Publisher's Synopsis
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. 1897 edition. Excerpt: ...sections have been found by projection. Taking the "cylinder" as the first curved-surfaced solid as our object, the problem is--Problem 49 (Fig. 148).--Given the plan of a cylinder with its axis. perpendicular to the 1IP, to find its elevation when its length is twice its diameter. Let the circle AB, No. 1, be the given plan; then its centre a will be the plan of the axis of the cylinder. Find by projection the elevation of this axis a' a. Assume the cylinder to be standing with one end on the HP; then, as its ends are in the same relative position as the sides of the rectangle which generated them--viz., parallel to each other--and one of them is on the HP, set off on the axis, from the IL, the length of the cylinder in the point a', and through it draw a line parallel to the IL. Now, in looking at the cylinder in the direction of the arrow in the plan No. 1, the visual rays will impinge upon its surface from A to B; at A and B the rays will be tangential, and being at the same time perpendicular to the plane of projection, or the VP, they will strike both sides of the cylinder in lines.drawn through A and B on its surface, perpendicular to the HP. Therefore through A and B, No. 1, draw the lines AC, BD, No. 2, and the required elevation is obtained. Now, let the cylinder be inclined to the HP, at an angle of 45--its axis being still parallel to the VP--and its plan when in that position be required. First draw in the elevation of the cylinder in the given position, as in No. 3. Its ends AB and CD are now inclined to the IL or HP, and will in plan become ellipses--as explained in Problem 40--because they are circular, but inclined to the plane--the HP--on which their projections are required. Now, in viewing the cylinder in...
