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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1913 Excerpt: ...to the distance from n to SPH, and dropping from the intersection of this arc with HPP a vertical line. »nBb is found at the intersection of this vertical line with VH. 101. The perspective of ab has been drawn from av, vanishing at fab. avbi on VH1 is made equal to the length of aHbu given in the plan of the card. A measure line through bx, vanishing at To'i", will determine the length of avbv. A line from bv vanishing at t/"d, and one from F vanishing at vab, will intersect at cv, completing the perspective plan of the card. 102. Even the vanishing points (vab and vad) for the sides of the card may be found without drawing a diagram. Since fn is drawn parallel to ab, it makes the same angle with HPP that ab makes. Similarly, since fh is drawn parallel to ad, it makes the same angle with HPP that ad makes. The angle between fn and fh must show the true angle made by the two lines ab and ad in the diagram. Therefore, having assumed SPH, we have simply to draw two lines through SPH, making with HPP the respective angles that the two sides of the cards are to make with the picture plane, care being taken that the angle these two Hnes make with one another shall equal the angle shown between the two sides of the card in the given plan. Thus, in Fig. 27, the two projections of the station point have first been assumed. Then through SPH, two lines (/« and//t) have been drawn, making respectively, with HPP, the angles (H and N) which it is desired the sides of the card shall make with the picture plane, care being token to make the angle between fn and fh equal to a right angle, since the card shown in the given plan is rectangular. Verticals dropped to VH from the points n and h will determine vab and Dad. Having found vab and vaA, »...
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