A NEW SYSTEM 
 
 OF 
 
 HAND-RAILING. 
 
 CUT SQUARE TO THE PLANK, WITHOUT THE 
 AID OF FALLING MOULDS, 
 
 A NEW AND EASY METHOD OF FORMING HAND-RAILS. 
 
 Containing a Large Number of Illustrations of Hand-Kails, 
 with Full Instructions for Working Them. 
 
 AN OLD STAIR-BUILDER 
 
 REVISED AND CORRECTED EDITION 
 
 NEW YORK: 
 THE INDUSTRIAL PUBLICATION COMPANY. 
 
 1885.
 
 * *
 
 CVUFORN 
 i A UAUBARA 
 
 PREFACE. 
 
 THE following method of getting out wreaths for hand-railing, while 
 i! ! ii -w. is nut generallyknowu in this country, which is to be regretted. 
 i;\ tlii.- method much time mid material arc saved, and the wreaths formed 
 with mon; accuracy than cutting them out by the older methods, as any 
 wreath whatever may be cut out square Iroin a plank of the same thickness 
 as the diameter of a circle, described round a section of the mil. 
 
 In the annexed cut, A B shows the thickness of the plank. But it is 
 1 . 1 1 ! for the learner to allow the plank 4 inch thicker, as it leaves more for 
 
 squaring the rail. By this method the same bevel applies to both ends of 
 the wreath an the square joint, for stairs with any number of winders in the 
 well: tills ensures the ra ll rising equally all round the weU, and not quick 
 in one part and flat in another. 
 
 All wreaths are cut out square from the plank with square joints, and 
 Boaoourate are the bevels and joints that the rail may he H,,med up ready 
 lor polishing before it leaves the bench. 
 
 Seven complete examples are RivRD, with ample illustrations and 
 .le,eri|,livetext .; these examples IHn- qu j ( .. slim ,.j.. n) 1o ,, nnh j c tho R t.He,u 
 to build a rail over any kind of a stair, no matter l,,,xv many winder* H.e.v 
 may l>e or how they are arranged. 
 
 New York, 1885.
 
 INTRODUCTION. 
 
 BEFOBE proceeding to describe the method of obtaining the lines 
 necessary for the correct formation of Hand-mils, it is deemed advis;ible 
 to present for the student's use a rudimentary treatise on what is known 
 as CAKPENTEK'S GEOMETKY, as it is felt that a knowledge of this will ma- 
 terially aid the young beginner to a better understanding of the principles 
 involved in the work. 
 
 The following treatise, which is taken from "Practical Carpentry," is 
 simple and easily understood. Let me add, however, that, although a 
 knowledge of geometry will greatly help the student to understand the 
 method herein described, such knowledge is not absolutely necessary, as 
 the Art of Hand-railing, as here presented, may be efficiently acquired 
 without it.
 
 TABLE OF CONTENTS. 
 
 PART I. 
 
 PAG!? 
 
 GEOMETRY. Straight Lines. Curved Lines. Solids. Compound Lines. 
 Parallel Lines. Oblique or Converging Lines. Plain' l-'iirmv<. 
 Angles. Right Angles. Acute Angles. Obtuse Angles. Right-angled 
 Triangles. Quadrilateral Figures. Parallelograms. Rectan _ 
 Squares. Rhomboids. Trapeziums. Trapezoids. Diagonals. Poly- 
 gons. Pentagons. Hexagons. Heptagons. Octagons. Circles. 
 Chords. -Tangents. Sectors. Quadrants. Arcs. Concentric and Ec- 
 centric Circles. Altitudes. Problems I. to XXIX. Drawing of Angles, 
 ('(instruction of (ieometrical Figures. Bisection of Lines.- Trisection 
 of Lines and Angles. Division of Lines into any Number of Parts. 
 Con>t ruction of Triangles, Squares and Parallelograms. Construction of 
 Proportionate Squares. Construction of Polygons. Areas of Polygons. 
 Areas of Concentric Rings and Circles. Segments of Circles. The u-e 
 of Ordinates for Obtaining Arcs ot Circles. -Drawing an Ellipse with a 
 Trammel. Drawing an Ellipse by means ot a String. Same by Ordin- 
 ates. Raking Ellipses. Ovals. Sixty-two Illustrations. Tangents and 
 Diagrams for Handrailing. -Three Illustrations, 7-34 
 
 PART II. 
 
 Lines and Method for Making the Wreaths for Stairs with Four Winders 
 in the Well. Four Illustrations and Diagrams. Lines and Method for 
 Making the Wreaths for Stairs with Six Winders in the Weil. Three 
 Illustrations and Working Diagrams. Lines and Methoi for Making the 
 Wivatlis lor Stairs with Three Winders in Half the Well, and Landing 
 in the Centre of the Well. Four Illustrations and Working Diagrams. 
 Method of Making the Wreaths for Stairs with One Riser iu the Centre of 
 Well. With Three Illustrations and Working Diagrams. How to Make 
 the Wreaths for Level Landing Stairs, the distance between Centre of 
 Rails across the Well being equal to the Tread of a Step, or nearly so. 
 With Three Illustrations and Working Diagrams. How to Make the 
 Wreath tor Quarter-landing Stairs, having One Winder in the Well. 
 Three Illustrations and Working Diagrams. How to Make a Wreath for 
 Quarter-landing stairs, with Itisers on the Line of Spring. Three Illus- 
 trations and Working Diagrams. - - :i.')-65
 
 HAND-RAILING. 
 
 PART I.-GEOMETRT. 
 
 1EFORE a knowledge of geometry can be acquired, it will 
 be necessary to become acquainted with some of the 
 terms and definitions used in the science of geometry, 
 and to this end the following terms and explanations are given, 
 though it must be understood that these are only a few of the 
 terms used in the science, but they are sufficient for our purposes : 
 
 1. A 'Point has position but not magnitude. Practically, it is 
 represented by the smallest visible mark or dot, but geometrically 
 understood, it occupies no space. The extremities or ends of lines 
 are points; and when two or more lines cross one another, the 
 places that mark their intersections are also points. 
 
 2. A Line has length, without breadth or thickness, and, conse- 
 quently, a true geometrical line cannot be exhibited ; for however 
 finely a line may be drawn, it will always occupy a certain extent 
 of space. 
 
 3. A Superficies or Surface has length and breadth, but no thick- 
 ness. For instance, a shadow gives a very good representation of 
 a superficies : its length and breadth can be measured ; but it has 
 no depth or substance. The quantity of space contained in any 
 plane surface is called its area. 
 
 4. A Plane Superficies is a flat surface, which will coincide with 
 a straight line in every direction.
 
 10 HAND-RAILING. 
 
 5. A Curved Superficies is an uneven surface, or such as will not 
 coincide with a straight line in all directions. By the term surface 
 is generally understood the outside of any body or object; as, for 
 instance, the exterior of a brick or stone, the boundaries of which 
 are represented by lines, either straight or curved, according to the 
 form of the object. We must always bear in mind, however, that 
 the lines thus bounding the figure occupy no part of the surface ; 
 hence the lines or points traced or marked on any body or 
 surface, are merely symbols of the true geometrical lines or 
 points. 
 
 6. A Solid is anything which has length, breadth and thickness ; 
 consequently, the term may be applied to any visible object con- 
 taining substance ; but, practically, it is understood to signify the 
 solid contents or measurement contained within the different sur- 
 faces of which any body is formed. 
 
 7. Lines may be drawn in any direction, and are termed straight, 
 curved, mixed, concave, or convex lines, according as they corres- 
 pond to the following definitions. 
 
 8. A Straight Line is one every part of which A B 
 lies in the same direction between its extremities, Flg- I- 
 and is, of course, the shortest distance between two pbints, as 
 from A to B, Fig. i. 
 
 9. A Curved Line is such that it does not lie in a straight direc- 
 tion between its extremities, but is continually changing by inflec- 
 tion. It may be either regular or irregular. 
 
 10. A Mixed or Compound Line is composed of straight and 
 curved lines, connected in any form. 
 
 11. A Concave or Convex Line is such that it cannot be cut by 
 a straight line in more than two points ; the concave or hollow 
 side is turned towards the straight line, while the convex or swell- 
 ing side looks away from it. For instance, the inside of a basin is 
 concave the outside of a ball is convex. 
 
 12. Parallel Straight Lines have no inclination, but are every- 
 where at an equal distance from each other; consequently they 
 can never meet, though produced or continued to infinity in either 
 or both directions. Parallel lines may be either straight or curved,
 
 HAND-RAILING. II 
 
 provided they are equally distant from each other throughout their 
 extension. 
 
 13. Oblique or Converging Lines are straight lines, which, if con- 
 tinued, being in the same plane, change their distance so as to 
 meet or intersect each other. 
 
 14. A Plane Figure, Scheme, or Diagram, is the lineal representa- 
 tion of any object on a plane surface. If it is bounded by straight 
 lines, it is called a rectilineal figure ; and if by curved lines, a 
 curvilineal figure. 
 
 15. An Angle is formed by the inclination of two lines meeting 
 in a point : the lines thus forming the angle are called the sides ; 
 and the point where the lines meet is called the vertex or angular 
 point. 
 
 When an angle is expressed by three letters, as A u c, Fig. 2, the 
 middle letter u should always denote the angular point: where 
 
 Fig. 3. 
 
 Fig. 4. 
 
 Fig. 5. 
 
 there is only one angle, it may be expressed more concisely by a 
 letter placed at the angular point only, as the angle at A, Fig. 3. 
 
 1 6. The quantity of an angle is estimated by the arc of any 
 circle contained between the two sides or lines forming the angle ; 
 the junction of the two lines, or vertex of the angle, being the 
 centre from which the arc is described. As the circumferences of 
 all circles are proportional to their diameters, the arcs of similar 
 sectors also bear the same proportion to their respective circum- 
 ferences; and, consequently, are proportional to their diameters, 
 and, of course, also to their radii or semi-diameters. Hence, the
 
 12 HAND-RAILING. 
 
 proportion which the arc of any circle bears to the circumference 
 of that circle, determines the magnitude of the angle. From this 
 it is evident that the quantity or magnitude of angles does not de- 
 pend upon the length of the sides or radii forming them, but 
 wholly upon the number of degrees contained in the arc cut from 
 the circumference of the circle by the opening of these lines. The 
 circumference of every circle is divided by mathematicians into 
 360 equal parts, called degrees ; each degree being again subdi- 
 vided into 60 equal parts, called minutes, and each minute into 60 
 parts, called seconds. Hence, it follows that the arc of a quarter 
 circle or quadrant includes 90 degrees ; that is, one-fourth part of 
 360 degrees. By dividing a quarter circle; that is, the portion of 
 the circumference of any circle contained between two radii form- 
 ing a right angle, into 90 equal parts, or, as is shown in Fig. 4, into 
 nine equal parts of 10 degrees each, then drawing straight lines 
 from the centre through each point of division in the arc ; the right 
 angle will be divided into nine equal angles, each containing 10 
 degrees. Thus, suppose B c the horizontal line, and A n the per- 
 pendicular ascending from it, any line drawn from B the centre 
 from which the arc is described to any point in its circumference, 
 determines the degree of inclination or angle formed between it 
 and the horizontal line B c. Thus, a line from the centre B to the 
 tenth degree, separates an angle of 10 degrees, and so on. In this 
 manner the various slopes or inclinations of angles are defined. 
 
 17. A Right Angle is produced by one straight line standing 
 upon another, so as to make the adjacent angles equal. This is 
 what workmen call " square," and is the most useful figure they 
 employ. 
 
 1 8. An Acute Angle is less than a right angle, or less than 90 
 degrees. 
 
 19. An Obtuse Angle is greater than a right angle or square, oi 
 more than 90 degrees. 
 
 The number of degrees by which an angle is less than 90 de- 
 grees is called the complement of the angle. Also, the difference 
 between an obtuse angle and a semicircle, or 180 degrees, is called 
 the supplement of that angle.
 
 13 
 
 20. Plane Figures are bounded by straight lines, and are named 
 according to the number of sides which they contain. Thus, the 
 space included within three straight lines, forming three angles, 
 is called a trilateral figure or triangle. 
 
 21. A Right-Angled Triangle has one right angle: the sides 
 forming the right angle are called the base and perpendicular; and 
 the side opposite the right angle is named the hypothenuse. An 
 equilateral triangle has all its sides of equal length. An isosceles 
 triangle has only two sides equal ; a scalene triangle has all its 
 sides unequal. An acute-angled triangle has all its angles acute, 
 and an obtused-angled triangle has one of its angles only 
 obtuse. 
 
 The triangle is one of the most useful geometrical figures for the 
 mechanic in taking dimensions; for since all figures that are 
 bounded by straight lines are capable of being divided into tri- 
 angles, and as the form of a triangle cannot be altered without 
 changing the length of some one of its sides, it follows that the true 
 form of any figure can be preserved if the length of the s : des of 
 the different triangles into which it is divided is kaown ; and the 
 area of any triangle can easily be ascertained by the same rr.le, as 
 will be shown further on. 
 
 Quadrilateral Figures are literally four-sided figures. They are 
 also called quadrangles, because they have four angles. 
 
 22. A Parallelogram is a figure whose opposite sides are parallel, 
 as A B c D, Fig. 5. 
 
 23. A Rectangle is a parallelogram having four right angles, as A 
 BCD, in Fig. 5. 
 
 24. A Square is an equilateral rectangle, having all its sidi-s 
 equal, like Fig. 5. 
 
 25. An Oblong is a rectangle whose adjacent sides are unequal, 
 as the parallelogram shown at Fig. TO. 
 
 26. A Rhombus is an oblique-angled figure, or parallelogram 
 having four equal sides, whose opposite angles only are equal, as 
 c, Fig. 6. 
 
 27. A Rhomboid is an oblique-angled parallelogram, of which the 
 adjoining sides are unequal, as D, Fig. 7.
 
 HAND-RAILING. 
 
 .28. A Trapezium is an irregular quadrilateral figure, having no 
 two sides parallel, as E, Fig. 8. 
 
 29. A Trapczoid is a quadrilateral figure, which has two of its 
 opposite sides parallel, and the remaining two neither parallel nor 
 equal to one another, as F, Fig. 9. 
 
 Fig. 7. 
 
 Fig. 8. 
 
 30. A Diagonal is a straight line drawn between two opposite 
 angular points of a quadrilateral figure, or between any two angular 
 points of -a polygon. Should the figure be a parallelogram, the 
 diagonal will divide it into two equal triangles, the opposite sides 
 and angles of which will be equal to one another. Let A B c D, 
 Fig. 10, be a parallelogram; join A c, then A c is a diagonal, and 
 the triangles A D c, A B c, into which it divides the parallelogram, 
 are equal. 
 
 31. A plane figure, bounded by more than four straight lines, is 
 called a Polygon. A regular polygon has all its sides equal, and 
 consequently its angles are also equal, as K, L, M, and N, Figs. 
 
 Fig. 11. 
 
 Fig. 12. 
 
 Fig. 13. 
 
 12-15. An i rre g u l ar polygon has its sides and angles unequal, as 
 H, Fig. 1 1. Polygons are named according to the number of their 
 sides or angles, as follows : 
 
 32. A Pentagon is a polygon of five sides, as H or K, Figs, u, 12. 
 
 33. A Hexagon is a polygon of six sides, as L, Fig. 13. 
 
 34. A Heptagon has seven sides, as M, Fig. 14.
 
 HAND-RAILING. 
 
 35. An Octagon has eight sides, as N, Fig. 15. 
 
 An Enneagnt has nine, a Decagon ten, an Undecagon eleven, and 
 a Dodecagon twelve sides. Figures having more than twelve sides 
 are generally designated Polygons, or many-angled figures. 
 
 36. A Circle is a plane figure bounded by one uniformly curv I 
 line, bed (Fig. 16), called the circumference, every part of which 
 is equally distant from a point "within it, called the centre, as a. 
 
 37. The Radius of a circle is a straight line drawn from the 
 centre to the circumference ; hence, all the radii (plural for radius) 
 of the same circle are equal, as b a, c a, e a,f a, in Fig. 16. 
 
 38. The Diameter of a circle is a straight line drawn through the 
 centre, and terminated on each side by the circumference; conse- 
 
 X 
 
 4 
 
 ig. 16. 
 
 Fig. }7 f 
 
 quently the diameter is exactly twice the length of the radius ; and 
 hence the radius is sometimes called the semi-diameter. (See b a c^ 
 Fig. 1 6.) 
 
 49. The Chord or Subtens of an arc is any straight line drawn 
 from one point in the circumference of a circle to another, joining 
 the extremities of the arc, and dividing the circle either into two 
 equal, or two unequal parts. If into equal parts, the chord is also 
 the diameter, and the space included between the arc and the di- 
 ameter, on either side of it, is called a semicircle, as bae'm Fig. 16. 
 If the parts cut off by the chord are unequal, each of them is called 
 a segment of the circle. The same chord is therefore common to 
 two arcs and two segments ; but, unless when stated otherwise, it 
 is always understood that the lesser arc or segment is spoken of, as 
 in Fig. 1 6, the chord c d is the chord of the arc c e d. 
 
 If a straight line be drawn from the centre of a circle to meet 
 the chord of an arc perpendicularly, as af y in Fig. 16, it will divide 
 the chord into two equal parts, and if the straight line be produced
 
 1 6 ( HAND- RAILING. 
 
 to meet the arc, it will also divide it into two equal parts, as 
 cfjd. 
 
 Each half of the chord is catted the sine of the half-arc to which 
 it is opposite; and the line drawn from the centre to meet the 
 chord perpendicularly, is called the co-sine of the half-arc. Con- 
 sequently, the radius, the sine, and co-sine of an arc form a right 
 angle. 
 
 40. Any line which cuts the circumference in two points, or a 
 chord lengthened out so as to extend beyond the boundaries of 
 the circle, such as g h in Fig. 17, is sometimes called a Secant. 
 But, in trigonometry, the secant is a line drawn from the centre 
 through one extremity of the arc, so as to meet the tangent which is 
 drawn from the other extremity at right angles to the radius. 
 Thus, vcb is the secant of the arc c e, or the angle c $ e, in Fig. 17. 
 
 41. A Tangent is any straight line which touches the circumfer- 
 ence of a circle in one point, which is called the point of contact, 
 as in the tangent line e b, Fig. 17. 
 
 42. A Sector is the space included between any two radii, and 
 that portion of the circumference comprised between them : c e v' 
 is a sector of the circle afc e, Fig. 17. 
 
 43. A Quadrant, or quarter of a circle, is a sector bounded by 
 two radii, forming a right angle at the centre, and having one- 
 fourth part of the circumference for its arc, as F/V, Fig. 7. 
 
 44. An Arc, or Arch, is any portion of the circumference of a 
 circle, as c d e, Fig. 17. 
 
 It may not be improper to remark here that the terms circle and 
 circumference are frequently misapplied. Thus we say, describe a 
 circle from a given point, etc., instead of saying describe the cir- 
 cumference of a circle the circumference being the curved line 
 thus described, everywhere equally distant from a point within it, 
 called the centre ; whereas the circle is properly the superficial 
 space included within that circumference. 
 
 45. Concentric Circles are circles within circles, described from 
 the same centre; consequently, their circumferences are parallel to 
 one another, as Fig. 18. 
 
 46. Eccentric Circles are those which are not described from the
 
 HAND-FAILING. iy 
 
 same centre; any point which is not the centre is also eccentric in 
 reference to the circumference of that circle. Eccentric circles 
 may also be tangent circles ; that is, such as come in contact in 
 one point only, as Fig. 19. 
 
 47. Altitude. The height of a triangle or other figure is called 
 its altitude. To measure the altitude, let fall a straight line from 
 the vertex, or highest point in the figure, perpendicular to the 
 base or opposite side ; or to the base continued, as at B D, Fig. 20, 
 should the form of the figure require its extension. Thus c D is the 
 altitude of the triangle ABC. 
 
 We have now described all the figures we shall require for the 
 purpose of thoroughly understanding all that will follow in this 
 book ; but we would like to say right here that the student who 
 has time should not stop at this point in the study of geometry, for 
 the time spent in obtaining a thorough knowledge of this useful 
 
 Fig. 18. Fig. 19. Fig. 20. 
 
 science will bring in better returns in enjoyment and money, than 
 if expended for any other purpose. 
 
 We will now proceed to explain how the figures we have de- 
 scribed can be constructed. There are several ways of constructing 
 nearly every figure we produce, but we have chosen those methods 
 that seemed to us the best, and to save space have given as few 
 examples as possible consistent with efficiency. 
 
 PROBLEM I. Through a given point c (Fig. 18 a), to draw a 
 straight line pat allel to a given straight line A B. 
 
 In A B (Fig. 1 8 a) take any point <t, and from d as a centre with 
 the radius d c, describe an arc c <?, cutting A B in f, and from c as a
 
 i8 
 
 HAND-RAILING, 
 
 centre, with the same radius, describe the arc //D, make dT> equal to 
 c e, join c D, and it will be parallel to A r. 
 
 PROBLEM II. To make an angle equal to a given rectilineal angle. 
 
 From a given point E (Fig. 19^), upon the straight line E K, to 
 make an angle equal to the given angle ABC. From the angular 
 point B, with any radius, describe the arc<?/ cutting B c and B A in 
 the points e and/ From the point E on E F with the^same radius, 
 
 Fig. 18 a. 
 
 Fig, 19 a. 
 
 /C 
 
 Fig. 20 . 
 
 describe the arc kg, and make it equal to the arc e f\ then from E, 
 through g, draw the line E D : the angle D E F will be equal to the 
 angle ABC. 
 
 PROBLEM III. To bisect a given angle. 
 
 Let ABC (Fig. 20 a) be the given angle. From the angular point 
 B, with any radius, describe an arc cutting B A and B c in the 
 points //and e; also, from the points ^/and e as centres, with any 
 radius greater than half the distance between them, describe arcs 
 cutting each other in/; through the points of intersection/ draw 
 B/D : the angle A B c is bisected by the straight line B D; that is, 
 it is divided into two equal angles, A B D and c B D. 
 
 PROBLEM IV. To trisect or divide a right angle into tJiree equal 
 angles. 
 
 Let ABC (Fig. 21) be the given right angle. From the angular
 
 HAND-RAILING. 
 
 point B, with any radius, describe an arc cutting B A and B c in the 
 points d and g\ from the points d a.ndg, with the radius B d or B g t 
 describe the arcs cutting the arc d gine and/; join i; e and B/: 
 these lines will trisect the angle A B c, or divide it into three equal 
 angles. 
 
 The trisection of an angle can be effected by means of elementary 
 geometry only in a very few cases; such, for instance, as those 
 where the arc which measures the proposed angle is a whole circle, 
 or a half, a fourth, or a fifth part of the circumference. Any angle 
 of a pentagon is trisected by diagonals, drawn to its opposite 
 angles. 
 
 PROBLEM V. From a given point c, in a given straight line A B, 
 to erect a perpendicular. 
 
 From the point c (Fig. 22), with any radius less than c A or c B, 
 describe arcs cutting the given line A B in d and e from these 
 
 Fig. 21. 
 
 Fig. 22. 
 
 points as centres, with a radius greater than c d or c e, describe arcs 
 intersecting each other in/: join c/, and this line will be the per- 
 pendicular required. 
 
 Another Method. To draw a right angle or erect a perpendicular 
 by means of any scale of equal parts, or standard measure of inches, 
 feet, yards, etc., by setting off distances in prooortion to the num- 
 bers 3, 4 and 5, or 6, 8 and 10, or any numbers whose squares cor- 
 respond to the sides and hypothenuse of a right-angled triangle. 
 
 From any scale of equal parts, as that represented by the line D 
 (Fig. 23), which contains 5, set off from B, on the line A B, the dis- 
 tance B e, equal to 3 of these parts; then from B, with a radius 
 equal to 4 of the same parts, describe the arc a .; also from e as a
 
 centre, with a radius equal to 5 parts, describe another arc inter- 
 secting the former in c ; lastly join D c ; the line u c will be per- 
 pendicular to A B. 
 
 This mode of drawing right angles is more troublesome upon 
 paper than the method previously given ; but in laying out grounds 
 or foundations of buildings it is often very useful, since only with a 
 ten-foot pole, tape line, or chain, perpendiculars may be set out 
 very accurately. The method is demonstrated thus : The square 
 of the hypothenuse, or longest side of a right-angled triangle, being 
 equal to the sum of the squares of the other two sides, the same 
 property must always be inherent in any three numbers, of which 
 
 o 
 
 Fig. 23. 
 
 the squares of the two lesser numbers, added together, are equal to 
 the square of the greater. For example, take the numbers 3, 4, 
 and 5 ; the square of 3 is 9, and the square of 4 is 16 ; 16 and 9, 
 added together make 25, which is 5 times 5, or the square of the 
 greater number. Although these numbers, or any multiple of 
 them, such as 6, 8, 10, or 12, 16, 20, etc., are the most simple, and 
 most easily retained in the memory, yet there are other numbers, 
 very different in proportion, which can be made to serve the same 
 purpose. Let n denote any number ; then * + i, * i, and 2//, 
 will represent the hypothenuse, base, and perpendicular of a right- 
 angled triangle. Suppose ;/ = 6, then n 2 + i =37, n 2 i =35, 
 and 2= 12: hence, 37, 35, and 12 are the sides of a right-angled 
 triangle. A knowledge of this problem will often prove of the 
 greatest service to the workman.
 
 HAND-KAI! 21 
 
 PROBLEM VI. 7l> Insut <r ghrn straight line. 
 
 Let A B (Fig. 24) be the given straight line. From the extreme 
 points A and B as centres, with any equal radii greater than half 
 the length of A r>, describe arcs cutting each other in c and D: a 
 straight line drawn through the points of intersection c and D, will 
 bisect the line A u in e. 
 
 PROBLEM VII. To divide a given line into any number of equal 
 parts. 
 
 Let A B (Fig. 25) be the given line to be divided into five equal 
 parts. From the point A draw the straight line A c, forming any 
 angle with A B. On the line A c, with any convenient opening of 
 the compasses, set off five equal parts towards c; join the extreme 
 
 Fig. 25. Fig. 26. Fig. 27 
 
 points c B; through the remaining points i, 2, 3, and 4, draw lines 
 parallel to B c, cutting A i; in the corresponding points, i, 2, 3, 
 and 4 : A B will be divided into five equal parts, as required. 
 
 There are several other methods by which lines may be divided 
 into equal parts ; they are not necessary, however, for our purpose, 
 so we will content ourselves with showing how this problem may 
 be used for changing the scales of drawings whenever such change 
 is desired. Let A B (Fig. 26) represent the length of one scale or 
 drawing, divided into the given parts A d, rf e, ef,fg, gh, and // B ; 
 and D E the length of another scale or drawing required to be di- 
 vided into similar parts. From the point B draw a line B C = D E, 
 and forming any angle with A r, ; join A c, and through the points 
 ^ ^/j C> an( l h, draw d k, el, ftn, g n, h <?, parallel to A c ; and 
 the parts c k, k /, I m, etc., will be to each other, or to the whole 
 line i; c, as the lines A d, d c, c f, dr.. are to each other, or to the 
 given line or scale A B. By this method, as will be evident from
 
 HAND-RAILING. 
 
 the figure, similar divisions can be obtained in lines of any given 
 length. 
 
 PROBLEM VIII. To describe an equilateral triangle upon a given 
 straight line. 
 
 Let A B (Fig. 27) be the given straight line. From the points 
 A and B, with a radius equal to A B, describe arcs intersecting each 
 other in the point c. Join c A and c B, and ABC will be the 
 equilateral triangle required. 
 
 PROBLEM IX. To construct a triangle whose sides shall be equal 
 to three given lines, F, E, D. 
 
 Draw A B (Fig. 28) equal to the given line F. From A as a 
 centre, with a radius equal to the line E, describe an are; then 
 
 Pig. 28. 
 
 Fig. 29. 
 
 from B as a centre, with a radius equal to the line D, describe an- 
 other arc intersecting the former in c ; join c A and c B, and ABC 
 will be the triangle required. 
 
 PROBLEM X. To describe a rectangle or parallelogram having one 
 of its sides equal to a given line, and its area equal to that of a given 
 rectangle. 
 
 Let A B (Fig. 29) be the given line, and c D E F the given 
 rectangle. Produce c E to G, making E G equal to A B; from G 
 draw G K parallel to E F, and meeting D F produced in n. Draw 
 the diagonal G F, extending it to meet c D produced in L ; also 
 draw L K parallel to D H, and produce E F till it meet L K in M j 
 then F M K H is the rectangle required.
 
 HA \D-1-: A 1 1, 1 NT,. 
 
 Equal and similar rhomboids or parallelograms of any dimen- 
 sions may be drawn after the same manner, seeing the comple- 
 ments of the parallelograms which are described on or about the 
 diagonal of any parallelogram, are always equal to each other; 
 while the parallelograms themselves are always similar to each 
 other, and to the original parallelogram about the diagonal of 
 which they are constructed. Thus, in the parallelogram c c; K i, 
 the complements c E F n and F M K H are always equal, while the 
 parallelograms K F H G and D F M L about the diagonal o L, are 
 always similar to each other, and to the whole parallelogram 
 c G K L. 
 
 PROBLEM XI. To describe a square equal to Iwo given squares. 
 
 Let A and B (Fig. 30) be the given squares. Place them so that 
 a side of each may form the right angle D c E ; join D E, and upon 
 this hypothenuse descnbe the square D E G F, and it will be equal 
 to the sum of the squares A and B, which are 
 constructed upon the legs of the right-angled 
 triangle D c E. In the same manner, any 
 other rectilineal figure, or even circle, may 
 be found equal to the sum of other two simi- 
 lar figures or circles. Suppose the lines c D 
 and c E to be the diameters of two circles, 
 then D F. will be the diameter of s. third, 
 equal in area to the other two circles. Or 
 suppose c D and c E to be the like sides of 
 any two similar figures, then D E will be the 
 corresponding side of another similar figure 
 equal to both the former. 
 
 PROBLEM XII. To descnbe a square equal to any number of given 
 squares. 
 
 Let it be required to construct a square equal to the three givt-n 
 squares A, B, and c (Fig. 31). Take the line n E, equal to the side 
 of the square c. From the extremity D erect D F perpendicular to 
 D E, and equal to the side of the square B ; join E F; then a square 
 described upon this line will be equal to the sum of the two given 
 squares c and B. Again, upon the straight line E F erect the per- 
 
 Fig. 30.
 
 HAND-RAILING. 
 
 Fig. 31. 
 
 pendicular F G, equal to the side of the 
 third given square A ; and join G E, which 
 will be the side of the square G E H K, 
 equal in area to A, B, and c. Proceed in 
 the same way for any number of given 
 squares. 
 
 PROBLEM XIII. Upon a given straiglit 
 line to describe a regular polygon. 
 
 To produce a regular pentagon draw 
 A B to c (Fig. 32), so that B c may be 
 equal to A B ; from B as a centre, with the 
 radius B A or B c describe the se'micircle 
 ADC; divide the semi-circumference 
 ADC into as many equal parts as there 
 are parts in the required polygon, which, 
 in the present case, will be five ; through the second division from 
 C draw the straight line B D, which will form another side of the 
 figure. Bisect A B at e, and B D at/ and draw e G and /G per- 
 pendicular to A B and B D ; then G, the 
 point of intersection, is the centre of a 
 circle, of which A B and D are points 
 in the circumference. From G, with a 
 radius equal to its distance from any of 
 these points, describe the circumference 
 A B D H K ; then producing the dotted 
 lines from the centre B, through the 
 remaining divisions in the semicircle 
 A D c, so as to meet the circumference of which G is the centre, 
 in H and K, these points will divide the circle A B D H K into the 
 number of parts required, each part behip' equal to the given side 
 of the pentagon. 
 
 From the preceding example it is evident that polygons of any 
 number of sides may be constructed upon the same principles, be- 
 cause the circumferences of all circles, when divided into the same 
 number of equal parts, produce equal angles; and, consequently, 
 by dividing the semi-circumference of any given circle into th^
 
 number of parts required, t\vo of these parts will form an angle 
 which will be subtended by its corresponding part of the whole 
 circumference. And as all Tegular polygons can be inscribed in a 
 circle, it must necessarily follow, that if a circle be described 
 through three given angles of that polygon, it will contain the 
 number of sides or angles required. 
 
 The above is a general rule, by which all regular polygons may 
 be described upon a given straight line; but there are other 
 methods by which many of them may be more expeditiously con- 
 structed, as shown in the following examples : 
 
 PROBLEM XIV. Upon a given straight line to describe a regular 
 pentagon. 
 
 Let A B (Fig. 33) be the given straight line; from its extremity 
 B erect B c perpendicular to A B, and equal 
 to its half. Join A c, and produce it till 
 c d be equal to B c, or half the given line 
 A B. From A and B as centres, with a 
 radius equal to B d, describe arcs inter- 
 secting each other in e, which will be the 
 centre of the circumscribing circle A B F 
 o H. The side A B applied successively 
 to this circumference, will give the angu- 
 lar points of the pentagon; and these 
 being connected by straight lines will complete the figure. 
 
 PROBLEM XV. Upon a given straight line to describe a regular 
 hexagon. 
 
 Let A B (Fig. 34) be the given straight line. From the extremi- 
 ties A and B as centres, with the radius A B describe arcs cutting 
 each other in g. Again from g, the point of intersection, with the 
 sa-r>e radius, describe the circle ABC, which will contain the given 
 side A B six times when applied to its circumference, and will be 
 the heA'agoiurequired. 
 
 PROBLEM XVI. To describe a regular octagon upon a given 
 sf might line. 
 
 Let A B (Fig. 35) be the given line. From the extremities A 
 and B erect the perpendiculars A E and B F; extend the given
 
 2 6 
 
 HAND-RAILING. 
 
 line both ways to k and /, forming external right angles with the 
 lines A E and B F. Bisect these external right angles, making each 
 of the bisecting lines A H and B c equal to the given line A B. 
 Draw H G and c D parallel to A E or B F, and each equal in length 
 to A B. From G draw G E parallel to B c, and intersecting A E in 
 E, and from D draw D F parallel to A H, intersecting B F in F. 
 Join E F, and ABCDFEGHis the octagon required. Or from D 
 
 Fig 34. 
 
 Fig. 35. 
 
 Fig. 36. 
 
 and G as centres, with the given line A B as radius, describe arcs 
 cutting the perpendiculars A E and B F in E and F, and join G E, 
 E F, F D, to complete the octagon. 
 
 Otherwise, thus. Let A B (Fig. 36) be the given straight line on 
 which the octagon is to be described. Bisect it in a, and draw 
 the perpendicular a b equal to A a or B a. Join A l>, and produce 
 a b to c, making b c equal to A b ; join also A c and B c, extending 
 them so as to make c E and c F each equal to A c or B c. 
 Through c draw c c G at right angles to A E. Again, through the 
 same point f, draw D H at right angles to B F, making each of the 
 lines c c, c D, c o, and c H equal to A c or c B, and consequently 
 equal to one another. Lastly, join B c, c D, D E, E F, F G, G H, H A ; 
 ABCDEFGH will be a regular octagon described upon A B, as 
 required. 
 
 PROBLEM XVII. In a given square to inscribe a given octagon. 
 
 Let A B c D (Fig. 37) be the given square. Draw the diagonals 
 A c and B D, intersecting each other in e ; then from the angular 
 points ABC and D as centres, with a radius equal to half the 
 diagonal, viz., A e or c e, describe arcs cutting the sides of the
 
 27 
 
 square in the points/ g, h, k, /, m, ;/, o, and the straight lines of, 
 g h, k /, and m n, joining these points will complete the octagon, 
 and be inscribed in the square A u c D, as required. 
 
 PROBLEM XVIII. To find the area of any regular polygon. 
 
 Let the given figure be a hexagon; it is required to find its area. 
 Bisect any two adjacent angles, as those at A and B (Fig. 38), by 
 the straight lines A c and B c, intersecting in c, which will be the 
 
 Fig. 37. 
 
 Fig. 38. 
 
 Fig. 39. 
 
 centre of the polygon. Mark the altitude of this elementary tri- 
 angle by a dotted line drawn from c perpendicular to the base 
 A B; then multiply together the base and altitude thus found, and 
 this product by the number of sides : half gives the area of the 
 whole figure. 
 
 Of otherwise, thus. Draw the straight line D E, equal to six 
 times, /'. e., as many times A B, the base of the elementary triangle, 
 as there are sides in the given polygon. Upon D E describe an 
 isosceles triangle, having the same altitude as A B c, the elementary 
 triangle of the given polygon ; the triangle thus constructed is 
 equal in area to the given hexagon; consequently, by multiplying 
 the base and altitude of this triangle together, half the product will 
 be the area required. The rule may be expressed in other words, 
 as follows : The area of a regular polygon is equal to its perimeter, 
 multiplied by half the radius of its inscribed circle, to which the 
 sides of the polygon are tangents. 
 
 PROBLEM XIX. To describe the circumference of a circle tlirough 
 three given points. 
 
 Let A, B, and c (Fig. 39) be the given points not in a straight 
 line. Join A Band B c; bisect each of the straight lines A B and 
 B c by perpendiculars meeting in D ; then A, B and c are all e?,ui-
 
 HAND-RAILING. 
 
 distant from n ; therefore a circle described from T), with the radius 
 D A, DB, or DC, will pass through all the three points as required. 
 
 PROBLEM XX. To divide a given circle into any number of equal 
 or proportional parts by concentric divisions. 
 
 Let ABC (Fig. 40) be the given circle, to be divided into five 
 equal parts. Draw the radius A D, and divide it into the same 
 
 Fig. 40. Fig. 41. Pig. 42. 
 
 number of parts as those required in the circle; and upon the 
 radius thus divided, describe a semicircle : then from each point of 
 division on A D, erect perpendiculars to meet the semi-circumfer- 
 ence in e,f,g, and h. From D, the centre of the given circle, 
 with radii extending to each of the different points of intersection 
 on the semicircle, describe successive circles, and they will divide 
 the given circle into five parts of equal area as required ; the centre 
 part being also a circle, while the other four will be in the form of 
 rings. 
 
 PROBLEM XXI. To divide a circle into three concentric parts, 
 bearing to each other the proportion of one, two, three, from t/ic centre. 
 
 Draw the radius A D (Fig. 41), and divide it into six equal parts. 
 Upon the radius thus divided, describe a semicircle : from the first 
 and third points of division, draw perpendiculars to meet the semi- 
 circumference in e and/ From D, the centre of the given circle, 
 with radii extending to e and/, describe circles which will divide 
 the given circle into three parts, bearing to each other the same 
 proportion as the divisions on A D, which are as i, 2 and 3. In 
 like manner circles maybe divided in any given ratio by concentric 
 divisions.
 
 RAND-RAILING. 29 
 
 PROBLEM XXII. To draw a straight line equal to any given are 
 of a circle. 
 
 Let A B (Fig. 42) be the given arc. Find c the centre of the arc, 
 and complete the circle A D B. Draw the diameter B D, and pro- 
 duce it to E, until D E be equal to c D. Join A E, and extend it so 
 as to meet a tangent drawn from B hi the point F ; then B F will 
 be nearly equal to the arc A B. 
 
 The following method of finding the length of an arc is equally 
 simple and practical, and not less accurate than the one just 
 given. 
 
 Let A B (Fig. 43) be the given arc. Find the centre c, and join 
 A B, B c, and CA. Bisect the arc A B in D, and join also c D; 
 then through the point D draw the straight line E D F, at right 
 
 Fig. 43. Fig. 44. 
 
 angles to c D, and meeting c A and c B produced in E and F. Again, 
 bisect the lines A E and B F in the points G and H. A straight line 
 G H, joining these points, will be a very near approach to the length 
 of the arc A B. 
 
 Seeing that in very small arcs the ratio of the chord to the double 
 tangent or, which is the same thing, that of a side of the inscribed 
 to a side of the circumscribing polygon, approaches to a ratio of 
 equality, an arc may be taken so small, that its length shall differ 
 from either of these sides by less than any assignable quantity ; 
 therefore, the arithmetical mean between the two must differ from 
 the length of the arc itself by a quantity less than any that can be 
 assigned. Consequently the smaller the given arc, the more nearly 
 will the line found by the last method approximate to the exact 
 length of the arc. If the given arc is above 60 degrees, or two- 
 thirds of a quadrant, it ought to be bisected, and the length of the
 
 30 HAND-RAILINC,. 
 
 semi-arc thus found being double, will give the length of the whole 
 arc. 
 
 These problems arc very useful in obtaining the lengths of veneers 
 or other materials required for bending round soffits of door and 
 window-heads. 
 
 PROBLEM XXIII. To describe the segment of a circle by means 
 of two laths, the chord and versed sine being given. 
 
 Take two rods, E B, B F (Fig. 44), each of which must be at least 
 equal in length to the chord of the proposed segment AC: join 
 them together at B, and expand them, so that their edges shall pass 
 through the extremities of the chord, and the angle where they join 
 shall be on the extremity B of the versed sine D B, or height of the 
 segment. Fix the rods in that position by the cross piece g /i, 
 then by guiding the edges against pins in the extremities of the 
 chord line A c, the curve ABC will be described by the point B. 
 
 PROBLEM XXIV. Having the chord and versed' sine of the seg- 
 ment of a circle of large radius given, to find any number of points i;i 
 tJie curve by means of intersecting lines. 
 
 Let AC be the chord and D B the versed sine. 
 
 Through B (Fig. 45) draw E F indefinitely and parallel to A c; 
 join A B, and draw A E at right angles to A B. Draw also A o at 
 
 right angles to A c, or divide A D and E B into the same number of 
 equal parts, and number the divisions from A and E respectively, 
 and join the corresponding numbers by the lines i i, 2 2, 3 3. 
 Divide also A G into the same number of equal parts as A L> or K B, 
 numbering the divisions from A upwards, 1,2, 3, etc.; and from the 
 points , 2 and 3, draw lines to B ; and the points of intersection of 
 these, with the other lines at //, k, /, will be points in the curve re- 
 quired. Same with r, c. 
 
 Anotlier Method. Let AC (Fig. 46) be the chord and D B the 
 versed sine. Join A B, B c, and through i; dra\v E F parallel to AC.
 
 HAND- RAILING. 3! 
 
 From the centre B, with the radius B A or B c, describe the arcs 
 A E, c F. and divide them into any number of equal parts, as i, 2, 
 3: from the divisions i, 2, 3, draw radii to the centre B, and divide 
 each radius into the same number of equal parts as the arcs A E 
 
 Fig. 46. 
 
 and c- F ; and the points g, //, /, ///, //, o, thus obtained, are points in 
 the required curve. 
 
 These methods, though not absolutely correct, are sufficiently 
 accurate when the segment is less than the quadrant of a circle. 
 
 PROBLEM XXV. To draw an ellipse with the trammel. 
 
 The trammel is an instrument consisting of two principal p.-irts, 
 the fixed part in the form of a cross E F GH (Fig. 47), and the 
 moveable piece or tracer kl m. The fixed piece is made of two 
 rectangular bars or pieces of wood, of equal thickness, joined to- 
 gether so as to be in the same plane. On one side of the frame 
 so formed, a groove is made, forming a right-angled cross. In the 
 groove two studs, k and /, are fitted 
 to slide freely, and carry attached to 
 them the tracer k I m. The tracer 
 is generally made to slide through a 
 socket fixed to each stud, and pro- 
 vided with a screw or wedge, by 
 which the distance apart of the studs 
 may be regulated. The tracer has 
 another slider ;//, also adjustable, 
 which carries a pencil or point. The instrument is used as fol- 
 lows : Let A c be the major, and H B the minor axis of an ellipse : 
 lay the cross of the trammel on these lines, so that the centre lines of 
 it may coincide with them ; then adjust the sliders of the tracer, so 
 that the distance between k and /// may be equal to half the major 
 axis, and the distance between / and >n equal to half the m'inr 
 
 Fig. 47.
 
 3 2 
 
 HAND-RAILING. 
 
 Fig. 48. 
 
 axis; then by moving the oar round, the pencil in the slider will 
 describe the ellipse. 
 
 PROBLEM XXVI. An ellipse may also be described by means of a 
 string. 
 
 Let A B ( Fig. 48) be the major axis, and D c the minor axis of 
 the ellipse, and F G its two foci. 
 Take a string EOF and pass it 
 over the pins, and tie the ends, 
 together, so that when doubled it 
 may be equal to the distance from 
 the focus F to the end of the axis, 
 B; then putting a pencil in the 
 bight or doubling of the string at 
 H and carrying it round, the curve 
 may be traced. This is based on 
 the well known property of the 
 ellipse, that the sum of any two 
 lines drawn from the foci to any 
 points in the circumference is the same. 
 
 PROBLEM XXVII. The axes of an ellipse being given, to draw 
 the curve by intersections. 
 
 Let A c (Fig. 49) be the major axis, and D B half the minor 
 axis. On the major axis construct the parallelogram AEFC, and 
 make its height equal to D B. Divide A E and E B each into 
 the same number of equal parts, and number the divisions from A 
 and E respectively; then join A i, i 2, 2 3, etc., and their intersec- 
 tions will give points through which the curve may be drawn. 
 
 Th points for a " raking " or rampant ellipse may also be found 
 by the intersection of lines as shown at Fig. 50. Let A c be the 
 major and E B the minor axis : draw A G and c H each parallel 
 to B E, and equal to the semi-axis minor. Divide A i), the semi- 
 axis major, and the lines A G and c H each into the same number 
 of equal parts, in i, 2, 3 and 4; then from E, through the divisions 
 i, 2, 3 and 4, on the semi-axis major A D, draw the lines E //, E k, E /, 
 and EOT; and from n, through the divisions i, 2, 3 and 4 on the 
 line A G, draw the lines i, 2, 3 and 4 is; and the intersection of
 
 HAND- KAN. I Mi. 
 
 Fig.. 49. Fig. 50. 
 
 these with the lines E i, 2, 3 and 4 in the points h klm, will be 
 points in the curve. 
 
 PROBLEM XXVIII. 7o describe with a compass a figure resem- 
 bling the ellipse. 
 
 Let A B (Fig. 51) be the given axis, which divide into three equal 
 parts at the points/^. From these points as centres, with the radius 
 /A, describe circles which intersect each other, and from the points 
 of intersection through /and g, draw the diameters CgE, c/n. 
 From c as a centre, with the radius c D, describe the arc D F, which 
 
 Fig. 51. Fig. 52. 
 
 completes the semi-ellipse. The other half of the ellipse may be 
 completed in the same manner, as shown by the dotted lines. 
 
 PROBLEM XXIX. Another method of describing a figure ap- 
 proaching the ellipse with a compass. 
 
 The proportions of the ellipse may be varied by altering the 
 ratio of the divisions of the diameter, as thus : Divide the major 
 axis of the ellipse A B (Fig. 52), into four equal parts, in the points 
 f g h. On/// construct an equilateral triangle /c //, and produce
 
 34 
 
 HAND-RAILING.
 
 HAND-RAILING. 35 
 
 - ; the sides of the triangle c /, c // indefinitely, as to D and E. Then 
 froiii the centres f and h, with the radius A f, describe the circles 
 A D g, B E g; and from the centre c, with the radius c D, describe 
 the arc D E to complete the semi-ellipse. The other half may be 
 completed in the same manner. By this method of construction 
 the minor axis is to the major axis as 14 to 22. 
 
 The following problems, which relate more particularly to hand- 
 railing, should be thoroughly mastered before passing to actual 
 work. 
 
 A tangent is a line touching a circle at right angles to the radius 
 as shown at Fig. 55. 
 
 To construct Fig. 53 : From the centre o with the radius o A, 
 describe a quarter-circle A p c; draw tangents A B and c B; join 
 A c; through the point B draw a straight line parallel to A c; with 
 the centre B, with the radius B A, describe the arcs A D and c E; at 
 the point E erect the perpendicular E F at right angles to n E to 
 any desired height; in laying down a hand-rail this height will be the 
 number of risers contained in the wreath; let F be the given height 
 (this being one pitch); join F D; extend o B to G; draw GH at right 
 angles to F D ; make G H equal to B i ; with the centre H and ra- 
 dius D G describe arcs cutting D F at K and L; draw H L and H 
 K, which are the tangents on the pitch, and which when placed in 
 position would stand plumb over ABC. 
 
 To construct Fig. 54, proceed as above until the height is 
 located. It will be seen that here B G is lifted higher, making the 
 pitch line and tangents F G and D G of unequal length. To obtain 
 the angle, continue B G to H, making B H equal to E F ; from H 
 draw the line H j to any distance at right angles to D G. With the 
 centre G and radius G F, describe an arc cutting the line H J at i ; 
 join i G and i D, and the angle is completed. 
 
 An easy way to prove these problems is to draw them on com- 
 mon thick paper ; then take a knife, and cut out the angle D E F, 
 place it perpendicularly over ABC, bringing D over A, and E over 
 c; then cut out the angle H L K, and if drawn correctly it will lie on 
 the pitch lines and fit the sides exactly. 
 
 To draw the curve line the quickest and most practical method
 
 HAND-RAILING. 
 
 Fig. 54.
 
 HAND-RAILING. 
 
 37 
 
 is to take B as a centre, and with a radius of B p, describe 
 an arc, touching the curve A p c, in the angle ABC, from H, as a 
 centre, with the same radius, describe an arc cutting H G ai M ; 
 then, take a thin flexible strip of wood of an even thickness, bend it 
 until it touches the points KIM; mark around it with a pencil, 
 and the curve is completed, and near enough to absolute accuracy 
 tor ;ill practical purposes. 
 
 Fig. 55.
 
 38 HAND-RAIUNG. 
 
 PART II. 
 
 TO MAKE THE WREATHS FOR STAIRS WITH FOUR 
 WINDERS IN THE WELL. 
 
 FIGURE i. Strike the quarter circle ABC, which is the centre 
 line of the rail, and also describe the inside and outside lines of rail 
 b b l>, and mark the risers at A B and C. Draw the lines A D 
 and CD. 
 
 FIGURE 2. Draw the straight line B K equal to the tread of one 
 step, and set up K A equal to the rise of one step. Draw the line 
 A C, and make A D and D C equal to the length A D and D C 
 in Figure i. Set up the height of two risers from C to F. Draw 
 the line A B, and mark the point G at one third the distance from 
 A to B, and draw the line F G. Continue C A to the point H, 
 and from D draw the perpendicular line D J. Make the point at 
 H square to G H, and ease off the rail from /to H, and draw the 
 top and bottom side, by setting half the thickness of the rail each side 
 the line / H. 
 
 FIGURE i. Draw the line C I through the centre X, and make 
 C I equal to D H in Figure 2, and draw HI. Set up C F equal 
 to C F in Figure 2, and C J equal to D J in Figure 2. Join 
 J I, and from the point fdraw the line f G parallel to J I, and 
 join G H. Draw ordinate lines a b b b at any convenient dis- 
 tance apart all parallel to G H, one of which to pass from D to X. 
 From the point /draw the line IN, square to J /, and from G 
 with the distance G H cut the line /, N at E, and draw G E. 
 Draw the perpendicular lines a c and from the points c c c draw 
 ordinate lines c d, all parallel to G E. Take the distance from a 
 to b b b, and apply them from c to d t d, d t and draw the face mould
 
 HAND-RAILING. 
 
 39 
 
 Fig. L
 
 HAND-RAILING. 
 
 Fig. 2.
 
 HAND-RAILING.
 
 42 HAND-RAILING. 
 
 through them. Take the distance X D, and mark it on the ordi- 
 nate from c to k, and draw the lines E K and K F. Make the 
 joint L M square to K F, and the end N O square to K E. 
 From /, with the distance I P, describe the arc P Q, and draw 
 Q If, and apply this bevel I Q Jf, to both ends of the wreath 
 through the centre O, as shown at / Q H in Figure 4. Take the 
 distance Q O in Figure 4, and apply it from / to ft in Figure i ; 
 draw J? S parallel to / J, and cut it with the line /, S. Apply 
 the distance R S from F\.o V, and F to IV; also from E to 7' and 
 E to U, and make the face mould the length from 7" to W. 
 
 FIGURE 3. The centre line E U F, and the joints are the same 
 as E U F'\\\ Figure i. Set the compasses to half the thickness of 
 of the plank C D in Figure 3^, and mark it each side the centre 
 line E U F, and make another face mould to this (Figure 3), mark 
 it on the plank, and cut it out square, and joint the ends square. 
 Now apply the face mould Figure i on the top side of wreath, 
 with the points V and T over the points Q Q at the ends of 
 wreath, and mark both edges; apply it on the underside with the 
 points W and U over the points H H, at the ends of wreath, and 
 mark both edges. Work the inside of the wreath with a round plane 
 to fit the rail on the plan, then work off the outside to the marks 
 on the top and bottom, and the bevels at the ends. To work 
 the top : At the mark K P on the face mould, take an equal por- 
 tion off the top and bottom side of the wreath that will leave it the 
 thickness of the rail, and work it to the bevels marked on the ends, 
 gradually twisting from end to end. The bottom part can be 
 gauged from the top side.
 
 HAND-RAILING. 43 
 
 TO MAKE THE WREATHS FOR STAIRS WITH SIX 
 WINDERS IN THE WELL. 
 
 FIGURE 4. Strike the quarter circle A C, which is the centre 
 line of the rail, and also describe the inside and outside lines of rail, 
 bob. Draw the lines A D, and D C. 
 
 FIGURE 5. Draw the straight line B K equal to the tread of 
 one step, and set up K A equal to the rise of one step. Draw the 
 line A C, and make A D and D C equal to A D and D C in 
 Figure 4; setup the height of 3 risers from C to F \ draw the line 
 A B, and mark the point G at one-third the distance from A to B, 
 and draw the line G F; erect the perpendicular line D J. Joint 
 the rail square at If, (which eases the rail better than if jointed at 
 the spring of the well at A). Ease off the rail from H to E, and 
 draw If I parallel to A C. 
 
 FIGURE 4. Draw the line C G through the centre X, and 
 make C I equal to L H'\\\ Figure 2, and draw I H. Set up C F 
 equal to I F in Figure 5, and C J equal to L J; join J /, and 
 from the point ^draw the line F G parallel to J I, and join G H. 
 
 Draw ordinate lines a b b b, at any convenient distance apart, 
 all parallel to G H, one of which to pass from D to X. From the 
 point /draw the line /^square to / J, and from G, with the dis- 
 tance G H, cut the line I N z\. E, and draw G E. 
 
 Draw the perpendicular lines a c, and from the points c c c, draw 
 ordinate line c to d d d, all parallel to G E. Take the distances 
 from a to b b ft, and apply them from c to d d d, and draw the face 
 mould through them. 
 
 Take the distance X D, and mark it on the ordinate from c 
 to X, and draw the line E K and K F. Make the joint L M 
 square to K F, and the end NO square to K E. 
 
 From /, with the distance / P, describe the arc PQ, and draw Q 
 If. Apply the bevel / Q H to both ends the wreath through the
 
 44 
 
 HAND-RAILING. 
 
 Kg. 4.
 
 HAND- RAILING. 
 
 45 
 
 Fitf. 5.
 
 HAND-P 'HUNG.
 
 HAND-RAILING. 47 
 
 centre O, as shown at / Q H in Figure 4. Take the distance Q 
 O in Figure 4, and apply it from / 7? in Figure i, and draw R S 
 parallel to G F, and cut it with the line / S. Apply the distance 
 R .9 from Flo V, and F io W, also from E to rand E to U, and 
 make the face mould from 7" to W. 
 
 FIGURE 6. The centre line E U F and the joints at the ends, 
 are the same as E U F in Figure i. Set the compasses to half the 
 thickness of the plank CD in Figure 7, and mark it each side the 
 centre line E U F. Make another face mould, (to Figure 6) mark 
 it on the plank, cut it out square, and joint the ends square. 
 
 Apply the face mould, Figure 4, on the top side the wreath, with 
 the points T^and Fover the points Q Q at the ends of the wreath, 
 and mark both edges. Apply it on the underside the wreath, with 
 the points W and U over the points // H at the ends of the 
 wreath, and mark it. 
 
 Work the inside the wreath with a round plane to fit the inside 
 curve on plan, then work off the outside to the marks on the top 
 and bottom and the bevels at the ends. 
 
 To work the top, at the mark K P on the face mould, take an 
 equal portion off the top and bottom side the wreath, that will 
 leave it the thickness of the rail, and work it from the bevels at 
 each end, through K P, gradually twisting from end to end. The 
 bottom part can then be gauged from the top side.
 
 48 HAND-RAILING. 
 
 TO MAKE THE WREATHS FOR STAIRS WITH THREE 
 WINDERS IN HALF THE WELL, AND LANDING 
 IN THE CENTRE OF THE WELL. 
 
 (THIS is ONE OF THE MOST DIFFICULT RAILS THAT CAN BE MADE.) 
 
 FIGURE 8. Strike out the semi-circle A B C, which is the centre 
 of rail, and also describe the inside and outside lines of rail. Draw 
 the lines C , A >, and D E. 
 
 FIGURE 9 {See Folding Plate). Draw the straight line OK equal 
 to the tread of one step, and set up KA equal to the rise of one step. 
 Draw the line A E, and make A D D B and B E equal to A D 
 D B and BE in Figure 8. Set up the height of 3 risers from E to 
 M, and six inches more from M to C, (this makes the rail 3 feet 
 high on the landing, or six inches higher than the rail is on the 
 stairs, which should be 2^ feet plumb with the face of the risers.) 
 
 Draw the line A O, and mark the point G about one-third the 
 distance from A to O, and join C G. Make the joint at Zf which 
 is a little above the spring of the well at A, and ease off the rail 
 as shown at G. 
 
 Draw the line H L /parallel to A B, and erect the perpendicu- 
 lar lines L J and I F. Now from the base line H L I and the 
 height L <7and If get the face mould and work the bottom wreath, 
 as shown and described in Figure 5, from the base l : ne H L I 
 and the heights L J"and IF. 
 
 To work the top wreath proceed as follows, viz. : 
 
 FIGURE 10. Lay down the plan of the rail, the centre line B 
 C F, being the same as B C F, in Figure 8, and make the straight 
 part about 5 inches from C to F. Draw the lines F A-A B-B E 
 and E F. Take the height N C in Figure 9, and apply it from F 
 to D, and join A D. Draw D G square to A D, and make D G 
 equal to F E.
 
 HAND-RAILING. 
 
 49 
 
 fig. 10. 
 
 . 8.
 
 5 
 
 HAND-RAILING. 
 
 M
 
 HAND-RAILING. 51 
 
 Draw ordinate lines a b b b parallel to A B, and draw the lines 
 a c, and from c c, draw ordinate c d d d, parallel to G D. Apply 
 the distance from a to b, b, b, on to the ordinates from c to d, d, d, 
 and draw the face mould through them. Make the joints at H 
 and D square, as shown. 
 
 FIGURE n. Shows the ends of the wreath. Take the bevel 
 F D A in Figure 10 and apply it at the top end of the wreath 
 through the centre O at F D A, and mark the size of the rail on 
 as shown. 
 
 FIGURE 12. Is the face mould to cut the wreath out square; the 
 centre line DIJf, and the joints are the saifie as the centre line D 
 I H'm Figure 3; put half the thickness of the,plank each side the 
 centre line D Iff, and cat out the wreath. Figure 8: Take half the 
 thickness of the plank from A to Z, draw L M parallel to A D, and 
 draw A M, apply the distance L J/to // J, and H K, and make 
 the face mould from J to D. 
 
 Apply Figure 10 face mould on the top side of wreath, wiih the 
 points D and J, over the points D J in Figure n, and mark it. 
 Apply it on the underside with the points D and .AT over the points 
 A and K in Figure n. 
 
 Work it as described in the preceding Figures 6 and 7,. to th 
 bevels at the ends as shown, and work the part from D to /straight; 
 then work the top gradually twisting from /to H in Figure 12. 
 
 . t
 
 HAN iv RAILING. 
 
 TO MAKE THE WREATHS FOR STAIRS WITH ONE 
 RISER IN THE CENTRE OF WELL. 
 
 FIGURE 13 (See Folding Plate}. Mark the plan of the rail as A B 
 CD, and E F G the centre line, the straight part E F being the width 
 of one step. Continue E F to H and draw G H. Draw D /paral- 
 lel to E F, and draw the lines A J and F 0. Make K equal to the 
 rise of one step, and draw J K, set up G L equal to the rise of two 
 steps. From the point L make the line L /parallel to J K and 
 join / E. Draw ordinate lines as i, 2, 3, 4, at any convenient 
 distance apart all parallel to / E, draw the perpendicular lines i, 
 5, etc., to intersect the line I L. From the point J, draw the line 
 J M, square to J K, and from 7, with the distance IE, cut J M at 
 the point N, and draw / N; draw ordinate lines 5, 6, 7, 8, etc., all 
 parallel to / N. Take the distances i, 2, 3, 4, and apply them 
 from 5 to 6, 7, 8, etc., and draw the inside and outside and centre 
 line of face mould through them. 
 
 Take the length of the ordinate H, and apply it on the ordinate 
 from P to Q, and draw the lines N Q, and L Q. Make the joint R 
 8 square to Q L, and the joint M N square to Q N. 
 
 FIGURE 14. The centre line N 7 7 7 L, and the joints at ends, 
 are the same as Figure 13. 
 
 Set the compasses to half the thickness of the plank, (as shown 
 at G D in Figure 15), and apply it on each side the centre line 
 L 7 7 7 JV, and draw the inside and outside lines. Mark this face 
 mould on the plank, and cut it out square, and joint Lhe ends 
 square. 
 
 To put the twist on the wreath, proceed as follows : 
 
 In Figure 13 from the centre J, describe the arc T U, and draw 
 U E. Continue the lines / E, and G H, until they meet at W. 
 Take the length G L and apply it from G to V, and draw V W. 
 From G draw G X square to V W, and describe the arc X Y, and 
 draw YI.
 
 HAND-RAILING. 
 
 53
 
 54 HAND-RAILTNG. 
 
 Take the bevel U J E, and apply it on the bottom end of the 
 wreath in Figure 15, at U J E, through the centre O, and mark 
 half the width of the rail on each side U E, also mark the top and 
 bottom line of rail square to U E. Apply the bevel H Y I'm Fig- 
 ure i to H Y I, at the top end of the wreath in Figure 15, through 
 the centre 0, and mark the size of the rail on as shown. Mark 
 the line Q P on both sides the wreath. Take the distance U 
 in Figure 15, and apply it from J to a in Figure 13, draw a b par- 
 allel to J K, and draw J b, apply the distance a b from N to c and 
 N to d. Apply the distance Y, in Figure 15, to X e in Figure 
 13, and draw ^/parallel to G V, and apply the distance X f to L 
 g, ami L h. Make a face mould to Figure 13, the full length 
 from c to h, and place it on the top of the wreath, with the point 
 ^over the point Y, in Figure 15, at the top end of the wreath, and 
 the point c over the point U, at the bottom end and mark both 
 edges, then mark it on the underside the wreath, with the point h 
 over the point / in Figure 15, and the point c over the^oint E. 
 
 First work off the inside of the wreath with a round plane that 
 will fit the rail on the plan, then work off the outside part. To work 
 the top at the mark Q P, take an equal portion off the top and bot- 
 tom side of the wreath that will leave it the finished thickness of the 
 rail, then work from the bevels marked at each end, through the 
 part taken off at Q P and the bottom part can be gauged from 
 the top.
 
 HAND-RAILING. 55 
 
 TO MAKE THE WREATHS FOR LEVEL LANDING 
 STAIRS. THE DISTANCE BETWEEN CENTRE OF 
 RAILS ACROSS THE WELL BEING EQUAL TO 
 THE TREAD OF A STEP, OR NEARLY SO. 
 
 FIGURE 16. A B C is the centre of tlie rail, A B being equal to 
 the tread of one step. Strike out the inside and outside of the rail b 
 b b b. Draw C D and D A. Set up E F equal to one riser, and 
 draw D F to G, and draw the line C G. Draw ordinate lines 
 a b b, at any convenient distance apart, and draw the lines tf, c. 
 From the points c c draw ordinate lines c d. square to the line D G. 
 Take the distance a b b, and apply them on the ordinates from c 
 to d d, and draw the face mould through them. 
 
 Take the thickness of the plank C A in Figure 18, and apply it 
 from D to / in Figure 16, draw I H parallel to D F, and cut it with 
 the line of D H. Take half of / H and apply it from K to L, and 
 K to J. Make the face mould the length from J to G, and mark 
 it on the top side' of the plank, and cut the two joints at J and G 
 square. 
 
 Take the bevel D J E in Figure 16, and apply it to the top joint 
 at D F E, as shown in Figure 18. The bottom end will be square, 
 as shown in Figure 17. 
 
 Apply the face mould underside the wreath with the point L 
 at the lower end of wreath. Cut it out bevel (as this kind ol 
 wreath is less labor cut bevel than having to bevel it if cut out 
 square). 
 
 Take an equal portion off the top and bottom side of the shank- 
 part of wreath to leave the rail the finished thickness as shown in 
 Figure 17, at the bottom end, then gradually work the top side 
 twisted to the bevel at the top end, so that when the wreath is 
 put up to its proper pitch the top side is level across. Gauge the 
 bottom from the top side.
 
 HAND-RAILING. 
 
 Fig. 16.
 
 HAND-UALL1NG. 
 
 57 
 
 I
 
 HAND-RAILING.
 
 HAND-RAILING. 59 
 
 NOTE. The distance across the well between centre of rails 
 should not exceed the tread of a step for level landing stairs, or the 
 wreath has the appearance of dropping. 
 
 If the distance is less than the tread of a step bevel ordinate 
 lines would be required, after the manner shown in Figures 13, 14, 
 and .15.
 
 60 HAND-RAILING. 
 
 TO MAKE THE WREATH FOR QUARTER-LANDING 
 STAIRS, HAVING ONE WINDER IN THE WELL. 
 
 FIGURE 19. From the centre strike the quarter circle A Jj, 
 the centre of the rail, and draw the lines A B B C and A 
 C, and draw the inside and outside curve of rail. 
 
 FIGURE 20 (See Folding Plate). Make D E equal to the tread 
 of one step, and-EA equal to one riser. 
 
 Draw the line A B and make A B and B C equal to A B and B 
 C in Figure 19. 
 
 Set up the height of 2 risers from B to F, and draw the line F 
 G, to pitch of the stairs. Join A D and mark the point H about 
 one-third the distance from A to Z>, then mark the point / about 
 one-third the distance from F to G, and draw the line I H. 
 
 Draw the perpendicular line C J, and continue A B to K. 
 
 Joint the end at K square to J K and ease off the rail at H. 
 
 Joint the top length at the point L and ease off the rail at /. 
 
 FIGURE 19. Make B K equal to C K in Figure 20. Make B 
 J equal to C Jin Figure 20, and B M equal to B M in Figure 20. 
 Draw the line J K, and from the point M draw M D parallel to 
 J K. Draw K J* 7 square to J K, and from the point D with the 
 distance D E cut it at the point G. Draw the ordinate a b b b 
 parallel to D E, draw the lines a c, and from c draw ordjnates c d d 
 d parallel to D G. 
 
 Apply the distances a b b b, on to the ordinates from c to d a d, 
 and draw the face mould through the points d d d. 
 
 Mark the length of the ordinate C, on to the ordinate from c 
 to H, and draw H G and H M. Joint the end at G square to 
 GH. 
 
 Take the distance M L in Figure 20, and apply it from M to T, 
 and joint the end at T square to H T. Apply the bevel K I E to 
 both ends the wreath as shown in Figure 22.
 
 HAND-RAILING. 
 
 61 
 
 \ 
 
 Fiji. 19.
 
 62 
 
 HAND-RAILING. 
 
 The face mould Figure 21, is got from the centre line of Figure 
 19, as before described, and the joints at T and G are the same as in 
 Figure 19. 
 
 Apply the distance 7, in Figure 22 to K L in Figure 19, draw 
 L N, and mark the length L N from G to P and G to Q, also 
 from T to S and T to R. Make the face mould Figure 19, the 
 length from P to R. 
 
 Fig. n. 
 
 Cut the wreath out square to Figure 21, face mould and joint 
 the ends. Apply Figure 19 face mould on the top side of wreath 
 with the points S P, over I 1 at the ends of wreath, then apply it 
 underside with the points R Q, over E E at the ends. 
 
 Work the inside the wreath first with a round plan to fit the rail on 
 plan, and work the top side straight from T to M and d to G in 
 Figure 21, and gradually work it twisted to the bevels at each end, 
 and take an equal portion off the top and bottom side at c H to 
 leave the rail the finished thickness.
 
 HAND-RAILING. 63 
 
 TO MAKE THE WREATH FOR QUARTER LAND- 
 ING STAIRS, WITH THE RISERS AS SHOWN IN 
 FIGURE 23. 
 
 FIGURE 23. From the centre strike the quarter circle A /?, 
 which is the centre of the rail, draw the lines A B B C 
 and .1 C, ai:d draw the straight part each end about 3 inches long 
 from A to D, and B to E, then draw the inside and outside lines 
 of rail. 
 
 FIGURE 24 (See Folding Plate] Make D E equal to one tread, 
 and E A equal to one riser. Draw the line A B, the parts A C 
 and C B, being equal to A C and C B in Figure i. Set up B F 
 equal to one riser and from F draw the line G F H, to the pitch of 
 the stairs, and draw (J G. 
 
 Join A D, and mark the point /about half way between A and 
 Z>, and draw / G. 
 
 Make K J" equal to A D in Figure 23, and joint the end at J, 
 square to 7 (7, draw the line J N. 
 
 Make F L equal to E B in Figure 23, draw L M and joint the 
 end at M square to F H. 
 
 FIGURE 23. Draw the line D J, set up B F equal to G in 
 Figure 24, and joint J F. 
 
 Set up B G equal to N F in Figure 24, and from the point G 
 draw G H parallel to J F. 
 
 From J, draw J I square to J F, and from H with the distance 
 // ft, cut J / at K, and draw If K. 
 
 Draw the ordinates a b b If, parallel to H D, draw the lines a c, 
 and from the point c, draw ordinate c d d d parallel to H K; 
 apply the distance a b b b, on to the ordinates c d d d, and draw 
 the face mould through the points d d d. 
 
 Mark the length of the ordinate a C, on the ordinate c L, and 
 draw L K and L <!.
 
 HAND- RAILING. 
 
 Fig. 23.
 
 HAND-RAILING. 
 
 Make G M equal to F M in Figure 24. Joint the end at K 
 squat -<j to A' />, and the end at M square to L M. 
 
 Get the level J D (or the bottom end, and jB # for the top 
 end of wreath, as described in Figure 15, and apply them at the 
 ends as shown in Figure 26. 
 
 The face mould Figure 25, is. obtained from the centre line of 
 Figure 23, as described in previous examples. 
 
 Fig. 26. 
 
 Apply the distance T U from K to V, and K to W, and the 
 length E Y apply from M to g, and M to /, and make Figure 23, 
 face mould the length from Fto/ 
 
 Cut the wreath out square, to Figure 25, face mould, and joint 
 the ends. 
 
 Apply Figure 23, face mould on the top of wreath, with the points 
 /and TFover the points S and at the ends of wreath, then mark 
 it underside with the points g and V, over the point H and D at 
 the ends of wreath. Proceed to work the wreath as before de- 
 scribed, to .the bevels at the ends, and take as much of the top and 
 bottom sides of the wreath at L a as will leave the rail the finished 
 thickness.
 
 THE LIBRARY 
 UNIVERSITY OF CALIFORNIA 
 
 Santa Barbara 
 
 THIS BOOK IS DUE ON THE LAST DATE 
 STAMPEDJSELOW.
 
 UC SOUTHERN REGIONAL LIBRARY FACIL'T 
 
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