

GEOMETRICAL PERSPECTIVE. 



9bJ 





I from PS as a centre. Now thin horizontal plane through tho 

 ye, K, is turned up upon the perpendicular or picture-plane ; 

 mark the course of the dotted arc from K to x*, therefore PS >' 

 !a equal to PS B. Thus, it will be seen, two planet are re- 

 duced to one. The result is shown al*> in fig. 22. Now we 

 must turn the ground-plane upon the picture-plane. Pint, let 

 us repeat a remark or two made in Lesson II., Fig. 5 (page 225). 

 Tho lino from X to A (Fig. 21) is a visual ray cutting the picture- 

 plano inn; then is the picture of tho point A. The line from A 

 to B p, on tho ground-plane, in the ground plan of this visual ray; 

 therefore a perpendicular line from c (where A BP cut* the base 

 of the picture) drawn to tho line x A, determines B, the picture 

 of A, the object. The pupil will now perceive there ia a plane 

 perpendicular to the ground, and also to the picture-plane, 

 upon which the distinctive points x, p s, B, c, A, and s p are 

 placed ; but first make c A 1 equal to c A, as shown in the arc 

 from A to A- ; this brings the ground-plane upon the picture- 

 plane (in the same way as we turned x to x *). Compare 

 Fig. 22 with Fig. 21 ; A 1 will bo seen in both figures. From c, in 

 Fig. -2, draw the arc A* A s . This will be recognised in Fig. 21 ; 



Piy. 25 



and tho line A* to D E will also be seen in both figures ; there- 

 fore the line E B A in Fig. 21 is turned round upon the picture- 

 plane, and represented by A 3 B DE', shown also in Fig. 22. 

 Thus the perspective projection B of the point A on the ground 

 is determined. We have remarked (see Lesson II.) that points 

 are the extremities of lines ; and if we can determine the 

 positions of points in the picture, we can represent straight 

 lines by uniting these points. Our pupils will also recollect 

 that we have said, " all lines which are perpendicular to the 

 picture-plane have the point of eight for their vanishing- 

 point." Let these observations be borne in mind as we proceed. 



PROBLEM VIII. (Fig. 23). A straight line, A B, 5 feet long, is 

 perpendicular to the picture-plane lying on the ground, and 1 foot 

 from it; height of eye, 5 feet, and distance from, the picture-plane, 

 10 feet. Scale, 4 feet to the inch. 



Draw the line CAB perpendicularly to PP, make c A 1 foot, 

 and A B 5 feet. Draw HL parallel with PP, and 5 feet from 

 it. Draw PS E perpendicularly to H L, and 10 feet long. 

 From PS, with distance PS E, describe the semicircle DE 1 X DB*. 

 From c, and with the distance c A, draw the arc A D. From 

 , again draw the arc B r, join F and D with DE*, also draw 

 the line c PS. Between the intersections of c PB with the 

 lines from F and D to DE" will be the perspective of A B, 

 Viz., a b. Let the line H lie 5 feet long, and at an angle of 50 



thri>iet,tr,:],lane. From H and I draw perpendicular* to th 

 I.n-tun-- plane, and proceed with each extremity M wa* done witls 

 tho lino A B ; h i will be tho perspective of the given line M i. 



Tho following remark upon the line K I will refer to all th 

 lines similarly drawn that ia, perpendicularly to P P ; became the 

 line K i is perpendicular to PP, therefore the perspective reprc- 

 tentation of that lino U drawn to the point of night, viz., K t Ft, 

 and somewhere upon that lino U the position of I in i found firwt 

 by drawing from the centre K the are I L, and joining L wit* 

 DE 1 ; this last line cutting K PB in i, fixe* the peiipeotite view 

 of i ; tho same may be repeated for h ; i h being joined give* th* 

 perspective view of H I. 



PROBLEM IX. (Fig. 24). Draw the perspective vie* of 

 pavement composed of square slabs, the edges of which shall meamrc 

 1*5 feet s height of eye and distance as before. 



Let A B be the total width of all the slabs, and A 1, 1 2, 23, 

 3 4, 4 B, each equal to 1*5 feet. Draw lines from each of thete 

 divisions to the point of sight : upon the horizontal line set off 

 from PB to DE 10 feet. From each of tho divisions on A B, viz., 

 A, 1, 2, etc., draw lines to DE; where these lines intersect thoM 



drawn from the given divisions A, 1, 2, etc., will be found the 

 angles from which ore drawn the opposite sides of the squares 

 viz., 5, 6, 7, 8, 9. We must here observe, as will be seen in Fig. 

 24, that all lines which retire at an angle of 45 with theVP have 

 the distance-point for their point of sight. For if one side of t 

 square is parallel with the P P, the other side will be at right 

 angles with the P P ; therefore the diagonal of the square wiB 

 be 45 with p p. 



PROBLEM X. (Fig. 25). Draw the perspective visw ofasqvan, 

 the sides of which are 3 feet in length, 2 feet from the PP. and 

 one side at an angle of 50 with the PP. 



Draw a 6 at the given angle. Find the point c according to 

 Figs. 13 and 14, Lesson HI., and Fig. 20. Construct the square 

 c d ef, and from each angle draw perpendicular line* to the PP, 

 and from thence vanishing lines to the PS. In these several 

 vanishing lines find the projected angle* of the square a* in 

 Problem VIII., Fig. 23 ; between these points respectively draw 

 straight lines which will produce the perspeetire representation 

 of the square. In the next problem we only give the proposition 

 and the diagram, trusting the pnpil will be able to work it, a* 

 the explanation would be a repetition of Problems VllL and X. 



PROBLEM XL (Fig. 26). Draw the perspective view of a paral- 

 lelogram 5 feet long, 3 feet broad, one edge at an angle of 40 tct& 

 the PP, and the nearest angle I foot within, or from the PP. 



