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SEC( 
 
 I 
 
 WI] 
 
PEESPECTIVE 
 
 AND 
 
 GEOMETKICAJ. 
 
 DRAWING 
 
 ADAPTKD TO THK t'8B OK CAND'DATKS POR 
 
 SECOND AND THIRD-CLASS TEACHERS' CERTIFICATES. 
 
 BY ^ 
 THOS. Irl. TVIcGTJIRL, B.A. 
 
 COMMRRCIAL MA8TRR, CoLIjINOWOOl) COLLKOIATK InhTITUTR. 
 
 
 TORONTO: 
 
 WILLIAM BRIGGS, 78 & 80 KING STREET EAST. 
 
 C. W. COATES, MONTRRAL, Que. S. F. HUESTIS, Halifax, N.S. 
 
 1887. 
 
Ms. 
 
 Entered, according to the Act of the Parliament nt r«««^ • *i. 
 
 one thousand eight hundred .„rf «C^ ! "'*' '" ^^^ ^^^ °' »"' ^r^ 
 
 u- ...'■, . 
 
 I 
 
PREFACE. 
 
 rd 
 d. 
 e, 
 
 Drawino having been at length recognized by the EJ :ca- 
 tion Depaitnient as an essential feature in High School edu- 
 cation, it is necessary that a work, at once simple and concise, 
 should be prepared on this subject. The incompleteness and 
 want of definiteness in the existing works on perspective, have 
 induced me to place this book before High School pupils. It 
 consists mainly of problems, etc., that have been given from 
 time to time in my own classes. To obviate the necessity of 
 copying problems from the blackboard, I have added a num- 
 ber in Geometrical Drawing, which will be found useful. 
 
 Believing that this work supplies a want long felt in our 
 schools, I have consented to place it before the public. 
 
 T. H. M. 
 
 Thf, I.vstitittk, March, 18S7. 
 
 ^1 
 
1^ 
 
 National Library 
 of Canada 
 
 Bibiiolheque nationals 
 du Canada 
 
CONTENTS 
 
 Introduction 
 
 Drawing to a Scale 
 
 The Point 
 
 Exercise on the Point ..... 
 
 The Line 
 
 Exercise on the Line 
 
 Surfaces— The Square 
 
 Exercise on the Square . , 
 
 The Oblong 
 
 Exercise on the Oblong . 
 
 The Triangle 
 
 Exercise on the Triangle . 
 
 The Hexagon 
 
 Exercise on the Hexagon 
 
 The Octagon 
 
 The Circle 
 
 Exercise on the Circle . . . 
 Solids 
 
 The Cube 
 
 The Plinth ^y. 
 
 Exercise on the Plinth 
 
 The Prism 
 
 The Cylinder 
 
 The Pyramid 
 
 The Cone 
 
 Exercise on the Prism and Cylinder 
 
 Exercise on the Pyramid and Cone 
 
 The Frusta 
 
 Exercise on the Frusta ug 
 
 PAOI. 
 
 9 
 15 
 16 
 22 
 23 
 27 
 28 
 36 
 37 
 38 
 39 
 45 
 47 
 
 52 
 
 54 
 
 57 
 
 63 
 
 64 
 
 66 
 
 68 
 
 70 
 
 72 
 
 75 
 
 78 
 
 80 
 
 81 
 
 82 
 
 83 
 
 87 
 
VI 
 
 CONTKNTS. 
 
 IMOR. 
 
 Soli ds — Coniinvr^l. 
 
 The Sphere 88 
 
 Exercise on the Hpltere 90 
 
 Foreshortening .... ' , 92 
 
 Synthetic Perspective 93 
 
 Perspective Kffect 94 
 
 Angular PEBSPKcmvE— Figures on Picture Plane 96 
 
 Figures within the Picture Plane 99 
 
 Kxercise in Angular Perspective 101 
 
 Miscellaneous Exercises 101 
 
 Geometrical Drawino lOfl 
 
 To draw a perpendicular to a given line from a point on the 
 
 line or away from it . . , 105 
 
 To describe a B(|uare on a given straight line 106 
 
 To describe a 8<iuare on a given diagonal 107 
 
 To construct an oblong of given dimensions 107 
 
 To bisect a given line IO7 
 
 To divide a given line int<i any number of ajual parts 108 
 
 To draw a line parallel to a given line from an external point. 108 
 
 To divide a line proportionally to another divided line 109 
 
 To construct a triangle of given dimensions 109 
 
 To bisect a given angle 110 
 
 To trisect a right angle 110 
 
 To inscribe a circle in a given triangle Ill 
 
 To draw a circle through three given points Ill 
 
 To find the centre of a given circle or arc 112 
 
 To draw a tangent to a circle from a point without or on the 
 
 circumference 113 
 
 To draw an isosceles triangle of given dimensions 114 
 
 To draw an equilateral triangle of given dimensions 114 
 
 To draw from a point an angle equal to a given angle . . 115 
 
 To constntct a triangle similar to a given triangle within a 
 
 given circle 115 
 
 To construct an equilateral triangle about a given circle 116 
 
 To construct a triangle similar to a given triangle about a 
 
 given circle 117 
 
 Within a circle to draw any number of equal circles, each 
 
 touching two others and the outer circle 118 
 
 To construct a polygon on a given line 119 
 
CONTENTS. 
 
 Vll 
 
 (jEomktrioal T)RA\\'\y(i— Covf hi nfd. paor. 
 
 To construct a regular polygon m a given circle 120 
 
 To construct a regular pentagon on a given line 120 
 
 To construct a regular hexagon on a given line 121 
 
 To couHtruct a regular octagon on a given line 122 
 
 To construct a regular octagon in a given si^uare 122 
 
 To construct an ellipse with axes and intersecting arcs 123 
 
 To construct an ellipse with concentric circles 124 
 
 To construct an ellipse when the longer axis only is given. . . . 125 
 
 To find the axes and foci of a given ellipse 125 
 
 To draw a tangent to an ellipse from a point in the curve .... 126 
 
 To draw a tangent to an ellipse from an external point 127 
 
 To draw an oval of a given width 127 
 
 To draw an involute to a circle 128 
 
 To draw a mean proportional (greater) to two given straight 
 
 lines 129 
 
 To draw a mean proportional (less) to two given straight 
 
 lines 130 
 
 To draw a circle touching two lines not parallel 130 
 
 To draw a circle touching another circle and a given line .... 131 
 Graded Exercise on Geometrical Drawing 132 
 
DEAWING 
 
 ilvj\ RAWIN(» may be defined as the representation of an 
 ^i2J/ object or collection of objects on a plain or level 
 surface. 
 
 There are two kinds of Drawing : Perspective, or the 
 representation of an object as it appears to the eye ; and 
 Ueometrical Drawin*;, as it actually is. 
 
 Our knowledge of the size of an object can be known oidy 
 by experiment. We must either see the object near the eye, 
 and observe its size, or know its distance from the eye. 
 
 From long practice we are enabled to tell the height of a 
 hill, breadth of a river, or capacity of a ship, though at a con- 
 sideraVile distance. If we hold a rule of definite length close 
 to the eye, and then withdraw it six or eight feet away, we 
 notice that it is apparently smaller. Experience teaches us 
 that it is not really smaller, the apparent diminution being 
 only the effect of distance. 
 
 Perspective aims, then, at measuring and representing 
 objects as they appear at a distance. 
 
 The horizon always bounds our vision. 
 
 If we look out upon a large lake we find that the sky and 
 water appear to meet ; this line of apparent union is called 
 2 
 
10 
 
 DRAWING. 
 
 i.r 
 
 the horizon. No matter hrw large an object is, if it recedes 
 far enough from us on « lake, it would at length appear 
 
 on the horizon as a poinL ^_ 
 
 If a person stands on a level plain he can see over a range 
 of 60 degrees in every direction without moving his head. 
 The point on the hoiizon directly in front of him is called 
 the Centre of Vision. 
 
 
 
 9As 
 
 
 Ji 
 
 c^^'cAi-^. 
 
 j-fon?o'ft. 
 
 Now, if we take this point as a centre, and join it with the 
 eye of the observer, and draw lines at an angle of 30° with it 
 from the observer's eye, we shall form a hollow cone, which 
 represents his range of vision : thus, if the observer's eye be 
 at the point X, and AB represent the horizon, and C the 
 point on it directly in front of X, if CX be joined, and 
 AX, BX be drawn at an angle of 30 degrees with OX, AB 
 
INTRODUCTION. 
 
 11 
 
 will be the horizontal range of the spectator's vision. If we 
 describe a circle from centre C, and distance CA or CB, such 
 circle will be called the base of the cone of visual rays ; for it 
 will be observed that from point X the spectator can see just 
 as far as the edge of this circle. The ground as a plane is 
 generally supposed to extend to the horizon, and is marked 
 for sake of abbreviation, G. P. The horizontal line is un- 
 limited in length, and is written H. L. (Fig. 1.) 
 
 All representations of objects are supposed to be made on a 
 plane (unlimited in extent), perpendicular to, and resting on, 
 the ground plane, and directly in front of the spectator. This 
 plane is called the Picture Plane, and is written P. P. 
 
 It must be distinctly borne in mind that the bottom of the 
 picture plane touches the ground plane, and this line is always 
 at right angles to that joining the spectator's eye with the 
 centre of vision. 
 
12 
 
 DRAWING. 
 
 The line joining the spectator's eye with the centre of 
 vision is usually denoted L. D. (line of direction, and some- 
 times length of distance). * 
 
 The position of the spectator's eye is called the point of 
 station, or P. S. (Fig. 2.) 
 
 If a spectator stand at O and observe a stick at AA, then 
 carry the stick back and place it at XX, parallel to A A, and 
 draw lines OX, OX ; it is plain the stick cannot appear as 
 large as AA, but will be represented in length by BB. If 
 again withdrawn to YY, it will appear as ee, and so on. 
 As we remove it the apparent length diminishes, or, in 
 other words, the angle mad^ by the line with O constantly 
 decreases. (Fig. 2.) 
 
 If the stick were removed to an infinite distance, its appar- 
 ent length would vanish to a point S, and the lines drawn to 
 O would coincide, forming an angle of O degrees. 
 
 Objects appear to diminish in size as they recede from the 
 eye, and vice versa. 
 
 At the centre of vision all objects have no apparent magni- 
 tude, and must be represented by a point, and steadily in- 
 creased in size till they approach the eye. This is why rails 
 on a straight piece of railway appear to meet in the distance, 
 though everywhere the distance between them is the same. 
 
 Now, if we stood on a straight piece of track, and if other 
 tracks were laid on each side and parallel with it, every track 
 would appear to vanish at the same point directly in front of 
 us, hence the rule : 
 
 Lines parallel to the line of direction will vanish at the 
 centre of vision. (Fig. 3.) 
 
 Thus, let C be centre of vision, and AA, BB the ends of 
 parallel lines : they will all meet at C, and all the lines drawn 
 from C to AB will really be right angles, however different 
 they may appear to be. Now draw DD parallel to AA, and 
 
INTRODUCTION. 
 
 13 
 
 B centre of 
 and some- 
 
 since tlie distance from A to A is always the same as from D 
 to D, we have the rule : 
 
 le point of 
 
 ■j AA, then 
 P AA, and 
 appear as 
 y BB. If 
 nd so on. 
 es, or, in 
 constantly 
 
 its appar- 
 drawn to 
 
 from the 
 
 lit magni- 
 eadily in- 
 why rails 
 distance, 
 same. 
 
 if other 
 Jry track 
 
 front of 
 
 h at the 
 
 ends of 
 s drawn 
 lifferent 
 ^A, and 
 
 Parallel lines drawn between vanishing lines are of equal 
 length, and, conversely, the lines joining the extremities of 
 equal parallel lines will vanish to a point. 
 
 ♦ Fig. 4. 
 
 Again, if the spectator stood on the ground at O, where the 
 picture plane rests, and looked towards C on the horizon, it 
 
14 
 
 DRAWING. 
 
 is clear that he could see objects lying on the ground any- 
 where between AB and HL. Hence, if we take AB as the 
 ground line, or base of picture plane, and OC as the height of 
 spectator's eye above the ground, the horizontal line will pass 
 through C and be parallel to AB. (Fig. 4.) 
 
 The ground line is denoted G. L. 
 
 If a stake AS be placed perpendicularly in the ground 
 plane at A, it will necessarily touch P.P. throughout its length. 
 If placed as TT, still erect, it is said to ^ e tvithin the picture 
 plane, and parallel to it. Now, if the stake be same height 
 throughout, TT will be equal to AS ; so also will V V be equal 
 to AS ; and being placed on the ground, their extremities will 
 lie in the straight lines SV, AV (being parallel). But lines 
 joining equal parallel lines will vanish, hence SV and AV, 
 being produced, will meet in C. 
 
 As nothing is ever supposed to be drawn 7iearer to the eye 
 tha,n the picture plane, we use the picture plane as a basis of 
 measurement, — for in the figure we can estimate the length 
 of TT or VV only by referring them to AS, which is drawn 
 on the picture plane. 
 
 Objects to the right of 00 are said to be to the right; 
 objects left of 00, to the left. Objects which are at an equal 
 distance on each side are said to be directly in front. 
 
 Perspective is of four kinds : — 
 
 1. Parallel, in which some side or face of the object is per- 
 pendicular or parallel to the P.P., and also to the G.P. 
 
 2. Angular, in which a side or face makes an angle less 
 than a right angle with the P.P., but is parallel to the Gr.P. 
 
 3. Oblique, in which the sides or faces make angles less 
 than right angles with both P.P. and G.P. 
 
 4. Aerial, or the perspective of distinctness* in a view. It 
 is related to shading and painting. 
 
THE POINT. 
 
 15 
 
 DEAWING TO A SCALE. 
 
 Unless objects were very small, and our drawing surface 
 large, we could not represent the size of the object as it is. 
 It is usual to draw the figure in miniature, or a certain num- 
 ber of times smaller. Every line in the drawing must bear 
 this fixed proportion to the corresponding one in the figure or 
 object. This is called drawing to a scale. 
 
 Usually, one quarter-inch for each foot is the proportion in 
 perspective, but any other ratio may be used. 
 
 Thus, if a line eight feet in length were to be drawn, it 
 would be represented by a line (on the picture plane) two 
 inches in length. 
 
 HL 
 
 C 
 
 
 HL 
 
 
 d 
 d 
 
 Wb 
 
 \ 
 
 
 
 \.\ 
 
 
 BL 
 
 \ 
 
 \ BL 
 
 
 D 
 
 A 
 
 
 B 
 
 
 
 Fig. 5. 
 
 Draw any horizontal line H.L., take any point O, draw OC 
 perpendicular to H.L. Let CD = height of spectator; tlirough 
 D draw B.L. parallel to H.L. If O be spectator, C wil) be 
 centre of vision, B.L. base line, and H.L. honzontal line. OC 
 will be line of direction and OD length of distance. (Fig. 5.) 
 
16 
 
 DRAWING. 
 
 I"; 
 
 I il 
 
 *l !,l| 
 
 :i 
 
 In B.L. take any points, A, B, and join them with centre 
 of vision (C.V.). If dab be drawn parallel to DAB, and 
 touching CI) and CB, it will be equal to DB, and da = DA 
 and ah = AB ; also, if the distance of B to the right of D be 
 known, the distance of h from d is known, for db = DB ; or, in 
 other words, every part of BC is the same distance from CD 
 that B is. 
 
 If we wish to find the position of a point within the plane, 
 we first find its distance on B.L. from D, and then join the 
 point marking this distance with C; the latter line would pass 
 through it. 
 
 Exercise I. 
 
 1. Find a point on the ground plane, at base of picture 
 plane, 4 feet to right. 
 
 2. Find a point 3 feet above G.P., 4 feet to right, and 
 touching P.P. 
 
 3. Find a point directly in front touching P.P., and 6 feet 
 above it. 
 
 4. Find position of a point 3 feet under G.P., 3 feet to" 
 right in P.P. produced. 
 
 (In foregoing examples assume spectator's height to be six 
 feet.) 
 
 All lines at right angles with the base line (B.L.)., or, 
 parallel with the direction we are looking (L.D) will vanish 
 at the centre of vision (C.V.) ; but lines parallel with the 
 base line (B.L.) or horizon (H.L.) will never vanish, but 
 always appear parallel. This is an important rule. 
 
 "to find the distance of a point within the plane. 
 
 In fig. 6, let AB DC represent the face of a cube resting 
 on the ground, and let the given face touch the picture plane. 
 The near lower ed^e will then coincide with base line AB. 
 Now, if we suppose the cube placed to the left of the spec- 
 
THE POINT. 
 
 17 
 
 tator, he will be able to see the side BD FE, or the top, if he 
 ]>e as high as the cube. 
 
 CV 
 
 We have said that all lines at right angles to AB will 
 vanish at point CV. ; and since all angles of a cube are right 
 
>■' 
 
 1i 
 
 18 
 
 DRAWING. 
 
 angles, the line BE, if produced, will reach C.V., and 
 so will line DF, for the same reason. But we want to find 
 the position of E, the extremity of the line BE. 
 
 Draw a line from C.V. perpendicular to AB, and produce 
 it indefinitely. A spectator stationed at N would see the 
 edge BE represented in size by the dotted line Be perpen- 
 dicular from B on EN. If he were placed at O, BE would 
 appear as B6, perpendicular to EO from B ; also, if placed at 
 P, BE would appear as Ba / and it will be noticed that, as 
 the spectator recedes, the apparent size of ,BE decreases, ?.e., 
 Ba is less than B6, and B6 less than Be, and so on. (Fig. 6.) 
 
 Now from B measure lengths of Be, B6, Ba on BE ; thus, 
 B/= Be, Be = B6, Bd = Ba, etc. 
 
 Take a point M on base line at a distance from B equal to 
 required length of BE ; this point may be on either side of B. 
 
 In given case, since BE = AB or BD, make BM = AB, and 
 suppose M to be drawn t( right of B. Join M/, Me, '^\d, etc., 
 and produce them backward to horizontal line touching it in 
 h, k, m, etc. 
 
 Now, if M be a fixed point for a distance from B, it will be 
 seen that as the spectator recedes from the object the points 
 hy k; m, etc., recede from C.V. ; hence, if the points h, k, wi, 
 etc., be given, and knowing position of M, we can find appar- 
 ent distance of BM within the plane, as shown by ^f, Be, etc., 
 and this is done by the following method : 
 
 Take C.V. as centre, and distances N, O, P, etc., of specta- 
 tors from object, as distance; and describe a semicircle cutting 
 horizontal line in m, k, or h on one side, and corresponding 
 points on the other. The points where the semicircle cuts 
 the horizontal line, are called the measuring points, and are 
 denoted by RMP, LMP, according as they are to the right or 
 left of C V. 
 
THE POINT. 
 
 19 
 
 To find the measuring points, having given the height of 
 
 spectator and his distance from the picture plane or base line. 
 
 In fig. 7, let H.L. be horizontal line, which is unlimited in 
 
 length ; take any point C.V. in it, and draw line L.D. from it 
 at right angles. Make C.V. = 6 feet, or whatever may be 
 the spectator's height, and through O draw B.L. parallel to 
 H.L.; produce C.V. O to S.P., so that O S.P. equals distance 
 
 f> 
 
20 
 
 DRAWING. 
 
 of spectator from base line. Then with centre C.V. and 
 distance C.V. S.P. describe a semicircle, cutting H.L. in LMP 
 and RMP, the right and left measuring points respectively. 
 
 In ilg 7, let any point A be taken, say 4 feet to the left, on 
 the base line ; join A C.V. 
 
 Suppose we wish to find a point the same distance to the 
 left that A is, but 4 feet within the plane : we know that the 
 point lies somewhere on A C.V., because every part of this 
 line is the same distance from C.V. O that A is. Now, we 
 proceed by measuring the required distance to right or left of 
 A, and joining the point thus found with the measuring point 
 opposite — that is, if point be taken to the right of A, as O, 
 we join O LMP ; if to the left, as E, we join E RMP. O 
 LMP and E RMP will always cut A C.V. at the same point 
 Bif EA = AO. 
 
 So also, if we take a lAsser distance, as AD, and make 
 AF = AD; join F RMP and D LMP; they will intersect in c. 
 Now AO or AE = AB, hence B is four feet within the plane 
 and four feet to the left. 
 
 In practice it is not necessary to draw to both measuring 
 points, one (the nearest) will answer every purpose. 
 
 Example 1. — Find position of a point on the ground 6 feet 
 directly in front, as seen by a spectator 6 feet in height, and 
 4 feet from picture plane. 
 
 In fig. 8, we draw H.L. and B.L. 6 feet apart, and C.V. 
 S.P. perpendicular to B.L. from C.V., and make O S.P. 
 the spectator's distance from base line. Draw a semicircle 
 to cut H.L. in LMP and RMP. Now, since point required 
 is on line between C.V. and O, we measure 6 feet either way 
 on B.L., as A or B, and join to measuring point as B LMP 
 or A RMP ; they will intersect in X, the point required. 
 
THE POINT. 
 
 21 
 
 Example ^.— Find a point 3 feet (3') to right, 4 feet (4') 
 within the P.P., and 5 feet (5') above it. Height of spectator 
 5 feet 6 inches (5' 6"), and his distance from P.P. 4 feet (4'). 
 
 BMP 
 
 Fit'. 9. 
 
 Draw H.L. and B.L. 5' 6" apart; draw C.V. S.P. cutting 
 B.L. in O ; make O S.P. = 4'. Find RMP (C.V. RMP must 
 always be equal to C.V. S.P.); make 0A = 3', from A measure 
 
 U\ 
 
22 
 
 DRAWING. 
 
 off AB oqual to the required distance of point within the 
 plane (4'). Join A C.V. and B IIMP, intersecting in C ; draw 
 AD perpendicular to B.L., and make AD equal to required 
 height (5'). Joii. D C.V. Draw CX parallel to AD and 
 meeting D C.V. in X. X is position required. For, since 
 AC = AB, and any point C in A C.V. is same distance to the 
 right that A is, then C is 3' to the right and 4' within P.P. 
 
 Now, D C.V. and A C.V. are vanishing lines, and CX and 
 AD are parallel lines drawn between them, then CX = AD ; 
 but point D is 5' high, then X is the same height, and is 
 vertically above C also ; therefore X is position of required 
 point. (Fig. 9.) 
 
 Note. — All measurements must be made on the P.P. 
 
 Exercise II. 
 
 (In the following examples take height of spectator 6 feet, 
 and his distance from P.P. 4 feet, and make scale ^ inch to 
 one foot.) 
 
 1. Find position of a point directly in front, and 8 feet 
 within P P. on G.P. 
 
 2. Find position of a point 4' to left (L), 4' within P.P. on 
 G.P. 
 
 3. Find position of a point 6' to right (R), 6' within P.P., 
 and 4' above G.P. 
 
 4. Find position of a bird flying 10' to R., 12' within P.P., 
 and 8' above G.P. 
 
 5. Find position of a fish resting in water 4' beneath G.P., 
 6' to R., and 8' within P.P. 
 
 6. In Ex. 1, show that however far the point be away, it 
 must always be nearer than the point C.V. 
 
THE LINE. 
 
 23 
 
 THE PERSPECTIVE LINE. 
 
 A straight line is the sliortest distance 'oetween any two 
 points. If we know the position of any two points we can 
 locate the line between them. 
 
 Example 1. — Draw a staff 8' high placed erect on ground 
 plane, 6' to R., and 4' within P.P.; disUnce 4', height of 
 spectator 6' (H = 6'), scale \" = 1 '. 
 
 i 
 
 Fig. 10. 
 
 Draw H.L. and B.L. 6' apart; draw C.V. S.P., making 
 O S.P. (LD) 4', and find RMP as before. Take A, 6' to right 
 of O, and erect perpeudioular AB 8' iu height; join B C.V. 
 
24 
 
 DRAWING. 
 
 and A C.V. ; take C 4' to left of A, and join C RMP, cutting 
 A C.V. in E. Through E draw ED parallel to AB ; ED is 
 line required. Now, since AE = AC, E is 6' to right and 4' 
 within P.P.; but ED = AB, then ED is 4' within P.P., 8' 
 high, and 6' to right. (Fig. 10.) 
 
 Example 2. — Draw a line 3' long lying on G.P., parallel to 
 P.P. (or base line), and 4' within it, near extremity of line 
 being 2' to left. H = 6 , L.D. = 4', scale Y = !'• 
 
 RMP 
 
 i!- 
 
 Draw H.L., B.L. and L.D. as before, and find also LMP. 
 From O mark off OB = 2' (distance of near extremity from O), 
 and make BA = 3' (length of line) ; join B C.V. and A C.V. 
 From B measure off BE = 4' (distance within P.P.) ; join 
 E LMP, cutting B C.V. in D ; through D draw DC to line 
 A C.V. and parallel to AB. CD is line required, for it is 
 equal to AB, and its near extremity D is same distance to 
 the left that B is, and BE = BD. (Fig. IL) 
 
 Example 3. — Draw a line 3' in length lying on G.P., per- 
 
 
THE LINE. 
 
 25 
 
 pendicular to P.P. (base line), 4' to R., and nearest extremity 
 2' within P.P. H = 6', L.D. = 4', scale ^ = 1'. 
 
 LMP 
 
 ^AfP 
 
 Draw H.L., B.L., L.D., as before, and mark point RMP. 
 Make OC = 4', join C C.Y. From C measure off CB = 2', and 
 from B mark off BA = 3' (length of line) ; join B RMP and 
 A RMP, to cut C C.V. in E, D. ED is the line required, for 
 DE is parallel to LD, and therefore perpendicular to BL. 
 EC = BC and DC = AC, then DE = AB, and point E is the 
 same distance from L.D. that C is. (Fig. 12.) 
 
 Example 4.— Draw a line 3' in length parallel to P.P. and 
 4' within it, and parallel to G.P., and 4' above it, line to 
 be drawn with near extremity 2' to left. H = 6', LD = 4', 
 scale i" = r. 
 
 In fig. 13 draw H.L., B.L. and L.D., and find LMP, 
 take B 2' to left and A 4' to left of B, also S 4' to ri<iht of 
 B. Erect AC and BD perpendiculars to AB, and each 4' in 
 height; join CD, C C.V., D C.V., B C.V., and S LMP. Let 
 S LMP cut B C.V. in G; draw GF parallel to BD, and FE 
 3 
 
fffiif 
 
 26 
 
 DRAWING. 
 
 parallel to CD : then EF is line required. For since BG = 
 BS, and BD = FG, then DF = BG, so also CE = DF. But 
 EF = CD, then EF is parallel to P.P., is same height above 
 G.P. as BD, and the same distance to the left as BD. 
 
 IV 
 
 l! 
 
since BG = 
 = DF. But 
 
 sight above 
 D. 
 
 THE LINE. 
 
 Exercise III. 
 
 27 
 
 (H = 6', L.D. = 4', scalei" = l'.) 
 
 1. Draw a line 4' in length on G.P., parallel to P.P. and 4' 
 within it, directly in front. 
 
 2. Draw a line 4' in length parallel to G.P. and 4' above it, 
 touching P.P. 4' to left, and perpendicular to it. 
 
 3. Draw a line 3' in length perpendicular to G.P., and 3' 
 above it ; line to be 4' to right. 
 
 4. Draw a line 4' in length parallel to G.P. and 5' above it, 
 and m contact with P.P. ; line to have one extremity 3' to 
 
 right. 
 
 5. A line 5' in length is drawn parallel to P.P. and 6' 
 within it ; it is parallel to the G.P., one extremity being 3' to 
 right, the other 2' to left. It is 4' above the G.P. 
 
 I 
 
DRAWING. 
 
 SUEFACES IN PEB8PECTIYE. 
 
 RECTANGULAR SURFACES. 
 
 m 
 
 J)' I 
 
 m 
 
 
 ' m 
 
 1 1 
 
 Thie Square. 
 
 A square is a parallelogram having two adjacent sides 
 equal, and the included angle a right angle. 
 
 We said previously that an object appears to decrease in 
 size as it recedes from the eye. If, therefore, a square is 
 placed on the ground plane with one side touching the picture 
 plane, it is clear that the side most removed will appear 
 smaller than that touching the picture plane, and so the 
 square may not appear to have even one right angle. 
 
 Let AB, BC, CD, DE be all taken of equal length, and let 
 C be on L.D., it is plain if lines parallel to AB And of equal 
 length be drawn on G.P. within the P.P., they will appear 
 shorter than AB. Find points M, N and join BM, CM, etc. 
 Then AF = AB, GB = BC = AB, etc. Hence FG = AB, GH = 
 BC, etc., also AF = BG, BG = CH, etc., and the squares AG, 
 BH, CK, and DL will all be equal, and the only real right 
 angles will be BCH and DCH ; all the other angles, FAB, 
 ABG, etc., though really right angles will not appear so. 
 (Fig. 14.) 
 
E. 
 
 ent sides 
 
 screase in 
 square is 
 le picture 
 11 appear 
 i so the 
 
 1, and let 
 of equal 
 11 appear 
 CM, etc. 
 B, GH = 
 ires AG, 
 3al right 
 ;s, FAB, 
 pear so. 
 
30 
 
 DRAWING. 
 
 To represent a square perpendicular to G.P. and P.P. 
 
 Let H.L. and B.L. be drawn, also L.D., and find X. Take 
 
 
 any points A, B, C on B.L. at equal distances, and at each 
 erect a perpendicular equal to AB or CD ; join EH, which 
 
THE SQUARE. 
 
 tl 
 
 will be parallel to AD; join A C,V., B C.V., C C.V. and" 
 D C.V. Take a point a at a distance from A equal to AB 
 (on left), and join AX to cut A C.V. in K ; draw KN parallel 
 to AD, intersecting the vanishing lines in L, M, N ;• at K, L, 
 M, N, erect perpendiculars to meet the vanishing lines in Q, 
 P, R, S ; join QS. Now, since AB = AE, AF is a square, 
 
 (a) AF is said to be drawn to right, perpendicular to G.P., 
 and touching or coincident with P.P. • 
 
 (h) Since Aa = AK = QE, then KE is a square drawn to 
 right, perpendicular to G.P., perpendicular to and touching 
 P.P. and parallel to L D. 
 
 (c) Since AK = AB = KL = LB, AL is a square drawn to 
 right, resting on G.P., perpendicular to and touching P.P. 
 
 (d) Since AB = AK = QE = PF, QF is a square drawn to 
 right, parallel to G.P. and raised above itf and perpendicular 
 to and touching P.P. and parallel to L.D. 
 
 The cube AP or BR will show the square in every position 
 in parallel perspective, provided it touch the picture plane. 
 
 (e) Similarly, square QL is drawn to right, perpendicular to 
 G.P., parallel to P.P., and within it. (Fig. 15 ) 
 
 Note. — If a figure be drawn parallel to the P.P. it will 
 always be drawn in its true shape, but smaller, hence QL 
 will be a true square. In above figure, take points a, 5, c, c?, 
 etc., at equal distances to the left ; erect perpendicular de at 
 dy and equal to c?c. Join e C. V. and d CY\, also cX, 6X, 
 rtX, to meet d C.V. in /, g, h. Through f, g, h draw 
 parallels to de, meeting e C.V. in /, m, n. Then, since de, fl, 
 etc., are parallels between vanishing lines, they are all equal 
 to each other and to dc, cb, etc., for de = dc. But df= dc, and 
 Jg — cb, etc., hence e/, Ig, mn are equal squares, and are said 
 to be drawn perpendicular to G.P. and also to picture plane. 
 One of them, e/, touches the P.P., the others are within it. 
 It will be seen that as the squares recede they appear smaller. 
 
 
 3' 
 
■r^ 
 
 DRAWING. 
 
 Example 1. — Draw a square, side 4', 4' to R., 4' within, 
 lying on G.R, ±_ to P.P. H - 6', L.D. = 4', scale J" = 1'. 
 
 Construct figure as before. Take AC one inch (4') and CD 
 one inch,* take AB to left one inch, join C C.V., D C.V., 
 B MP, A MP, cutting C C.V. in F, E. Draw EK and FG, 
 parallel to AC, and meeting D C. V. in K, G : then FK is 
 square required, forFG = EK = CD, EC = CA, and EF = AB. 
 (Fig. 16.) . 
 
 - Ck 
 
 Fig. 16. 
 
 to 
 
 Example 2. — Draw a square, side 4' and 6' to left, _ 
 P.P. and 3' within it, and _]_ to G.P. and 1' above it. 
 Other specifications as in last example. 
 
 Construct figure as before. Take C H" (6') to left of A, 
 take B 1" (4') to right of A, erect CE ±^ to AC at C and 1 J" 
 (5') in height, and mark off D Y (!') above C; join E C.V. 
 D C.V., C C.V., A MP and B MR Through F, G draw FL 
 and GS parallel to CE, and meeting vanishing lines in L, S 
 and H, K respectively : LK is the figure required. For 
 
THE SQUARE. 
 
 « 
 
 LS = HK = FG = AB = 4', SK = LH = ED = 4' and HF = DC 
 
 = 1', also FC = AC = 6'. (Fig. 17.) 
 
 Fig. 17. 
 
 Example 3. — Draw a square 4' side directly in front, paral- 
 lel to G.P. and 3' above it, and J_ to P.P. and 3' within it. 
 Same specifications as before. 
 
 Construct figure and find M as before. From A measure 
 off Y both ways to B and C (BC = 4') ; erect perpendiculars 
 BF and CG, each j" (3'); join FG, B C.V., F C.V., and G C.V. 
 From B measure ofT BD \" (3'), and from D measure off DE 
 1" (4') ; join DM and EM, cutting B C.V. in H, K. Draw 
 HL, KN parallel to BF, and meeting F C.V. in L, N; through 
 L, N draw U and NP parallel to FG : then LP is the 
 square required. For FL = BH = BD = 3', and NP = LJ = FG 
 = BC = 4', L]Sr = HK = DE = 4', andLH = FB = 3'; alsoSL = 
 SJ = XF = XG = AB = AC = 2'. (Fig. 18.) 
 

 mw 
 
 m 
 
 '*' 
 
 , ! 
 
 .34 
 
 DRAWING. 
 
 . / 
 
THE SQUARE. 
 
 36 
 
 Example Jf.. — Draw a square 4' side, 4' to right, lying on 
 G.P., _L to P.P. and 4' within it, and within this square 
 place centrally a square whose sides are 2'. 
 
 Fig. 19. 
 
 Note. — Figures are said to be placed centrally when their 
 centres coincide and their like sides are parallel to each other. 
 Concentric circles are always placed centrally with respect to 
 each other. 
 
 Draw H.L., B.L., L.D., as before, and find M. Measure oflF 
 aA, ae, and AE, each one inch ; bisect AE and ae in C and 
 c, and bisect CE, CA, ce and ca in D, B, d, b respectively. 
 Join A C.V., B C. v., C C.V., D C.V., aM, 6M, cM, c^M, and 
 eM, cutting A C.V. in/, g, n, k, I; and through these latter 
 points draw parallels to AE, cutting E C.V. in t, s, r, q and j. 
 It will form the outer square, and XX the inner square. 
 For «c = BD = 2', and kg = dh = 2', and /A = Aa and/< = AE, 
 etc. (Fig. 19.) 
 
 '; 
 
T"^ 
 
 36 
 
 DRAWING. 
 
 Note.— If aM. be joined, it will always pass through ,/ if 
 tlie figure be a scjuare. It will also pass thiough X and X. 
 
 i ! 
 
 Exercise IV. , 
 
 (H = 6', L.T).-4', scale 1" = 1'.) 
 
 1. Draw a square 4' side, 3' to left, resting on G.P., J_ to 
 and touching P.P. 
 
 2. Draw a square 5' side, directly in front, lying on G.P., 
 I to and touching P.P. 
 
 3. Draw a square 6' side, _L ^-o Gr-P-) _L *^ P.P-, touching 
 both, and 6' to left. 
 
 4. Draw a square 3' side, parallel to G.P. and 2' above it, 
 4' to R., perpendicular to P.P. and 2' within it. 
 
 5. Draw a square 4' side, parallel to P.P. and 4' within it, 
 resting on ground plane, left corner touching L.D. 
 
 6. Draw a square 4' side, parallel to G.P. and 2' above it, 
 right corner 1' to right; square to have side _J_ to P.P. and 
 r within it. 
 
 7. Draw a square 4' side, _|_ to G.P. and 1' above it, 3' to 
 right, _J_ to P.P. and 2' within it. 
 
 8. Draw a square 6' side, directly in front, J_ to P.P. and 
 2' within it, parallel to G.P. and 4' above it ; and place a 
 square of one-fourth its area centrally within it. 
 
 9. Draw a square 3' side, 4' to right, _[_ to P.P. and 3' 
 within it, _l_ to G.P. and just its own height below it. 
 
 10. Draw a cube (edge 4') touching P.P. 4' to left, resting 
 on G.P. 
 
 11. Draw a cube (4' edge) directly in front, and to rest on 
 G.P., one edge parallel to P.P. and 2' within it. 
 
THE OBLONG. 
 
 37 
 
 The Oblong;. 
 
 An oblong is a figure wlio.sc^ opposite sides are eijual and 
 parallel, and whose angles are right angles. 
 
 The drawing of the oblong differs little from that of the 
 square, care being required only to distinguish the sides. 
 
 Jig. 20. 
 
 Example 1. — Draw an oblong 3' x 2', lying on G.R, _\_ to 
 P.P., side 3', parallel to, and 2' within it; oblong to be 4' to 
 right. Specifications as before. 
 
 ! 
 I 
 
 m 
 
■Mm. DRAWING. 
 
 Construct figure as before. Find M, and measure off AC 4' 
 = (distance to right), CD = 3' (length of side), and measure off 
 BE 2' (breadth); measure off BC = 2' (distance within P.P.). 
 Join EM, BM, C C.V., D C.V.; through H, F draw HK, FG 
 parallel to CD, meeting D C.V. in G, K : then FK will be 
 oblong required. For FG = HK ^-- CD --= 3', GK = FH = EB = 
 2', and HC = CB = 2'. (Fig. 20.) 
 
 i i 
 
 Exercise V. . 
 
 (H = 6', L.D. = 4', scale i" = V.) 
 
 1. Draw an oblong 4' x 2' J_ to G.P., and 4' to left, J to 
 P.P. and 3' within it ; oblong to rest on end. 
 
 2. Draw an oblong 7' x 5', standing on end on G.P., parallel 
 with P.P. and 2' within it, directly in front. 
 
 3. Draw an oblong 3' 6" x 4' 6", parallel to G.P. and 2' 6" 
 above it, right corner 3' 6" to left ; end of oblong to be paral- 
 lel to P.P. and 6" within it. 
 
 4. An oblong 6' x 4' is buried in the ground to a depth of 
 2' ; it is parallel with LD and 4' to right, and 2' within P.P. 
 
 5. An oblong 6' x 4', with side resting on G.P. 4' to left, 
 anc' _\_ to it, intersects a square of 4' side, resting on G.P. 
 and parallel £o P.P. ; the oblong divides the square into two 
 equal portions, and the square divides the oblong into parts of 
 4' and 2' respectively, the greater portion being nearest. The 
 square is 4' within the P.P. 
 
 6. Draw an oblong 4' x 2', end touching P.P. 4' to R. ; 
 oblong lying on G.P. 
 
 7. Draw an oblong 5' x 3', lying on G.P. directly in front; 
 end parallel with P.P. and 2' within it. 
 
 8. A wall, whose height is 8', begins at a point 10' to left, 
 and stret^ihes inwards indefinitely ; at distances of 10' and 20' 
 doors 5' x 3' are made in it. Drtiw it. 
 
 9. Draw an oblong 6' x 4', lying on ground, sides parallel 
 to P.P. and 2' from it, oblong 3' to right ; and within it place 
 an oblong 4' x 2' centrally. 
 
 10. Draw an oblong 3' x 2', lying on G.P. 4' to left, end 
 parallel to P.P. and 2' from it; and about it draw an oblong 
 6' X 4' centrally. , .. 
 
to 
 
 TRIANGLE. 
 
 39 
 
 Tine Triang-ie. 
 
 A triangle is a figure enclosed 
 by three straight lines ; these 
 may be of uniform length, but 
 the length of any two taken 
 together must be greater than 
 the third. 
 
 The triangle cannot be con- 
 veniently drawn alone, but is 
 drawn with reference to an ob- 
 long. 
 
 Triangles are equilateral, with 
 three equal sides ; 
 
 Isosceles, with two equal sides ; 
 
 Scalene, with unequal sides. 
 
 Right angled, or containing a 
 right angle ; 
 
 Obtuse angled, or containing 
 an obtuse angle ; 
 
 Acute angled, or containing 
 three acute angles. 
 
 To draw a triangle we first 
 draw a plan, showing the tri- 
 angle and its bounding oblong. 
 
 Thus, 1 would be a plan for 
 an equilateral triangle ABC, in 
 which D A = AE = BF = FC. 
 
 2 would be a plan for a right- 
 angled triangle, in which the 
 angle bac is th eright angle, with 
 he as hypothenuse ; the segments 
 bfj fc of the hypothenuse w6uld 
 determine the position of a. 
 
 ii^^ 
 
 
 w 
 
 '1 
 
 ■ I 
 
DRAWING. 
 
 In 3, the triangle DAC being given, CE would be drawn 
 perpendicular to DA, and CB parallel to DA ; also DB and 
 AF would be drawn parallel to CE : then position of the 
 angles at A and would be easily determined. 
 
 Example 1. — Draw an equilateral triangle, 3' to a side, 
 lying on ground, one side touching P.P. near left angle, 4' to 
 right. Height 6', distance 4', scale J" = r. 
 
 f; \ 
 
 '■■I 
 
 \ t 
 
 I! 
 
 Draw H.L., B.L. and L.D. as before. Take A 4' to right of 
 S, and B 3' to right of A ; join A C.V. and B C.V.; on AB 
 describe equilateral triangle ABE (below AB). Draw AD, 
 BF at right angles to AB, and through E, draw DF parallel 
 to AB, and meeting AD in D and BF in F ; draw EG paral- 
 lel to AD. Join G C.V., and with centre A and distance 
 AD, describe arc DL, cutting B.L. in L. Join LM, cutting 
 
THE TRIANGLE. 
 
 41 
 
 ght of 
 n AB 
 AD, 
 
 rallel 
 paral- 
 
 tance 
 iitting 
 
 A CV. in N ; through N draw NK parallel to AB, cutting 
 G CV. in C and B C.V. in K. Join CA and CB : then 
 ABC is the triangle required. For AB = AE and AD = AL 
 = AN = CG = height of triangle ; also NC = CK = AG = GB = 
 DE = EF. Then AC = AE = EB = BC, for they are diagonals 
 of equal oblongs. (Fig. 22.) 
 
 In this figure the vertex C is directed away from the eye. 
 
 If we joined NG and KG we would have a similar triangle, 
 but with vertex G directed towards the observer. 
 
 Example 2. — Draw an isosceles triangle whose sides are 
 2', 3' and 3' respectively, lying on ground plane, vertex 
 directed towards the spectator, base parallel with P.P. and 5' 
 away from it ; near angle 2' to left. 
 
 Draw B.L. and H.L. as before, find M. Take K 2' to left 
 of X, and G 2' to left of K ; bisect GK in H ; draw GR, HS 
 and KT at right angles to B.L., and 5' in length ; join RST. 
 Take TP 3', and with centre T and distance TP describe arc 
 PN, cutting HS in N ; join NT and NR : NRT will be the 
 plan of the triangle. 
 
 Througli N draw UNQ parallel to RT, cutting GR hi U 
 and KT in Q ; with centre K and distance KQ describe arc 
 QL, cutting B.L. in L ; and with same centre and distance 
 KT describe arc TV, cutting B.L. in V. Join G C.V., 
 H C.V. and KC.V. ; and join also LM and VM to cut 
 K C.V. in E and C. Through E draw EAF, and tlirough 
 C draw CDB, each parallel to B.L,, and cutting H C.V. in A 
 and D respectively ; join BA and AC : then ABC will be the 
 triangle required. For CK = KV = KT = 5', and EC = LV = 
 QT = NS = height of triangle, and EK = KL = KQ. Hence 
 AD = NS and AH = HN. But since triangle is isosceles, GH 
 is made equal to HK ; hence BD = DC = GH = HK, then CA 
 = NT = NR = BA. (Fig. 23.) 
 
 ;■ ( 
 
?II 
 
 r 
 
 DRAWING. 
 
 ' I ti 
 
 Fig. 23. 
 
 I I 
 
 E.icainple 3. — Draw an equilateral triangle, each side 4', one 
 side on ground plane parallel to L.D. ; triangle placed _]_ to 
 ground plane and touching P.P. 3' to right. 
 
 Draw H.L., B.L. and L.D., and find M. Take point D 3' 
 to R., and B 3' to right of D ; on PB describe the equilateral 
 triangle DGB. Bisect DB in E, join EG ; through G draw 
 FH parallel to B.I;., and through D, B draw DF, BH parallel 
 to EG, cutting FH in F, H. Join B C.V., draw EM, DM, 
 cutting B C. V. in N and A. With centre B and distance BH 
 describe arc HK, cutting BK, a perpendicular on B.L., at K ; 
 
T- 
 
 THE TRTANOLE. 
 
 48 
 
 join K C.V. Through A and N draw parallels to BK, meet- 
 ing K C.V. in L and C ; join AC, CB. Then ABC is the 
 triangle required. For LA = CN = KB = BH = EG = altitude 
 of the triangle ; and AN = NB = DE = EB, and AB = DB = 
 
 length of side ; then C corresponds to G, and AC, CB to DG 
 and GB respectively. (Fig. 24.) 
 
 Example Jf.. — Draw a right angled-triangle whose hypothe- 
 nuse is 4' and one of the other sides 3' ; the right angle is 
 
44 
 
 DRAWING. 
 
 directed away from the observer, and is 5' to the left, and 
 the hypothenuse is 3' to the left. The triangle lies on the 
 ground with the hypothenuse in contact with the B.L. 
 
 Fig. 26. 
 
 Draw H.L., B.L. and L.D. as before. Find M, and take a 
 point D, 5' to left, C, 3' to left, and B, 4' to left of C. On 
 BC describe the semicircle BGC. (The angle BGC in a semi- 
 circle is always a right angle.) Join DG, B C.V., D C.V. and 
 C C.V.; through G draw EF parallel to BC, and through B, C 
 draw BE, OF perpendiculars on EF, at E and F. With 
 centre C and distance CF, describe arc FL, cutting B.L. in L ; 
 join LM, cutting C C.V. in H ; through H draw KAH 
 parallel to BO, cutting D C.V. in A ; join AB and AC : then 
 
THE TRIANGLE. 
 
 45 
 
 ABC is the triangle required. For AB = BG and AC = CG, 
 and AD = DG = CF = CL, and K A = BD and AH = DC ; then 
 KH = BC, and A corresponds to G ; then angle BAC cor- 
 responds to angle BCG. (Fig. 25.) 
 
 Exercise VI. 
 
 (H = 6', LD = 4', scale ^" = 1'.) 
 
 1. Draw an equilateral triangle, 3' side, lying on G.P., one 
 side parallel to L.D., vertex directed to the left and distant 
 from the L.D. 4' ; the triargle touches P.P. 
 
 2. Draw an equilateral triangle parallel to P.P. and 3' from 
 it, _L to G.P., near angle 2' to left, triangle 3' to side, one 
 side on ground. 
 
 3. Draw a triangle whose sides are 4', 5', 6', respectively, 
 6' side on ground, _[_ to P.P. and 2' within it ; triangle 3' to 
 right. 
 
 4. Draw an isosceles triangle whose base is 3' and each 
 equal side 4', lying on ground plane directly in front, vertex 
 directed towards P.P. and 1' from it. 
 
 5. Draw a right-angled triangle whose hypothenuse is 5', 
 and the perpendicular on it from the right angle divides it 
 into segments of 3' and 2'. The triangle is parallel with G.P. 
 and 4' above it, and the vertex is directed away from P.P. at 
 a distance of 4'. The right angle is directly in front, and the 
 larger segment is to the right. 
 
 6. Draw a triangle, each side being 3', _]_ to G.P. and 1' 
 above it, with a side parallel to it. The triangle is 4' to left, 
 J_ to P.P. and 2' within it. 
 
 7. Draw an isosceles right-angled triangle, the equal sides 
 being 3' ; one equal side is parallel to G.P. and 2' above it. 
 The triangle is 2' to right, parallel to P.P., and 2' within it. 
 
 8. Draw an isosceles triangle, base 4', equal sides 3' each, 
 directly in front, parallel to P.P. and 4' from it ; vertex touches 
 ground, and base is parallel to it. 
 
 9. An equilateral triangle, each of whose sides is 4', lies on 
 the ground, vertex directed away ; one side parallel to P.P. 
 
Sf ? 
 
 46 
 
 DRAWING. 
 
 and 2' from it. Triangle 2' to right. Within this place cen- 
 trally a similar triangle whose sides are 2'. 
 
 10. An equilateral triangle, each of whose sides is 4', is _L 
 to ground and also to P.P., which it touches at a point 5' to 
 left ; the vertex of the triangle is directed downwards, and 
 one side is horizontal The triangle is buried one-fourth in 
 the ground. Draw it. • 
 
 
 ;^l .: 
 
 
 >i '. 
 
 \, ! ! , 
 
 • 1 ,it 
 
 1 '■ 
 
 .1 ^.^! •.. 
 
THB HEXAGON. 
 
 47 
 
 Thie Hexagon. . 
 
 A hexagon is a figure bounded by six sides. Wlien the 
 sides and angles are all equal the hexagon is regular. Only 
 regular hexagons will be considered here. 
 
 In drawing a hexagon, a plan is made somewhat similar to 
 that for the triangle. The following is the general form of 
 the plan. 
 
 K 01 D AC C £ 
 
 - G^ 
 
 Let B.L. = base line, and AB given side of hexagon, and in 
 given position. Bisect AB in C, niake AD and BE each equal 
 BC ; on DE describe equilateral triangle DFE, and construct 
 oblong DL. Bisect DK in G ; draw GH parallel to KL ; join 
 AG, GM, BH and HN. AVith centre D and distances DG 
 and DK describe arcs to cut B.L. in X and P respectively. 
 Like triangles, hexagons may lie flat on the ground, witii a 
 side parallel to the P.P. or to L.D., or they may be drawn 
 perpendicular to the ground plane, with a side or an angle 
 touching it ; and a side or an angle may touch the P.P. or be 
 within it. (Fig. 26.) 
 
 If a side touch the P.P., the extremity of it determines the 
 
 f 
 
^. 
 
 I ! 
 
 48 
 
 DRAWING. 
 
 
 i 1 
 
 ( 
 
 J! [. 
 
 .; i 
 
 ; i 
 
 distance of the hexagon to the right oi- left When an angle 
 touches the P.P., that point determines tiie distance ; so also 
 when the figure is within the P.P. 
 
 Example 1. — Draw a hexagon, each side 2', lying on G.P., 
 one side coincident with P.P. and 4' to R. H = 6', L.D. — 4', 
 scale J" =r. 
 
 Fljf. s 
 
 Let H.L., B.L. and L.D. be drawn. Take C.V., find O and 
 MP; take AB 4' to R. of Q, make AB = 2'. On AB con- 
 struct equilateral triangle ABH, and produce to Y, Z, making 
 HZ and HY each equal BH ; join AY and YZ, and produce 
 YZ both ways. Through H draw GHW parallel to YZ ; 
 through C draw CE parallel to AY ; draw also DWF parallel 
 to AY. Join CC.V., A O.V., B C.V. and D C.V. ; also 
 L MP and K MP, cutting C C.V. in N and P. Through 
 N, P draw NT and PV : parallel to AB ; join SP, PA, TV 
 and VB, then STVBAP is the hexagon required. For Pf; = 
 
 I! 
 
THE HEXAGON. 
 
 49 
 
 CK = CG, and NP = LK = GE. Then SP = GY, also PA = 
 AG=GY; similarly, TV = VB = BW = WZ = AB, and QA 
 = 4'andAB=2'. (Fig. 27.) 
 Example 2. — Draw a hexagon same as in Example 1, but 
 i to G.P., and with an angle touching P.P. 4' to L. 
 
 Fig. 28. 
 
 Here take E 4' to left, and for plan of the hexagon we 
 may proceed as follows : — Bisect EQ (4') in Z, also EZ, QZ 
 in A, B respectively ; with centre A and distance A 1 > 
 describe arc BC ; similarly, draw arc AD. Draw EC and 
 QD, perpendiculars to AB at E and Q respectively, meeting 
 arcs in C and D ; produce EC to F, making CF = EC, 
 Erect perpendicular EN and make it equal EF, and make 
 
oO 
 
 DHAWINCJ. 
 
 ES-EC; join N C.V., ^^ C.V. and E O.V., also AMP, 
 B MP and Q MP, cutting E C.V. in K, L, M. Through 
 K, L, M draw parallels to EN, cutting N C.V. in O, P, R ; 
 join OS, SK, LT and TP : then OPTLKS will be hexagon 
 required. For ES = EC and EN = EF, also K L = AB and EM 
 = EQ. Then OS = SK = AB ; and PT = TL = BD. (Fig. 28.) 
 Example 3. — Draw a hexagon, 2' side, resting on G.P. _J_ 
 to P.P., having one side coincident with it and 4' to right. 
 H = 6', L.D. = 4', scale J" =1'. 
 
 \ < 
 
 i 
 
 1 
 
 
 ■•] 
 
 
 
 
 r " 
 
 Fijc. 29. 
 
 We proceed as follows : — After drawing H.L. and B.L. and 
 finding positions of O, MP and C. V., take A 4' to right of S ; 
 at A erect perpendicular AC = 4', and measure off AE = 1' and 
 DE = 2'. Produce CA to B, making AB = AC and FX = DE ; 
 make FG = FX and GH = GA ; join HKMP, GLMP, A C. V., 
 
THK HEXA(K)N. 
 
 61 
 
 I) C. v., E C.V. and i) O.V. At K, L erect perpendiculars to 
 meet vanishing lines in N and M ; join MR, Ml), LE and 
 LP : then RMDELPR will be hexagon reciuired. For AK = 
 AH and AL = AG = FX = 2', and DE = RP and ML = NK = 
 ^ A. Then RM = MD = DE = 2', etc. (Fig. 29.) 
 
 Example ^. — Draw a hexagon, 2' side, lying on ground, 
 near side parallel vvitii P.P. and 2' within it; hexagon to be 
 directly in front. H = 6', L.D. = 4', scale \" = 1'. 
 
 ,)' 
 
52 
 
 DRAWING. 
 
 Draw H.L., B,L., and find O C.V. and MP as before. 
 Make SV and ST each = 1', and make TX and VA each = 1' ; 
 make AF-2', and draw it J_ to AX; draw XP, TY, VZ 
 parallel to AF. Make RG = RL or TV, and GZ also = RL, 
 and complete the oblong MH. Make Aa = AF, A6 = AG and 
 AE = AH; join EMP, 6MP, aMP, and A C.V., V C.V., 
 T C.V. and X C.V. Through intersecting points D, C, B, draw 
 parallels to AX, meeting X C.V. in e, J, g, respectively ; join 
 cf, fk, dC and Cm, completing the hexagon : then AB = Aa = 
 2', and BC = CD = rt6 = &E = GH, and hence Cm = GZ and dC 
 = YG = 2', etc. (Fig. 30.) 
 
 I 
 
 I 4 
 
 Exercise VII. 
 
 In these examples take H = 6', L.D. = 4', and scale ;J' = 1' ; 
 but a scale of ^" = 1' may be used if thought more convenient. 
 
 1. Draw a hexagon, side 2', lying on ground plane, one side 
 perpendicular to P.P., and an angle touching it at a point 4' 
 to right. 
 
 2. Draw a hexagon, side 3', standing on edge, _l^ to ground 
 plane and P.P., and having an angle touch the P.P. 3' to left. 
 
 3. Draw a hexagon, side 2', standing on edge, parallel to 
 P.P. 1^ to ground plane, directly in front, and 3' away. 
 
 4. Draw a hexagon, 2' side, lying on G.P., one side parallel 
 to P.P. and 3' away ; hexagon 4' to left. 
 
 5. Draw a hexagon, 3' side, resting on an angle _[ to 
 G.P. and P.P., one side parallel to P.P. 4' to right and 4' 
 within it. 
 
 6. Draw a hexagon whose edge is coincident with that of a 
 s(juare, and lying in same plane. The square is 2' to the side, 
 and is placed I to P.P. and G.P. 2' to right and 2' within. 
 
 7. Draw a hexagon, 3' side, parallel to G.P. and 4' above 
 it, 4' to right, 3' within P.P., and one side parallel to P.P. 
 
 8. Draw a hexagon 3' to side, 4' to right, one angle touch- 
 ing P.P.; hexagon to be _l to P.P. and parallel to G.P. 
 
THE HEXAGON. 
 
 .5.S 
 
 9. Draw a hexagon 3' to a side, directly in front, lying on 
 ground plane, one angle touching P.P., and sides _j_ to it. 
 
 10. Draw a hexagon about an equilateral triangle lying on 
 G.P., vertex directed away from observer ; the triangle is 3' 
 to a side, and one side is parallel to P.P. and 3' within it. 
 The vertex of the triangle is 4' to left. 
 
 11. Draw a hexagon, 2' side, placed _l to P.P. and G.P., 
 4' to right, lower side parallel to ground plane and 4' above 
 it. 
 
 12. Draw a hexagon, 4' side, lying on G.P., near side paral- 
 lel to P.P. and 2' within it ; hexagon to be 4' to right. 
 Within this draw (centrally) another hexagon, whose sides 
 shall be 2' in length. 
 
 13. Draw a hexagon, side 2', parallel to P.P. and 2' within 
 it, lying on G.P. directly in front. 
 
 14. Represent a hexagon, side 3', half buried vertically in 
 the ground, one side parallel to G.P. ; hexagon I to P.P. 
 and 3' within it, 4' to left. 
 
 to 
 4' 
 
 )ve 
 
^1 
 
 54 
 
 DRAWING. 
 
 Tlie Octagon. 
 
 The drawing of an octagon differs but little from that of 
 the hexagon, we shall, therefore, merely show the plan. The 
 following are methods of drawing the plan: — 
 
 Plfir.81. 
 
 Let AB = given side (on B.L.) ; bisect AB in O, and draw 
 OG _[ to OA and equal it, and describe semicircle AGB ; 
 
THE OCTAGON 
 
 
 join AG and GB ; with centres A, B, and distances equal to 
 AG, describe arcs to cut AB produced in C and D ; on CD 
 describe square CDFE. With centre A and distance AB 
 describe arc to cut CE in H ; find P similarly; with centres 
 H and P and distances equal to HA, describe arcs to cut CE 
 and DF in K and N respectively, and with same distances 
 describe arcs to cut EF in L and M ; join AH, KL, MN and 
 BP, which will complete the hexagon. (Fig. 31.) 
 Another way : 
 
 t 
 
 . 1 
 
 Fig. 32 
 
B 
 
 ^m 
 
 56 
 
 DKAWJNG. 
 
 I It 
 
 Let AB = given side ; take centre O, and with distance AO 
 describe circle ACD ; draw diameter CD J to AB. Join 
 CA and produce it. With centre A and distance AB describe 
 arc to cut CA produced in E ; then draw EF _l_ to BA pro- 
 duced, and produce FE, making EG = EA, etc. (Fig. 32.) 
 
 Note. — In parallel perspective a hexagon or an octagon 
 must be supposed to have a side parallel or perpendicular to, 
 the picture plane. 
 
TMK cikclp:. 
 
 AO 
 Join 
 ;ribe 
 pro- 
 
 ) 
 
 xgon 
 r to, 
 
 Xlie Circle. 
 
 Hitherto we have been dealing exclusively with straight 
 lines, in so far as the appearance of figures is concerned ; we 
 now proceed to represent curved lines in perspective. It is 
 evident that a curve cannot be correctly represented, without 
 the aid of straight lines. 
 
 There is only one position in which a circle will appear true 
 to the eye, and that is, when the eye is in a line exactly per- 
 pendicular to its plane, at its centre. In all other positions it 
 will appear an ellipse, varying from a circle to a line. If, for 
 instance, ve place a hoop on the ground, and look at it 
 directly, it will appear true, but if turned on an imaginary 
 axis it will assume the form of an ellipse. The height or 
 diameter of the hoop corresponding to the imaginary axis will 
 remain the same, while the diameter at right angles bo it, or 
 the revolving axis, will diminish, till at length it is a mere 
 point. Hence, to know the appearance of a circle not viewed 
 directly, we must know the angle the eye makes with its 
 plane, or its appearance in relation to some figure easy of 
 representation, contained by straight lines. 
 
 Now, a square answers admirably for this purpose, for if we 
 draw the diameters of a square, and then draw a circle so as 
 to touch its sides at the extremities of the diameters, we can 
 without much difficulty represent the circle, for we will have 
 four points as guides. If, however, the diagonals also, of the 
 square be drawn, the four points where they cut the circum- 
 ference of the circle will furnish additional points, so that we 
 will have altogether, eight points for guidance in drawing the 
 
 circle. Thus, — .•...' 
 
 5 , 
 
58 
 
 DRAWING. 
 
 J? I 
 
 Let KLMO=- given square, draw diameters and diagonals, 
 and inscrilje circle cutting the diagonals in A, C, G, E; 
 join AC and GE, and product} them to meet KL in S, S ; join 
 
 H* 
 
 
\ 
 
 / 
 
 THE CIRCLE. 
 
 59 
 
 lals, 
 
 E; 
 join 
 
 K C. v., S C. v., etc. ; also make KR = KB and RX = BO or 
 KB. Join X MP and R MP, and where they meet K C.V. 
 draw parallels to KL : then hf will represent the diameter 
 BF, and it will be easily seen that a, 6, c, c?, e,y, g, h will cor- 
 respond with A, B, C, D, E, F, G, H respectively, and the 
 curve traced between them will represent the circle (in this 
 case) lying on the ground plane. (Fig. 33.) 
 
 Example 1. — Draw a circle, diameter 4', lying on G.P., cen- 
 tre 4' to right and 2' within P.P. H = 6', L.D. = 4', scale 
 
 cy 
 
 Fijf. 34. 
 
 Here ED = 4', EA = 2', DB = 2'. Describe semicircle ASD, 
 draw AG and BM _L to AB, and draw GSM through S, 
 
60 
 
 DRAWiNr;. 
 
 parallel to AB ; make AE = AG and AF = AB; join F MP 
 and E MP, also A C.V., D C.Y,, B C.V. Join DG and DM, 
 cutting curve in H and K. Diaw HC and KL parallel to 
 AG ; join C C.V. and L C.V. ; complete square NABP, draw 
 diagonals; then on the eight points thus shown draw the curve 
 required. (Fig. 34.) 
 
 Example 2. — Draw a circle touching P.P. 4' to left, stand- 
 ing on G.P. and J_ to it and P.P. ; circle to be 4' in diameter. 
 H = 6', L.D. = 4', scale J" = 1'. 
 
 Fig. 35. 
 
 Here take A 4' to left, bisect it in G ; describe semi- 
 circle, and complete oblong APQB ; join GP, GQ, cutting 
 curve in 11, S ; draw RF, SH J to AB ; join A C.V. Erect 
 at A the perpendicular AC = AB, and bisect it in X ; join 
 C C.V., X C.V., also F MP, G MP, H MP, B MP. At points 
 of section K, L, M, E draw parallels to CA; draw diagonals 
 
 
I. 
 
 THE CIRCl.K. 
 
 61 
 
 CE, AD, and trace curve between the eight nmrked points. 
 (Fig. 35.) 
 
 ^a:am;)/e J.-Draw a circle, dian^eter 4', directly in front 
 lying on G.P., centre 4' within P.P. H = 6', L.D. = 4', scale 
 
 4 •*• • , 
 
 J 
 
62 
 
 r)RAWfNG. 
 
 Here take OA, OB, each = 2'; upon AB describe square 
 ACDB; bisect BD in E, and draw EM parallel to CD; 
 describe semicircle and find points F, G, as already shown. 
 Make BK = BD, BH = BE and KL = DE; join L MP, 
 K MP, etc.; also A C.V., G C.V., etc.; and on eight points 
 thus formed describe circle required. For BX = BH = BE 
 = 2', and XZ = HL = BD = 4, etc. 
 
 Example Jf. — Draw a circle, diameter = 4', J^ to G.P., 
 parallel with P.P. and 2' within it; centre of circle 5' to 
 right. H = 6', L.D. = 4', scale J" = 1'. 
 
 Here take A 5' to right of O ; erect the perpendicular AB. 
 2' = radius of given circle; join B C.V. and A C.V. ; take C 
 2' to left of A; join MP, cutting A C.V. in D. Draw DE 
 parallel to AB ; then with centre E and distance ED describe 
 circle required. For AD = AC = BE ; then E is 2' within 
 P.P. and 5' to right, also DE = AB = 2', etc. (Fig. 37.) 
 
THK CIRCLE. 
 
 6d 
 
 Exercise VIII. 
 
 (H = 6', L.D. = 4';scale J" = l'.) 
 
 1. Draw a circle, diameter 4', resting on ground plane and 
 touching P.P. at a point 4' to right. 
 
 2. Draw a circle, diameter 4', resting on (-J. P., centre 4' to 
 left and 4' within P.P. 
 
 3. Draw a circle, diameter 4' its plane perpendicular to 
 G.P. and touching it at a point 4' to left; the circle is per- 
 pendicular to P.P. and touches it. 
 
 4. Draw a circle, diameter 4', lying on ground plane directly 
 in front, centre 5' within P.P. 
 
 6. Draw a circle, diameter 4', parallel to G.P. and 2' above 
 it, placed with centre 4' to right and 4' within P.P. 
 
 6. Draw a circle, diameter 4', coincident with P.P. and 
 touching G.P., centre 4' to right. 
 
 7. Draw a circle, diameter 4', plane perpendicular to G.P. 
 and P.P.; the centre of the circle is 5' to left and 3' within 
 P.P. 
 
 8. Draw a circle, diameter 4', parallel to G.P. and 9' above 
 it, directly in front, centre 4' within P.P. 
 
 9. Draw a circle, diameter 4', parallel to P.P. and 6' within 
 it ; centre of circle 1' to right and 3' below G.P. 
 
 10. Draw a circle, diameter 6', lying on G.P., centre 4' to 
 right and 4' within P.P., and within it draw a concentric 
 circle of 3' diameter. 
 
 11. Draw a quadrant, radius 2', lying on G.P., vertex 
 directed away, and placed 4' to left and 4' within P.P. ; the 
 radii make an angle of 45° with P.P. 
 
 12. Draw a circle, diameter 4', buried vertically in the 
 ground to a depth of 1'; the circle is perpendicular to P.P., 
 and its centre is placed at a point 5' to left and 3' witliin 
 P.P. 
 

 \l 
 
 i 
 
 i 
 
 64 
 
 HRAWINa. 
 
 SOIvIUS. 
 
 It is expected that the pupil will have drawn all the figures 
 mentioned in the exercises. Unless the problems have been 
 thoroughly understood, comparatively little progress can be 
 made in the perspective of solids. 
 
 Solids may be classified thus : . ' 
 
 I. Those contained by plane surfaces. 
 
 II. Those partially or wholly contained by convex surfaces. 
 They are sometimes classified as solids with Developable, or 
 
 with UndevelopaV)le surfaces. 
 
 Those belonging to Class I. are Cubes, Plinths, Parallelo- 
 pipeds. Prisms, Pyramids, "Wedges, and Frusta. 
 
 Of those contained by convex surfaces in part, are Cones, 
 Cylinders, Hemispheres, and frusta of Cones. 
 
 Those contained wholly by convex surfaces are Spheres, 
 Spheroids, Ellipsoids, Cylindroids, Spindles, etc. 
 
 All the latter have undevelopable surfaces, i.e., they cannot 
 be straightened out to a plane surface. 
 
 Solids contained by plane surfaces may be subdivided into : 
 
 1. Those rectangular throughout. ^ 
 
 2. II partly rectangular. 
 
 3. II wholly oblique. 
 
 The latter class of solids cannot be readily drawn in per- 
 spective, and will not be treated of, here. 
 
 Of (1) are Cubes and Plinths or Parallelepipeds. 
 
 A cube is a solid contained by six equal squares, and all 
 its angles are right angles. 
 
 A plinth is a solid contained by three pairs of equal and 
 similar oblongs. Each pair of surfaces may be equal or 
 
SOLIDS. 
 
 66 
 
 unequal to one or both of the otlier pairs, but the angles are 
 right angles. 
 
 (2) A prism is a solid contained by two regular polygons 
 whose planes are parallel to each other, and whose like sides 
 are joined by rectangular planes. 
 
 A pyramid is a solid formed by joining the angles of a 
 triangle, square, etc., with some external point. If the exter- 
 nal point be vertically above the centre of the pyramid, the 
 pyramid is said to be riyht ; if in any other position, oblique. 
 
 A frustum of a pyramid is the part remaining after a 
 smaller pyramid is cut off by a plane parallel to the base. 
 
 Of solids with convex surfaces : — i 
 
 A sphere is formed by the revolution of a semicircle around 
 the diameter, which remains fixed. 
 
 A cone is formed by the revolution of a right-angled 
 triangle around one of the containing sides, which remains 
 fixed. 
 
 A cylinder is formed by the revolution of an oblong around 
 one side, which remains fixed. 
 
 A spheroid is formed by the revolution of a semi-ellipse 
 around one of the axes, which remains fixed. 
 
 If the fixed axis be inajor^ the spheroid is prolate ; if 
 minor, oblate. , 
 
! fr 
 
 66 
 
 DRAWING. 
 
 ill 
 
 if 
 
 n 
 
 , The Cube. 
 
 The diuwing of the cube is so simple that a single example 
 will suffice for its explanation. 
 
 be 
 
THE CUBE. 
 
 67 
 
 Example 1. — Draw a cube, edge i', placed on G.P., one side 
 parallel to P.P. and 2' within it, near angle 3' to right. 
 
 H = 6', L.D. = 4', scale i" = l'. 
 • Draw B.L., H.L., and 0, C.V. and find MP as before. 
 Take A 3' to right, and B 4' to right of A ; take C 2' to left 
 of A, and D 4' to left of C ; join A C. V., D MP and 
 C MP, cutting A C.V. in F and E. On AB describe square 
 ASMB; join S C.V., M C.V. and B O.V.; through PJ and F 
 d aw parallels to AS, meeting S C.V. in H and G ; through 
 G, H draw parallels to SM, meeting M C.V. in K and L. 
 Draw LN from L J_ to MB, and EN from E parallel to AB. 
 This will complete the required cube. For XA = 3', and 
 AE = AC = 2', AF = AD and EF = CD = AB = 4'; also FG = 
 EH = AS-4', and GK = HL = EN = AB= 4': then EF = GH 
 = KL = 4', etc. (Fig. 38.) 
 
 ii» 
 
' 
 
 68 
 
 DKAWIXfJ. 
 
 Tlrie Plintln. 
 
 : I, 
 
 111 
 
 The plinth differs from 
 a cube only in the rela- 
 tion of its dimensions; 
 the principle employed 
 in drawing them is the 
 same, but a particular 
 side of the plinth is men- 
 tioned in reference to 
 the P.P. or G.P. In the 
 cube this is quite un- 
 necessary, as all the sides 
 are equal. 
 « Example i.— Draw a 
 .^ plinth whose dimensions 
 are 4' x 3' x 2' (4' long, 
 3' wide and 2' thick), 
 the side 4' x 3' rests on 
 the ground plane, and 
 side 3' X 2' is parallel to 
 P.P. and 2' from it ; the 
 plinth is 2' to the left. 
 H = 6',L.D. = 4',scale 
 
 I" 
 
 4 
 
 1'. 
 
 Draw H.L., B.L. and 
 
 tind C.V.,Oand MP as 
 before. Take B, 2' to 
 left, and A, 3' to left 
 of B; also M, 4' to 
 right of O. On AB 
 
THE PLINTH. 
 
 (id 
 
 
 construct the oblong ABDC, a'xii'; join C 0. V., D (J V 
 AC. v., B C.V. ; also O MP and M MP. From points K h' 
 where B C.V. intersects O MP and M MP, erect KN and HG 
 paralle] to BD ; and through N, G draw NE and GF paral- 
 lel to CD. Draw EL J_ to EN and KL | to EL, which 
 will complete the plinth. For BK = BO and KH = OM • then 
 EF ^ NG = KH = OM = 4', and FG = EN ^ LK = AB = .3' GH 
 = NK=:DB = 2',etc. . ' 
 
 Bxample ^.-Draw a flight of four steps, each step 4' x 1' 
 X r. The ends of the steps are coincident with the P P and 
 4' to right. H = 6', L.D. = 4', scale y=r. 
 
 Fijf. 40. 
 
 Here take A, 4' to i-ight of K, and 15 4' tu ri^dit of A • on 
 AB describe square ABDG, and divide it into sixteen equal 
 squares ; join each angle, as shown in ligure, with the C Y ' 
 .)oin also h: MP, cutting A C.V. in F; draw FG parallel to 
 AC, GS L to GF, etc. Then FG = AK --^ 1 ', and GS = KN 
 = r, etc.; and AF=:AE = AB = 4'. (Fig. 40.) 
 
TW 
 
 11 
 
 70 
 
 DRAWING. 
 
 Example 3. — Draw same, with ends perpendicular to P.P., 
 one step being coincident with it and 4' to left. 
 
 Fig. 41. 
 
 Draw H.L., B.L., and find C.V. and MP as before. Take 
 A, 4' to left, and B, 4' to left of A ; on AB construct square 
 ABCD ; join O MP ; and on AE, complete square ADFE ; 
 then draw steps similar to preceding exauiple. 
 
 Exercise IX. 
 
 (H = 6', L.D. = 4', scale ^' = 1'.) 
 
 1. Draw a cube, edge 4', 2' within P.P. parallel to it, and 
 4' to right. 
 
 2. Draw a cube, edge 3', directly in front, at a distance of 
 3' from P.P., resting on G.P. 
 
 3. Draw a cube, edge 5', parallel to G.P. and 2' above it; 
 cube 3' to right, parallel to P.P. and touching it. 
 
 4. Place two cubes, each 4' edge, on a line parallel to L.D., 
 cubes to be 4' apart, and nearest 4' from P.P. and 4' to left. 
 
 5. Draw a cube, edge 4', touching P.P. and 4' to right; and 
 place a cu>)e, 2' edge, centrally upon it. 
 
THE PLINTH. 
 
 71 
 
 6. Draw a plinth 6' x 4' x 2', side 4' x 2' on ground, side 
 6' X 2' parallel with P.P. and 4' to left; figure to be 2' within 
 P.P. 
 
 7. Draw a slab 4' x 2' x 2', lying on ground directly in 
 front, side 4' x 2' parallel to P.P. and 4' from it. 
 
 8. Draw a slab 5' x 5' x 1' lying flat on G.P., side 5' x 1' 
 parallel to P.P. and 3' from it; slab to be 4' to left. PLace 
 centrally on this slab a cube whose edge is 3'. 
 
 9. Draw a cube, 2' edge, on each side of L.D., 2' from it, 
 and touching P.P.; on these cubes place a slab 6' x 2' x 1' 
 coincident with the cubes. 
 
 10. A wall 8 high and 2' thick starts from a point on the 
 P.P. 4' to the h^ft, and runs straight forward to the horizon ; 
 at distances of 6' and 12' doors 5' x 3' are placed. 
 
 11. Draw a cross whose beams are 7' x T x 1' and 5' x 1' x 1' 
 respectively ; the cross-beam is placed at a height of 3'. The 
 cross stands erect, its cross-beam parallel to P.P. and 4' within 
 it ; the foot of the cross is 4' to right. 
 
 12. Draw same, with end of cross-beam coincident with 
 P.P., 4' to left. 
 
 13. Draw same, lying on G.P., cross-beam J_ to P.P. and 
 its end coincident with it, 3' to right. 
 
 14. Draw same, lying on ground directly in front, cross- 
 beam directed away, end of main beam coincident with P.P. 
 
 15. A circular table 4' in circumference is supported by 
 four legs 2' high, which proceed from the edge of the table ; 
 the legs form a square whose side is parallel to P.P. and 3' 
 within it. The centre of the table is 4' to right. Thickness 
 of neither table nor legs, taken into account. 
 
 16. Draw a set of four steps, each 4' x 1' x 1', ends parallel 
 to P.P. and 2' within it ; to be 5' to left, facing toward right. 
 
 17. Draw same, ends perpendicular to P.P. and 2' within 
 it, and 3' to right. 
 
 1 8. Draw same, with steps descending as they recede ; back 
 coincident with P.P. and 2' to left. 
 
 19. Draw same, directly in front, steps ascending as they 
 recede, and 2' within P.P. 
 
 ! ! 
 
72 
 
 DRAWINf}. 
 
 Ttie Prism. 
 
 Prisms are square, triangular, hexagonal, etc., according to 
 their ends or bases. 
 
 The square prism may be considered as a mere modification 
 of the plinth. 
 
 To draw a prism, we have only to draw the two surfaces 
 forming its ends, and join similar angles. 
 
 Example 1. — Draw a triangular (equilateral) prism, length 
 6', side of base 2', lying on G.P., one end perpendicular to 
 P.P. and 3' to left; prism to touch P.P. H = 6', L.D. = 4', 
 scale Y =V. 
 
 MP CV 
 
 Take B, 1' to left. A, 3' to left, and D, 6' to left of A. On 
 AB describe equilateral triangle ABC; draw CL J_ to AB 
 
THE PRISM. 
 
 M 
 
 and bisecting it. Draw AE ± to AB and equal to CL; simi- 
 larly draw DF; join F C.V., E C.V. and A C.V, also B MP 
 and L MP ; through M draw MG parallel to AE, meeting 
 E C.V. in G; draw GH parallel to EF; join HD, AG and 
 GK, completing the prism. Then AK = AB = AC = BC = 2' 
 Hence AG == KG = AK, and GM = EA = LC = required height, 
 and HG = FE = DA = 6', etc. (Fig. 42.) 
 
 Example ^.— Draw a hexagonal prism (edge of base 2') 
 whose length shall be 4', one side touching P.P. 4' to right; 
 prism to stand on end. H = 6'; L.D. = 4'; .scale, 1" = 1. 
 
74 
 
 DliAWING. 
 
 Draw tlie plan in proper situation, as already explained; 
 then draw the hexagons, one on G.P., the other 4' above it; 
 then join similar angles in each, forming the required hexagon. 
 (Fig. "43.) 
 
 I 
 
THE CYLINDER. 
 
 n 
 
 The Cylinder. 
 
 The drawing of the cylinder differs from that of the prism, 
 only in the plan. Draw the circles, forming the ends, in the 
 proper positions, and then draw tangents to them, forming 
 the cylinder. 
 
 Example 7.— Draw a cylinder, length 4' and diameter 4', 
 lying on ground plane parallel to P.P. and 2' within it; the 
 end qf cylinder 4' to left. 
 
 -Fig. 44. 
 
 1 [ 
 
 H 
 
76 
 
 DRAWING. 
 
 Example 2. — Draw a cylinder lying on G.P., 4' to ricflit, 
 having end perpendicular to P.P., and touching it; cylinder 
 8' long and 3' in diameter. 
 
 Take A, 4' to right and E, 4' to left; make AB = 1|', and 
 describe circle; join B C.V. and A C. V., also E MP, and from 
 C, draw CD parallel to AB, and describe smaller circle; then 
 draw common tangents, completing the cylinder (Fig. 45.) 
 
i.cjht, 
 nder 
 
 THE CYLINDEK. ff 
 
 Example -^.— The figure shows how to draw a common pail, 
 showing staves and hoops. 
 
 II 
 
 Fig. 
 
 and 
 from 
 then 
 
 45.) 
 
 I i 
 
w 
 
 78 
 
 DRAWING. 
 
 y 
 
 m 
 
 Ttie Pyramid. 
 
 « 
 
 We now come to consider solids, which are not wholly rec- 
 tangular; they are cones and pyramids and their frusta. 
 
 In speaking of the height of a pyramid or cone, we mean 
 the distance from the vertex perpendicularly to the base. 
 Tiiis is important, especially in frusta, where the slant height 
 migiit be mistaken for the real height of the solid. 
 
 Ml ample 1. — Draw a pyramid, 8' high with square base, 
 each side of which is 4', and touches P.P. 4' to right. H = 6', 
 L.D. = 4 , scale \" = '. 
 
 K L W 
 
 Take A, 4' to right and B, 4' to right of A; complete the 
 square ABDC; draw diagonals intersecting in O'; join O with 
 
THE PYHAMID. 
 
 C.V. and profluce it backward to iiK^ot hase 1 
 erect 
 
 7» 
 
 ine in E; at K 
 
 perpfMidicular, 8' in lioi^dit to M; Join M C.V.; tlnou^d. 
 O, draw OX, parallel to KM; join XA, XR, XC, completing 
 the pyramid. Now, OX = EM ^-8', and this represents the 
 vertical liei«,djt. (Pig. ij.) 
 
 It IS not absolutely necessary to join O with C.V. We 
 may draw it to any point on the H.L., as N or S., and pro- 
 duce it backward to F or A, and erect a perpendicular from 
 either of these points; but it must be carefully remembered, 
 that the so-found point K or L, must be joined to S or N 
 respectively. Such lines, KS, LN, M C.V,' etc., will all pass 
 through same point X, which may be considered as a locus 
 for all such lines. For convenience, however, the line O C.V. 
 Hhould be used, unless tlie solid be directly in front. 
 
 , ^4 
 
■NH 
 
 SO 
 
 DRAWING. 
 
 i 
 
 Tl::Le Cone. 
 
 The di awing of the cone does not differ materially from 
 that of the pyramid. The circle forming the base being 
 drawn, and the position of the vertex found, it is only neces- 
 
 »-•. 
 
THE CONE. 
 
 81 
 
 sary to draw tlie tangents from it to tlie circle. We give a 
 particular example : — 
 
 Draw a cone whose base — 6' in diameter and slant height 6'. 
 The cone lies on its side; plane of base, I to P.P., and the 
 line joining the centre of base with the vertex is parallel to 
 the P.P. and 4' from it. The cone is 4' to the right. H = 6', 
 L.D. = 4', scale i" = l'. 
 
 Take A, 4' to right and C, 6' to right of A; on AC describe 
 equilateral triangle ADC; drav/ DB perpendicular to AC; 
 join A C.V., B C.V., C C.V.; take V and E, 3' from O; join 
 E MP, O MP and V^ MP; through F, G, F draw parallels to 
 AB; through S, K, K draw parallels to DB; join PF, NG, 
 MH; then in square PH describe circle; produce GR to L, 
 and from L draw tangents LX, LX to circle, completing the 
 cone. (J'ig. 48.) 
 
 'l! 
 
 Exercise X. — On the Prism <md the Cylinder 
 
 (H = 6', L.D. = 4', scale i" = r.) 
 
 1. Draw a prism G' in length, triangukir base, ea-^h side of 
 which is 2'. The prism stands on end 4' to right, with one 
 side coincident with P.P. 
 
 2. Draw same, 3' to left, one side perpendicular to P.P. 
 
 3. Draw same, directly in front, one side parallel to P.P. 
 and 2' from it, vertex away. 
 
 4. Draw same, lying on ground plane, perpendicular to P.P., 
 3' to right and 3' within P.P. 
 
 5. Draw a hexagonal prism 6' high, each side of base 2', 
 standing on end directly in front, one i.ide touching P.P. 
 
 6. Draw same, lying on ground parallel lo P.P. and 2' within 
 it; one end projecting 2' to right, and opposite ui 4' to left. 
 
 7. Draw same, lying on groimd perpendicular to P.P., 3' to 
 right and 3' within P.P. 
 
 8. Draw a cylintler, diameter of base 4', height 6', lying on 
 ground, parallel to P.P. and 4' within it; left end just in line 
 with L.D. 
 
 •I I 
 
 .1 
 
 !l 
 
f 
 
 82 
 
 J)UA\V1N(;. 
 
 Kp 
 
 9. Draw a cylindrical v«!8sel 4' feet in heiglit and 4' in 
 diani(^t(!i', standin<^ on end, t()uchin<^ P.P. 4' to left; show 4 
 hoops at distances of 1' from each otlua-. 
 
 10. Draw ;i hollow cylinder, 4' in length, outer diameter 4', 
 inn(!r diameter ."V, lying on ground jK^rpendicular to P.P., 2' to 
 right and touching P.P. 
 
 KxKiUMHK XL- On the Pi/r<nvid <i7i(l Cone. 
 
 (11 = 6', L.D. = 4', scale i"-l'.) 
 
 1. Dniw a s(|uare pyramid 5' high, each side of base 3', 
 standing on ground, directly in front, touching P.P. 
 
 2. Draw sanu^, 4' to light, parallel to P.P. and 4' within it. 
 
 3. Draw same, standing on a 'V cube, parallel to P.P. and 
 2' within it, 4' to left. 
 
 4. Draw same, 3' above ground plane and parallel with it, 
 4' to right and touching P.P. 
 
 5. Draw same, with v(;rt(!X flownwards, base parallel with 
 ground and P.P., vertex, 4' to left and 3' within P.P. 
 
 6. Draw a cone, height 5', diameter of base 4', standing on 
 ground 4' to right and 4' within P.P. 
 
 7. Draw same, touching P.P., 3' to left. 
 
 8. Draw same, directly in front, 3' from P.P. 
 
 9. Draw same, sta)iding on a cylinder 4' in diametei- and 4^ 
 high, 3' to right and touching P.P. 
 
 10. Draw sanu^, placed centridly on a cylinder of 5' in diam- 
 eter, directly in front and touching P.P. 
 
 11. Draw a cone, base 4', slant height 4', lying on side; 
 \tHMo perpendicular to P.P. and touching it, 4' to left; vertex 
 directed toward left. 
 
 
FRL>WA. 
 
 HI] 
 
 The dimensions of a frustum may be given by stating dinuMi 
 sions of eacli end, and vertical lieigiit. 
 
84 
 
 DRAWING. 
 
 
 I 
 
 ExmapLf 1. — Draw the fru':tuin of a scjuaro pyramid, whose 
 })aHes are 6' and 4' square, respectively ; the frustum touches 
 P.P. 4' to right, height 4'. (Fig. 49.) 
 
 From the ahove tlie method of drawing may be easily 
 understood. 
 
 Example 2. — A pyramid with square base, each side of 
 which is 4', stands on th(^ ground plane 4' to the left, touch- 
 ing P.P. The pyramid is 8' in height ; 3' from the vertex 
 the pyramid passes through a square plinth 4' x 4' x ] ' placed 
 parallel to the ground plane. The pyramid cuts the plinth 
 centrally. H = 6', L.D. = 4', scale \' = 1 '. (Fig. 50.) 
 
 B,V. 60. 
 
 By a careful observation of the linen drawn above, the 
 method may be easily seen. 
 
FHUSTA. 
 
 8.-) 
 
 EXERCISK XII. 
 
 (H. - 6', L.D. = 4', scale ^ = !'•) 
 
 1. A pyramidal frustum with square l)ase, and vortical 
 height 4', touches P.P. 4' to right, resting on ground i)]ane ; 
 the sides of the square are .'V and .5' respectively. 
 
 2. Draw a pyramidal frustum same as No. 1, :V to left, and 
 3' within P.P. 
 
 3. Draw a triangular frustum (eijuilatcial), sides of base 
 5' and 3' respectively, height 4'; on (I.P. 4' to right and 4' 
 within P.P., vertex away, one edge parallel to P.P. 
 
 4. Draw a square pyramidal frustum, height 4', sides of 
 square 2' and 4' respectively, standing on CI. P. reversed, 
 directly in front, 1' within P.P. 
 
 5. Draw a conical frustum, height 5', diameters 5' and 3' 
 respectively, on G.P., and touching P.P. 4' to right. 
 
 6. Draw No. 5, 6' to left and 3' within P.P. 
 
 7. Draw same, directly in front, 3' within P.P. and 3' 
 above G.P. 
 
 8. Draw a hexagonal frustum, lieight 5', sides of bases 3' 
 and 2' respectively, touching P.P., resting on G.P. 4' to right. 
 
 9. Draw same, 4' to left and 4' within P.P. 
 10. Draw an octagonal frustum, height 5', edges of bases 
 2' and 1' respectively, reisting on G.P., and touching P.P. 
 4' to left. ^ 
 
w 
 
 I Ml 
 
 S(J 
 
 DHAWFNC;. 
 
 \ 
 
 Fi^. 51 shf)ws a riK^thod of layiri^r out a plan for a friistu 
 of a cone. 
 
 m 
 
 I 
 
 i 
 
 
ituni 
 
 KllirsTA. 
 
 EXKKCIHK XTir. 
 
 87 
 
 (H = 6', L.D.-- r, scale l"-!'.) 
 
 1. 4' to ri-hf jui(i -I' witl.iu P.P. (Iniw a fru.stuiii r.f a 
 square pyramid, r(ljr(.s of siiuarcs 4' and 2' rospectively, lici^dit 
 4 . 
 
 L'. Draw saiiH!, touching p.p. o' to left. 
 
 :5. Draw sani(;, dir(3ctly in front, parallel to (J. P. .md IV 
 above it. 
 
 4. Draw a conical frustum, liei<,dit T)', diameters IV and 2' 
 respectively ; frustum rests on .irn.nnd, with centre of base 
 :V to ii<,dit and IV within P.P. 
 
 5. Draw a frustum of a trian<rular pyi*amid, edges of ends 
 -V and L>' respectively, height 4'; it rests on (;.P. with one 
 edge coincident with P.P., and IV to left. 
 
 6. Draw a fiustum of a Ix^xagonal prism, edges of bases 
 :V and 2' respectively, height \' ; one edge of frustum is per- 
 pendicular to P.P., and an angle touches it at a point IV to 
 the right. ' 
 
 7. A cone, whose height is 8' and dian»eter of base 4' 
 touches the P.P. 4' to left; it is encircled by a rectan^rular 
 collar whose dimensions are 4' x 4' x T, placed centrally^over 
 It, 4' above the ground. The cone rests on the ground.' 
 
 A frustum of a squan* })yi'amid, whose bases ajt; 5' and 
 3' respectively, and whose height is 4', suppo.ts a cone placed 
 centrally upon it ; the diameter of the cone is iV and its height 
 3'. The edge of the })ase touches th(! P.P. 2' to right. 
 
 1 
 
88 
 
 DRAWINC;. 
 
 ■ I 
 
 Thie Sphere. 
 
 Tho perspective of tlie sphere must necessarily be repre- 
 sented by a true circle, and little difficulty will be experienced 
 in drawing a complete sphere. However, when a hemisphere 
 
TiiK sniKRi:. 
 
 m 
 
 is to be represented, an apparent fallacy appcjars, owing to 
 the representation of the circle that shows the section of the 
 sphere. A sphere must always he supposed to be drawn with 
 the radius of the circle as distance. Howevei-, as the per- 
 spective of the circle, viewed in any ol)li(jue position, sjjows 
 diameters of varying length, care must be used in drawing the 
 curve of the hemisphere at the f/reatest apparent diameter, 
 and this diameter cannot i^e dcjinitely determined in perspec- 
 tive, if drawn in any but a direct view. The sphere rests on the 
 ground at a point directly bent^ath the centre, and it touches 
 P.P. at a point perj)endicular to the vertical, from the centre*. 
 Example /.— ])riiw a s/uere, radius 2', rwiting on ground 
 at a point :V to right and 2' within P.P. H = 6', L.D .4' 
 scale J" =1'. 
 
 Here FA = 3', BA = AD = 4' and AC - KD -- 2'. (Fig. 52. ) 
 Example 2.~0n centre of the top of a culx' of 4' edge, placed 
 4' to left and touching P.P., place a sphere of radius U'. 
 
90 
 
 DRAWING. 
 
 Iloro (JA = AiJ-AD = AE = 4', KF=FIVI=2', r,F=li' 
 lienco K is centre at intersection of diagonals and HK— GF. 
 
 (Fig. 5;^.) 
 
 NoTK. — Th(( sphere will not touch th(^ V.V. unless HK = 
 KF. 
 
 Example 'J. — Draw a cube, edge 4', touching P.P. 4' to 
 right, and in this place a sphere whose radius = 2'. Here the 
 sphere will touch the centre of each side. 
 
 Draw cube, and diagonals of those sides, perpendicular to 
 P.P., join intersections, and bisect this horizontal line as 
 shown. The circle drawn on this line as diatneter will repre- 
 sent the sphere, and touch the centre of each side. (Fig. 54.) 
 
 EXEItClSK XIV. 
 
 (H = 6', L.D. = 4', scale I" = 1.) 
 
 1. Draw a sphere, radius 2', placed centrally on a cylinder 
 (on end), touching P.P. 4' to left, cylinder 3' high and diame- 
 ter 4', 
 
THE SPIIKKI. 
 
 91 
 
 2. Draw a sphere, diameter .T, 8' hi<rli, 6' to ri.rht and fi' 
 within P.P. 
 
 3. Draw a sphere, diameter 4', directly in front, tourhinf^ 
 
 1. Draw a sphere, radius 2', huried coniphjtely heneath the 
 ground, centre of spliere 0' to i-ight and fi' within P.P. 
 
 f). Place a sphere in a cubical box of 4' edge, diameter of 
 sphen. 4'; cube to be 4' to left, 4' within P.P., parallel to, and 
 2' above (».P. 
 
 G. A cylinder whose height is 4' and dianu'ter IV stands on 
 end, touching P.P. 3' to right; this cylinder passes centrally 
 through a sphere whose diameter is 4' and whose centre coin- 
 cides with that of the cylindei'. 
 
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 w 
 
92 
 
 DRAWING. 
 
 Koresliorteniog. 
 
 i !l 
 
 Foreshortening consists in re- 
 presenting the apparent length of 
 a visible object. It depends on 
 the distance of the object, and its 
 position with regard to the eye. 
 Thus, a lead-pencil may he so 
 turned as to show only the end, 
 or it may be placed so as to show 
 its whole, or greatest length. 
 Again, if AB represent a line of 
 definite length, and O, the ob- 
 server, the apparent length of 
 AB as seen from O will be AD. 
 (Fig. 5.5.) 
 
 This representation of a line 
 AB by AD, which is always le88 
 than AB, is called "foreshorten- 
 ing." 
 
 Note. — AD is always perpen- 
 dicular to the longest side OB. 
 
SYNTHETIC PERSPECTIVE. 
 
 98 
 
 Taking an object " out of " perspective means, that when an 
 object is drawn, and the position of the observer's eye given, 
 the size and position of the object may be determined. 
 
 Fig. 56. 
 
 Here, the figure BX only, would be given, and it would be 
 assumed to touch the picture plane. We first produce the 
 vanishing lines DX and AC to meet in C.V., then draw 
 C.V. S perpendicular to B.L., and make it equal to height of 
 spectator; then draw H.L. parallel to B.L. through C.V. ; 
 next take a point MP at a distance to the left equal to the 
 height of spectator and his distance away, combined ; then 
 join MP with C, and produce it to base at E. Then, scale 
 being given, find BA, AD and AC, the dimetunona of the solid, 
 and AS will show its distance to the left. (Fig. 56.) 
 
 
94 
 
 DRAWING. 
 
 Perspective Effect. 
 
 This consists in showing merely the appearance of an object 
 when placed in a certain position. The dimensions and dis- 
 tance of the object are not ta.ken into account. 
 
PERSPECTIVE EFFECT. 
 
 bject 
 [ dis- 
 
 7^ 
 
 It will be remembered that if an object of less height than 
 the observer, he placed on the ground, the observer will be 
 able to see the upper side of it, and if placed above him he 
 will see the under side ; if placed on his right, he will see 
 the front and left sides ; if placed directly in front, he will 
 see front side and upper or lower sides, according to the 
 height of the object. 
 
 Take a cube, for instance, placed on 
 the ground parallel with the P.P. 
 
 Now, if the cube be lower than the 
 observer's eye, the upper face will be 
 visible ; if raised above, the lower face, 
 and so on. 
 
 Figure 57 will illustrate perspective 
 effect. 
 
 (1) Shows object above and left of eye. 
 •(2) Above and directly in front of eye. 
 
 (3) Above and to right of eye. 
 
 (4) Level with and to left of eye. 
 
 (5) Level with and directly in front. 
 
 (6) Level with and to right. 
 
 (7) Below and to left. 
 
 (8) Below and directly in front. 
 
 (9) Below and to right. 
 If an object, as for instance a cube, is 
 
 to be drawn, say to right and above the 
 ^ ^. eye, draw first a square, then take a 
 ^ 2 point to left and below, draw the van- 
 ishing lines to this point, and mark off' 
 C ^ lines for thickness, etc. 
 
96 
 
 DRAWING 
 
 ANGULAR PERSPECTIVE. 
 
 I 
 
 We now come to consider the rules pertaining to angular 
 perspective, or the perspective of two vanishing points. If a 
 rectangular object, as a cube, rests on the ground parallel to 
 P.P., it is evident that its sides, if produced, will appear to 
 vanish directly in front, at the poii t called the centre of 
 vision. If we move the cube by even a small amount 
 fS*om tlie parallel position, its sides will no longer vanish at 
 the centre of vision, but at a point to the right or left of it, 
 and at a distance from it, depending on the angle which the 
 sides make with the P.P. Now, in parallel perspective we 
 deal with only one vanishing point — the centre of vision ; but 
 there are really two: for all lines parallel to the P.P., if pro- 
 duced to an infinitely great distance, will appear to meet at a 
 point to right or left. Hence in parallel perspective only one 
 vanishing point is of practical utility. However, when the 
 cube is moved out of its parallel position, this apparently- 
 hidden vanishing point appears, and strikes the horizon at a 
 distance from the centre of vision, depending on its angle, as 
 already explained. Thus : — 
 
 In Fig. 59, AB on left side shows the base of a cube in 
 parallel perspective, while in Fig. 60 AB has been moved 
 arcund to position of AD, and AC will not now vanish to 
 C.V., but to a point Y to right of it; so also AD will not 
 vanish at a point parallel to AB, but at X, a point in the 
 
ANGULAR PEUSPECTIVE. 
 
 97 
 
 Fig. 60. 
 
 h6rizontal line to left of C.V. We will now proceed to 
 ascertain the positions of these points. 
 
 Eocample 1. — Draw a square (side 4') lying on ground; sides 
 make an angle of 45° with picture plane, and the angle 
 touches the P.P. at a point 4' to the right; scale Y = 1'. 
 
 h 
 
ANGULAR PERSPECTIVE. 
 
 99 
 
 CO « 
 
 Here draw H.L. and B.L. as before, and take P.S. at given 
 distance; then draw a straight line through P.S. parallel to 
 B.L. and on each side of L.l>. ; lay off the sides at required 
 angle (in this case 45 ); produce these lines till they meet 
 the horizon in R V.P. and L V.P. (the vanishing points) ; with 
 L V-.P. Jis a centre, and P.S. as distance, describe an arc to 
 cut H.L. in R MP.; similarly find L MP. These are called 
 measuring points. Now take A, 4' to right and draw A L V.P. 
 and A R V.P.; take 4' on each side of A, namely, B and C, 
 and draw B R MP and C L MP to cut vanishing lines in E 
 and D. Draw D L V.P. and E R V.P. to cut in F. Then 
 ADFE will be the square required. For AD and EF are 
 parallel to P.S. R V.P. placed at given angle, and AE and 
 DF are parallel to P.S. L V.P. also placed at given angle; 
 and DF = AE = AB = 4', and EF = AD = AC = 4'. (Fig. 61.) 
 
 JExample 2. — Draw a cube, edges 4', right face at an angle 
 of 60° and left face at an angle of 30° with P.P. H = 6', 
 L.D. = 4', scale Y =1'. Cube to have an angle 4' to left and 
 2' within P.P. 
 
 Draw H.L., B.L. and 00 as before; find also C.V., R V.P., 
 L V.P., R MP, L MP and MP (parallel perspective), as already 
 shown. Take A, 4' to left ; join A C.V.; erect AK = 4'; join 
 K C.V.; join also B M.P,; through E draw EL parallel to 
 AK; join E RMP and E LMP, and produce them backwards 
 to meet B.L. in Z and X. Mark ofi* XC = 4', also ZD = 4'; 
 join D RMP and C LMP, also L R V.P. and LLV.P.; 
 through F draw FM, and through G draw GH, parallel to 
 EL; join M L V.P. and H R V.P. to meet at N, completing 
 the required cube. For FM = EL = HG = AK = 4', and EG = 
 EF = BC = AD = 4' ; so also MN = HL = GE, etc. (Fig. 62.) 
 
 Note. —The point E njust always be determined by parallel 
 perspective; hence necessity for finding MP. 
 
100 
 
 DRAWING. 
 
 I / 
 
 ^■J 
 
' 
 
 MISCELLANEOUS EXEUCiSES. 
 
 101 
 
 »i 
 
 Exercise XV. — Figures in Angufnr I'ertiperiive. 
 
 (In the following consider H = 6', L.D. = 4', scale J" = 1.) 
 
 1. A square whose sides are 4', lies on ground; an angle 
 touches P.P. 2' to right; angle 45°. 
 
 2. Draw same, 3' to left and 2' within P.P.; right side makes 
 angle of 60". 
 
 3. Draw same, touching P.P. directly in front; angle 45'^. 
 
 4. Draw a cube, of 4' edge, touching P.P. at a point 2' to 
 right; angle 45°. 
 
 5. Draw same, 4' to left and 1' within P.P.; angle 45°. 
 
 6. Draw a square pyramid, edge of base 4', height 8', angle 
 45°; touches P.P. 4' to left. 
 
 7. Draw same, 4' to right and 2' within P.P.; angle 45°. 
 
 8. Draw same, 2' to left and 2' within P.P.; left and right 
 angles, 30° and 60° respectively. 
 
 9. Draw a triangular prism, each edge of base 3' and 5' 
 long; standing on end, angle touching P.P. 4' to left; angles 
 60° and 60°. 
 
 10. Draw a square pyramid, edge of base 3', placed cen- 
 trally on a cube of 4' edge; angle 45°; touches P.P. 3' to left. 
 Height of pyramid 4'. 
 
 11. Show perspective effect (angular) of a pyramid to left 
 and above the eye. 
 
 12. Show angular perspective effect of a square pyramid 
 placed centrally over a cube of snmller base, to right and lielow 
 eye. 
 
 Miscellaneous Exercises. 
 
 (Unless otherwise stated, consider H = 6', L.D. = 4', and 
 scale f = l'.) 
 
 1. Two circles, whose diameters are 4', and intersect at 
 right angles, having their common diameter perpendicular to 
 ground and touching it at a point 4' to left and 4' within P.P. 
 
 2. Draw an equilateral triangle, lying on ground plane, side 
 3', vertex directed away, one side parallel to P.P. ; vertex 4' 
 to right and 4' within P.P. 
 
 3. Draw a hexagon, each side 2', standing on ground, _[_ to 
 P.P., one side touching it 4' to left. 
 
102 
 
 DRAWINO. 
 
 4. Draw a circle, diainettjr 4', toucliin;^ G.P. .'iiul P.P., and 
 perpendicular to both, 4' to right. 
 
 5. A rod is placed obliquely in the ground, and its outside 
 length is 5' ; it makes an angle of 30'' with the ground and 
 60° with the P.P. The rod descends toward the left, and 
 lower point is 6' within P.P. and 6' to right. Draw it. 
 
 6. Draw an octagon, side 2', lying on ground, touching 
 P.P. 4' to left. 
 
 7. Within a circle, diameter 4', lying on ground plane, 4' to 
 right and 4' within, describe a squai'e wliose side slmll be 
 parallel to PP. 
 
 8. Draw a triangular prism, length G', edges 2', parallel to 
 P.P. and' 3' within it, one end 4' to right, other 2' to left. 
 
 9. Draw a cone, diameter 4', height 4', standing on ground 
 plane, touching P.P. directly in front. 
 
 10. Draw a pyramidal frustum (square), edges 2' and 4', 
 height 4', touching P.P. 4' to left. 
 
 11. Draw same, in angular perspective, angle 45°, 4' to left, 
 touching P.P. 
 
 12. Draw a sphere, diameter 4', half buried in ground, 6' 
 to right and 6' within P.P. 
 
 13. Draw a hemisphere, plane directed towards right and 
 perpendicular to P.P. and G.P., touching each ; hemisphere to 
 be 4' diameter and 4' to left. 
 
 14. Draw a cylinder on end, diameter 4', height 4', touch- 
 ing P.P. 4' to right, and on this place a hemisphere centrally, 
 4' diameter, convex surface upward. 
 
 15. Draw a sphere touching sides of a cubical box of 4' edge, 
 box on ground parallel to P.P., 4' to left and 4' within P.P. 
 
 16. Draw a pyramid, base 4' square, 4' high, 4' to right 
 and 3' within P.P. 
 
 17. Draw an equilateral triangle, sides 3', in angular per- 
 spective; angle 60°; 4' to left, 3' within P.P., on G.P. 
 
 18. A square, sides 4', stands on ground plane perpendicular 
 to it, making an angle of 45° with P.P. and 2' from it at 
 nearest lower point ; it is 3' to left. 
 
 19. Draw a triangular pyramid, height 8', each side of base 
 4', presenting an angle of 60° to the P.P. 4' to left. 
 
 20. A square prism, length 4', edge 2', stands on end, an 
 angle touches P.P. directly in front; sides at 45°. This prism 
 
MISCELLANEOUS EXKUriSKS. 
 
 lo;^ 
 
 per- 
 
 Icular 
 it at 
 
 base 
 
 ail 
 )risni 
 
 supports a pyramid placed evenly upon it, of e<|ual base and 
 4' in height. 
 
 21. Draw a plintii 6' x 4' x 2', side C x 4' on ground, placed 
 at angles of 60" and HO", 4' to left, touching P.P. 
 
 22. Draw an ordinary Roman cross, beams 0' and 4' in 
 length, and 1' square at ends, at an angle of 45", 4' to right 
 and 4' within P.P. 
 
 23. Place a cube of 4' edge on top of a cylinder (on end), of 
 4' diameter, centres coincident; cylinder touches P.P. directly 
 in front. 
 
 24. Draw middle zone of a sphere whose radius is 4', height 
 of zone 2', plane parallel to ground plane, centre of zone 4' to 
 right, 4' within P.P. and 4' above it. H = 6', L.D. = 6', scale 
 
 25. Draw four pyramids, each in contact at bases, 4' square 
 and 4' in height, standing on ground plane at an angle of 45° 
 with P.P., 4' to left and 2' within P.P. H = 6', L.D. = 6', 
 scale i" = l'. 
 
 26. Draw a frustum of a cone, height 5', diameters 3' and 
 2' respectively, touching P.P. 4' to right. 
 
 27. A cone whose slant height is 6' and diameter of base 
 6', rests on ground plane, slant touching ground, parallel to 
 P.P. and 6' within it, vertex directed towards left ; cone to 
 be 4' to left. 
 
 28. A cube of 4' edge contains a cylinder of equal diameter 
 and height; the cube makes an angle of 45" with P.P., and 
 touches P.P. 4' to right. The cylinder is vertical. 
 
 29. A pyramid, whose base is 4' square and whose height is 
 6', presents an angle to the P.P. 4' to left, the inclination of 
 the sides being 45°. This pyramid passes centrally through a 
 plinth 4' X 4' X 1', placed horizontally upon it at a height of 3'. 
 
 30. A cube, whose edges are 4', is suspended from an angle 
 so as to just touch the ground directly in front, while another 
 angle touches the P.P. directly in front. 
 
 31. Draw .* stove-pipe elbow, diameter of ends 6", length of 
 each half (outside) 1'. The elbow rests on the ground, one 
 end touching P.P. 2' to left, tiie other bending towards the 
 right and parallel to P.P. H = 6', L.D. - 4', scale 1 " = 1 '. 
 
 32. Show perspective effect of a pipe lying on ground paral- 
 lel to P.P. and to left. 
 

 104 
 
 DRAWING. 
 
 \£ 
 
 
 33. Show same, standing on end to right and below eye. 
 
 34. Show angular perspective effect of a square pyramid to 
 right aid ^^elow eye. 
 
 35. 'how perspective effect of a cylinder on end, to right 
 and beloA^. 
 
 36. Show perspective effect of a pail with threfihoops, below 
 and directly in front. 
 
 37. Show perspective effect of a water pitcher, to left and 
 below; lip to right. 
 
 38. Show perspective effect of a chair, straight back, directed 
 away, angle to right and below the eye. 
 
 39. Show hollow pipe lying on ground, perpendicular to 
 P.P., to right. 
 
 40. Show perspective effect of an ink bottle (conical). 
 
 41. Show perspective effect of an ink bottle in form of 
 pyramidal frustum, angle to left, below the eye. 
 
 42. Show perspective effect of a plinth to left and below, 
 angular. 
 
 43. Show perspective effect of a teacup below the eye. 
 
 44. Show perspective effect of a sphere placed centrally on 
 a cube, directly in front, below the eye. 
 
 45. Show angular perspective effect of a table below and to 
 right. 
 
 46. Show perspective effect of a triangular prism on end, 
 one side perpendicular to P.P., below and to right. 
 
 47. Show perspective effect of a reversed cone, below the 
 eye. 
 
 48. Show perspective effect of a hexagon on one side, per- 
 pendicular to ground, to right. 
 
 49. Show perspective effect of a hollow conic frustum lying 
 on ground parallel with P.P., larger end to right, smaller end 
 to left. 
 
 50. Show perspective effect of a door in three different 
 positions, revolving round an axis through the hinges : 
 
 (a) When shut, parallel to P.P. 
 
 (6) When opened at an angle of 45° with P.P. 
 
 (c) When opened perpendicularly to P.P. 
 
 I'ir 
 
GEOMETRICAL DRAWING. 
 
 106 
 
 GEOMETEICAL DEAWING. 
 
 To constiftct the following figures, pupils should provide 
 themselves with a pair of good compasses with pen attach- 
 ment, and a ruler, with marks for inches and fractions of an 
 inch. No proof is necessary, but it will be well to investi- 
 gate the methods as far as possible, many of which are but 
 modifications of the Euclidian. 
 
 ' No. 1. — To draw a perpendicular to a given line (a) from a 
 point on the line. 
 
 Fig. 64. 
 
 Fig. 63. 
 
 Let AB be the given line, and let B be a point at which 
 the perpendicular to AB is to be drawn. In first, take any 
 point C above, and with distance CB describe circle, cutting 
 AB in E; join EC and produce to meet circumference in D; 
 join DB, which will be the perpendicular j*equired. In second, 
 take any point C above, and with B as centre and BC as dis- 
 tance describe arc, cutting AB in A; then with C as centre 
 and CB as distance describe arc, cutting former arc in E; 
 with E as centre anjd same distance describe arc, cutting in D ; 
 8 
 
 r'il 
 
106 
 
 DRAWING. 
 
 join DB, which will be perpendicular to AB, at B. (Figs. 63 
 and 64.) 
 
 (6) From a point above or below AB. 
 
 »' 
 
 fi4 
 
 k<i<n,/:.". 
 
 .2^ 
 
 y 
 
 B 
 
 Fig, 65. 
 
 Let AB be the given line and C given point above; with 
 centre C describe an arc to cut AB in D and E; with centre 
 D and distance greater than half of DE describe an arc ; with 
 centre E and same distance describe an arc to cut former arc in 
 F; join GF, which will cut AB at right angles at G. (Fig. ^5.) 
 * No. 2. — To describe a square (a) on a given line. 
 
 Fig. 66. 
 
 In first, let AB be the given line; erect at A a perpendicular 
 and make it equal to AB; with centre C and distance CA 
 describe arc AD; with centre B and distance BA describe 
 
T 
 
 GEOMETRICAL DRAWING. 
 
 107 
 
 le 
 
 arc to meet former arc in D; join CD and BD, completing 
 the square. (Fig. 66.) 
 
 (6) On a given diagonal AB. 
 
 With centre A and distance greater than half of AB,' de- 
 scribe arc CD, and with centre B and same distance describe 
 an arc to cut former arc in C and D; join CD and produce 
 both ways; with centre E and distance EA or EB describe a 
 circle cutting diameter in F and G; join FA, FB, GA and 
 GB, completing the square. (Fig. 67.) 
 
 No. 3. — To construct an oblong of given dimensions. 
 
 Fig. 68. • »; 
 
 Let AB represent the greater side ; erect AC perpendicular 
 to AB at A, and with centre C and distance equal to AB 
 describe an arc ; with centre B and distance equal to AC 
 describe an arc cutting former arc in D; join DC and DB, 
 completing the required oblong. (Fig. 68.) 
 
 No. 4. — To divide a given line into (a) two equal parts. 
 
 B 
 
 Fig. 69. 
 
 I 
 
108 
 
 DRAWING. 
 
 '% Let AB be the given line; with centre A and distance equal 
 to more than half of AB describe an arc, and with centre B 
 and same distance describe an arc cutting former in C and D ; 
 join CD, cutting AB in E into two equal parts. (Fig. 69.) 
 
 (6) Into any number of equal parts. 
 
 Note. — Before this can be done it is necessary to show how 
 to draw a line parallel to another from an external point. 
 
 0/ D 
 
 *.,..- ^,. Figr. 70. . . - 
 
 • Let AB be the given line and C the external point; at any 
 point B and distance BC describe an arc to cut AB; with C 
 as centre and CB as distance describe an arc, and with B as 
 centre and distance equal to AC describe an arc cutting 
 former arc in D ; join CD, which will be parallel to AB. 
 (Fig. 70.) 
 
 (b) To divide a line into any number of equal parts. 
 
 C 
 
 
 
 Fig. 71. 
 
 
 
 
 
 . Let AB be the 
 
 given 
 
 line; draw 
 
 DE 
 
 parallel 
 
 to AB on 
 
 
 either side, and on 
 
 this line set off the 
 
 r 
 
 required 
 
 number of 
 
 
GEOMETRICAL DRAWING. 
 
 lOd 
 
 on 
 of 
 
 equal distances (in this case five); then join each point of 
 section with A, which will divide AB into the same number 
 of equal parts. (Fig. 71.) 
 
 (c) To divide a given line proportionally to another line. 
 
 »••-' i^j 
 
 .. . . . Fijf. 72. 
 
 Let it be required to so cut AB, that the smaller part 
 shall be to the greater, as the greater is to the whole line. 
 Let any line CD not equal to AB, be cut in G, so that 
 CG:GD::GD:CD; place CD parallel to AB, and join CA 
 and DB; produce them to meet in E; join EG to cut AB in 
 F; then will AF:FB::FB:AB. (Fig. 72.) .-..; 
 
 Ifo. 5. — To construct a triangle of given dimensions. 
 
 Fijf. 73. 
 
 Let AB, CD and EF be the given sides, no two of which, 
 taken together, are equal to, or less than, the third. Take 
 one of them AB, and with centre B, and distance equal 
 
 ji 
 

 110 
 
 DRAWING. 
 
 to CD describe an arc; with centre A and distance equal to 
 EF describe an arc cutting former arc in G; join GA and 
 GB, completing the triangle. (Fig. 73.) 
 No. 6. — To bisect a given angle. 
 
 Let BAG be a given angle ; with centre A and any distance 
 AB describe an arc BC ; with centre B f^d any distance less 
 than half of BC describe an arc; with centre C, and same dis- 
 tance, describe an arc to cut former arc in D; join AD, which 
 will bisect the angle. (Fig, 74.) 
 
 Ifo. 7. — To trisect a right angle. 
 
 \ , 
 
 Fig. 76. 
 
 Let ABC be the given right angle; with B as centre and 
 at any distance BA describe the quadrant AC ; with centre A 
 and distance equal to AB describe an arc to cut AC in E; and 
 with centre C, and distance equal to CB or BA describe an 
 
GEOMETRICAL DRAWING. 
 
 Ill 
 
 1 to 
 and 
 
 arc to cut arc in D. Then D, E will be points of trisection, 
 and lines from D and E to B will trisect the angle. (Fig. 75.) 
 No. 8. — To inscribe a circle in a given triangle. 
 
 mce 
 
 less 
 
 dis- 
 
 hich 
 
 and 
 
 'e A 
 
 and 
 
 3 an 
 
 '""j 
 
 Pig. 76. 
 
 Let ABC be the given triangle; bisect the angle at A by 
 AD, and the angle at B by BD, cutting AD in D ; draw DE 
 perpendicular to AB at point E; then with centre D and 
 distance DE describe the circle. (Fig. 76.) 
 
 No. 9. — To draw a circle through three given points, which, 
 however, cannot be in the same straight line. 
 
 ^i e 
 
 Fig. 77. 
 
 iJ 
 
li 
 
 112 
 
 DRAWING. 
 
 Let A, B, C be the given points; join them to form a tri- 
 angle; with centre A, and distance greater than half of AC, 
 describe an arc; with centre C and same distance, describe an 
 arc to cut former; join points of section of arcs; this line will 
 pass through the centre of the circle; draw similar arcs on 
 AB, and join points of section to meet in D, the centre; then 
 a circle drawn with centre D, and distance DA, will pass 
 through A, B and C respectively. (Fig. 77.) 
 
 No. 10. — To find the centre of a whole or part of a circle. 
 
 ». 
 
 Fig. 78. 
 
 i 
 
 Let BACD be a circle or arc; draw any two chords AB, 
 CD, and draw arcs EF and GH, bisecting the chords, respec- 
 tively; join EF and HG and produce them to meet in K; 
 then K will be the centre, and if K and D be joined, KD 
 will be a radius, and a circle may be thus described with it 
 with the centre thus found. (Fig. 78.) 
 
GEOMETRICAL DRAWING. 
 
 113 
 
 No. 11. — To draw a tangent to a given circle (a) from a 
 point in the circumference. 
 
 Fig. 79. _;„ 
 
 ^ y 
 
 Let C be a given point in the circumference; let A be the 
 centre; join CA; at C erect the perpendicular CE, which, will 
 be tangent required. Also if H be the given point, join All 
 and produce it, making HG equal to part produced ; bisect 
 this line by KL, which will also be a tangent at point H. If 
 AC be produced, the line FC, perpendicular to the tangent at 
 the point of contact C, is called a "normal." (Fig. 79.) 
 
 (b) To draw a tangent to a circle from an external point. 
 Join point with centre, and at point where it cuts the circum- 
 ference draw a perpendicular upon it ; thus, if D be given 
 point, HL will be a tangent. 
 
114 
 
 DRAWING. 
 
 
 '4 ^ 
 
 fill 
 
 iTo. i^. — To construct an isosceles triangle of a given 
 altitude. 
 
 A H 
 
 Let AD be the given altitude ; at A and D draw perpen- 
 diculars, and with A as centre and any distance describe an 
 arc EF ; with D as centre and any distance less than DA 
 describe an arc to cut arc in E and F ; join AE and AF, and 
 produce them to base, forming isosceles triangle ABC. (Fig. 
 80.) 
 
 iVb. 13. — To construct an equilateral triangle of a given 
 altitude. 
 
 £ A F 
 
 \.i'. 
 
 V^^.--,. -; 
 
 :. i.'i'. i'j.'' 
 
 Fig. 81. 
 
 Let AD be the given altitude ; through A, draw EAF per- 
 
GEOMETRICAL DRAWINO. 
 
 115 
 
 pendicular to AD ; with centre A describe any semicircle 
 EGHF, and with centre F and distance FA, describe arc to 
 cut arc GH in H ; also with gentre E and same distance de- 
 scribe arc to cut GH in G. Join AG and AH, and produce 
 them to meet BC in B and C : then ABO will be the eqxii 
 lateral triangle required. (Fig. 81.) 
 
 iTo. 14' — (a) To draw, from a given point in a straight line, 
 an angle equal to a given angle. 
 
 per- 
 
 Let BAG be given angle and let the given point be at D ; 
 with centre A and any distance less than a side describe an 
 arc CB ; with centre D and distance equal to AB describe arc 
 EF ; with centre E and distance equal to BC describe an arc 
 to cut EF in F ; join DF : then the angle EDF will be equal 
 to the angle BAG. (Fig. 82.) 
 
 (6) Within a given circle to construct a triangle similar to 
 another triangle. (Triangles are similar when the angles in 
 one are equal to the angles in the other, each to each. They 
 are similarly situated when the sides of one are parallel to the 
 sides of the other, each to each. These are called hojiioloyous 
 sides.) 
 
 Let ABC be the given circle and LHK, the given triangle. 
 At any point C, draw a tangent DE, and describe a semicircle 
 DE ; with centre H, and any distance HM, describe an arc, 
 and with centre K and same distance describe an arc. Make 
 EG = NO and DF = MO ; join CF and CG, and produce them 
 
i 
 
 Ml 
 
 
 116 
 
 DRAWING. 
 
 to the circumference in A and B ; join AB, completing the 
 triangle required. (Fig. 83.) 
 
 Fig. 83. 
 
 iTo. 15. 
 circle. 
 
 — To construct an equilateral triangle about a given 
 
 f- 
 .it 
 
 • .;. if 
 
 Fig. 84. 
 
GEOMETRICAL DRAWING. 
 
 117 
 
 the 
 
 Let BCD be given circle ; at any point A in the circumfer- 
 ence, with distance equal to radius, describe an arc, cutting 
 circle in B and C ; with centre C, and same distance descril)e 
 an arc, cutting circle in D; describe siniilar arcs with centres 
 B and D; these arcs intersect in points E, F and (I; join 
 these, completing the triangle required. (Fig. 84.) 
 
 No. 16. — About a given circle to construct a triangle simi- 
 lar to a given triangle. 
 
 
 Fig. 85. 
 
 Fig. 86. 
 
 Let DEF be the given circle and TON the given triangle ; 
 find the centre G, and draw any radius GD ; at D, draw a 
 
118 
 
 DRAWING. 
 
 \iiii 
 
 '. m 
 
 tangent to the circle. With cent»'P N and any distance MN, 
 describe an arc MS, and with centre O, and same distance 
 describe an arc RP ; with centre G, and distance equal to 
 MN or OP, describe a circle cutting GD in H. Make arc 
 HK = RP and HL = MS ; join GK and GL, and produce them 
 to the circumference in E and F respectively. Draw tangents 
 at E and F to meet the other tangent in A and C ; then 
 triangle ABC will be similar to TON. (Figs. 85 and 86.) 
 
 1^0. 17. — .Vithin a circle to draw any number of equal 
 smaller circles, each touching two others and the outer circle. 
 
 .1! 
 
 
 
 
 
 \ 
 
 /o 
 
 
 
 
 \ 
 
 V 
 
 - . 
 
 / 
 
 \ 
 
 
 7 1 
 
 
 
 \ 
 
 \g/ 
 
 ^ 
 
 ^ 
 
 V 
 
 / 
 
 
 i 
 
 \ 
 
 ^' 
 
 % 
 
 > 
 
 
 
 \ff/ 
 
 X:' 
 
 
 P 
 
 '^^ 
 
 ^. 
 
 ^ 
 
 
 ^ 
 
 M 
 
 Let KME be the given circle, and divide it (in this case) 
 into six equal parts. Take centre O, and join any two, as OA, 
 OB ; bisect the angle AOB by OE, and at E draw a tangent 
 CD ; produce OA and OB to meet the tangent in C and D. 
 
GEOMETRICAL DRAWING. 
 
 119 
 
 Bisect the angles at C and D by OF and DF, meeting at F ; 
 then with centre O and distance OF describe a circle ; also 
 with centre F and distance FE describe a circle : this will be 
 •one, and the remaining five may be similarly drawn. (Fig. 87.) 
 No. 18. — To construct a regular polygon {a) on a given line. 
 
 Fig. 88. 
 
 Let AB be the given line, produce it both ways; then with 
 centre A and distance AB, describe the semicircle DEB, 
 describe pIso a similar semicircle AFC. Divide the circumfer- 
 ence DEE into as many equal parts as the polygon is to have 
 sides (in this case five), and join A with the second point of 
 division ; make the arc FC = DE ; join BF and with centres 
 E and F and distance EA and FB describe arcs to intersect 
 at G ; join GE and GF,'completing the polygon. 
 
 Note. — This method will be clear if it be remembered that, 
 if from a point within a polygon straight lines be drawn to 
 the angles, the figure will be divided into as many triangles 
 as it has sides, and each triangle will contain two right 
 angles, but the angles around the common point within, to- 
 gether make four right angles. Then if N represent the 
 number of sides, the number of degrees in the angle of a 
 
 regular polygon will be 
 
 ^- ''^^^1^, that is, 'i^(" --i). 
 n n 
 
 Now, 
 
120 
 
 'jrjyj.-i DRAWING. 
 
 in the above figure the line DB may be called 180°, or two 
 right anglesi Then the angle EAB will be represented by 
 
 5-2 
 
 — — , or f of 180°; hence it is always necessary to draw 
 
 through the secono? point of division. . .__ .;j'. 
 
 (b) In a given circle. 
 
 Let ABE be the given circle; draw any diameter FC, and 
 divide it into as many equal parts as the figure is to have 
 sides (in this case five) ; with centre C and distance CF 
 describe arc FG, and describe similar arc CG, intersecting at 
 G. Draw GA from A, through second point of division ; join 
 FA, and continue this around the circumference, completing 
 the polygon. (Fig. 89.) . „ ^. _, . , , 
 
 
 'H- 
 
 
 i 
 
 ^ 
 
 y 
 
 y^ 
 
 ; 
 
 ■t. .. 
 
 
 3 
 
 k 
 
 r 
 
 4 
 
 L 
 
 J 
 
 
 B"^ 
 
 
 
 ■\ 
 
 Fig. 89. 
 
 No. 19. — To construct a regular pentagon on a given line 
 by a special method. / /' * 
 
 Let AB be the given line ; describe arcs CAD and CBD, 
 with radius AB ; join CD ; with centre D and distance same 
 as AB describe arc EABF, cutting former arc in E and F. 
 
 i't 
 
 i 1 
 
GEOMETRICAL DRj^WING. 
 
 121 
 
 Join PG and EG, and produce them to meet arcs in H K- 
 jom AH and BK; with centre H and distance HA defcrit' 
 arc; and w.th centre K and distance KB, describe arc cutt"n^ 
 former arc in L Join T h „ ^ t t^ , t**^ ouinng 
 
 gon. (Fig. 900 ' ''""^ ^^ ^^ '''' P^"*^- 
 
 h ■ 
 
 Fig. 90, 
 
 Tfo. 20.- 
 line. 
 
 -To construct a regular hexagon on a given straight 
 
 9 
 
122 
 
 DRAWING. 
 
 n 
 
 11 
 
 ■•r ^ 
 
 Let AB be the given line ; with centre A and distance AB 
 describe arc ; with centre B and distance BA, describe arc 
 cutting at C; join CA, CB. With centre C and distance CA, 
 describe circle cutting in D, B and A ; join DC and produce 
 to E ; produce AC to CI and BC to F ; join AE, EF, FG, 
 GD and DB, completing the hexagon. 
 
 No. 21. — To construct a regular octagon (a) on a given 
 straight line. j ' ' •' 
 
 ■fa ;|' 
 
 m 'I 
 
 
 t-ic 
 
 i - • i ■• 
 
 -.« 
 
 >.-• 
 
 Let AB be the given line ; produce it both ways, and 
 describe the semicircles CKB and AMD. Erect a perpen- 
 dicular at A ahd at B; bisect the right angles CAG and DBH 
 by AK and BM, respectively, and erect perpendiculars KE 
 and MF, at K and M, respectively. With centre K and dis- 
 tance KA describe arc EA cutting KE in E; draw similar 
 arc BF cutting MF in F; with "centre E and distance EK 
 describe arc to cut AG in G ; similarly find H ; join EG, GH 
 and HF, completing the octagon. (Fig. 92.) . 
 ,.(6) In a given square. 
 
 Let BCDE be the given square; draw diagonals intersecting 
 at A. With centre B, and distance BA, describe arc cutting 
 ^ides of square in H and P; similarly find points K, N,^ F, M 
 
>•"■', 
 
 GEOMETRICAL DRAWING. 
 
 m 
 
 and G L; join FG, PN, ML and HK, completing the octagon. 
 (*ig. 93.) 
 
 B e H _c 
 
 ' ' S'.f 
 
 i.yty"', 
 
 •r! >,.^ 
 
 ■A 
 
 N M 
 
 Fig. 93. 
 
 No. 22.— To draw a perfect ellipse by means of the foci 
 and intersectmg arcs, axes being given. 
 
 Let the axes, AB and CD, be pla.,ed centrally, at right 
 angles to each other ; then measure from the distance from 
 A to D, and describe arc cutting AB in F and F: these are 
 the foci. Between and F, take any number of points, 1, 2 
 etc—the more the better; then with centre F (left) and dial 
 
^i; 
 
 1^4 
 
 DRAWING. 
 
 tance equal to distance from 1 to B, describe arcs E, E ; and 
 with centre F (right) and same distance describe arcs E, E. 
 Then with centre ¥, (left) and distance equal to that from 1 
 to A, describe arcs cutting the former, so also describe arcs 
 from centre F (right). Thus for each point between O and F, 
 we get four points. Having thus found a number of points, 
 join tliem, or rather draw a curve through them : this curve 
 will be an ellipse. (Fig. 94.) 
 
 J^o. 23. — To draw an ellipse (a) by means of concentric 
 circles and intersecting perpendiculars. 
 
 
 ^' 
 
 I 
 
 
 >• •. y 
 
 A '■*' • 
 
 Let the concentric circles EF and AB be drawn ; draw 
 diameters AB and CD at light angles to each other, divide 
 each quadrant into the same number of equal parts, and join 
 opposite points ; draw perpendiculars from the outer points 
 and horizontals from the inner points to meet them ; thus 
 i>fi.w perpendicular from H and horizontal from a to meet 
 ii> J similarly find 2, 3, 4, etc., all around the circle. Draw 
 f 'iXiTYd through the points of intersection thus found, which 
 will form an ellipse. (Fig. 95.) 
 
GEOMETRICAL DRAWING. 
 
 125 
 
 (b) When the major axis (transverse diameter) only is given. 
 
 , » 1. : . 
 .v".7l,<V' 
 
 IW 
 
 de 
 )in 
 its 
 lus 
 
 kt 
 
 IW 
 
 3h 
 
 Let AB be the given diameter; divide it into four equal 
 parts in G, M, D ; with centre D and distance DG describe 
 arc EF, and with centre G and same distance describe arcs to 
 intersect in E and F. Join FG, FD, EG and ED, and produce 
 them to the circumference in C, H, K and L respectively. 
 Then with centre F and distance FC describe arc CH, and 
 with centre E and distance EK describe arc KL, completing 
 the elliptical curve. (Fig. 96.) 
 
 Note. — No part of a true ellipse is an arc of a circle. 
 
 J)^o. 2^. — An ellipse being given, to find axis and foci. 
 
 Draw any two parallel chords AB and CD; bisect each 
 and join points of section FE, and produce each way to 
 meet the circumference in G and H; bisect GH in K, and 
 with centre K describe a circle to cut the ellipse in four 
 points N, O, R and P; join these to form a rectangular paral- 
 lelogram; bisect each side and join the opposite points of sec- 
 tion, and produce both ways to meet circumference in L, M, 
 T and V; then LM and TV will be the axes; and if the dis- 
 
126 
 
 DRAWING. 
 
 tance TK be taken with centre L or M, the arc will cut TV 
 in S, S, which will be required foci. (Fig. 97.) 
 
 I'f"^- 
 
 
 •■■} 
 
 Fig. 97. 
 
 N'o. 25. — To draw a tangent to an ellipse (a) from a point 
 in the circumference. 
 
GEOMETRICAL DRAWING. 
 
 127 
 
 Let B, C be the foci and A, given point on the curve; join 
 BA and CA and produce them to D and F; bisect the angle 
 BAD by AK, and the angle DAF by AG. Then will KA be 
 a tangent and GA a perpendicular or normal to it, at the 
 point of contact, A. (Fig. 98.) 
 
 (6) From an external point. "'"^.^ " . ' ^ 
 
 Fig. 99. 
 
 "»-. 
 
 ■ If 
 
 Let A be an external point; draw major axis BC, and on 
 it describe the semi-circle BFC; draw a tangent AF to the 
 circle at F; draw FE perpendicular to BC, cutting curve in 
 G; join AG, which will be a tangent, to the curve. (Fig. 99.) 
 
 1^0. 26. — To draw an oval of a given width. 
 
 Let AB be the given width ; bisect it in C, and on it describe 
 the circle ADB; draw CD at right angles to AB; with centre 
 B and distance BA describe curve AE; similarly describe 
 
128 
 
 DRAWING. 
 
 curve BF; join BD and AD, and produce them to meet curve 
 in E and F; with centre D and distance DE describe curve 
 EF, completing the oval. (Fig. 100.) 
 
 
 iVi- 
 
 iVb. 27. — To construct the involute of the circle. 
 
 Divide the circle into any number of equal parts and draw 
 the radii, numbering them 1, 2, etc. Draw the tangents, mak- 
 ing the first the length from 1 to 2, the second twice this 
 length, the third three times, and so on. When all the tan- 
 gents have been drawn thus, begin again at 1 by producing 
 it, and so get a second series of points. Then draw a curve 
 through the points, commencing with 8, or last, and joining 
 it with 1, then 2, etc. (Fig. 101.) 
 
GEOMETRICAL DRAWING. 
 
 129 
 
 3--.— - 
 
 Fig. 101. 
 
 M. 28.- -To find (a) a mean proportional between two given 
 
 lines 
 
 Fig. 102. 
 
 ^I*t AC and BC be the given lines; place them in a 
 straight hne AB; bisect AB in D, and on AB describi 
 
130 
 
 DRAWING. 
 
 r 
 
 II '; 
 II 
 
 semicircle AEB; through C draw CE at right angles to AB; 
 then will AC : CE : : CE : CB. (Fig. ^) 
 (h) To draw a third proportional (greater). 
 
 Fig 103 
 
 Let AB and BC be the given lines ; draw AE greater than 
 AC, making any angle with AC ; make AD = BC ; join BD, 
 and through C draw CE parallel to BD ; then will AB : BC : : 
 BC:DE. (Fig. 103.) 
 
 (c) (Less). Make AE greater than BC and le;ss than AC. 
 
 ■ Kg. 104. 
 
 I 
 
GEOMETRICAL DRAWING. 
 
 131 
 
 No. 29. — To draw a circle of given radius wliich shall "touch 
 both lines of a given &Dgle. 
 
 Let BAG be the given angle and ST the given radius; bisect 
 angle by line AF; erect on either line a perpendicular DE 
 equal to ST ; through E draw EG parallel to AD, and cutting 
 AF in G ; then a circle drawn with centre G and radius equal 
 to DE or ST, will touch the sides AB and AD. (Fig. 104.) 
 
 No. 30. — To draw a circle of given radius which shall touch 
 another given circle and a given straight line. 
 
 Let FG be a given line, BX a given circle, and DE a given 
 radius; draw a line GH perpendicular to FG and equal to 
 DE; through H draw HK parallel to FG; draw any radius 
 AB and produce it, making the part produced equal to DE; 
 then with centre A and distance AC, describe an arc cutting 
 HK in K: the circle drawn with centre K, and at a distance 
 equal to GH or DE will touch the circle BX and the line 
 FG. (Fig. L05.) 
 
132 
 
 'r?fi?i 
 
 DRAWING. 
 
 t . .' 
 
 ? - 
 ■ /•*■ 
 
 ■ •■■•\ !• • 
 
 Graded. Exercises. 
 
 ■ t I 
 
 'M 
 
 1. Construct a square whose side is 2". ' ' \' ["'' 
 
 2. Construct a square whose diagonal is 4". 
 
 3. Construct an oblong whose sides are If" and 2^" respec- 
 tively. 
 
 4. A rectangular field is 900 yds. long and 400 yds. wide; 
 divide it into four eCj^ual fields each 400 yds. long. Scale 
 100 yds. = 1". 
 
 5. Divide a line 3|" long into two parts in the ratio 3 : 4. 
 
 6. Draw a line parallel to and between two other parallel 
 lines 2^" apart, the line to be twice as near one as the other. 
 
 7. Construct an equilateral triangle whose side is 2". 
 
 8. Centrally within the triangle in No. 7 construct a tri- 
 angle whose side is 1|". 
 
 9. Construct a triangle whose sides are 2|" 3^" and 3|" 
 respectively. 
 
 10. Inscribe a circle in a triangle whose sides are same as 
 No. 9. 
 
 11. The diagonal of a parallelogram is 4' and one side is 2'. 
 Draw it. ' ' 
 
 12. Find the extent of an angle of 22J°,of 37 J°, of 41 J°. 
 
 13. Construct an isosceles triangle whose base is 2" and 
 vertical angle 37|°. 
 
 14. Construct an isosceles triangle whose base is 2" and 
 altitude 3^". ^ • • .- ■ :- : 
 
 15. Construct an equilateral triangle about a circle whose 
 diameter is 4'. 
 
 1 6. Construct a triangle whose sides are in the ratio 3:4:5 
 about a circle whose diameter is 4". 
 
 17. Describe a circle about a square whose side is 3*. 
 
 18. Within a triangle whose sides are 3", i" and "5" respec 
 tively, inscribe a circle. 
 
 19. Construct a regular pentagon whose side is l^*. 
 
 20. Construct a regular heptagon in a circle 4' diameter. 
 
 '•>! 
 
 5 1">-i'.' i 
 
GRADED EXERCISES. 
 
 133 
 
 21. Draw a hexagon whose side is 2". 
 
 22. Draw a hexagon within an equilateral triangle of 3". 
 
 23. Construct a regular octagon whosv^ side is 1^". 
 
 24. Construct a regular octagon in a square, side 3". 
 
 25. Construct a regular octagon in a circle of 3" diameter. 
 
 26. Inscribe seven equal circles in a circle of 3" diameter. 
 
 27. The diameters of an ellipse being 3^" and 2 |" respec- 
 tively, draw it. 
 
 28. Draw an elliptical curve on a transverse axis of 3". 
 
 29. Draw an elliptical figure from two squares, diagonal of 
 each 3". 
 
 30. A circle and an ellipse touch the angles of an oblong 
 3" X 5", find axes and foci of the ellipse. 
 
 31. Draw an oval whose shorter axis is 2^". • : 
 
 32. Construct an involute to a circle whose diameter is |". 
 
 33. Find a mean proportional between two lines 2^" and 
 3f " respectively. 
 
 34. Draw a third proportional (greater) to two lines 2^" 
 and 3|" respectively. ■ 
 
 35. Draw same as No. 34, less. . 
 
 36. Lay out a circular garden whose radius is 30 yds., which 
 shall just touch a fence on one side, and another garden whose 
 radius is 50 yds. on the other. Scale 20 yds. = 1". 
 
 37. Lay out a circular garden, radius 30 yds., which shall 
 touch two fences not parallel. Scale 20 yds. = 1 ". 
 
 38. Lay out a circular garden, radius 30 yds., which shaU 
 touch two fences not parallel, and whose edge shall just touch 
 a tree in a given position. Scale 20 yds. = 1". 
 
 39. The axes of an elliptical flower garden are 40 and 60 
 yds. respectively, and a point is taken 60 yds. to left of the 
 shorter (produced), and 40 yds. above the longer (produced). 
 Draw a path from this point to touch the elliptical garden. 
 Scale 20 yds. = 1". Longer axis horizontal. 
 
 40. Construct a triangle whose angles are 75°, 45° and 60° 
 respectively, on a line 2" in length. , , ,; ., . .,»l' 'i 
 
 41. Two circles, whose diameters are each 3|", intersect, the 
 intercepted arc of each being one-fifth of the whole circum- 
 ference. 
 
 42. The base of a right-angle triangle being 2", and the 
 
134 
 
 DRAWING. 
 
 .) 
 
 
 perpendicular on the hypothenuse from the right angle being 
 1^", construct the triangle. 
 
 43. Two lines meet at a point. Find a point between them 
 that will be 2" from one and 3" from the other. 
 
 44. Show that the number of degrees in the angle of any 
 
 regular polygon is represented by ^ , where n = 
 
 num- 
 
 n 
 
 ber of sides. 
 
 45. Show that a circle is but a particular form of an 
 ellipse. 
 
 46. If an ellipse be drawn with a string around two fixed 
 pins as foci, show that the sum of the distances from any 
 point in the curve to th€ foci is the same. 
 
 47. From the point found in No. 43 draw two equal 
 straight lines to the given line. 
 
 48. Draw a parallelogram 3|" x 5^", and within this draw 
 one of half the size similar and similarly situated. 
 
 49. If one hexagon be inscribed in, and another inscribed 
 about, a circle, show that their areas are in ratio 3 : 4.' ' : 
 
 50. Three circular gardens, diameters 20 yds., 30 yds. and 
 40 yds. respectively, are to be placed with walks of 5 yds. 
 between them. Scale 20 yds. = 1". 
 
 51. Divide a triangle whose sides are 3", 4" and 5" respec- 
 tively, into four equal and similar triangles, 
 
 52. From the vertex of a scalene triangle draw a straight 
 line to the base which shall exceed the less side as much as it 
 is exceeded by the greater. 
 
 53. One of the acute angles of a right-angled triangle is 
 three times as great as the other; trisect the smaller. 
 
 54. One side of a right-angled triangle is 4", and the differ- 
 ence between the hypothenuse and the sum of the other two 
 sides is 2" ; construct the triangle. - •' ■"- 
 
 55. The altitude of an equilateral triangle is 2"; construct it. 
 
 56. Place a straight line, 3" in length, between two straight 
 lines, each 2" in length, which meet, so that it shall be equally 
 inclined to each of them. 
 
 57. Describe an isosceles triangle upon a given base, having 
 each of the sides double of the base. 
 
 58. Draw a square equal in area to two unequal oblongs. 
 
GRADED EXERCISES. 
 
 186 
 
 59. Given the base, the vertical angle and the perpendicular 
 of a plane triangle, construct it. 
 
 60. Cut off two-thirds of an isosceles triangle by a line 
 parallel to the base. 
 
 61. Describe two circles with given radii which shall cut 
 each other and have the line between the points of section 
 equal to a given, limited line. 
 
 62. Describe a circle with a given centre cutting a given 
 circle in the extremities of the diameter. 
 
 63. Describe a circle which shall pass through a given point 
 and which shall touch a given straight line in a given point. 
 
 64. Describe a circle which shall touch a given straight 
 line at a given point and bisect the circumference of a given 
 circle. 
 
 65. Two circles are described about the same centre; draw 
 a chord to the outer circle, which shall be divided into three 
 equal parts by the inner one. What are the limits of the 
 diameters ? 
 
 66. The perimeter of an oblong is 20", and the sides are in 
 ratio 3:2; construct it. 
 
 67. Construct an isosceles triangle of given vertical angle 
 and given altitude. , 
 
 68. Draw a square equal in area to an oblong 3" x 2". 
 
 69. Describe a circle of given radius to touch two points. 
 What limits the position of the points ? 
 
 70. Show how to dtuw a similar triangle within another. 
 
 71. Two equal ellipses, axes 5" and 3", cut each other at 
 right angles, and their centres are coincident ; draw them. 
 
 72. Trisect a given line. . ./ 
 
 73. Show how an angle may be trisected (mechanically). 
 
 74. Divide a square into three equal parts by lines drawn 
 from an angle. 
 
 75. Divide an oblong into three equal parts by lines drawn 
 from an angle. 
 
 76. Find the number of degrees in the angle of a regular 
 
 diiodecagon. 
 
 Ci" 
 
 ni^- 'i. 
 
 ^■'