Problem 21.

Given the traces of a line, to find its projections.

Let AB' be the traces of the given line; ie., A and B' are the points where a line in space meets the planes of projection.

The line joining A and B' will be the line in space.

First, let us find its horizontal projection. Draw B'B' at right angles to xy and join AB'; then AB is the projection upon the horizontal plane of the line in space.

Secondly, let us find its vertical projection. Draw AA' at right angles to xy and join A'B'; then A'B' is the projection upon the vertical plane of the line in space.

NOTE 1.—Neither plan nor elevation expresses the real length of the line.

NOTE 2.-The vertical trace B' shows that B' is elevated above the horizontal plane a distance equal to B'B', and that A shows that A' is removed from the vertical plane a distance equal to A'A.

Problem 22.

Given the projections of a line, to find its length.

Let AB and A'B' be the projections of the given line. It is required to find its length.

If we conceive a vertical plane as passing through AB, this plane will have AB for its horizontal trace, and BB' for its vertical. Imagine then this plane to revolve upon AB, until it coincides with the horizontal plane. In order to illustrate what is meant by the

plane revolving upon AB, let a triangle be placed with the bevelled edge on AB, and keeping this edge in contact with the surface of the paper, let the triangle be turned down, until it becomes horizontal. It will thus assume the position of the triangle ABC. In order to construct this triangle, draw BC perpendicular to AB, and make it equal to BB' (because BB' expresses the height of B above the horizontal plane of projection) and join AC. Then AC the hypotenuse is the real length of the line.

AC may be regarded as the elevation of AB, when viewed at right angles to the plane, passing through it at right angles to the horizontal plane, i.e., in the direction of CB.

The construction may also be made in the vertical plane as follows:--make B'D' equal to AB, and join B'D'. Then B'D' is the real

B

18

BY

length of the line. The triangle B'D'B represents the vertical plane conceived to pass through AB, after it has been made to coincide with the vertical plane of projection, by being moved through the arc AD'.

We thus have this practical rule for finding the real length of a line, whose projections are given-viz., upon the given horizontal projection construct a right-angled triangle of which the altitude or perpendicular is equal to the difference of the altitudes of the extremities of the line above the plane of projection. With reference to the vertical plane, we should make the vertical projection of the line the base of a right-angled triangle of which the perpendicular is equal to the difference of the distances of the extremities of the line from the vertical plane of projection.

Problem 23.

Given the projections of a line, to find the angles which it makes with the planes of projection.

Let AB and A'B' be the projections of the given line. It is re

quired to find the angles which it makes with the planes of projection.

From a consideration of the preceding, it will be readily seen that the angle made with the horizontal plane is CAB. The angle made with the vertical plane is A'B'C', which is found as follows:-From point A', A'C' is drawn at right angles to A'B', and equal to À'A. B'C' is then joined.

NOTE 1.-The angle which a line makes with its plan is its inclination to the horizontal plane, and the angle which a line makes with its elevation is its inclination to the vertical plane,

NOTE 2.-When the actual number of degrees is not required, the inclination of a line to the horizontal plane is usually indicated by the Greek letter 0, and to the vertical plane by p..

Problem 24.

The traces of two planes being given, to find the projections of their common intersection.

Let AB and AC be the horizontal traces of the two planes, meeting in A; and D'B, D'C', the vertical traces of the two planes, meeting in D'. It is required to find the projections of their common intersection.

Since the points A and D' are common to the two planes, the line joining them will be the line in which the planes intersect; and the projections of this line will fulfil the conditions given.

Now as A and D' are the traces of a line in space, its projections can be found by Pr. 21. Hence, draw D'D at right angles to xy,

and join AD; then AD is the horizontal projection. Similarly, to

B

y

and

find the vertical projection. Draw AA' at right angles to xy, join A'D'; then Â'D' will be the required vertical projection. NOTE. When the horizontal traces AB and AC are parallel, the hori

zontal projection of their common intersection, DE, will be parallel to AB, AC; and F'D', its vertical projection, will be parallel to xy.

To determine the angle contained by two straight lines, AB and BC, given by their projections.

In this case, if the horizontal traces of the lines be joined, a third line will be formed, which, with the two given lines, will form a triangle, the vertical angle of which it is required to determine. If the triangle be constructed into the horizontal plane, its base being the axis of rotation, its true shape will be determined, and therefore the required angle between the lines AB and BC.

Find d and e, the horizontal traces of AB and BC, and join de; then dBe will be the triangle mentioned. In constructing it into the horizontal plane, d and e will be fixed, and the point B will travel in a vertical plane, at right angles to de.

Through b draw bf at right angles to de and produce it beyond b. The actual distance of B from f is the length of the hypotenuse of a

right-angled triangle, of which bf is the base, and hb' the perpendicular.

Along xy, from the point h, set off hf' equal to bf. Join b'f', and make fB in the plan equal to f'b'. Join Bd and Be, and the angle dBe is that between the two given lines.

Problem 26.

The traces of two parallel planes being given, to find the distance between them.

Let A'A', AA, and B'B', BB, be the traces of the given parallel planes. It is required to find the distance between them.

Draw CD at right angles to the horizontal traces of the planes, and DE' perpendicular to xy; CD and DE' will be the traces of a plane at right angles to the horizontal plane of projection.

Now this third plane will cut the given planes in two straight lines, which will be parallel to each other; for if two parallel planes be cut by another plane, their common sections with it are parallel. The plane CD, DE, cuts AA', A'A', in the line CD, and BB', B'B', in the line DF. Now we proceed to find the angles which the given planes make with the horizontal plane of projection; we shall then obtain the distance required. Thus, make DF,