PROBLEM 40.

A plane making an angle of 40° with the horizon contains on its surface a line 1.25" long inclined to the horizon at an angle of 30°; show plan and elevation.

Draw V'g at an angle of 40° to X Y in the elevation, and g h ༡ at right angles to X Y in plan.

Then V' g h are the required traces of the plane.

In the vertical plane draw line A B inclined at an angle of 30° and 1.25" in length.

From A draw A b' parallel to X Y, meeting V' g in b'.

Then g b' is the elevation of the required line.

With centre g and radius g b' describe arc meeting X Y in c. At c erect a perpendicular c b" to X Y.

From a point a in the horizontal draw a b′′ 1.25" long, and from b" draw b" b parallel to X Y, intersecting the projector from b' in the point b.

Join a b.

Then a b is the required plane of the Ine.

The position of the point a may be fixed by the conditions of problem.

The line A B may be inclined in either direction, as all we require is the height A above the horizontal.

The plan a b can also have another position, viz., that of being inclined from X Y.

PROBLEM 41.

To determine the plan of a square 1.25" side when lying in a plane forming an angle of 30° with the horizontal, and one of its sides inclined at an angle of 15°.

Obtain a b, a' b', and A B, as in last problem.

On A B, the true length of the side of the square, construct a square A B C D.

From each of the angular points erect perpendiculars to X Y, and revolve the points o p ƒ thus obtained into the vertical trace by arcs o c', p d', f b', having a' as centre.

From b', d'and c' drop perpendiculars to X Y, which produce in plan.

From C D and B draw lines parallel to X Y until they meet the projectors from b' d' c' in the points b c d.

Join a b, bc, c d.

Then a b d c is the plan of the square required.