Obs. 1°. There is a limit to the conditions of the problem. The angle contained by the two planes (the angle m xy), cannot be less than the angle (the angle a cb or a b c), which the required line is to make with the given plane (cm, mm'). If the two angles are equal, only one line can be drawn; if the angle contained by the planes is greater, two lines can be drawn, as shown in the figure.

Obs. 2°. It is not strictly necessary, in the solution of the problem, to determine the ellipse which is the plan of the base of the cone. To draw an ellipse neatly is by no means an easy operation, and where accuracy can be attained without it, it will be advisable to dispense with it.

(a) The problem, then, may be solved thus:-It will be observed, that the solution consists in finding the points d', d". Now, the transverse axis of the ellipse, which is the projection of the base of the cone, will be equal to the original diameter of the circle, as has been explained (51). Thus, the plan of the diameter represented by o, will be equal in length to the original; and if a number of ordinates be taken in the base of the cone parallel to this diameter, each ordinate will be shown in plan equal in length to the original. Now, we want to find the length of the ordinate represented by d, that is to say, we want to find the distance between the two points in which the plane ad cuts the circle which is the base of the cone. With centre o, and radius o b or o c, describe a semi-circle (we have shown only a part of this semi-circle in the figure). right angles to cm, meeting the circumference of the semicircle in y. From h, where the vertical from d intersects

From d, draw d g at

a' c', set off hd, hd", each equal to d g; d' d" is the ordinate expressed by d. Join a'd', a' d'"; a' d', a' du are the lines which fulfil the conditions of the problem.

Fig. 63,

Obs. 3°. By a modification of the construction explained at (a), Obs. 2°, the problem may also be solved thus::-"Construct" the plane cm, Fig. 63, with the base of the cone, and the line in which the plane ad cuts the base.

This line is shown at g h in the base of the cone 66 constructed," and its

projection is d' d". The points d', d", joined to a', the plan of the apex of the cone, give the lines required.

53. If a plane pass over the surface of a right cone, it will cut the base in a straight line, which is a tangent to the circle, representing the plan of the base. If the base of the cone is situated in a horizontal plane, this tangent will be the horizontal trace of a plane, whose inclination is equal to that of the generator of the conical surface. In other words, if any circle whose plane is horizontal, be taken, with its centre, as the plan of a right cone, a tangent to any point of the circle will be the trace of a plane whose inclination is

equal to that, which the generator makes with the plane of the base. If at the point d (See Fig. 61) a line be drawn a tangent to the circle de cf, this line will be the trace of a plane whose inclination is expressed by the angle a cb. The generator a' d, at right angles to this tangent, is the line on the conical surface over which the tangent plane passes.

Now, this line, a' d as referred to the tangent plane, is the same as di, Fig. 26, Prob. 14, and alluded to at Obs. 2°, of the same problem. In fact, the principles just enunciated are employed in the construction shown at Fig. 26, and in several of the subsequent figures.

The line be, Fig. 26, tangential to the base of the cone whose generator is a c, is the horizontal trace of a plane inclined at an angle a c d, and containing the line d h.

By a modification of Prob. 14, we are now enabled to solve the following problem.

PROBLEM 38.

Required a plane which shall make a given angle with a given plane, and contain a line not lying in the given plane.

It will be observed, that, in this problem, we have to do with the given plane, what is done at Prob. 14, Fig. 26, with the horizontal plane.

Let m m, m m' be the traces of the given plane; and a b,

a b', the plan and elevation of the given line. Make any point a', in a'b', the apex of a right cone, having its base in m m, the generator a' c or a d making with mm the angle, which the required plane is to make with it. Now, the given line intersects m m in b'. "Construct the given

plane with the base of the cone, and the point b'. We thus obtain the circle, and the point g'. circle, touching it at f Find of the plan of o,

From g' draw g' fa tangent to the From ƒ draw ƒ h, and transfer h to o. and join o' ; ob will be a tangent to the base of the cone, when its base is situated in the plane m m, m m', and, as has been explained, this tangent will represent a plane passing over the surface of the cone, and having the same inclination as that which the generator of the conical surface makes with the given plane. It will be observed, that we have also shown the construction by means of the ellipse, which is the projection, on the given plane, of the base of the cone.

Obs. There is a limit to the conditions of the problem. The given line cannot make with the given plane a greater angle than the required plane is to make with it. (See Obs. 1°, Prob. 14.)

EXAMPLES.

Note. Those questions marked with an asterisk (*) are taken from the Reports on the Military Examinations.

1. A line in plan measures 1 inch, the indices of its extremities being 8 and 3 inches; find the real length of the line, and its inclination to the plane of projection. Scale -inch to 1 inch.

Make a b, Fig. 24, 1 inch long. Draw the projector b b' at right angles to a b, and make it 5 inches (8-3), the difference of the indices of the extremities of the line, and join a b'. Since the scale is 4-inch to 1 inch, b b will be made 5 times 1-inch 11 inches, or 1.25 inches.

=

The real length of the line is a b', and its inclination to the plane of projection is expressed by the angle b' a b.

The same result would be obtained by drawing projectors from a, b, and setting off upon them, 3 times and 8 times 4-inch respectively. For an explanation of the principle involved in the foregoing solutions, see Art. 38.

2. (1) Draw a plane inclined at 50°, and in it place a line inclined at 39°. (2) Draw a second line in the plane, at an angle of 60° with the first line, and find the real angle contained by the lines.

For the solution of (1) See Prob. 14, Fig. 26. (2) Assuming dh as the line inclined at 39°, from any point in it, as d, line as d i, making the angle h d i, 60°. Then, taking