cone may be considered as being a pyramid of an infinite number of faces. Method. 1. Pass a series of auxiliary planes perpendicular to one of the coordinate planes and cutting elements from the cone. 2. Each auxiliary plane, save the tangent planes, will cut two elements from the cone and a right line from the given plane. The intersections of this line with the elements give two points in the required curve. Construction. Fig. 160. Let it be re- quired to determine the curve of intersection between plane N and the oblique cone with its circular base parallel to H. Pass a series of auxiliary cutting planes through the cone, containing its apex, a, and perpendicular to H. Plane X is one sach plane which cuts two ele- ments, ah and ac, from the cone, and the line B from plane N. B intersects element ac in point e and element ah in point /, and these are two points of the required curve of inter- section between plane iVand the cone. Simi- larly determine a sufficient number of points to trace a smooth curve. The planes passing through the contour elements of each pro- jection should be among those chosen. As in the case of the pyramid, the work of construction may be slightly shortened by observing that since all of the auxiliary planes are perpendicular to H^ and pass through the apex of the cone, they all contain the line passing through the apex and perpendicular to H. This line pierces plane iV^in (f, a point common to all lines of intersection between N and the auxiliary planes. The true size of the cut section is deter- mined by revolving each of its points into V about FiVas an axis. 123. To determine the development of any- oblique cone. Principle. When a conical surface is rolled upon a plane, its apex will remain stationary, and the elements will successively roll into contact with the plane, on which they will be seen in their true lengths and at their true distances from each other. DEVELOPMENT OF CX)NE 87 Construction. Since in Fig. 160 the base of the cone is parallel to J?", the true distances between the elements may be measured upon the circumference of the base ; therefore, to develop the conical surface upon a plane, through any point a, Fig. 161, representing the apex, draw a line ac, equal in length to a'^Cj, the revolved position of element oc. Fig. 160. With a as a center, and with a radius equal to the true length of the next element, ak^ draw an arc of indefinite length. With c as a center, and with a radius equal to (^V"^ draw an arc intersecting the first in^. This process must be repeated until the complete DESCRIPTIVE GEOMETRY development has been found. The accuracy of the development depends upon the number of elements used, the g;reater number giving greater accuracy. The development of the curve of intersection is obtained, as in the pyramid, by laying off from the apex, on their corresponding ele- ments, the true lengths of the portions of the elements from the apex to the cut section, and joining the points thus found. 124. To determine the curve of intersection between a plane and any cylinder. Principle. A series of auxiliary cutting planes parallel to the axis of the cylinder and perpendicular to one of the coordinate planes will cut elements from the cylinder and right lines from the given plane. The intersections of these elements arid lines will determine points in the required curve. Construction. Fig. 162. Given the ob- lique elliptical cylinder cut by the plane iV. Plane X is one of a series of auxiliary cutting planes parallel to the axis of the cylinder and perpendicular to H. Since it is tangent to the cylinder, it contains but one element, A^ and intersects the given plane in the line (r. Since lines G- and A lie in plane X, they in- tersect in a, one point in the required curve of intersection between the cylinder and the given plane N. Likewise plane Z intersects the cylinder in elements C and D, and the plane iV^in line K, the intersections of which with lines C and D are c and c?, two other points in the required curve. The true size of the cut section has been determined by revolving it into V about VN as an axis. 125. To develop the cylinder. Principle. When a cylinder is rolled upon a plane to determine its development, all the elements will be shown parallel, in their true lengths, and at their true distances from each other. Since in an oblique cylinder the bases will unroll in curved lines, it is necessary to determine a right section which will develop into a right line, and upon which the true distances between the elements may be laid off. This line will be equal to the periphery of the DEVELOPMENT OF CYLINDER 89 right section, and the elements will be per- pendicular to it. The ends of these perpen- diculars will be at a distance from the line equal to the true distances of the ends of the elements from the right section. A smooth curve may then be drawn through the ends of the perpendiculai-s. Method. 1. Draw a right line equal in length to the periphery of the right section. 2. Upon this right line lay off the true dis- tances between the elements. 3. Through the points thus obtained draw perpendiculars to the right line. 4. On these perpendiculars lay off the true lengths of the corresponding elements, both above and below the right line. 5. Trace a smooth curve through the ends of the perpendiculars. Construction. Figs. 162 and 163. The secant plane Noi Fig. 162 has been so chosen as to cut a right section from the cylinder, that is, the traces of the plane are perpen- dicular to the projections of the axis of the cylinder (Art. 62, page 44). Element D has been revolved parallel to V to obtain its true 90 DESCRIPTIVE GEOMETRY Fig. 163. DEVELOPMENT OF CYLINDER 91 length i)j, and d" is projected to d^, thus oh- taining the true lengths of each portion of D. Since all the elements of a cylinder are of the same length, 2)j represents the true length of each element, and their segments are ob- tained by projecting the various points of the cut section upon it, as at Jj, intersecting in 2Q. C perpendic- ular to H\ D parallel to GrL. E and F not intersecting. E perpendicular to F; F inclined to Fand H. Both in 4 Q. 10. Draw the H-pr. and V-pr. of the fol- lowing lines (Arts. 8 to 15, pages 7 to 10). A and B parallel, and inclined to V and JT, in 5^. and D intersecting in 1 Q. C inclined to Fand ^; ^ parallel to T^only. E and F intersecting in 3Q. E parallel to H and inclined to V; F parallel to CrL. 11. Required the Hpr.^ V-pr., and P-pr. of the following lines (Arts. 21-23, pages 14, 15). State the ^'s in which they appear, and the direction of inclination (Art. 17, page 10). a, — 6, — 4, 8. , I c, 6, 4, 9. 5, -2,-4,0. ""^[j, 2, -4,0. e, - 2, 4, 10. ah (/, -8, 4, 0. PROBLEMS 127 12. Required the R-pr.^ V-pr.^ and P-pr. of the following lines (Arts. 21-23, pages 14, 15). State the ^'s in which they appear, and the direction of inclination (Art. 17, page 10). a, — 2, 6, 7. , I c, 6, 1, 10. h, — 5, 1, 0. ef «, - 6, - 4, 9. I/, -2,4,0. 13. Required the H-pr. and V-pr. of the following triangles (Art. 21, page 14). a, -8,-3, 11. {d, -2, 2,8. abc\ J, - 1, _ 10, 7. def. e, - 2, 9, 4. c, 0, -3,0. |/,-2, 2, 0. 14. Required the H-pr. and V-pr, of the following triangles (Art. 21, page 14). a, -6, -6,8. f(f, -1,-1, 11. ahcl b, -1, - 2, 6. defi e, 6, 4, 0. e, 0, -1,0. [/, 4, -2i6. 15. Three points, a, J, and c, lie in P, and in 5^. a and 6 have their V-prs. in the same point, and b and I8 Il4'll 12 4,< C3/ 131 8' 6 2^-' 2 2V 1 / Determine the true size of the polygon by revolving it into V or H. (Art. 43, page 31.) Note. — In Problems 7-12 the traces of the plane of the polygon should first be determined. Unit of measure, | incb. Space required for each problem, 5x7 inches. Angles between GL and traces of planes, multiples of 15=. Measurements from GL, in lig^ht tjrpe, and from right-hand division line, in heavy type. PLATE 6 Draw the prs. of an equilat- eral triangle on plane N. Its center is point a, -6, -4. One side of the triangle is 7 units long and parallel to V. Dra-w the prs. of a regular hexagon on plane R. Point b is one extremity of a long diameter coinciding with line A. Side of hexagon is 5 units long. Draw the prs. of a square on plane S. Its center is point c, 8, 5. One side is 8 units long and at an angle of 60° with HS. Dra'w the prs. of a regular octagon on plane T. Its cen- ter is point d, -8, -5. Its short diameter is 8 units long and parallel to H. '20 Dra-w the prs. of a circle tangent to lines A and B. Its diameter is 8 units. Draw the prs. of a circle tangent to lines A and B. Its diameter is lO units. Draw the prs. of a circle tangent to the traces of plane M. Its diameter is 1 2 units. Draw the prs. of a circle tangent to the traces of plane W, and lying in 3Q. Its di- ameter is 16 units. Dra^ the projections of the figures indicated. (Art. 45, page 35J.) Unit of measure, } inch. Space required for eacli problem, 2J x 3 inches. Angles between GL and traces of Dl ATC A planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. "^-A I C. O Plane S is perpendicular to V Determine the prs. of the line of intersection between planes N and S. (Arts. 49, SO, page 35, Art. 52, page 37.) Unit of measure, } inch. Space required -for each problem, 2J x 3 inches. Angles betw^een OL and traces of PI ATE 7 lianes, multiples of 15^. Meaaurenaenta from GL, In light type, and from right-hand divlaion line, In heavy type. HM H8 VN VN 9 8 VS vs HH HS S passes through IQ and 3QatC0°with V S passes through 2Q and 4Qat30' with H Determine the prs. of the line of intersection ber«-een planes N and S. (Arta. 53-S6. pages 37-40.) Problems 1-4. Solve by Case 2. (Art. 51, page 36.) Problems 5-12. Solve by Case 3. (Art. 54, page 38.) Unit of measure, J Inch. Space required for each problem, 2.^x3 inches. Angles between GL and traces of qi a^c O planes, multiples of 15°. Measurements from GL, In light type, and from right-hand division line, in heavy type. "LA I C. O ^^1 14; loj 6 9^ HN \<^^ Il8 ^ 2 8 \^ VN \<^«. 10 VN HN 9 12 -6 22 16 Determine the pra. of the point in which line A pierces plane N. Indicate the prs. of the point by c^ and c*^. (Arts. 57-59, page 42.) Unit of zneasnre, | inch. Space required for each problem, 2^ x 3 inches. Angles between GIj and traces of Dl ATP Q I )lape8, multiples of 15-. Meaaurementa from GL, in light type, and from right-hand diviaion line, in heavy type. ' ^'^ I t l7 a -3 VN 6' '2 !l2 6*1 2 1 HM — 5 O ^8 b'l^ VN HM i|2 a -2 b'U a'-r9 bi-9 6^8 r* 12 a -2 a* 4 6*6 b'ls HM 6t2 a^8 1 -^ 8 1 61- — 1 <, J^> .6 ! 'i^ -^^ 6, ;« ^k 0' 1 1 I3 1 3| ^^ 1 1 ^■^ 1 1 1 1 8' B» ^e! Problems 1-8. Determine Problems 9-12. Determine the prs. the prs. of the point in -which line ab pierces plane N. (Art. 60. page 42.) of the point In which line A pierces the plane of lines B and D. (Art. 61, page 44.) Unit of measure, | Inch. Space required for each problem, 2^x3 Inches. Measurements from QL, in light PIAfP 1A type, and from right-hand division line,, in heavy type. t^^-^^ io[\ \ B 8 \ 2 ^^ C 191 18 112 6 1 2 ^1 B 6' ' 6 >i^ 1 1 1 2^ 120 14 !l2/ /^,2 1 1 14 1 k 6/0 10 6 2 17 1^ 7 6 ^0 K ' 6| Q \ v|<, 9 II Problems 1-4. Determine the prs. of the point in •which the line pierces the polygon. Problems 5-15. Determine the prs. of the points in which the line pierces the object. (Art. 61, page 44.) (Art. 61, page 44.) Unit of measure, | Inch. Space required for each problem, 2^ x 3 inches. Angles between OL and traces of Dl A^C 11 planes, multiples of 15-. Measurements from GL, in light type, and from right-hand division line, in heav>' type. • LM I & II 20^ \ aj6 >^y^ y^ J ^^ 2 afs 'l4 5 6 3 17 ^\lO 4- 1 1 a''8 ^ J^y^ 1 "-^^ 1 ^ fl*-4 1 .' a-8 ^^ VH -8 5 HN 8 6 h ar8 y" . 7 I6X 12 8 a'j6 1 Hit k a-r6 1 1 1 !l2 1 1 '12 *1 a-8 1 1 ^* i/l2 5 1 0x8 1 .! a-8 12 ^ 9 11/ /" ws -8" //S 12 1 b\2 f^ A 12 -4 12 Problems 1-8. Determine the prs. and true length of the line, ab, measuring the shortest distance from point a to plane N. (Arts. 62-65, pages 44, 45.) Problems 9-12. Determine the prs. of a perpendicular, ab, to plane S at point b on S. Line ab to be 8 units long. (Art. 36, page 24: Arts. 62-65, pages 44, 45.) Unit of measure, \ inch. Space required for eacb problem, 5 type, and from right-band division line, in heavy type. 7 inches. Measurements from GL, in light PIATF 10 K -28- /'^^^ '4-v"r TV'I Problems 1, 2, 5-7. Determine the prs. of the shado-ws of the object on itself and on V and H. (Arts. 66-69, pages 46-48.) Problems 3, 4. Determine the prs. of the shadow of the chimney on itself and on the roof. Problem 8. Determine the prs. of the shadoTV of the bracket on itself and on V. planes Unit of measure, | inch. Space required for each problem, 2V x 3 inches. Angles between OL and traces of Dl ATCT 1 *) nes, multiples of 15=. Measurements from GL, in light type, and from right-hand division line, in heavy type. • LM I C I O \ \ «*6 1 \^ a'rl 1 1 1 1 2 -9 3 -6 HN 4 \ a +3 a'j3 \ !i« - \ i|2 a.\ I6\^ 16 \i .12 il2 ^\^ 1^ a^6 \^ .•u VH 9 aVe 5 6 V 7 <.'t» ^ 1 ? f^a'-A Xc'-* a'|2 8 - 16 / 1 / '8 20! 16 20 8 16 14 / 7 2 .•U 'K >< / - 9^ C" 6't6 1 3 j 6 9 5 • il8 10 1"' -9^ « II 6' le 7 2 -1 12 |I6 8 ..i A. 1 ' 1 5\ ;i2 6 \l9 1 d' 1 O 6 12 -3 8 Problems 1-6. Determine the traces of the plane containing point a and parallel to plane Problems 7-12. Determine the traces of the plane containing point b and perpendicular to N. (Art. 71, page 50.) line C. (Art. 72, page 50.) Unit of measure, | inch. Space required for each problem, 2| x 3 inches. Angles between GL and traces of PIATE 14 planes, multlplea of 15°. Me asurements from GL, in light t ype, and from right-hand division line.ln heavy type. 2 9^8 9 \ \ %j^ ^5 18 '",<'■ |8 L- J 6 6 8. 1 \ 1 1 / ^l .Y ^2 61 \'=2 18 12 1 \ 1 12 'r\ <\ [l8 n \8 I0| \ 2*^^^ \^ 1 ^ Problems 1-4. Through point b pass a plane parallel to lines A and B. (Art. 73, page 51.) Problems 5-8. Through line A pass a plane parallel to line B. (Art 74, page 52.) Problems 9-12. Through line A pass a plane perpendicular to plane N. (Art. 75, page 62.) Unit of measure, J inch. Space required for each problem, 6x7 inches. Measurements from GL, in light qi atC 1R le. and from risrht-hand division line, in heavv tvne. ~ imf\ I C> t \f type, and from right-band division line, in heavy type fiil4 4 1 ■ J^ 19 \^'P 27 I8\|4 8 \^«.; ; 1^8 J 1 « / I 7 6 2* a •'-16 Determine the prs. and true length of the line measuring the shortest distance bet'sveen lines A and B. (Art. 79, page 54.) Note. — The problenti is best solved by passing the auxiliary plane through line A, Unit of measure, } Inch. Space required for each problem, 2^x3 inches. Angrles bet-ween GL and traces of qi atc 1 A planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. • ^'^ • C. I O 1 / 2 3 9n /4 2oX ^^ 14 20/ ||6 -4i 1 !6 - I6|\ J4 6 |22 iVii 1 1 1 1 1 1 ^ e^"""^ 1 7 s'^ ^^ ^\ 81^ ^^8 ^ 6 8r\ HN ^ HN /!% 6 f 6^ /9 7 afs V\^ 8 1 ^^^^v^ ^ j20 ^^1 > 16 - 110/ a- a" 2^ ^ |4 1 1 1 ^ ^ < 12 ■3 ^ 12 4 ■6 1^ VN ^^^'S a" /l9 9 •71-^.^^ >^° 10 =N li a" 9 12 je !6 K ^;. - 1 1 . \ 6 i^ U 1 2L 1 16 6 18 ^\ N2 a\2 ^■^^^ 1 ^^4 a ^"^6 6"^ 7 Problems 1-8. Determine the true size of the angle between line A and plane N. (Art. 80, page 54.) Problems 9-12. Determine the true size of the angle bet-^reen line B and V and H. (Art. 81, page 55.) Note. — Iietter the angle with V as X, and -with H as y. Unit of measure, } Inch. Space reqviired. for each problem, 5x7 inches. Angles between GL and traces of planes, multiples of 15^. Measurements from GL, in light type, and from right-hand division line, in heavy type. PLATE 17 33 28 13 8 Problems 1-4. Determine the tme size of the diedral angle between planes M and T. (Arts. 83-85 pages, 57, 58.) Problems 5-8. Determine the true sizes of the diedral angles of the objects. (Arts. 83-35, pages 57, 58.) Unit of measure, J inch. Space required for each problem, 2.^ x 3 inches. Angles between OL and traces of pi AXP 1ft planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. r L_M It 1 O \l4 2 4 ^^->^ \^ Jp^ 7 5 / 12/ 6 % 7 8 7 16/ 7 ^y^ ^y^ 2 9 V8 , 10 II ys 3 12 "^ 3 "^ n ^8 , 1 Determine the angle between plane S and V and H.. (Art. 86, page 68.) Note. — Letter the angle with V as x, and -with H as y. Unit of measure, I inch. Space required for the problem, 7 x lO inches. PLATE 19 HIP RAFTER. Determine length — down cut- heel cut — side cut — top bevel. JACK RAFTER. Determine side cut. PURLIN. Determine down cut — tide cut- angle between face and end. Determine the bevels, cuts, and lengths of roof members, aa above. (Art. 87, page 60.) Unit of measure, } inch. Space required for the problem, 7 x lO inches. PLATE 20 >./^. R. — Plane of web of Hip Rafter. P. ^ Plane of web of Purlin. S. — Plane perpe/idicular to line of in- tersection between R and P. Determine angrle of cut on top of purlin (A). Bevel on -web of purlin (B). Angle befween plane of -web of hip rafter and purlin, or bend of gusset (C). Angle between top edges of gusset (D). (Art. 87, page 61.) Unit of measure, j inch. Space required for each problem, 2i x 3 inches. Angles between GL and traces of Ol ATC 01 planes, multiples of 15~". Measurements from GL, in light type, and from right-hand division line, in heavy type. • L.r\ I ^ A I " ^ 2 60°with H. ^ TS'with H. * \ 45°with H. 30°with V. HH 7 5 / 6 / ^ 7 8 ^ 60°with H. 60°vyith H. eo'with V. 30°with V. 9 10 II 12 eo'with V. 45'with H. 75°w;th H. 45°with V. 30°v»ith H. eo'with V. 90°with H. 45'with V. Problems 1-8. Problems 9-12. Determine one position of the missing trace of plane N. (Arts. 88, 89, page 62.) Determine the traces of plane S making the g^iven angles t^tb V and H. (Art. 90, page 62.) Unit of measure, J Inch. Space required for each problem, 7 X lO inches. PLATE 22 In belt transmission one condition must always be obtained, namely : the point on a pulley from which the belt is delivered musr lie in the mid-plane of the pulley to which it is delivered. If the shafts are at right angles and placed as in the figure, they will run in the direction indicated, point b lying in the mid-plane of the large pulley, and point c lying iri the mid plane of the small pulley But if the direction be reversed a guide pulley would be necessary to compel the belt to fulfill the above conditions. The point at which the direction of the belt Is changed by the guide pulley is governed by con- venience. Problem I is that of a steering gear and requires one guide pulley. Prob- lem 2 illustrates a condition which necessitates two guide pulleys, although but one is to be determined. Problem 1. Draiv the projections of a guide pulley 8 units diameter, 2 units face, and determine the shaft angle with V and H. (Art. 45, page 32.) Problem 2. Dra-w the projections of one guide pulley 8 units diameter, 2 units face, and determine the shaft angle with V and H. (Art. 46, page 32.) Unit of measure, | inch. Space required for each problem, 7 x lO Inches. PLATE 23 Locate GL 38 units from lower margin line. Locate point b on GL 16 units from right-hand margin line. Draw the prs. of a regular hexagonal prism in IQ, resting on a plane, the vertical trace of which makes an angle of IS-" with GL, and the horizontal trace, an angle of 45^ with GL. The center of the base of the prism is a point of the plane, 17, 4 The sides of the hexagon are 6 units and two of the sides are perpendicular to the horizontal trace of the plane Altitude of prism is 12 units. The traces of the plane intersect GL in point b. Determine the shadow of the pnsm on the plane. Locate GL 38 units above lower margin line. Locate point b on GL 20 units from righi-hand margin line. Draw the prs. of a regular pentagonal pyramid in IQ, resting on a plane, the vertical trace of which makes an angle of 15° with GL, and the horizontal trace, an angle of 45° with GL The pyramid rests on its apex at a point 17, 4 Axis of pyramid is perpendicular to the plane and is 12 units long. Circumscribing circle of base of pyramid is 12 units in diameter. One side pf pentagon to be parallel to H. The traces of the plane intersect GL in point b. Determine the shadow of the pyramid on the plane. Locate GL 38 units from lower margin line. Locate point b on GL 16 units from right-hand margin line. Draw the prs. of a regular hexagonal pyramid in IQ, resting on a plane which makes an angle of 20' with H and 75° with V. The pyra- mid rests on its apex at a point in the plane, 18, 4. Axis of pyramid is perpendicular to the p'ane, and is 12 units long. The short diameter of the base is parallel to H; the long diameter is 12 units in length. The traces of the plane intersect GL in point b. Determine the shadow of the pyramid on the plane. Locate GL 38 units from lower margin line. Locate point b on GL 60 units from right-hand margin line. Draw the prs. of a cube in IQ, resting on a plane which makes an angle of SO'' with H and 65° with V. The center of the base of the cube is 10 units from the horizontal trace of the plane and 8 units from V. One diagonal of the base is parallel to H. Edge of cube is 8 units. The traces of the plane intersect GL in point b. Determine the shadow of the cube on the plane. Dra'w the prs. of a solid resting on an oblique plane, and determine the prs. of its shadow on the plane. Unit of measure, | inch. Space required for each problem, 5x7 Inches. Measurements from GL, in light type, pi AfC OA and from right-hand division line, in heavy type. Problems 1, 2, 8. Draw the traces of a plane which is tangent at point a of the surface. (Arts. 104, 105, pages 72-74.) Problems 3, 4, 7. Dra-w traces of planes w^hich are tangent to the surface and contain point a. (Art. 106, page 74 ; Art. 108,page75.) Problems 5, 6. Draw^ traces of planes which are tangent to the surface, and parallel to line B. (Art. 107, page 75 ; Art. 109, page 76.) Unit and from of me&sure, | inch. Space required for e»cii problem, 5x7 inches. Measarements from Gt>. in light typ«, qi A'^C f^ti im right-hand division line, in heavy tvpe. r ^^^ I & Axl Problems 1-4. Dra^e the traces of a plane •which is tangent at point a of the snrface. (Art. Ill, page 78.) Problems 5, 6. Draw traces of planes containing point a and tangent to the surface at given parallel. (Art. 112, page 79.) Problems 7, 8. Dra-n- the traces of planes tangent to the sphere and containing line A. (Art. 114 page 80.) Note. Problems 5-8. Determine the points of t»ngency. Unit of measure, J inch. Space required for each problem, 7 x 10 inches. Angles bet-ween GL and traces of planes, multiples of 15°. Measurements from GL, In light type, and from right-hand division line, In heavy type PLATE 26 Determine the Intersection of the plane -with the solid, and develop the surface. Problem 1. (Arts. 120, 121, pages 83, 84.) Problem 2. (Art. 127, page 92.) Problem 3. (Arts. 124, 125, page 88.) Problem 4. (Arts. 122, 123, page 86.) Unit at measure, } inch. Space required for eacb problem, 7 x lO inclies. PLATE 27 The generatrix makes an angle of 15^ with H, and one revolution. -31- Pitch 12 units ; length of longest elefnent, '9 units. Problems 1, 2. Dra-tr and develop an taelical convolute. (Arts. 129-131, pa^es 93-05.) Problems 3, 4. Delvelop one quarter of the lamp shade. Problem 3. (Art. 126, page 91.) Problem 4. (Art. 123, page 86.) Unit of measure, J inch. Space required for each problem, 5x7 inches. Angles bet\veen GL and traces of qi a "re Oft planes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type, i 1_M I H ^O A is the generatrix of an hyperboioid of revolu- tion. Determine the intersection of the plane N with the double-curved surface of revolution. (Art. 132, page 96.) Unit of measare, J inch. Space reqtdred for each problem, 7 x lO inches. Measnrementa from OL, in U^ht type, qi A"T"P OQ and from right-hand division line, in heavy type. r l-^ I C ^7 Determine the intersection between the solids. Unit of measure, J Inch. Space required for each problem, 7 x lO inches. Measurements from GL, in light type, and from right-hand division line, in heavy type. PLATE 30 37 28 16 35 28 Determine the intersection betiveeu the solids. t— -22— , Unit of measure, J inch. Space required for eacb problem, 6x7 inches. Measurement3 from GL, in light type, and from right-hand division line, in heavy type. PLATE 31 29 1 4— «r 7 - 1 ^ 'IS T 1 V V 1^ in Determine the intersection bet^veen the solids. The unit of measurement is 1/16 inob. Space required 7 x lO inches. PLATE 32 Oeveloprnent of I Dome drawn here. A sectional and end view of a portion of a boiler is shown. It is reqi/ired to develop one half of the dome, and that part of the slope sheet lying between the elements ef and cd, adding the amount necessary for flanging or lap _ Make two full views of the boiler, indicating the thickness of plate by a single line. Determine the projection of the two curves of the flange ab, observing that the radial elements ot the flange make angles varying from 0^ to y^. The elements of the slope sheet will be parallel to each other and to the vertical plane, thus appearing in their true lengths in vertical projection. The development may be obtained by either of the followingf methods, the first being that used in practice. First method: the elements being parallel to V, the development may be obtained .as in Art. 126, page 91, the amount indicated for the lap being added to the sheet. Second method : obtain a right section VX by Art 118, page 82, the development will be a right line, mn. To obtain the distance between the elements revolve the plane of the right section to P, and measure the required distances in the profile projection, or end view. Develop as in Art. 125, page 8S Draw and develop the dome and connection sheet, or slope sheet, of a locomotive boiler. Unit of measure, | Inch. Space required for each problem, 5x7 inches. Angles between GL and traces of pi A-rC QO planes, multiples of 15^. Measurements from GL, in light type, and from right-hand division line, in heavy type. ~ *~'^ ' ^ '**' Problems 1, 2. I>Taw an element of fhe -warped surface through point a. (Art. 145, page llO.) Problems 3, 4. Draw an element of the warped surface through point a. (Art. 146, page 111.) Problems 5, 6. Draw an element of the warped surface parallel to line C of the plane director, N. (Art. 147, page 112.) Problems 7, 8. Draw an element of the hyperbolic paraboloid through point a, of a directrix. (Art. ISO, page 116.) Unit of measure, | Inch. Space required for each problem, 6x7 Inches. Angles between GL and traces of ni ATP "^A planes, multiples of 16°. Measurements from GL, in light type, and from right-hand dlvison line, in heavy type. ' ^" ' ^ "^ A is the generatrix of an hyper- boloid of revolution. A Is the generatrix of an hyper- boloid of revolution. A is the generatrix of an hyper- boloid of revolution. A is the generatrix of an hyper- boloid of revolution. / 12 Z-. 1 y 12 3e~- to / 10 22 C "^ \ A \^c M nor Axis 12 X Problems 1, 2. Draw an element of the hyperbolic paraboloid through point a. (Art. 161, page 116.) Problems 3, 4. Draw^ an element through point a. (Art. 156, page 120.) Problems 6, 6. Draw a plane tangent to the surface at point a. (Art. 159, page 121.) Problems 7, S. Througli line A pass a plane tangent to the surfaca. (Art. 160, page 122.) .»i^. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. 5Hov'48)C 23Mar'54MC Ubhary use AUG 2 2 J957 200cV5^-VD REC"D LD RECD LD SEP26'65-4PM to ?;*o, YC 22387 laoT^'/ Q.A5-C 4 4 UNIVERSITY OF CALIFORNIA LIBIL\RY