This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1899 edition. Excerpt: ...of the end elevation will now become visible in the plan, and if this plan be turned (as at Fig. 4) at an angle (say 45) to I L, perpendicular lines from the points of the plan intersected by horizontal lines from the elevation, will give the projection of the object at a compound angle. Plate XVI. Fig ...
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1899 edition. Excerpt: ...of the end elevation will now become visible in the plan, and if this plan be turned (as at Fig. 4) at an angle (say 45) to I L, perpendicular lines from the points of the plan intersected by horizontal lines from the elevation, will give the projection of the object at a compound angle. Plate XVI. Fig. 1 shows the plan and elevation of four such prisms meeting at a point, a figure which often occurs in designing or drawing roofs of houses, churches, &c. The plan is formed of two figures similar to the plan of the last prism, crossing each other at right angles, and from this the elevation is easily projected. Fig. 2 shows the projection of the object when placed at an angle to the vertical plane. And Fig. 3 is the development cf one of the four parts of which the model is composed. To construct this development on a straight line, set off three spaces equal in width to the sides of the prism, a' b f a, and erect perpendiculars from the points. Make these perpendiculars equal to the lines similarly lettered in the plan. Now it will be clear that when two parallelograms, like those forming the plan of the prism, cross each other, they will form four right angles at the centre. Therefore, at d' and c construct angles of 45 which will meet in c, and form the required right angle, and this will complete the under side. Draw c h and c i at right angles to e c and d c, and equal to the altitude of the triangle gc in Fig. I. Join d' i and h e. Then the right-angled triangles, d' c i and e c h, turned up at right angles to a b c d e, will form the upright sides of the mitre; z and h will then come together. The triangular end, which is represented as bent down, being now turned upward at right angles to the under side, the two upper sides are...
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