Further, to complete the projection of the cone when seen in this new position ; if we look at No. 1 in the direction of the arrow k, which is at right angles to the section plane, it will be seen that by using this plane as the plane of projection, all the points necessary for the figure may be obtained upon it. For example, the base of the cone A'B', whose true form is a circle, being at an angle with this plane, its projection will be an ellipse. From B', C', and A', draw projectors to meet the line mn in P', &c. From P', take the distance P'b', and set it off from b' to P". Now the point R will represent the centre of the base. Take P'R', and set it off from P" to R", and the same distance from R" to s'; then P" and S will be the projections of the minor axis of the ellipse. Through R', draw a line parallel to xy and make it equal to A'B' for the major axis. An ellipse described about these axes is the base of the cone, and lines drawn tangential to the two ellipses will complete the figure.

Lastly, to obtain the plan of this figure, it will be viewed in the direction of arrow a. From a', draw a line at right angles to mn. If we conceive this line to represent an edge view of the plan of projection, projectors drawn from the points of the figure will represent on this line the projections required, as shown in No. 1.

Take any point, R, on the line a'P" produced, and through it draw a line parallel to wy; this will represent the major axis of the ellipse as seen in the figure. To obtain the minor axis, set off from R, the distances 1 2, and 2 3, and complete the ellipse by any of the ordinary methods. It will be seen that the plan of the section plane in this position becomes a straight line, whose breadth is determined by dropping projectors from p and p.

Problem 61. To find the projection of a cone standing on its base, when the section plane is parallel to one side of the cone, and perpendicular to the vertical plane; also the true form of the section.

By following the method laid down previously, we can readily obtain any number of points, as F, G, K, W, &c., in the curve representing the horizontal projection of the section specified. It must be remarked that the horizontal plane passing through M' gives only one point M (which is the vertex of the curve required), because the circle which denotes the section that it makes with the cone is a tangent to the given plane.

In order to determine the actual outline of this curve, let us suppose the plane mn to turn, as upon a pivot at M', until it has assumed the position M'B', and transfer M'B' parallel to itself, to M"B". The point F will thus have first described the arc FE till it reaches the point E', which is then projected to E"; so that, if we

conceive the given plane, now represented by M"B", to turn upon that line as an axis, until it assumes a position parallel to the vertical plane, we shall find that the point Ė", which is distant from the

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axis M"B" by the distance FV, No. 2, will now be projected to F", No. 1. The same distance FV, set off on the other side of the axis M"B", gives another point G' in the curve required, which is that called the parabola.

Section VII.


THE preceding sections being thoroughly understood, the following problems will present no difficulty to the student. All three relate to cylinders, and are of an elementary character.

Problem 62.

To draw the curve of penetration of two right cylinders, whose axes are at right angles to each other.

Let ABCD be the plan of the horizontal cylinder, and the circle E the plan of the vertical cylinder. Find the vertical projection of the two cylinders. Thus we get four points in the curve of penetration required, viz., a', a', 6', 6. Draw mn parallel to DC, and let it represent a section plane at right angles to the horizontal plane, cutting both cylinders. Now, this plane cuts the vertical cylinder in a rectangle whose elevation is c'"e"é', which is found by raising perpendiculars from the points c and e, where the plane mn cuts the circle E.

The plane mn also cuts the horizontal cylinder in a rectangle, the elevation of which is m'n'p'r'. It is found thus ;-On DA a semicircle is described, representing half the base or end of the cylinder ABCD. Produce mn to meet the circumference of the semicircle in d. The ordinate md represents half the width of the rectangle, which is the section of the horizontal cylinder by the plane mn. Hence, from D' (D'C" is the elevation of DC) set off D'm', D're, each equal to md, and through m'r' draw m'n', r'p' parallel to D'C". The rectangle d'"e"e' intersects the rectangle m'r'p'n' in the points s's' and t't. We have thus found two more points in the required curve.

Lastly, taking DC as another section plane, this plane cuts the vertical cylinder in a rectangle, whose elevation is h'h"k"K', found by erecting perpendiculars from the points h, k, where DC intersects the circle. Now, this rectangle intersects D'C' in f' and g', the elevations of f and g points in DC. The curve described through a's'f's'a

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and b't'g't'b' is that in which the horizontal cylinder ABCD intersects the vertical cylinder E.

Problem 63. To draw the curve of penetration of two right cylinders, whose diameters are equal, and whose axes are at right angles to each other.

From the following remarks it will be seen in what respects the present problem differs from the foregoing. The diameters of the cylinders being equal (Pr. 63), the curves of penetration are projected vertically in straight lines perpendicular to each other. For, if we proceed to apply the method before given, we shall soon discover that the various points in these curves are situated in two planes at right angles to each other, and to the vertical plane, the sections formed by them being, in fact, ellipses equal and similar to each other. It is not necessary to enter into any details in illustration of this case, other than to call attention to the figure, where the

projections of some of the points are indicated, both in elevation and plan, by the same letters of reference.

Problem 64. To draw the curve of penetration of two right cylinders, whose diameters are equal, and whose axes are at right angles to each other, one of the cylinders being inclined to the vertical plane.

The two preceding figures being drawn (Prs. 62 and 63), we may easily find the projection c, of any point such as c', by observing that it must be situated in the perpendicular cc; and that, since the

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