GEOMETRICAL PER8PK< Tl V K. 



when its rays arc parallel with tho picture ; secondly, when the 



in front of tho i>irtiitv ; thirdly, when it u 

 .1 tlm picture. 



; its rays are parallel with tlie picture. The un is 

 :luT on thu right huud or on tho loft; its rays, although 

 at an inclination, with thu ground, uro parallel with tho pirtim?- 

 pl&no. 



When the nm is before or in front of the picture ; that 

 is, \\in-ii it is behind the spectator, or whon the spectator u 



n thu HUH und tho object. 



:~.r<\. U .' ,i the sun is behind the picture. By thu U meant 

 when tho object upon which tho light folia is between tho sun 

 and tho spectator. Our first examples will be to illustrate the 

 first of these positions. 



PROBLEM XLV. (Fig. 75). A block of stone 3 feet high, 4 feet 



.id 5 feet long, has its end parallel with the picture plane, 



to the right of the eye and 1 foot within the picture. 



Height of tlie eye, 5 feet, and 10 feet from Die picture plane. The 



angle of the inclination of tho rays, or the sun's elevation, it 50 



the line of contact r in tn and t, and are eosrthnied on th* 

 face of the wall to n ; from r to e is 2 feet, to out off boa e 



the nearest angle of the wall within the picture ; from c to I' U 

 2*5 foot, the portion of wall on this side the opening. Linec 

 drawn from k t/ o (equal to fc v o) to the l>r* will cut the beer 

 of the wall for the perpendiculars of the opening ; between Uiest 

 perpendiculars the parallelogram I k t o and the diagonals most 

 be repeated ; the corresponding points will be easily recognised, 

 and through them the perspective of the arch must be drawn 

 by hand. For the shadow draw any line a b, as in the last 

 problem, at an angle of 45 with the PP, and draw lines parallel 

 to it through e, b, c, to meet lines on the ground drawn from the 

 bases of the perpendicular line* e b and e, and parallel to the 

 PP in the points d, e,fi draw the arc def by hand. Th* 

 shadows of the angles of the wall n, o are found as d and e in 

 the last problem. 



PROBLEM XLVII. (Fig. 77). The block of Problem 45 hat 

 a pole 10 feet long laid across it horizontally at an angle of 40 

 with the picture plane. The nearest portion of the pole which if 



with the horizon, and to the right of the eye. Project the shadow 

 of tlie block. 



Anywhere upon the PP draw an indefinite line a b, at an 

 angle of 50 with tho PP. Through tho angles of tho block c 

 and d draw lines parallel to a b, until they meet other lines 

 drawn from / and e parallel with the PP in m and n.. The side 

 of tho block cdfe will bo the broad shadow, that is, the 

 shadow on the object; e/ urn will bo tho cast shadow on tho 

 ground, that is, tho shadow caused by tlie object. It will bo seen 

 that tho edge of the shadow on the ground from tho upper edge 

 of tho block retires to the PS, the same vanishing point to which 

 the block retires, because it is parallel with the block. 



PROBLEM XLVI. (Fig. 76). Tlie face of a wall pierced by 

 <Mt opening having a semicircular arch; retires at right angles 

 with the PP, nearest end 1 foot within tlie picture. Height of wall, 

 9 feet. Horizontal length, 10 feet, and 5 feet to tlm left of the eye. 

 Breadth of opening, 5 feet, and height 7 feet. Height of eye, 5 fiet; 

 distance 10 feet. Sun's elevation, 45, and its rays parallel *cith 

 the picture plane. The thickness of the wall is purposely omitted. 



To draw the perspective elevation of the arch, its elevation 

 must be constructed parallel with the PP. At the given 

 height of the spring of the arch from the ground at o, draw k o 

 equal to the width of the arch ; draw the diagonals v I and v t ; 

 also the horizontal lines p m and It; p in must intersect 

 the diagonals where they intersect the arch ; these lines meet 



tn contact with the block is 1 foot from the right-hand corner of 

 the block, and 2 feet of the pole as it approaches the picture plant 

 hang over the side. Project the shadows of the block and the pole, 

 Sun's inclination 50. 



Project tho shadow of the block as in Problem XLY. To 

 determine tho perspective position of the polo, mark the point a 

 2 feet from 6; this will include the distance of the block from 

 the PP, and rule it towards DE', cutting b PS in c. Draw the 

 perpendicular c d (d marks the edge of the block over which 

 the pole projects). Through c and d draw indefinite lines 

 towards vp (tho vanishing point for the pole) ; the lower line 

 through c will be the plan of tho pole. Draw a line from c to 

 directed by the DVP, and make ef equal to 2 feet; draw a line 

 from / towards DVP to meet the plan of the pole in o ; draw the 

 perpendicular o tn ; d m will then be that part of the pole which 

 projects 2 feet over the side of tho block ; make / g oqnal to 10 

 feet, and draw from g to UVP, cutting the plan of the pole in 

 h ; draw the perpendicular h i ; then the portion of tho line 

 between m and i will be the perspective representation of the 

 pole in tho position given. To project the shadow, draw lines 

 from the end of the polo parallel to the sun's inclination, and 

 from h draw a line h k parallel to the PP to cut the inclined 

 line ; from this intersection will be traced the shadow of the 

 pole in the direction of vr, appearing only beyond the shadow 

 of tho block. 



