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EXERCISE 106 (Vol. II., page 156). 

 1. We cannot get through this forest. 2. I received these letters 

 this morning through a kind acquaintance. 3. Along the river they 

 saw the enemy's glittering arms. 4. One should be ready to sacri- 

 fice everything for a friend. 5. For this behaviour the father 

 punished the boy. 6. They directed the cannons against the town. 7. 

 You are not now nearly so frank towards me as was the case formerly. 

 8. The enemy shot all the prisoners without exception. 9. Without 

 the father's interposition, the children might have caused a great mis- 

 fortune. 10. Without doubt my friend will arrive here to-day. 11. In 

 order not to increase the mother's fear still more, he did not tell her 

 the truth in all respects. 12. Maria Theresa, empress of Austria, 

 waged war for seven years with Frederick the Second, king of Prussia, 

 about the possession of Silesia. 13. For this disease there is no 

 medicine. 



GEOMETRICAL PERSPECTIVE. VII. 



PROBLEM XX. (Fig. 40). Upon the board of tlie last question 

 (see Vol. III., page 73) describe a circle, the circumference touch- 

 ing tlie edges. 



Fig. 39, Problem XIX., must be repeated: then upon / b de- 

 scribe a semicircle, and about the semicircle the parallelogram 

 / h g b ; from z, the centre, draw x h, z g ; through the points 

 where these last lines cut the circumference draw lines parallel 

 to / h or b g, to i, Tt ; next, through the points t, z, k, draw 

 straight lines parallel to/ a' c or b d; draw the diagonals a' d, 

 c' b, which intersect the parallel lines dropped from the incline 

 k z i; the points of intersection numbered 1, 2, 3, 4, 5, 6, 7, 8, 

 will bo those through which the plan of the inclined circle must 

 be drawn. 



Before proceeding with the perspective delineation of the board 

 and circle, as shown in the lower part of Fig. 40, we must detain 

 the pupil whilst we examine the principles upon which the plan 

 of the inclined circle is represented. The explanation given 

 with Figs. 38 and 39 will be sufficient to clear all difficulties 

 with respect to the board only. As the circle is lying on the 

 board or inclined plane, the end or profile of which is / b, we 

 must ascertain the wliereabouts of Hie points through which the 

 circle is drawn upon the incline. Let the pupil draw a square 

 upon a separate piece of paper, and describe within it a circle, 

 then hold the paper at an angle with the horizon, the inclined 

 edge being opposite the eye, he will first see how from an in- 

 clined line we can represent the whole of a square, as illustrated 

 by Figs. 38 and 39 ; but in this case we have the addition of a 

 circle within the square, therefore the points through which the 

 circle is drawn must be brought to the edge of the inclined 

 square represented by the line / b (Fig. 40) ; a semicircle 

 will be sufficient to help us in this, as the opposite portions of 

 the circle and the several points through which it passes 

 correspond; therefore the method of construction above given 

 will enable us to produce upon the plan of the board the plan 

 of the circle also. 



To proceed with the perspective representation, let the pupil 

 draw visual rays from all the points in c' d and a' b, to cut 

 the respective sides of the perspective projection of the square ; 

 draw lines between the corresponding points on the opposite 

 sides of the perspective square, and also the diagonal lines of the 

 square : the points through which the circle is to be drawn by 

 hand will be those which are found to answer to the same in the 

 ground plan. 



PROBLEM XXI. (Fig. 41). A truncated pyramid has a square 

 base of 1*5 inch side, the top is of 1 inch side, the lieight 2 - 5 inches. 

 Give a perspective representation of the pyramid resting on a hori- 

 zontal plane with the plan of the picture inclined to one of the 

 edges of the base at an angle of 15. The line of sight to be f of 

 the height of the pyramid. 



After placing the line c D (an edge of the base) at the given 

 angle, 15 with the PP, draw the plan according to the instruc- 

 tions given in Problem VI. (Vol. II., page 297). Here is an instance 

 where the use of one VP only will be absolutely necessary ; 

 there are two sets of retiring lines, viz., c D and its parallels, 

 and c B and its parallels ; if we were obliged to determine 

 the VP for c D and its parallels, we should find by drawing 

 from the station point a parallel to c D that the VP would be 

 at a very considerable and inconvenient distance out of the 

 paper ; therefore produce the parallels to c B, viz., A D, a d and 



b c, to the PP in the points E F G ; determine the VP for these 

 lines only, and follow the instructions given with reference to 

 Figs. 27, 28, and 29 (Vol. III., page 9) in drawing the perspective 

 of the base ; the points of contact E and C will be the points to 

 be brought down to E' and c' for the base. The lines of contact 

 from F and G must also be brought down to the base of the 

 picture upon which to measure the height of the pyramid F'M and 

 G' N. Divide F' M or G' N into three equal parts, and through 

 the second from the base draw the line of sight parallel with 

 the PP. Find the VP, to which draw lines from the points of 

 contact E' and c' ; these lines cut by visual rays from D and B ir* 

 the plan will decide the extent of the base in h i and k. For the 

 top lines must be drawn from M and N to the VP, and cut by 

 visual rays from the plan of the top, as was done with the base ; 

 draw the inclined edges m h, n i and ok; this will complete the 

 subject. 



PROBLEM XXII. (Fig. 42). Supposing an equilateral triangle, 

 having its side 2'5 inches, to be the base of a pyramid 2'5 inches 

 high, draw a perspective representation of the pyramid. Assume 

 one side of the base to be inclined at an angle of 20 with tlie 

 picture plane, the nearest edge of the pyramid to be i inch from 

 the picture plane, and tlie observer's eye to be 5 inches from the pic- 

 ture plane, and 1'5 inch above the Iwrizontal plane on which the 

 pyramid stands, and opposite a point 2 inches to the left of the 

 angle of the pyramid nearest the picture plane. (From a Military 

 Examination Paper.) 



Draw a line, A x, at an angle of 20 with the PP, determine 

 the point B | inch from the PP, and make A B equal 2'5 inches, 

 upon which describe an equilateral triangle, the base of the 

 pyramid. The centre of the triangle must be found by bisecting 

 two of the angles (or by bisecting two of the sides, because the 

 figure is a regular one, having equal angles and equal sides) ; the 

 intersection of the bisecting lines will be the centre at G, which 

 is the plan of tlie apex of the pyramid. Produce the line c G to 

 D, and draw from A and B parallel lines to meet the PP in E and 

 F. From E, D and F draw perpendicular lines to the base of the 

 picture BP. Place the station point (SP), and draw the HL 

 according to the given distance stated in the question ; find the 

 VP for D G c, which will also be the VP for the other parallel 

 lines drawn from the plan to the PP: visual rays drawn from 

 A, B, c, and cutting other lines drawn from H i K to the VP, will 

 at their intersections give the perspective positions of the several 

 angles of the triangle, which must be completed by straight 

 lines forming these angles. Thus far there is no particular 

 difference in the rule for drawing the perspective of the base from 

 the one given for the last problem and several others gone before ; 

 but we wish especially to draw the attention of our pupils as to 

 which of the lines of contact E H, D I, or F K must be the one 

 upon which the elevation or height of the pyramid is to be set off. 

 It will be easily understood, when we consider that the vertex 

 of the pyramid is over the centre of the base, that the line of contact 

 connected with the centre must be tlie one, viz., D i. Therefore 

 upon D i mark the height of the pyramid, viz., i L ; from L draw 

 a line to the VP, and a visual ray from G cutting this line in M 

 will give the position of the vertex of the pyramid. Draw from 

 M lines to meet the angles at the base, which will complete the 

 representation required. Suppose the three inclined faces had 

 not been equal, and that the plan of the vertex had been at g, 

 then g d must be drawn parallel to G D, the line of contact 

 brought down, and from the height measured to t a line drawn to 

 the VP, and the visual ray from g to cut this line, to find the 

 vertex from which intersection the edges drawn to the angles at 

 the base as before will represent the pyramid. 



Suppose the solid to be a regular tetrahedron, that is, a figure 

 with four equal faces, each face would then be an equilateral 

 triangle ; the height in this case would have to be found. This 

 obliges us to have recourse to geometrical or orthographic pro- 

 jection. Upon a little reflection the pupil will see that the dis- 

 tance of the vertex from the ground will be less than the length 

 of the edge of the pyramid ; first, because a straight line drawn 

 from an angle of the equilateral triangle to the centre of the 

 opposite side is less than the side ; and again, it would be further 

 reduced because the triangular face is inclined. Now, how much 

 the height may be less than the edge can be determined by 

 the following mode of proceeding : Let ABC (Fig. 43) be the 

 plan of the pyramid at the base, and D the plan of the vertex. 

 Now it is understood that all the faces of this solid are equal, 

 and that they are equilateral triangles. Again, we have the full 



