368 



THE POPULAR EDUCATOR 



GEOMETRICAL PERSPECTIVE. XI. 



PROBLEM XXXIII. (Fig. 55). A stone slab is inclined at an 

 angle oj 35 with the ground; it is resting on an edge of one of its 

 ends, which is paral- 

 lel with the picture 

 plane; tlie edge on 

 the ground is 4 feet 

 within ; length of 

 alab 9 feet ; breadth 

 8 feet; thickness 2 - 5 

 feet. Height of the 

 eye 4 feet ; and dis- 

 tance from the PP 8 

 feet. 



Although the slab 

 is on an incline, yet 

 its ends are parallel 

 with the PP, there- 

 fore it may be con- 

 sidered a case of 

 parallel perspec- 

 tive ; had it been 

 flat on the ground, 

 its vanishing point 

 would have been tho 

 PS ; but being an 

 incline of 35 its VP 

 is perpendicularly 

 above the PS, found 

 by drawing a line 

 at 35 with the HL 

 from DE' (we hope 

 our pupils now clear- 

 ly understand that 

 the vanishing lines 

 for inclinations are 

 always drawn from 

 the DP of the VP, to 

 which they would 

 retire if they were 

 horizontal ; we beg 

 them to turn back 

 and examine the 

 figures of the pre- 

 vious problems to 

 confirm this). Tho 

 first thing to be 

 done is to find the 

 point b within the 

 picture ; make c a 

 and c d each equal 

 to 4 feet (these distances are 

 gether equal to the width of 

 slab) ; and r- ' is 4 feet within 

 will use the point a again, and rule 

 a line to DE 1 ; where this line cuts 

 the line c PS gives the point b : join 

 d PS and a PS ; through b draw e f. 

 Now, as the face and end of the stone are at 

 right angles with each other, it is very evident 

 that if the vanishing line of the face is at VP 1 , 

 the vanishing line of the end must form a right 

 angle with it and terminate at VP 2 . Find VP% 

 and its distance point DVP 2 ; through d draw 

 the perpendicular h d g in i. Now we must draw 

 the end of the stone o e p f thus : because it 

 vanishes at VP 2 , therefore we must draw a line 

 from DVP 2 through e to the measuring line in h ; 

 mark off the thickness of the stone 2'5 feet from 

 h to g, and rule g back again to DVP*, this gives 

 the perspective thickness of the slab o c directed 

 to VP 2 . To obtain the opposite corners p, f will 

 present no difficulty ; from e draw to VP 1 , and from the distance 

 point of VP 1 draw a line through e to in on the measuring line ; 

 make m i equal to the length of the slab ; a line from i to DVP 1 

 will cu<; the vanishing line e to VP 1 in n ; draw n s towards VP 2 , 

 and s t meeting a line fsom y> to the VP 1 , this will complete the 



VP 



perspective of the slab. We will now draw tha same slab not 

 having any of its sides parallel with the PP. 



PROBLEM XXXIV. (Fig. 56). The slab of the last problem, 

 having the same dimensions; the position only different; one of its 



edges is at an angle 

 of 40 with the PP ; 

 the remaining con- 

 ditions as before. 



Draw the HL, BP, 

 distance E and semi- 

 circle. Find the 

 VP for the end, viz. 

 VP 1 , by a line from 

 E at an angle of 40 

 with the tangent 

 line. VP 2 is found 

 by drawing a line at 

 a right angle with 

 E VP 1 ; draw E PS b; 

 b c is the distance 

 of the nearest point 

 within, determining 

 a (remember DE is 

 the distance point 

 for cutting the line 

 b PS to find a). Tho 

 vanishing point for 

 the face of the slab 

 will be VP 3 , found 

 by drawing a line 

 from DVP 2 at an 

 angle of 30; VP 

 will be the vanish- 

 ing point for the 

 end. For their dis- 

 tance points, draw 

 from VP 4 , with a 

 radius to DVP 2 , an 

 arc to DVP 4 . From 

 VP 3 , with a radius 

 to DVP 2 , draw an 

 arc to DVP 3 . To 

 draw the horizontal 

 edge a o, draw a 

 line from a to VP 1 ; 

 draw from DVP 1 

 through a to g; 

 make g h equal to 

 the width ; from h 

 rule back again to 

 DVP', giving the re- 

 quired length of the 

 end a o. Through a draw from 

 DVP 4 to e on the measuring line, 

 e f is equal to the thickness of the 

 slab ; draw from / to DVP 4 , and 

 through a directed to VP 4 draw an; 

 draw from n to VP 1 . Through o 

 draw o p directed to VP 4 , this will 

 be the end of the slab. Draw from a to vp 3 ; 

 through a from DVP 3 draw a line to i ; make i m 

 equal to the length of the slab ; draw from m 

 back again to DVP S , this will produce a r. Draw 

 a line from n to VP 3 for the inclined edge. A 

 line directed from VP 4 through r. meeting the 

 line from n to VP 3 in s, will be the termination 

 of the length. From s to t , directed to VP', will 

 be the upper edge of the face of the slab. 



If these two problems upon the same slab, in 

 the same position, and having the same dimen- 

 sions, but viewed from different points, are well 

 v | r ~ studied, with regard to that especial reason 



which suggested their introduction viz., the 

 principle of finding vanishing points for inclined lines and 

 planes, and the method of treating them according to the 

 characters and proportions of the objects, and the view we 

 have of them they will help to make our future problems) pos- 

 sessing more details easy to be understood. 



