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. 1874 edition. Excerpt: …A Bis 1 from the horizontal plane, and A is 5 from the vertical. Find also distance of B from the vertical. In Fig. 4 draw a’ b ‘ 3 long, at a distance of one inch from X Y parallel to it. From a b ’ drop indefinite perpendiculars to X Y. At a distance 5 from X Y, draw a b at an angle of 30 to X Y, and produce it until it meets the perpendicular from b ‘ in the point b. Then a b is the required plan of the line. The distance from i to X Y is the required distance of the point B from the vertical plane. Problem 5. Given the plan of a line a b, with the distances of the ends of A B above the horizontal plane, to draw the elevation. Suppose the heights of a’ and 6’ are.5 and 175 respectively: Erect perpendiculars from a and b to a’ b’, so that a’ shall be ‘5 above X Y; 6’ 175 above X Y. Join a’ b’. Then a’ b’ will be the elevation of the line required. This is Evident–first, because the elevation is, from what has previously been given, within the perpendiculars produced from the ends of a b to X Y; secondly, because all distances from the horizontal plane are measured on the vertical and vice versd. Note.–That a b being inclined to the vertical does not at all iffect the measurements on the vertical for the elevation. Problem 6. Given the traces a b’ of a line A B, to find its projection. The traces of a line are, as before stated, the points in which a line meets the planes of projection. The line joining a and 6’ will be the line in space. From a and 6’ draw a a’ and 6 6’ at right angle to X Y. Join a’ b’ and a 6. Then a’ b ‘ will be the elevation of the line and a b the plan. Note.–The traces a and b’ are simply points. The line joining a & is not the real length of the line in space….
