T 
 369 
 
 73 
 
 UC-NRLF 
 
 BACKBONE OF 
 PERSPECTIVE 
 
 TAYLOR 
 
LIBRARY ~,_ 
 
 OF THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Class 
 
BACKBONE 
 
 OF 
 
 PERSPECTIVE 
 
 NET BOOK This Book is supplied 
 to the trade on terms which do not 
 admit of discount. 
 
 THE MYRON C. CLARK PUBLISHING CO. 
 
 CHICAGO 
 
 THE MYRON C. CLARK PUBLISHING COMPANY 
 1910 
 
BACKBONE 
 
 OF 
 
 PERSPECTIVE 
 
 By T. U. TAYLOR 
 
 C. E., University of Virginia; M. C. E., 
 
 Cornell University ; Professor of Civil 
 
 Engineering, University of Texas ; 
 
 Member Am. Soc. C. E. 
 
 CHICAGO 
 
 THE MYRON C. CLARK PUBLISHING COMPANY 
 1910 
 
TJ6? 
 
 T3 
 
 Copyright, 1910, 
 
 By 
 THE MYRON C. CLARK PUBLISHING Co. 
 
PREFACE 
 
 These notes have been given in the form of lec- 
 tures and drawing-board exercises for many years. 
 They are here reduced to print to save time in note- 
 taking on the part of the student. A reader that 
 finds no errors in these pages should read again. 
 I am indebted to E. E. Howard, a former instructor, 
 for much valuable assistance. Prof. E. C. H. 
 Bantel, and Instructor S. P. Finch have rendered 
 substantial aid. The chapter on Axometric Projec- 
 tions is a modification of notes taken under Dr. W. 
 M. Thornton of the University of Virginia, when I 
 was a student there. 
 
 T. U. TAYLOR. 
 
 Austin, Texas. 
 
 June, 1910. 
 
 212042 
 
( UNIVERSITY ) 
 
 >IUFORjfifeX 
 
 BACKBONE OF PERSPECTIVE 
 
 CHAPTER ONE. 
 
 Primary Methods. 
 
 i. Plan and Projection. When a perpendicu- 
 lar is dropped from any point P to a plane, the point 
 of intersection p of the perpendicular and the plane 
 is called the foot of the perpendicular and p is called 
 the projection of P. If from two points, A and B, 
 perpendiculars are dropped on any plane, and their 
 feet a and b be joined, the line ab is called the pro- 
 jection of AB on the given plane. The projection 
 of a point on a horizontal plane is called the plan of 
 
 Figure 1. 
 
 the point, and its projection on a vertical plane is 
 called its elevation. We shall denote the horizontal 
 plane by H in these pages. 
 2. Perspective. The perspective of a point P 
 
2 BACKBONE OF PERSPECTIVE. 
 
 ( Fig. i ) with reference to any other point E and a 
 plane MN is the point of intersection of the line PE 
 and the plane. Thus if P be any point, the per- 
 spective of P with reference to the point E and the 
 plane MN is the point P' where EP cuts the plane 
 MN. Similarly the perspective of Q with reference 
 to the point E and plane MN is the point Q' where 
 EQ cuts MN. 
 
 The point E is the eye of the observer. The per- 
 spective of any point P with reference to any plane 
 MN is the intersection of the line of sight PE with 
 the plane. The plane MN is called the perspective 
 plane. 
 
 3. Perspective of a Line. The perspective of 
 a line will be found by joining all points of the line 
 to the eye and rinding their intersections with the 
 perspective plane. The perspective of a straight 
 line will be a straight line. Since by joining P and 
 Q with E we have a triangle EPQ (Fig. i), and 
 the plane of the triangle will cut the perspective 
 plane MN in a straight line P'Q', it is evident that a 
 line joining any point in PQ with E will cut the 
 plane MN somewhere in P'Q', as such a line lies in 
 the plane EPQ, and as the plane cuts the perspective 
 plane MN in P'Q'. It is sufficient in determining the 
 perspective of a definite part PQ of a straight 
 line, to find the perspective of the two points P and 
 Q and join these points by a straight line. 
 
 4. Point of View. A perspective is defined 
 with reference to the point of view, which is the eye 
 of the observer. If we stand in a room and look 
 through a window glass at a point A, the intersec- 
 tion of the line of sight with the plane of the win- 
 
BACKBONE OF PERSPECTIVE. 3 
 
 dow glass will be the perspective of the point with 
 reference to that particular location of the eye. 
 The point A will have a perspective for every posi- 
 tion of the eye in the room. 
 
 5. Perspective of a Point. Let P (Fig. 2) be 
 any point and E the eye and MN the perspective 
 plane which is taken as vertical. Let the plane 
 MN cut the horizontal plane in GL. Project P 
 and E on the horizontal plane in p and e. Join pe, 
 cutting GL in D. The plane PpeE will cut plane 
 MN in a vertical line, as the plane PpeE is itself 
 vertical. The line of intersection of the two ver- 
 tical planes PpeE and MN will be parallel to Pp 
 and Ee. The perspective of P will lie somewhere 
 on the vertical line Dt, and as it must lie on EP it 
 will be the point where EP intersects this vertical. 
 
 Figure 2. 
 
 It is our object to develop some method of finding 
 this point of intersection. 
 
 Draw the perpendiculars from P and E to MN, 
 
4 BACKBONE OF PERSPECTIVE. 
 
 and let the feet of these perpendiculars be p' and e' 
 respectively. The projection of PE on MN will be 
 p'e'. The perspective t of the point P will be on 
 EP and it will lie on e'p', because PE will intersect 
 the plane MN at some point on p'e'. Hence the per- 
 spective of the point P will lie on p'e', and on Dt, 
 and therefore at their intersection. 
 
 6. Perspective Plane in H. If in Fig. 3 we 
 revolve the plane MN about GL as an axis, each 
 point in the plane will describe a circle whose cen- 
 ter lies on and whose plane is perpendicular to GL. 
 If the plane is folded to the horizontal position, the 
 points p' and e' will fall as far from GL as the 
 points P and E are above H. Thus p' will de- 
 scribe a circle whose radius is Qp' and whose center 
 
 Figure 3. 
 
 is Q. When MN is folded into the horizontal posi- 
 tion p' will fall at P!. Similarly e' will fall at E x 
 and t will fall at T x . 
 
 7. Example. In Fig. 3 let SQ=4", Qp=z", 
 
BACKBONE OF PERSPECTIVE. 
 
 Figure 4. 
 
 Find the perspective of 
 the point P. Draw as in 
 f? Fig. 4 a line GL and 
 make SQ = 4", and 
 make Se = 3", pQ \ = 
 2". Join pe, cutting 
 GL at D. When MN 
 is folded into the hori- 
 zontal position p' and e' 
 (Fig. 3) will fall at P, 
 and at E l5 respectively. The folded position of the 
 perspective will lie on P^ and the folded position 
 of the vertical line Dt of Figs. 2 and 3 will lie on 
 Dt'. 
 
 PROBLEM i. If SQ=2^", Qp=2, QP o, eS= 
 2^2", and SE =i", find the perspective of P. 
 
 PROBLEM 2. If SQ=2", Qp=2", QP 1 =i", eS= 
 3", and SEj 2", find the perspective of P. 
 
 8. General Method. The method just out- 
 lined is perfectly general, and can be used to find 
 the perspective of any structure, however compli- 
 cated. The plane of the paper upon which the con- 
 struction is made represents H, and the perspective 
 plane (which in all practical cases is vertical) is 
 folded into H. The horizontal projection p of the 
 point P is located on the paper, and the perspective 
 is found to be at t, which (after folding) is in the 
 same plane as p. The points p' and e' are the pro- 
 jections of P and E on the perspective plane, and 
 P! and Ej are the folded positions of these points. 
 
 9. Translation of the Perspective Plane. The 
 horizontal projection of points and the folded posi- 
 
BACKBONE OF PERSPECTIVE. 
 
 tion of their projections on the perspective plane 
 are generally above GL, as the perspective plane is 
 taken between the eye of the observer and the ob- 
 ject. Where there are many points the figure may 
 become complicated. To prevent confusion and to 
 keep the plan and perspective of the points separate, 
 the perspective plane before folding is brought from 
 position GL in Fig. 5 to position G'L', while the 
 point e and the plan p of the point remain 
 fixed. It is clear that Ej and P x will occupy exactly 
 the same positions with respect to G'L' that they 
 did with respect to GL; that is, S'E"=SE 1 , Q'P"= 
 QP r The perspective T of the point P lies on the 
 perpendicular at D to GL and on P^; and it is 
 clear that in bring- 
 ing GL to G'L' we 
 have simply 
 brought all the 
 points except the 
 horizontal projec- 
 tion p into the po- 
 sition P'^E", by a G ' 
 motion of transla- 
 tion which has not 
 altered their relative positions. 
 
 10. True Height Line of a Point. From Fig. 
 5 we see that the line pP x is perpendicular to GL, 
 and that Q'P" is equal to the true height of the 
 point above H, and that the point P" is joined to 
 E". This can be expressed in the following con- 
 structive rule: 
 
 ( i ) Join the plan of the point to the plan of the 
 
 Figure 5. 
 
BACKBONE OF PERSPECTIVE. 7 
 
 eye, cutting GL in D, and at D draw DD' perpen- 
 dicular to the ground line. 
 
 (2) Drop a perpendicular from p, the plan of the 
 point, on the new ground line (G'L r ) and from G'L' 
 lay off on this perpendicular the true height of the 
 point=Q'P". Join the point thus located to E" , and 
 where the line E"P" cuts DD' is the perspective of 
 the point P. 
 
 In the following articles the perspective of the 
 point P will be marked P', that of K will be marked 
 K', etc. 
 
 ii. Perspective Triangle. Two points P and 
 K whose heights above H are O'P" and Q'K" re- 
 spectively lie on a common vertical line. Let the 
 height of E S'E" and e and p be located with re- 
 spect to GL and G'L' as in Fig. 6. Join pe, cutting 
 
 G3 
 
 GL at D and on the true height line Qp of P and K 
 lay off Q'K" equal to the height oF K and Q'P" 
 equal to the height of P. Take S'E" equal to 
 height of E and join E"P" and E"K", cutting DD' 
 
8 BACKBONE OF PERSPECTIVE. 
 
 in K' and F. The triangle E"FK' is called the 
 perspective triangle, and it is the perspective of the 
 triangle EPK. 
 
 PROBLEM 3. Given SQ=3", Q'P"=i", Q'K"= 
 3", S'E"=2", find the perspective of PK. 
 
 12. Special Case. When the plan of a point 
 lies on or near eE", the foregoing method of article 
 10 is indeterminate or lacks exactness. In the first 
 case the former method can not, and in the latter 
 case should not, be used. In the first case the 
 perspective triangle resolves into the straight 
 line eE", and in the latter case its sides cut DD' at 
 such sharp angles that the solution is not definite. 
 
 Figure 7. 
 
 In Fig. 7 let p and e be the plans of the point P 
 and of the eye E. If the trapezoid PEep be folded 
 around pe into H, PE will fall at P 2 E.,, and E 2 e, 
 Dtj and P 2 p will be perpendicular to pe, and Dt t is 
 
BACKBONE OF PERSPECTIVE. 9 
 
 equal to the height of the perspective of P above 
 GL. 
 
 In Fig. 8 draw lines at e, D, 
 and p perpendicular to pe and 
 make eE 2 equal to the height 
 of the eye, and P 2 p equal to the 
 height of the point. Join P 2 E 2 , 
 cutting Dtj in t . Lay off D'P' p' s\ 
 equal to Dt^ The point P' is 
 the perspective of P. 
 
 PROBLEM 4. If Fig. 5 SQ=o, 
 Qp=2", Q'P"=i" eS=2", and Figure 
 
 S'E 2 =3", find the perspective of P. 
 
 PROBLEM 5. If SQ=o, Qp=i", Q'P"=o, 
 2", and S'E 2 =i", find the perspective of P (Fig. 5). 
 
 PROBLEM 6. The plan of a point whose height 
 above H is 3" is 2" behind GL, and the plan of the 
 eye whose height is I inch is 5 inches from GL. If 
 pe is at right angles to GL, find the perspective of 
 P. 
 
 13. Perspective of a Block. Let ABCD (Fig. 
 9) be the base of a rectangular block, whose side 
 and end views are shown by ABFT and ADHT. 
 The block rests on the horizontal plane and has its 
 edge in the perspective plane. We have AB= 
 15', ADr= 9 ', height of blocfc=io', baL=3O. The 
 eye is taken 25'. in front of the perspective plane and 
 at a height of 6'. 
 
 To find the perspective of AT join ea, cutting GL 
 at a and from this point drop a perpendicular to 
 G'L', cutting it at A'. As AT lies in the perspec- 
 tive plane, it will be its own perspective, and all we 
 have to do is to lay off A'T'=io', thus determin- 
 
10 
 
 BACKBONE OF PERSPECTIVE. 
 
 ing the perspective of AT. Then drop a perpendic- 
 ular from b on G'L', cutting it at Q'. Now as Q'b 
 is the "true height" line for all points that lie on 
 
 H T 
 
 Figure 9. 
 
 Q'b lay off the height of B and F on Q'b. The 
 height of B is zero and that of F=io'=Q'R'. Join 
 eb, cutting GL at N. The points B' and F' where 
 E"Q' and E"R' cut the perpendicular to GL at N 
 are the perspectives of B and F, respectively. The 
 perspectives of DH and CK are found as that of 
 BF. The whole perspective is easily completed, and 
 appears as A'B'F'T'H'D'C'K'. 
 
 PROBLEM 7. Given (in Fig. 9) AB=i2', AD= 
 8', ea=i6', height of block==8', height of eye=4'. 
 The perspective plane GL passes i' below a and 
 AB makes 30 with GL. Construct the perspective 
 of the block. 
 
 PROBLEM 8. Find the perspective of the block of 
 Fig. 9 when AB makes 45 with GL. 
 
BACKBONE OF PERSPECTIVE. 
 
 II 
 
 14. True Height Line of a Horizontal Line. 
 
 If the line HK (Fig. 9) be extended to the per- 
 spective plane it will intersect this plane directly 
 above G (or G') at a distance equal to the height of 
 HK above the horizontal plane. The true height 
 of HK above the horizontal plane can be laid off on 
 GG' equal to G'J'. The point J' is the perspective 
 of the point of the line HK that lies in the per- 
 spective plane. Now, if we join J'H' we have the 
 perspective of HK extended to the perspective 
 plane. The perspective of K lies somewhere on this 
 line and can be found in two ways : 
 
 First. Join ek, and from the point where it cuts 
 
 Figure 9. 
 
 GL drop a perpendicular, cutting J'H' at K'. This 
 determines^ the perspective of K. In the same way 
 we can find the perspective of any point not in eE". 
 
12 
 
 BACKBONE OF PERSPECTIVE. 
 
 Second. Produce kf to cut GL at L. Then LL' 
 is the "true height" line of all horizontal lines that 
 lie in the face of BFKC. Lay off L'S' equal to the 
 height (10') of line KF. Join S'F' and produce it 
 to cut J'H' at K'. 
 
 One of these methods should always be used for 
 points near eE", for the construction lines of the 
 usual method intersect at such sharp angles that 
 their point of intersection is not sufficiently definite. 
 15. Horizontal Squares. A square abed, 6'x6', 
 lies on H and its side dc, Fig. 10, is parallel to and 
 i' from GL. The eye (whose height is 6') is 10' in 
 front of the perspective plane and lies in a per- 
 pendicular to GL 2' to right of c. Find the per- 
 spective of abed and of the nine squares into which 
 it is divided. 
 
 SOLUTION. The true 
 height line of ad is H'd, 
 and as D and A lie on 
 the horizontal plane, 
 their true heights are 
 zero. We therefore join 
 ea and ed cutting GL in 
 K and P, respectively. 
 The perspective of D 
 will lie on E"H' verti- 
 cally below P, and that 
 of a will lie on E"H' 
 / vertically below K. The 
 perspectives of a and d 
 t A' 
 
 H 
 
 Figure 10. 
 
 D'. In the same way we find the perspectives of b and 
 c at B' and C'. Join A'B', B'C, CD' and D'A', 
 
BACKBONE OF PERSPECTIVE. 13 
 
 forming A'B'C'D', the perspective of abed. The 
 perspective of any point f can be found by the usual 
 method. 
 
 1 6. Perpendiculars and Parallels. It will be 
 observed from the preceding example that the per- 
 spective of all lines perpendicular to the perspective 
 plane pass through a common point E". When the 
 perspective of a series of parallels passes through a 
 common point as E", they are said to vanish at E", 
 and E" is called the vanishing point of this system 
 of parallel lines. It will be shown later that all sys- 
 tems of parallel lines have a vanishing point. 
 
 The point E" (the vanishing point of perpendicu- 
 lars) is called the center of the picture. The per- 
 spectives of all horizontal lines parallel to the per- 
 spective plane are parallel to GL. 
 
 PROBLEM 9. Find the perspective of the square 
 in Fig. 10 when eE" lies on BC produced, other data 
 remaining the same. 
 
 PROBLEM 10. Find the perspective of the same 
 when eE'' bisects AB. 
 
 PROBLEM n. Find the perspective of the squares 
 when the eye is 2' to the left of AD. 
 
 PROBLEM 12. Find the perspective of a regular 
 hexagon of 2' side that lies on H when one of its 
 sides is parallel to and its center is 3' from GL, 
 when the eye (height 4') lies 6' in front of the per- 
 spective plan and in a perpendicular from the center 
 of hexagon. 
 
 PROBLEM 13. Find the perspective of hexagon 
 in problem 12 when one side is perpendicular to 
 GL. 
 
I 4 BACKBONE OF PERSPECTIVE. 
 
 PROBLEM 14. A circle of diameter 4' lies on H 
 and has its center 3' from the perspective plane. 
 Find the perspective of the circle when the eye 
 (height 5') lies in a perpendicular through center to 
 GL and 5' in front of GL. 
 
 PROBLEM 15. Find the perspective of a circle 
 whose plane is parallel to the perspective plane and 
 at a distance of 3' from it, the height of its center 
 above H being 5', when the eye (height 5') lies in 
 
 Figure 11. 
 
 the perpendicular to the perspective plane from the 
 center of the circle and 6' in front of the perspec- 
 tive plane. Diameter of circle=4'. 
 
BACKBONE OF PERSPECTIVE. 15 
 
 17. Perspective of Steps. A series of four 
 steps (Fig. ii ) six feet long, each having a tread 
 and rise of i', has one end parallel to the perspec- 
 tive plane at a distance i' from it. Find the per- 
 spective of the steps when the eye (height 8') is 6' 
 to the right of lowest step and 9' in front of the 
 perspective plane. Height of base fbc=i'. 
 
 On aA" the true height line of ad lay off the 
 heights of the different steps from G'L' equal to I, 
 2, 3, and 4 feet at N', R', S', T and Q'. Join ea 
 and ed, cutting GL at P and I. Drop perpendicu- 
 lars from P and I, cutting E"Q' in A' and D', 
 the perspectives of a and d. The perspective of f 
 lies on E"N' and vertically below P. Join R', S', 
 and T' to E", cutting A'F' at M', etc., through 
 which draw lines parallel to G'L'. To find the per- 
 spective B'X' of an edge BX, join eb, and from the 
 point where it cuts through GL drop a perpendicular 
 to cut the horizontals through F' and M' at B' and 
 X'. Join B' and X' to E" and drop a perpendicular 
 from point of intersection of ec and GL, cutting 
 E"B' at C', the perspective of c. In the same way 
 we can find the perspective of all other edges. 
 
 PROBLEMS. 
 
 PROBLEM 16. Find the perspective of the steps 
 in Fig. ii when height of eye is 9' and lies 4' to 
 the right of the lowest step, other dimensions re- 
 maining the same. 
 
 17. Find the perspective of a hollow box 4'x6', 
 height 5', that rests on H with the long face per- 
 pendicular to the perspective plane. The open end 
 next the perspective plane is 2' from it, and the eye 
 
16 BACKBONE OF PERSPECTIVE. 
 
 (height 3') lies on a line bisecting the plan of the 
 box and at a distance of 10' from the perspective 
 plane. 
 
 18. Find the perspective in problem 17 when the 
 height of the eye is i', other dimensions remaining 
 the same. 
 
 19. One corner of the box in problem 17 lies 
 in the perspective plane and the short end makes an 
 angle of 30 with it. If the eye (height 2^') lies in 
 a perpendicular to GL at the nearest corner, find the 
 perspective of the box. 
 
 20. The base of a monument is 4'x6' by i' high. 
 Upon the base rests a rectangular block 3^5' by 3' 
 high. The block is capped by a pyramid whose 
 base is the upper base of the block and whose height 
 is 2'. The longest face of the base makes 30 with 
 the perspective plane and the nearest corner of 
 base is i' from the perspective plane. The eye 
 (height 2') lies in a perpendicular from the nearest 
 corner of base and is 8' in front of the perspective 
 plane. Find the perspective. 
 
 21. Find the perspective in problem 20 when the 
 height of the eye is 5', other dimensions remaining 
 fixed. 
 
 22. Find the perspective in problem 20 when the 
 long face of base makes 45 with the perspective 
 plane, other dimensions remaining as in problem 20. 
 
 23. Find the perspective in problem 20 if the 
 height of the eye is i', other dimensions remaining 
 the same. 
 
 24^ Eight cubes are the corners of a larger cube 
 of 12' edge. One face of the cubes is parallel to the 
 perspective plane and 2' from it. The eye (height 
 
BACKBONE OF PERSPECTIVE. 17 
 
 6') lies in a perpendicular to GL from the central 
 line of the plan and is 12' from the perspective 
 plane. Draw the perspective of the cubes if edge 
 =2'. 
 
 25. A framework whose outer dimensions form 
 a cube 12' edge is composed of pieces i'xi' along 
 each edge of the cube. One corner of the cube lies 
 in the perspective plane and the plane of one face 
 makes 30 with it. If the eye (height 6') lies in a 
 perpendicular to GL through the corner in the per- 
 spective plane and is 15' from it, draw the perspec- 
 tive of the framework. 
 
 26. Draw the perspective of the framework in 
 problem 25 when the eye is moved 2' to the right, 
 other dimensions remaining the same. 
 
 27. Draw the perspective of the framework in 
 problem 25 when the height of the eye is 4', other 
 dimensions remaining the same. 
 
 28. Draw the perspective of the framework in 
 problem 25 when the front face of the cube lies in 
 the perspective plane and the eye (height 8') lies in 
 the central line of plan that is perpendicular to GL. 
 Other dimensions are the same as in problem 25. 
 
CHAPTER TWO. 
 
 Vanishing Point Method. 
 
 1 8. Vanishing Points. To find the perspective 
 of a line AB in Fig. 12 whose height above H is h, 
 we can find the perspective of A and B as before. 
 But if one of the points is on eE" or near it the solu- 
 tion by the former method lacks exactness. It is 
 
 Figure 12. 
 
 advisable to find the perspective of points on AB 
 favorably located. One of these favorable points 
 is K where AB cuts the perspective plane at a height 
 h above GL. The second favorable point should be 
 taken as far from eE" as the limits of the paper will 
 permit. Let Q be such a point. The perspective 
 
 18 
 
BACKBONE OF PERSPECTIVE. ig 
 
 Q' of Q is found as follows : Drop a perpendicular 
 from q on G'L', cutting it at P' ; lay off P'Q" equal 
 to the true height of point Q. Join Q" and E" ; join 
 eq cutting GL at D and draw DD' perpendicular to 
 GL, cutting E"Q" at Q', the perspective of Q. K'Q' 
 is the perspective of the line desired. The perspec- 
 tives of A and B are found by joining A and B to 
 e and by dropping perpendiculars from the points 
 where these lines cut GL to cut the line K'Q'. 
 
 The line E"F is drawn parallel to GL and G'L'. 
 
 In the similar triangles E"SQ' and E"FQ", 
 
 FQ" E"F* 
 But E"S=OD', E"F=OP'. 
 
 = 
 
 FQ" OP 7 ' 
 
 Let t = height of eye above H = OE". 
 h = height of AB above H = P'Q". 
 x = OP' = distance from eye to the true 
 
 height line of Q. 
 
 z = OD' = distance from eye line to the ver- 
 tical through the perspective of the 
 point. 
 
 y = SQ' = height of perspective of point 
 from horizontal through E", called the 
 horizon. 
 
 Now, let Q move along AB to an infinite distance 
 from K. The line eq that is drawn from q to e 
 
20 BACKBONE OF PERSPECTIVE. 
 
 will then be parallel to AB and will cut GL at L, 
 and D' will move to L'. But x becomes infinity, 
 
 Therefore the perspective of the point on AB that 
 is at infinity is zero distance from E"V in LL'. The 
 line E"V is called the horizon. 
 
 Corollary : The perspectives of all lines parallel to 
 AB pass through V, and are said to vanish at V. 
 
 From these results we derive the following rule 
 for obtaining the vanishing point of a system of 
 parallel lines: 
 
 Rule. Through the plan (e) of the eye draw a 
 line parallel to the lines zvhose vanishing point is 
 desired. From the point where this line cuts the 
 perspective plane, GL, drop a perpendicular to cut 
 the horizon at V. 
 
 Thus the vanishing point is absolutely indepen- 
 dent of the height of the line above H and of the 
 position of its plan. The only controlling factors 
 that locate the vanishing point are the angle the 
 system of lines makes with GL and the position of 
 the eye. 
 
 eg. Perspective of Cross. Given the side and 
 end views of a cross and base as shown in Fig. 
 13. Let abed be the plan of the base and let the 
 perspective plane pass through the corner a. 
 Through e draw lines parallel to the sides of the 
 base, cutting GL at G and L. From these points 
 draw perpendiculars to GL, cutting the horizon 
 line at V and V. As the corner of the base is in 
 the perspective plane, lay off A'F' equal to the 
 
BACKBONE OF PERSPECTIVE. 
 
 21 
 
 height of base and join A' and F' to V and V. 
 Join b and d to e, cutting GL at i and 4, and from 
 these points draw perpendiculars to GL, cutting 
 F'V and F'V at B' and D'. The base is thus 
 defined. 
 
 To draw the perspective of the arms of the cross 
 produce one side to cut the perspective plane at M. 
 
 Figure 13. 
 
 Then MM' is the true height line for all lines in 
 this face of the arm. Lay off M'J' and M'H' equal 
 to the true heights of the upper and lower surfaces 
 of the arm. Join J'V and H'V. Join 5 and 6 to e 
 and from the points where these lines cut GL draw 
 perpendiculars to GL. The intersection of these 
 perpendiculars with J'V' and H'V'will define the 
 front face of the arm. 
 
 To draw the upright, lay off the height of the 
 base equal to M'Q' and height of top=M'R' on M'M 
 
22 BACKBONE OF PERSPECTIVE. 
 
 from M'. Join these to V and join s, k and t to e, 
 and from the points of intersection of these lines 
 with GL draw perpendiculars to GL, cutting R'V 
 in K' and S'. Draw K' V and join t to e and from 
 point where et cuts GL drop perpendicular to cut 
 K' V at T 1 . The rest can be drawn in the same 
 way. 
 
 PROBLEM 29. Construct the perspective of the 
 cross in Fig. 13 when side of base = 7', end of 
 base = 5', length of crossarm = 6', other dimen- 
 sions as in Fig. 20. 
 
 20. Perspective of a House. Let one corner a 
 of the plan abed (Fig. 14) be in the perspective 
 plane. Through e draw lines parallel to ab and ad, 
 cutting GL in L and G. From these points drop 
 perpendiculars, cutting the horizon in V and V, 
 the vanishing points for the systems of lines parallel 
 to ab and ad, respectively. Draw a line from a to e. 
 The point A' where it cuts G'L' will be the perspec- 
 tive of a. Join A'V and A'V and from the points 
 where eb and ed cut GL drop perpendiculars cut- 
 ting A'V and A'V in B' and D', the perspective of 
 the points b and d. Lay off the true heights of all 
 points in the visible sides of the house on vertical 
 A'a from G'L', and the perspective of these points 
 will lie on the line joining their proper height to 
 their vanishing point. To draw the window pq, lay 
 off the height of the top and bottom of window 
 from A' on aA' at K' and T'. Join T' and K' to V, 
 and from the points where ep and eq cut GL drop 
 perpendiculars to cut K'V and TV in P', M', Q', 
 and O'. To locate the roof, extend the comb to cut 
 the GL at U, and drop a perpendicular from U on 
 
BACKBONE OF PERSPECTIVE. 23 
 
 G'L', cutting it at U'. Lay off on UlT from U' the 
 true height of the comb, equal to U'x, and join x to 
 V. Join the ends of the comb to e, and where these 
 lines cut GL, drop perpendiculars, cutting the line 
 xV in Z' and Y', the perspective of the ends of the 
 ridge. To find the perspective of the eaves, extend 
 the plan of the eave line to cut GL say at N, and 
 
 Figure 14. 
 
 from N' lay off true height of eave line equal to 
 N'F', and join F' to V. From the points where er 
 and es cut GL, drop perpendiculars, cuting F'V in 
 R' and S' the perspectives of r and s. In the same 
 way the perspectives of the other eave-lines can be 
 found. 
 
 PROBLEM 30. Construct the complete perspective 
 of the house in Fig. 14 when AB = 6', AD = 4', 
 height of eaves = 4', height of comb = 6j^', height 
 of eye = 2', distance from eye to perspective plane 
 = 12', other dimensions remaining as in Fig. 14. 
 
24 BACKBONE OF PERSPECTIVE. 
 
BACKBONE OF PERSPECTIVE. 25 
 
 21. Architectural Perspective. In finding the 
 perspective of a house, it is convenient to let one 
 corner of the house pass through or lie in the per- 
 spective plane. Thus, if is it desired to construct 
 the perspective of the house whose side and front 
 elevations are marked "front" and "side" (Fig. 15), 
 it will shorten the work if we let a corner common 
 to the two views given lie in the perspective plane 
 GL. It is always best to take e as low as the size 
 of the drawing sheet will permit. After e is locat- 
 ed draw lines through e parallel to the sides of the 
 house, cutting GL in A and B, and then drop per- 
 pendiculars from A and B to the horizon line 
 through E", cutting the horizon in V and V. The 
 corner that lies in the perspective .plane will be the 
 true height line for all points lying in the planes of 
 the two faces or sides seen in the two views. These 
 heights are laid off from G'L' on verticals through 
 E" and joined to V or V. To find the perspective 
 of any horizontal line like the ridge or comb pq, 
 we produce pq to cut the perspective plane at F 
 and from F drop a perpendicular cutting G'L' at F'. 
 On line F'F lay off the true height of the ridge equal 
 to F'D. Join D and V, and join e and p, cutting 
 GL at G, and from G drop a perpendicular, cutting 
 DV at P', which is the perspective of p, or the left 
 end of ridge. In the same way we can find T', the 
 perspective of t. Join T' to V and where it cuts 
 DV will be the perspective of q. Other horizontal 
 lines can be found in the same way. If a line is 
 not horizontal in the building it is best to find the 
 perspective of each end separately by the projective 
 method. 
 
CHAPTER THREE. 
 
 Axometric Projections. 
 
 22. In order to show the different parts, con- 
 nections and relations of a framework, it is often 
 desirable to take its projection on some plane not 
 parallel to any of the plane faces. The basal planes 
 of most frameworks are composed of a series of 
 surfaces each of which is at right angles to the 
 other two. Three axes each at right angles to the 
 other and parallel to the edges of the framework 
 can be drawn, and all lines can then be located with 
 references to these axes. 
 
 23. Axes. Let PA, PB, and PC in Fig. 16 be 
 the axes in space each at right angles to the other 
 two, and let these axes be _ 
 projected normally on 
 
 any plane ABC in KA, 
 KB, and KC. Let PAK 
 = a, PBK = b, PCK = 
 c. Let AK, BK, and CK 
 be produced to intersect 
 BC, AC, and AB in D, E, 
 and F. The lines KA, 
 KB and KC are the axes 
 of projection. 
 
 Then as PA and PK are at right angles respec- 
 tively to the planes BPC and ABC, the plane of 
 these two lines is at right angles to the intersection 
 of the two planes BPC and ABC. That is, APD 
 is at right angles to BC. Hence AD and PD are 
 
 26 
 
BACKBONE OF PERSPECTIVE, 27 
 
 the altitudes of the triangles ABC and BPC. Sim- 
 ilar relations are true of BE and CF. 
 
 The right angles BPC, APC, and APB are pro- 
 jected in the angles BKC, AKC, and AKB, which 
 we represent by x, y, and z. 
 
 24. Reduction Cosines. The axis PA has been 
 projected in KA and its length has been reduced 
 from PA to KA. But in the right triangle PAK 
 we have 
 
 KA = PA cos a. 
 Similarly 
 
 KB = PB cos b. 
 KC = PC cos c. 
 
 Thus each axis is reduced" in the ratio of the 
 cosine of the angle it makes with the plane of pro- 
 jection. All lines that are parallel to PA, PB, and 
 PC will, when projected on the plane ABC be 
 reduced in the ratio of the cosines of a, b, and c, and 
 these are therefore called the reduction cosines. 
 
 25. Fundamental Formula. In the right tri- 
 angles PBK and PCK of Fig. 16 we have 
 
 PK h 
 
 PK __ h_ 
 ~ PC ~PC 
 
 where PK = h 
 
 Squaring and adding, 
 
 h 2 h 2 PB 2 +PC 2 
 
 ' ' 
 
 In the right triangle PBC we have 
 PB 2 PC 2 = BC 2 
 
28 BACKBONE OF PERSPECTIVE. 
 
 Also as expressions for the area M of PBC we 
 have the following : 
 
 PBxPC PDxBC 
 
 or PB 2 PC 2 = PD 2 BC 2 
 Substituting in (A) gives 
 
 BC 2 h 2 
 
 = h2 pD^B^ = PD* ..... (B) 
 
 But in the right triangle APD, DPK = PAK=a, 
 and cos DPK = = cos a> Putting this value of 
 
 (B) we get 
 
 sin 2 b + sin 2 c = cos 2 a 
 . . sin 2 a -j- sin 2 b + sin 2 c == i 
 or, cos 2 a + cos 2 b + cos 2 c = 2 (i) 
 
 26. Angles between Axes. In the triangle 
 CKD of Fig. 16 we have 
 
 (C) 
 
 Now DKC 180 y, and in the right triangles 
 PKD and PKC, 
 
 DK = PK tan DPK = h tan a 
 
 CK PK h 
 
 tan PCK tan c 
 
 Substituting these values of PK and CK in (C) 
 we get 
 
 cos ( 180 y ) = tan a tan c 
 . . cos y = tan a tan c 
 
OF 
 
 ^BACKBONE 
 
 OF PERSPECTIVE. 
 Similarly 
 
 cos x = tan b tan c 
 cos z = tan a tan b 
 PROBLEM 31. In the right triangle DKC, sin 2 y = 
 DC 2 PC 2 -PD 2 . h _^ h 
 
 \^t . Jr LJ = 
 
 sin c cos a 
 
 (2) 
 
 sin b 
 
 KC 2 KC 2 
 
 and KC = h cot c, then show that cos y 
 
 cos a cos c 
 
 PROBLEM 32. Given cos a = 2/3, cos b = y 2 , 
 find the cosines of x, y, and z. 
 
 27. Relation of Reduction Angles. It is as- 
 sumed that the axes make 
 some definite angles with 
 the plane of projection. 
 These angles must lie be- 
 tween o and 90. As 
 neither of the cosines can 
 be o or unity, it follows 
 that the square of either 
 cosine must be less than 
 unity; and therefore that 
 the sum of the squares of Figure 17. 
 
 the other two cosines must be greater than unity. 
 If a = 32, then 90 a = 58. Now if we assume 
 that b = 58, we have 
 
 cos 2 a + cos 2 b = cos 2 32 + cos 2 58 = 
 
 cos 2 32 + sin 2 32 = i. 
 
 But as cos 2 a -f cos 2 b + cos 2 c =2, 
 
 then cos 2 c = i , . . cos c = I , . . c = a 
 This is contrary to hypothesis. Now as the cosine 
 of an angle decreases as the angle increases, we see 
 that for possible values b must be less than 58. It 
 
BACKBONE OF PERSPECTIVE. 
 
 can have any value between o and 58. Thus none 
 of the angles a, b, or c can be equal to or greater 
 than the complement of either of the others. 
 
 Case I : If a = 90 b, then cos 2 a + cos 2 b = 1 
 
 .'. cos 2 c = 1 .'. c =0. 
 
 One axis is parallel to 
 the plane of projection 
 and the plane of the oth- 
 er two perpendicular to 
 the plane of projection. 
 This is the case of ordi- 
 nary plans and eleva- 
 tions, and is not treated 
 here. Figure IT. 
 
 Case II: b>90-a. Then cos'a + cos 2 b<l 
 ."-. cos 2 c>l. c is impossible. 
 
 Case III: If b<90 a, then cos 2 a + cos 2 b>l 
 .'. cos 2 c<l, c is possible. 
 
 28. Application. If we assume the angles a, 
 b, and c, then each dimension of the structure paral- 
 lel to PA, PB, and PC will be reduced in the ratio 
 of the cosine of a, b, and c. 
 
 EXAMPLE. Let cos a = Jv's", cos b = Vf , then 
 
 cos c = v/|' . 
 sin a = J, sin b = Vf , sin c =V^ . 
 
 tan a = ^ , tan b=v / r, tan c =v i . 
 
 cos x = tan b tan c = V^. 
 cos y = tan a tan c = v^. 
 cos z = tan a tan b = v~f . 
 
BACKBONE OF PERSPECTIVE. 31 
 
 log cos x = l / 2 [log 5 log 21 ] 
 
 = l /2 [0.698970 I.3222I9] 
 
 = l /2 [19.376751 20] = 9-6883755 
 .-.X= 119 12' 21" 
 
 log cos y = y 2 [log i log 21 ] 
 = ' l /2 [01.322219] 
 
 = l /2 [18.67778120] = 9.3388905 
 
 ..y=i0236'i6" 
 
 log cos z = y 2 [log 5 log 9] 
 
 = l /2 [0.698970 .954243] 
 = V* [19744727 20] =9.8723635 
 . -.2=138 II' 2 3 " 
 
 Check : x + y + z = 360 
 
 Now, all dimensions of the structure will be re- 
 duced in the ratios of cos a (= .8660), cos b 
 (=. .6124), and cos c ( = -9354), and a separate 
 scale will have to be made for each axis. 
 
 29. Reduction Ratios. Instead of assuming 
 the angles, we can assume their ratios. Thus 
 
 Let cos a = lp, cos b = mp, cos c = np. 
 
 cos a : cos b : cos c : : 1 : m : n. 
 
 Then from (i) 
 
 J2 p 2 + m 2 p 2 + n 2 p 2 = 2 
 
 2l 2 ,, 2m 2 
 
 cos 2 a = r^ ; ; , cos 2 b = 
 
 2n 2 
 
 COS 2 C = 
 
 From these we can find tan a, tan b, and tan c, and 
 then x, y, and z. 
 
32 BACKBONE OF PERSPECTIVE. 
 
 30. Practical Application. We can, instead of 
 using the incommensurable fractions represented by 
 the cosines of a, b, and c, use their ratios, 1, m, and 
 n, and thus get the relative dimensions. The object 
 of axometric projections is to show the connections 
 and relations more fully and completely than ordi- 
 nary projections or perspectives would do. It is 
 therefore allowable to vary these ratios proportion- 
 ally. It will be far more expeditious and far more 
 satisfactory to assume the ratios 1, m, and n, and find 
 the values of x, y, and z from these. 
 
 31. Systems. There are three general sys- 
 tems: 
 
 Isometric : 1 : m : n, where 1 = m n ; all equal. 
 Dimetric : 1 = m or n, or m = n ; two equal. 
 Trimetric : 1 : m : n ; all different. 
 In any system the ratios must fulfill formulas ( i ) 
 and (2). 
 
 32. Isometric System. The ratios are 1 : m : n 
 i : i : i. 
 
 2 
 
 P = 12 .2 . -2 = ? ' 
 
 cos 2 a = 2 / 3 , cos 2 b = 2 / 3 , cos 2 c = ft. 
 
 sin 2 a ft, sin 2 b = ft, sin 2 c = ft. 
 
 tan 2 a = y 2 , tan 2 b = ft, tan 2 c = ft. 
 
 cos x = tan b tan c = ft, . . x : 120. 
 
 cos y = tan a tan c = ft, . ' . y = 120. 
 
 cos z = tan a tan b ft, . ' . z = 120. 
 
 33. Isometric of Stone Cap. In the isometric 
 system, the line OX is drawn vertically, and then 
 
BACKBONE OF PERSPECTIVE. 
 
 33 
 
 the angles XOA and XOB are made equal to 120, 
 thus making AOB 120. Let OH equal the total 
 length of a stone cap, OB the total width and HG 
 the total height. Make OB = its true width, and 
 OC equal to its true length, and draw the parallelo- 
 gram OBDC. Then draw CF and DE parallel to 
 axis OA and make CF equal 
 to its true length, and draw 
 FE parallel to axis OB. To 
 locate the top GMLK, 
 through G draw lines paral- 
 lel to the axis OA and OB, 
 and make GM and GK equal 
 to their true lengths in the 
 stone cap, and draw lines 
 GK, ML, and KL. Then 
 join KF and EL, forming 
 the parallelogram FKLE. To draw the back face 
 AGMR, make OA equal to its true length and join 
 AG. Draw BR parallel to OA and AR parallel to 
 OB, intersecting at R. Join R with M, thus com- 
 pleting the figure. 
 
 PROBLEM 33. Draw an isometric of a cube 4' 
 edge that has a square hole (i'xi') connecting each 
 pair of parallel faces. 
 
 34. One-Half Dimetric System. In the dimet- 
 ric system two of the ratios are equal, but they can 
 bear a variety of relations to the third. 
 
 The one-half dimetric : (1 : m : n) ( I \y 2 : I ) 
 
 Figure 18. 
 
 ! 2 +m 2 
 
 __ 8 
 
34 
 
 BACKBONE OF PERSPECTIVE. 
 
 cos 2 a = |, cos 2 b = |, cos 2 c = | : 
 sin^a = J, sin 2 b = }, sin 2 c =]. 
 tan 2 a = J, tan 2 b = |, tan 2 c = . 
 .'. cos x = tan b tan c = J\/7~ 
 cos y = tan a tan c = J 
 cos z = tan a tan b = 
 
 As an example of the application of this system 
 
 Figure 19. 
 
 draw the dimetric of the monument in problem 20. 
 
 Draw in Fig. 19 a vertical line and make OA = i, 
 and draw AM at right angles to OA. With O as 
 a center and 8 as a radius, cut AM at M. Join OM 
 and produce in OC. 
 
 Now cos MOA = - = 
 
 /. cos COA = -J /. COA = y. 
 
 As x and z are equal, bisect the angle COA and 
 produce in OB, then COB = x, and BOA = z. Full 
 dimensions are laid off on OA and OC and one-half 
 
BACKBONE OF PERSPECTIVE. 35 
 
 of the real dimensions are laid off on OB. Thus 
 Oa and Oc are made equal to i' and 6' respectively 
 the height and length of the side of the base of 
 the monument, while Ob is laid off equal to one- 
 half of the real length (4') of the end of the base. 
 Through a, b, and c draw lines parallel to the axes 
 intersecting in e and f. Lay off ag and ke equal to 
 i' to the full scale, while ah and fn are laid off to 
 the half scale. Through g and k draw lines parallel 
 to the axis OB, and through h and n lines parallel 
 to OC, intersecting the former lines in p, q, and r. 
 Through p draw a vertical (parallel to OA) and 
 make ps = 3' (the height of the block). Then 
 through s draw lines parallel to the axes and make 
 su=^5', and st = 3' on the half scale. Complete 
 the parallelogram tsuv, the top of the block. Draw 
 the diagonals and from the point of intersection 3 
 lay off 34 = 2' and join 4 to s, t, u, and v. 
 
 35. Three-Fourths Dimetric. In this system 
 the same scale is used on two of the axes, while a 
 scale equal to three-fourths of the first is used on 
 the other one. 
 
 In the three-fourth system we have, 
 (1 : m : n) = (1 : J : 1) 
 
 cos 2 a = ff, sin 2 a = 9 T , tan 2 a = ^; 
 C os 2 b = if, sin 2 b = -|f , tan 2 b = \ f ; 
 cos 2 c = ff , sin 2 c = T 9 T , tan 2 c = -fa. 
 cos y = tan a tan c = A- 
 
 In Fig. 20 draw OA = 9 parts on some scale, 
 draw AM perpendicular to OA, and with O as a 
 
36 BACKBONE OF PERSPECTIVE. 
 
 center and a radius equal to 32, cut AM at M. Join 
 OM and produce in OC. Then bisect angle AOC 
 and produce the bisector in the line OB. Now, full 
 dimensions are laid off on the axes OA and OC and 
 three-fourth dimensions on OB. On axis OC make 
 O2 = 7", the full length of the side O2. Now, the 
 side 03 is equal in the original figure to eight 
 inches, but we lay off only three-fourths of this on 
 
 03 ; that is, we make 03 = 6". Draw lines 04, 25, 
 and 36 parallel to axis OA each equal to one inch, 
 and through 4, 5, and 6 draw lines parallel to the 
 axes, marking out the top of the base. The shaft 
 of the cross is i" by i" '. Make distance 48=3", 
 and draw 8g three-fourths of 3.5". Then through 
 g draw lines parallel to the axes and lay off dimen- 
 sions equal to one inch on axis parallel to OC and 
 to three-fourths of one inch on axis parallel to axis 
 OB. Make gc = 4", be = i", ba = i" , and through 
 these points draw lines parallel to the axes. Make 
 ax = i", xy = $4", be = 2^", etc. 
 
BACKBONE OF PERSPECTIVE. 37 
 
 PROBLEM 34. Draw the one-half dimetric of a 
 box without top, whose outside dimensions are 
 2'x3'x2^', the thickness of the material being 2". 
 
 36. The One-Third Dimetric. (1:1/3:1). We 
 have 
 
 n 2 = 
 
 cosa 
 
 2 a = j- 1 , sin 2 a = ^g , tan 2 a = 
 cos 2 b = T * 9 , sin 2 b = } j, tan 2 b = 
 cos 2 c =-}!, sin 2 c = 1*9 , tan 2 c = 
 cos x = tan b tan c = 
 cos y = tan a tan c = 
 cos z = tan a tan b = 
 
 The axes can be laid off as in Fig. 19, except that 
 while OA equals i, OM is 18. The drawing of a 
 structure in the one-third dimetric is made in all 
 respects like the one-half dimetric, except that one- 
 third is measured along the OB axis instead of 
 one-half. 
 
 PROBLEM 35. Show in the (m : i : m) dimetric 
 
 that cos y=--_. 
 
 PROBLEM 36. Show in the ( i : m : i ) dimetric 
 that cos y = . 
 
 PROBLEM 37. In the system ( I : % i ) find the 
 cosines of x, y, and z, and draw the cross in Fig. 13 
 in this system. 
 
 PROBLEM 38. In the system ( i : % : i ) show 
 that cos y = 2/9, and draw the monument in 
 
38 BACKBONE OF PERSPECTIVE. 
 
 Problem 20. 
 
 37. Trimetric System. (l:m:n). In this sys- 
 tem the ratios are all different, but must fulfill the 
 conditions of formula (3). 
 
 Find the angles between the axes for the ( l /z : % 
 : i ) system. 
 
 P 2 = H - * 
 
 cos 2 a = ^, cos 2 b = i, cos 2 c = f . 
 
 As no cosine can be greater than unity, the sys- 
 tem is impossible. 
 
 For the system (9 : 10 : n) we have 
 2= 2 2 
 
 81 + 100 + 121 
 cos 2 a = J||, cos 2 b 
 sin 2 a = Jf $, sin 2 b 
 tan 2 a = iff, tan 2 b 
 cos x == tan b tan c 
 cos y = -tan a tan c = -VH 
 cos z = tan a tan b = - 
 
 Log COS X=:^ [log IO2+log 60 log 2OO log 242] 
 log 102 = 2.008600 log 200 2.301030 
 
 log 60 = 1.778151 log 242 = 2.383815 
 
 3.786751 4.684845 
 
 .-. Logcosx = y 2 [3.786751 4.684845] 
 
 = J4 [19.101906 20] =9.550953 
 
 Log cos y=^2{log i4O+log 60 log 162 log 242] 
 log 140 = 2.146128 log 162 = 2.209515 
 
 log 60=: 1.778151 log 242 =^2.38381 5 
 
 3.924279 4-593330 
 
BACKBONE OF PERSPECTIVE. 39 
 
 Log cos y = y 2 [3-924279 4.593330] 
 
 = */2 [ I9.330949 20] = 9.6654745 
 
 .-.y= 117 34 25" 
 
 Log cos z= */< [log 1 40+ log 1 02 log 162 log 200] 
 log 140 = 2.146128 log 162 = 2.209515 
 
 log 1 02 2.008600 log 200 2.301030 
 
 4.154728 4.5*0545 
 
 Log cos z= y 2 [4.154728 4.5!0545] 
 
 = y 2 [19.644183 20] =9.8220915 
 
 2=131 35' 48" 
 Check: x + y + z = 360. 
 
 38. Example. Draw the perspective of a 4" 
 cube that has a hole i" square connecting each set 
 of opposite faces, the axis of the hole coinciding 
 with the axis of the cube, in the (4:5:6) system. 
 
 9 
 
 16 + 25 + 36 
 
 cos 2 a = f|, sin 2 a = 4f , tan 2 a = ||. 
 cos 2 b = -?4, sin 2 b = 14, tan 2 b = fj. 
 cos 2 c = , sin 2 c = - 5 , tan 2 c = -. 
 
 cos x = -tan b tan c = -Vfiff^ ' x = l l $' W 
 cos y - -tan a tan c = -VfUfj -'. Y == 108 12 r 36 /x 
 
 cos z - -tan a tan b - -VfUfj * z = iSO 37 r 27 X/ 
 Check : x + y + z = 360 0' 0". 
 
 This system is equivalent to (4/6:5/6:1). 
 Draw a vertical line for the axis OA (Fig. 20) and 
 draw OB making angle AOB = 108 12' 30", and 
 angle AOC= 150 37' 27". The angle BOC is then 
 = 101 9' 57". Now as full sizes are laid off on 
 OC and the dimension on OA must be reduced 
 
40 BACKBONE OF PERSPECTIVE. 
 
 by multiplying by 4/6, or 2/3, we can use the 4 o 
 scale on OC and the 6 o scale on OA, as this makes 
 the reduction for us. Similarly we could use the 
 5 o scale on OC and the 6 o on OB. The cube is 
 then easily drawn. 
 
 PROBLEM 39. Draw a cube in the (3 14 15) sys- 
 tem. 
 
 PROBLEM 40. Find the angles x, y, and z in the 
 (5 : 6 17) system. 
 
 PROBLEM 41. Find the angles x, y, and z in the 
 (10 : ii : 12) system. 
 
 PROBLEM 42. Find the angles x, y, and z in the 
 (8 : 15 : 17) system. 
 
CHAPTER FOUR. 
 
 Shades and Shadows. 
 
 39. Shade. An opaque body exposed to light 
 will be illuminated on the side next the light and 
 dark on the opposite side. Thus if an opaque 
 
 Figure 21. 
 
 sphere be exposed to a source of light P, a set of 
 rays of light from the source P will be tangent to 
 the sphere. If the source of light is a point P the 
 rays of light that are tangent to the sphere will 
 form the surface of a cone whose vertex is the 
 source of light P and which touches the sphere in 
 the circle of contact, ABC. The part ABCK of the 
 body, that does not receive any light, is in shade and 
 the line of the contact ABC is the curve of shade. 
 That part of space from which the source of light 
 cannot be seen is called the shadow in space. 
 
 40. Shadow. If light is excluded from a sec- 
 ond body MN, that part of the surface of the second 
 body from which the source of light cannot be seen 
 is said to be in shadow, and the dark part is called 
 shadow. The line of contact of the second body 
 
 41 
 
42 BACKBONE OF PERSPECTIVE. 
 
 with the cone of rays tangent to the first is called 
 the curve of shadow. 
 
 Thus in Fig. 21, P is the source of light, ABCK 
 is an opaque sphere and MN is another body that 
 intersects the shadow cone in the curve of shadow 
 DEFG. The area DEFG is called the shadow area 
 of the sphere on the body MN. 
 
 41. Drawings in Projection. The ordinary 
 drawings of structures consist of a side and an end 
 view called elevations, and a top view called the 
 plan. If the structure is a house, we ordinarily 
 have the end elevation, side elevation, and a rear 
 elevation. 
 
 In 'addition to this we have the ground plan and 
 roof plan and these elevations and plans define the 
 structure completely, from which we get a concep- 
 tion of the structure as a whole. 
 
 Projective drawings are referred to a horizontal 
 plane (called H) and a vertical plane (called V), 
 while their intersection is called the ground line 
 (G. L.). The phrases "horizontal projection" and 
 "vertical projection" will hereafter be abbreviated 
 into hp and vp respectively. 
 
 42. Shadow of Points. The shadow of a point 
 on a plane is where the ray of light through the 
 point cuts the plane. If the point is a material point 
 it is supposed to intercept the light and its shadow 
 will appear as a dark spot on the plane. 
 
 43. Directions of the Rays of Light. The con- 
 ventional direction of any ray of light is such that 
 the elevation and plan of the ray make 45 with 
 the ground line. The ray of light is supposed to 
 come over the left shoulder as we face the drawing 
 
BACKBONE OF PERSPECTIVE. 43 
 
 in such a way that its projections make 45 with 
 the ground line (G. L.). 
 
 44. Angle with H or V. In Fig. 22, let M be 
 a plane perpendicular to H and V cutting H in LA 
 and V in LB. 
 
 PQ is a ray of light whose hp is pQ, vp is p'Q 
 and whose projection on M is LR. The lines pQ 
 
 Figure 22. 
 
 and p'Q make 45 with GL. .-. CQp'=45 = 
 CQp. But if angles CQp and CQp'=45> then 
 
 Let CQ = a, then Cp = a 
 
 *- PQ 
 Tangent 
 
 .'. PQp = 35 16' 
 
 Now R is the projection of P on M, and PR is 
 parallel to GL and perpendicular to M. Hence LR 
 is the projection of the ray PQ on M. Draw RA 
 and RB J. to H and V. As GL is J- to M, pA and 
 p'B will be parallel to GL. 
 
 Cp' = LB 
 But CQ = Cp = Cp' 
 
 .-. LA = LB or LA = RA 
 
 .-. The angle RLA is 45 
 
44 
 
 BACKBONE OF PERSPECTIVE. 
 
 Hence the projection of the ray of light on a plane 
 M J- to H and V makes an angle of 45 with lines 
 j. to H and V, or with new ground line LA. 
 
 45. Shadow of Vertical Rod. a. Shadow all 
 on H. Let cd-c'd' Fig. 23 (a) be the projec- 
 tions of a vertical rod. To find the shadow of a 
 point C on H, draw rays through c and c', making 
 45 with GL. Find the horizontal trace at C'. 
 
 
 c' 
 
 
 
 \ 
 
 c' 
 
 
 \ 
 
 J\ 
 
 P' 
 
 3 d 1 
 
 X 
 
 > \ c 
 
 \ J^ 
 
 
 V # 
 
 D' 
 
 
 \ 
 
 \ 
 
 d 
 
 b (/ 
 
 
 . 
 
 (a) P 
 
 r x7^ 
 
 
 C 
 
 / 
 
 
 Figure 24 
 
 
 Figure 23. 
 
 The point D is in H and is its own shadow. The 
 shadow of the rod will be the heavy line D'C'. 
 
 b. Shadow on H and V. Through cc' draw 
 projections of a ray making 45 with GL and find 
 the vertical trace at C'. Find the shadow P' of 
 some point P on the rod. Join P'c and produce to 
 cut GL at 3 ; then join C'3. The broken line C'3C 
 is the shadow. The part C'3 on V makes 90 with 
 GL, while the part C3 on H makes 45. 
 
 46. Horizontal Lines. a. Shadow on V. 
 Let cd-c'd' be the projection of a line perpendicu- 
 lar to V. Fig. 24 (a). Draw rays through c and 
 
BACKBONE OF PERSPECTIVE. 
 
 45 
 
 c' and find shadow of C at C'. In the same way 
 find the shadow of D at D'. Connect C' and D', 
 giving the shadow CD'. 
 
 b. Shadow on H and V. Fig. 24 (b). Find 
 the shadow of point C at C' and D at D', also find 
 the shadow of P at P'. Join C'P' to cut GL at 3 
 and then join 3D'. The broken line D'3C' is the 
 shadow required. 
 
 47. Shadow of Any Line. Given a line ab-aV. 
 
 i . Find the 
 shadow of A at 
 A', and shadow 
 of B at B'. 2. 
 Take any inter- 
 mediate point P 
 on the line AB 
 
 P 
 
 Figure 25. 
 
 Join A' and K'*. 
 
 and find its shad- 
 ow at P'. Produce 
 B'P' to cut GL at 
 
 K'. Join A' and K'-. The broken line A'K'B' is the 
 
 shadow of the line AB. 
 
 48. Shadow of a Shelf. Let a'b'c'f be the ver- 
 tical projection of the shelf and aebc the horizontal 
 projection. 
 
 1. Find shadow of the line 
 cb-c'b'at C'B'; 
 
 2. Find shadow of line ab- 
 a'b' at A'B'; 
 
 3. Find shadow of a line 
 cf-c'f in c'C'. 
 
 .-.Area e'A'B'C'c'b'e' is 
 the shadow of the shelf on V as shaded in Fig. 26. 
 
 49. Shadow of Block on H. Let abed be the 
 
46 BACKBONE OF PERSPECTIVE. 
 
 plan of the top and a'b'c'f be the elevation of a 
 rectangular block resting on H. 
 
 1. Find shadow of line 
 bc-b'c' at B'C' ; 
 
 2. Find shadow of line 
 cd-c'd' at CD' ; 
 
 3. Find shadow of line 
 df-d'f at DM ; 
 
 . 4. Find shadow of line 
 be-b'e' at B'e. 
 
 The area bB'C'D'dcb 
 
 (shaded) is the shadow of 
 
 block on H. Fi * ure 27 - 
 
 50. Shadow of a Circular Disc on H. Given 
 the projections ef-e'f of the circular disc whose 
 plane is parallel to H and perpendicular to V. 
 
 The rays of light that 
 touch the circle form a 
 cylindrical surface of rays 
 which is cut by H parallel 
 to plane of circle. Hence 
 the section cut by H from 
 cylinder of rays is similar 
 to that cut by the plane 
 of circle. Thus the shad- 
 ow is a circle. To find this 
 shadow, find the shadow 
 of the center of the circle 
 at C'and with C as a cen- 
 Figure 28. ter and radius equal to 
 
 that of the circle draw the circle A'B'D', which is 
 
 the required shadow. 
 
 51. Shadow of a Horizontal Circular Disc. 
 
 
 
BACKBONE OF PERSPECTIVE. 
 
 47 
 
 cf b f 
 
 Given a circle abcd-a'b'c'd' whose plane is horizon- 
 tal and perpendicular to V. Find the shadow of a 
 series of points as 
 a-a', b-b', c-c, d-d/ 
 at A', B', C', D', etc. 
 Through these 
 points draw a curve. 
 In this case the 
 shadow falls on V 
 and forms an ellipse. 
 
 52. Construct the 
 shadow of a hori-- 
 zontal circular disc 
 where its center is 
 equidistant from H 
 and V. Let abcd-a'b' d 
 c'd' be the projec- 
 tions of the circular 
 
 d' c' o' a' b' 
 
 Figure 29. 
 
 the 
 the 
 
 disc. Find 
 shadow of 
 two semi-circu- 
 lar halves abc 
 and adc. The 
 part of shadow 
 that falls on H 
 is a circular 
 curve, while that 
 on V is an ellip- 
 tical curve. First 
 find the shadow 
 of center o-o' at 
 O' and with cen- 
 ter O' and radius equal to oa draw semicircle B'A' 
 
48 BACKBONE OF PERSPECTIVE. 
 
 K'. Then construct elliptical part B'C'K' by points. 
 53. A circular disc 2" in diameter touches H at 
 a point 3" in front of V and its plane is perpen- 
 dicular to V. Construct its shadow. Let abed- 
 
 Figure 31. 
 
 a'b'c'd' be points on the circumference of the cir- 
 cular disc. Find the shadows of these points at 
 A'B'CD', etc., and sketch a curve through these 
 points. The shadow will be an ellipse. If we have 
 vp (b') of a point B, the hp can be found by draw- 
 ing the circle O. Draw b'B" parallel to GL to cut 
 circle and diameter at B" and K. Lay off KB" 
 
BACKBONE OF PERSPECTIVE. 
 
 49 
 
 from a on ac making ab KB". In the same way 
 the projections of the other points can be found. 
 
 54. The plane of a circular disc 2" in diameter 
 is parallel to V and perpendicular to H. Its center 
 is 3" from V and i" from H. Construct its shadow. 
 
 (See Fig. 32.) Locate points abcd-a'b'c'd' and find 
 their shadow at A', B', C', D', etc. Draw the curve 
 through them, defining the ellipse of shadow. 
 
 55. Shadow of a Chimney. In Fig. 33 let 
 epqf be the plan of a hip roof and p'e'f's' the eleva- 
 tion, while abc and a'b'c' represent the plan and 
 elevation of a chimney. The line ab-a'b' is perpen- 
 dicular to V and its shadow on H will be in the line 
 ef, which is found by drawing a'e' through a' at 
 
50 BACKBONE OF PERSPECTIVE. 
 
 45 with GL. Produce e'e to cut the eaves of roof 
 at e and f. If the line AB were indefinite in extent, 
 its shadow on the roof would pass through e and f. 
 The shadow of AB will cut the comb of the roof at 
 o-o' where a'e' cuts the comb s't'. Then the shadow 
 
 Figure 33. 
 
 of AB will lie in the lines oe and fo. Through a 
 and b draw rays, ag and bh, making 45 with GL 
 cutting oe at g and fo at h. The broken line goh 
 will be the shadow of ab-a'b' on the roof. The 
 shadow of the line bc-b'c' will be parallel to GL and 
 hence will be mh; while the shadow of the vertical 
 corner ak-a'k' will be ag ; and that at C will be cm. 
 The shaded area agohmcba will be the plan of the 
 
BACKBONE OF PERSPECTIVE. 51 
 
 shadow, and k'g'o's't' will be the elevation of the 
 shadow. 
 
 56. Shadow of Cap on Cylinder. Given a ver- 
 tical cylinder 2" diameter and a cap 3^' ', required 
 
 Figure 34. 
 
 to find shadow of cap on the cylinder. See Fig. 34. 
 Let abc-a'b'c' be the projections of the cap and 
 efg-e'f'g' the projections of the cylinder. Draw 
 diameter eh parallel to GL, and from e draw ray ea 
 at 45 with GL. Mark a' vertically above a. Draw 
 a' A' at 45 with GL, cutting a vertical through e at 
 A'. Draw any intermediate ray through bb', cutting 
 surface of cylinder at f. Locate b' vertically above 
 b, and draw b'B' at 45 with GL to cut vertical 
 through f at B'. The extreme point of shadow 
 
52 BACKBONE Ob PERSPECTIVE. 
 
 curve will be above g where Og makes 45 with 
 GL. Draw gc perpendicular to Og, cutting rim of 
 cap at c. Find c' vertically above c and draw c'C 
 at 45 with GL to cut vertical through g at C'. 
 The line A'B'C' is the shadow line. The rest of the 
 visible part of the cylinder to right of vertical 
 through g is in shade as indicated by area h'g'. In 
 the same way the part of cap to right of vertical 
 through k is in shade. 
 
 57. Shadow of a Straight Line with the Pro- 
 jections Perpendicular to GL. Let ab-a'b' be 
 
 the projections of a straight line where ab and 
 a'b' are perpendicular to GL, in Fig. 35. Find the 
 shadow of A at A' and B at B'. If the line AB is 
 revolved around its vertical projection into the ver- 
 tical plane, it will appear at AB and its horizontal 
 trace and vertical traces will be D and E respectively. 
 If these points D and E are taken back to their true 
 
BACKBONE OF PERSPECTIVE. 
 
 53 
 
 positions they will appear at D' and E' and each 
 point will be its own. shadow. Join D'A' and pro- 
 duce to cut GL at C and then join CB'. The broken 
 line A'CB' will be the required shadow. The shad- 
 ows of all lines parallel to AB will be parallel to 
 A'C and B'C. 
 
 58. Shadow of Wall on Steps. Let i, 2, 3, 4, 
 5, 6, 7 be the end view of a series of steps and ABC 
 
 i ! ! 
 j -B-H-- i- +-H 
 
 / I 
 
 that of a wing-wall, in Fig. 36. The vertical and 
 horizontal projections are shown above and below 
 GL. Now the shadow of a line AB is found as in 
 Art. 57 in the broken line E'KD'. That part D'c 
 is on H and at point c the shadow strikes the verti- 
 
54 BACKBONE OF PERSPECTIVE. 
 
 cal face of the first step. The shadow on the 
 vertical face of the first step will pass through c' 
 and be parallel to KE' ; i. e., c'e' will be this shadow. 
 Through e, the horizontal projection of e' draw ef 
 parallel to D'K. The shadows on the other steps 
 will be parallel to c'e' and ef respectively. The 
 shaded areas represent the shadows on the vertical 
 and horizontal planes of the steps. 
 
 59. Shadow on a Semicylinder. Let abc-a'b'c' 
 represent the horizontal and vertical projections of 
 
 a semicylindrical sur- 
 face. (See Fig. 37.) 
 Required to find the 
 shadow of the rim on 
 the cylindrical surface. 
 Through a draw ad at 
 45 with GL to cut the 
 cylindrical surface at d 
 and through a' draw 
 a'A' at 45 with GL to 
 cut vertical through d 
 at A'. To find the 
 shadow of any inter- 
 mediate point C, through c draw C3 at 45 with GL 
 to cut surface of cylinder at 3. Through c' draw 
 c'C' at 45 with GL to cut vertical through 3 at C'. 
 If Ob makes 45 with GL, b' is a point in shadow 
 line. Thus the shadow of rim is A'C'b'. The area 
 a'b'A'd'G is the required shadow. 
 
 60. Shadow of Half-Cylinder on H. Given 
 abed and a' c', in Fig. 38, the horizontal and ver- 
 tical projections of a half-cylinder whose axis is 
 parallel to GL. It is required to construct the 
 
BACKBONE OF PERSPECTIVE. 
 
 55 
 
 shadow cast by the cylinder on its curved surface 
 and on H. The shadow of the semi-circle ab-aV 
 will be a semi-ellipse A'OB' which is found as in 
 Art. 53. That of cd-c'd' is similarly found, but 
 
 only that part C'E' will be seen. The shadow of 
 the left end on the interior surface of the cylinder 
 will be found as in Art. 59. The shaded area D' 
 Oad is on the cylinder while A'gOB'C'E'b is on H. 
 61. Shadow of Cone. Let Fig. 39 represent a 
 right cone whose axis is vertical with base resting 
 on the horizontal plane. Find the shadow of the 
 vertex at A' and from A' draw A'b and A'c tan- 
 gent to the circle bee. The area A'cebA' will be 
 the shadow that the cone casts on H. The area 
 cebac is the horizontal projection of the shade cast 
 by the cone on its own surface, while a'b'f is the 
 
56 BACKBONE OF PERSPECTIVE. 
 
 vertical projection of that part of the shade that is 
 visible. Fig. 40 shows how the shadow of the cone 
 falls when the shadow of the vertex falls on the 
 vertical plane. Draw the rays of light through the 
 vertex at 45 with GL. The horizontal trace of the 
 ray will be at A" and its vertical trace at A'. From 
 
 Figure 39. Figure 40. 
 
 A" draw lines A"b and A"c tangent to the circle. 
 Join the points where A"b and A"c cut GL to the 
 vertical trace A'. The shaded area A'ceb will be 
 the shadow of the cone on the horizontal and ver- 
 tical planes. The areas ceba and a'b'f will be the 
 horizontal and vertical projections of the shade 
 the cone casts on its own surface. 
 
OF ** ***** 
 
 THIS BOOK ON THE DATE * n"** T ' 
 WILL 'NCREASETO 50 CE N?S n E PENAL TY 
 DAY AND TO $, OO ON TM* THE FOU *T H 
 OVERDUE. ' THE SEVENTH DAY 
 
 21-loo m -8,'34 
 
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 VD ! r-o 
 
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