72 



THE POPULAE EDUCATOR 



GEOMETRICAL PERSPECTIVE. VI. 



PROBLEM XV. (Fig. 34). A square pyramid 4 feet side andQ 

 feet high, one edge of the base at an angle of 60 with the pic- 

 ture plane, and the nearest angle 1 foot within the picture. The 

 base of the pyramid must be treated as the square in Problem 

 X., Fig. 25, Vol. II., page 361 ; after which draw the diagonals in 

 the perspective projection, namely, a b and c d ; their intersection 

 will determine the perspective of the centre e, from which erect a 

 perpendicular e f; this will 

 be the axis of the pyramid. 

 From DE 1 draw a line through 

 the centre of the base to the 

 picture plane at g, from 

 which draw the line of con- 

 tact, and upon it from g set 

 off the given height of the 

 pyramid g h. Draw the re- 

 tiring line from h to DE I , and 

 at the point where this inter- 

 sects the axis e f will be found 

 the perspective height of the 

 pyramid, namely, at i. Draw 

 lines from the angles a, b, c, d 

 to i, to represent the inclined 

 edges of the pyramid ; this 

 will complete the problem. 

 If we place the ground plan 

 beyond the picture plane (Fig. 

 35) wo must proceed as fol- 

 lows: Place the plan with 



give the perspective height of the pyramid. Complete the in- 

 clined edges as in Fig. 34. (Observe, we might have drawn e f 

 not parallel to the side of the square ; in that case we should 

 have been compelled to find another vanishing point ; therefore 

 to save extra work we draw ib parallel to the side of the square, 

 so that we may make use of the VP of that side. Definition 

 13, Vol. II., page 162.) 



PROBLEM XVI. A cone 4 feet diameter and 6 feet high. This 

 will be done from almostthe same directions as the pyramid. Look 



back to Problem XII., Fig. 

 31, Vol. III., page 9, where we 

 have the perspective of a circle. 

 Now the base of the cone, 

 being a circle, must be treated 

 in the same way. To draw 

 the elevation, draw a perpen- 

 dicular line, the line of con- 

 tact from d or b (Fig. 31) ; 

 mark off upon this line the 

 given height, and from that 

 point draw a line to the re- 

 spective VP ; thus, if the line 

 of contact is from d, DE 2 will 

 be its VP; a perpendicular 

 line drawn from the centre of 

 the circle to cut this vanish- 

 ing line will be the axis, and 

 the point of intersection will 

 mark the apex, from which 

 draw lines to o and p for the 

 sides of the cone. 



the PP, as required by the question, and produce two of the 

 parallel sides to the PP, to be continued perpendicularly to the 

 base of the picture, from which draw the retiring lines to the 

 VP. Visual rays from each angle of the plan, cutting the retiring 

 lines, will give the positions of the angles of the square. This 

 method of using only one VP has been fully explained in Les- 

 son V. ; we trust the pupil will make himself master of it, as 

 we shall have to employ it very frequently. For the elevation, 

 diagonals may be drawn and the subject completed as in 

 Fig. 34 ; but we will show another way, and draw the diagonals 

 in the ground plan which give the centre e. Draw e f parallel 

 to the side of the square, and the perpendicular line/;;, the 

 line of contact. Upon / g mark the given height of the pyra- 

 mid g h, and from h draw a line to VP ; a visual ray drawn from 

 the centre e of the plan, cutting the line from h to the VP, will 



PROBLEM XVII. (Fig. 36). A cylinder 4 feet diameter and 8 

 feet high stands on its end ; the eye is opposite half the height 

 of the cylinder. In working this problem we prefer placing 

 the plan beyond the PP, it being necessary to draw a circle for 

 each end of the cylinder, therefore the same perpendicular lines 

 drawn from the plan will answer for both. It will be seen that 

 when these perpendiculars have reached the base of the picture 

 other lines are drawn from them to the PS, and the circle is 

 drawn by hand as in Fig. 31, Vol. III., page 9. For the upper circle, 

 a b is drawn horizontally across the perpendiculars according to 

 the height of the cylinder, and the same process with regard to 

 the circle is followed as in the one for the base ; lastly, lines 

 c, d, drawn tangential to the outer edges of the circles, will give 

 the sides of the cylinder. 



PROBLEM XVIII. (Fig. 37). To draw the perspective represen- 



