GEOMETRICAL PERSPECTIVE. 



I 



purpose of placing boforo our pupils one or two canes to show how 



:i put int. i |>ia.-i !'', will simplify tho work in 



many ipi- nmis which may ap]"'ai- to In- ditHiuilt. Lttt tho pupil 

 turn t.) 1'nili. I., Ki>f. 7 (Vol. II., p. 225) of course wo presume 

 that this problem has boon learnt, and, for the sake of practice, 

 l:iir-t have I" ,n placed at other angles and worked out also 

 jinil.lfiu contains the kev to tho whole theory of 

 ground-plan re. Now, if tho pupil consider for a 



moment tho prim ![ ,!> exemplified in that problem of p"' 



'V, ho will see that it is nothing nioro 



than tin.inii: tii.- p/-itioim of o and B. For as a straight lino 

 li.'twi .MI tin- two points in the plan, therefore, when tho 

 perspective of those two points is determined by drawing a lino 

 between them, wo shall have tho perspective of the line. In 

 order that wo may bo fully prepared to go further into tho 

 subject, wo will take a point only, independent of any line, and 

 place its plan anywhere beyond the p p, Fig. 27. Let A be the 

 point. From A draw a lino at any angle with tho P P, and meet- 

 ing it in a. Find tin- vanishing point of a A, bring down tho 

 point of contact a perpendicularly, to b, and join 6 v P; a visual 

 ray from A cutting this lino will give the perspective position of 

 A at c. Wo will now mako an applica- 

 tion of this principle, and show how, 

 with only one vanishing point, wo may 

 work out a problem which in plan is 

 composed of a great number of lines, 

 and none of them parallel with each 

 other. We will propose a very simple 

 case (Fig. 28), composed of three lines 

 a b, b c, and c d. Draw a line a e from 

 a to the P P, and from the extremities of 

 each line draw others parallel with ae; 

 find the v r for 

 one of these 

 lines, which will 

 be tho V P also 

 for the rest, be- 

 cause parallel 

 retiring lines 

 have the same 

 V P ; bring down 

 the points of 

 contact e, f, g, h, 

 to B P, tho base 

 of the picture, 



and then join them with the v r. 

 Upon these last lines respectively 

 will bo the perspective positions of 

 the points or extremities of tho lines 

 abed found by tho visual rays as in 

 Fig. 27, viz., a' b' c' d'; unite thesu 

 points by straight lines corresponding 

 with tho plan, and then will be pro- 

 duced the perspective view of that represented by the plan. The 

 pupil must practise this by giving himself a similar combination 

 of lines, and of greater number. We will add another position of 

 lino to this figure. A line, a b, is drawn across the arrangement 

 (Fig. 29). There is no necessity to repeat that which has been 

 already shown in Fig. 28, so wo will proceed to explain the 

 method of representing tho line a b. It will be understood 

 that the perspective of c d is c 1 d' and of e f, e' f. Now 

 the given line a b crosses these lines in the plan through 

 in and n; it will be necessary, then, to find the projection of 

 these points m and n, which will be done if visual rays are 

 brought down from those points to cut tho perspective lines 

 c' d' in o, and d f in p ; then, to complete tho problem through 

 these points o and p, tho line a' b' must be drawn to the extent 

 each way as determined by visual- rays from the extremities 

 a and b. These figures may in themselves have no particular 

 meaning, we merely employ them to illustrate a principle, be- 

 cause the perspective of any other object of an intelligible 

 character may be worked out by this method. 



Fig. 30 is the plan of a hexagon (see Lessons in Geometry 

 XIX., Problem LI., Vol. II., page 192), having one of its edges 

 at an angle of 20 with the p p. We will only give the figure, 

 and recommend the pupil to work it out ; afterwards he should 

 mako other regular or irregular polygons for practice also. Re- 

 member that a regular polygon is one that has all its sides and 



Fig. 33. 



all its angle* equal; an irregular polygon has nneqoal lides and 

 unequal an 



PIIOIJLKM XII. (Fio. 31). To draw the perspective of a circle. 

 Diameter of the circle, 4 feet ; height of eye, 5 feet / distance oj 

 the eye from the p P, 10 feet, and opposite the centre of the circle. 

 Scale, 4 inch to the foot. Draw tho B P and also the H L 5 feet 

 above it ; anywhere (at a) draw tho semicircle bed, with a 

 radius of 2 feot ; from a erect a perpendicular to B r, and from 

 tho point where this line cuts the L in P 8 describe tho semi- 

 circle D E 1, E, o E 2, with a radius equal to the distance 

 of tho eye from the picture plane. About the semicircle 

 cribo tho rectangle d efb ; draw a e, af, and through the 

 point -i in tho semicircle cut by tho lines a e and a /, draw tho line* 

 <j i ami h k parallel to o c or d e. From the points d, k, a, i, b, draw 

 lines to tho p 8. Draw a lino from d to D K 2, and also one from 

 6 to D E 1, and call them distance lines; notice where these lost 

 lines cut bra and dps, between these intersections tho line 

 m n must bo drawn. We have now a square in perspective, in 

 which tho circle is to bo drawn. Through the intersection of 

 tho distance lines draw o p parallel to m n. With the hand draw 

 through Che points t, u t p, q, a, r, o, s, to t again, the perspec- 

 tive circle required. This latter part 

 of tho process that is, drawing the cir- 

 cle will be tho most troublesome; as it 

 does not admit of the use of compasses, 

 it must be done by the hand only, and 

 will require much practice to dp it 

 neatly. 



PROBLEM XHI. (Fio. 32) is the 

 first example of elevations. A row oj 

 7 rods, perpendicular to the ground, 6 

 feet hiyh, and 1 foot apart; the plan 

 55 with the pic- 

 ture plane. Their 

 distance apart 

 must be ar- 

 ranged in the 

 plan 1, 2, 3, 4, 

 5, 6, 7; their 

 heights most be 

 set off on the 

 line of contact, 

 be; therefore a 

 line drawn from 

 c to the v P will 



represent their retiring position on 

 the ground, and a lino from b to the 

 V P will represent the perspective of 

 their heights; their widths apart 

 will be found by the visual rays 

 drawn from 1, 2, 3, etc., on the plan. 

 PROBLEM XIV. (Fio. 33). A cube 

 4 feet edge has one of tte faces at 

 the P P ; its nearest angle touches 



with 



an angle of 40 

 the P P. Draw the plan of the cube abed, making d c 

 at an angle of 40 with PP. Find the vanishing points of 

 the sides a d and d c, viz., v p 1 and v P 2. Proceed as in 

 Prob. VII., Fig. 19, for the extent of the base. For the 

 elevation, as tho cube touches the p P at d, therefore a per- 

 pendicular line from d brings down tho point of contact to e, 

 making the lino d e the line of contact. As all heights are 

 measured upon the line of contact, therefore the height of 

 the cube must bo represented by e f equal to d c (as all the 

 edges of a cube are equal). Draw lines from / and e to eacli 

 vanishing point ; then tho visual rays from o and c will de- 

 termine the outer and perpendicular edges of the cube g h and 

 i k. Again, as parallel retiring lines have the same vanisliing 

 point (and opposite sides of a cube are parallel), draw a line 

 from i to V P 1, and another from g to V P 2, where these lines 

 intersect at m will complete tho upper horizontal face of the 

 cube, viz., g f r,i i. Mako all these lines dark as in the diagram. 

 To find the perspective of the centres of any of the faces of 

 tho cube, it will bo only necessary to draw the diagonals on the 

 faces of the perspective projection viz., from /to fc and t to e, 

 or from / to b and g to . If a line is required to be drawn 

 across the centre of a face, in a horizontal position, the above 

 i central point must first bo found, and then the line must be 

 I drawn through this point to the v p of the face respectively. 



