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DC', equal respectively to DF, DC, and join GF, HC. Then ab at
right angles to these lines is the required distance between the planes.
Problem 27. To determine by its traces a plane parallel to a given plane, and at a given distance from it.
Let the given plane be A'BC, and ab the given distance.
It is required to determine by its traces a plane parallel to A'BC.
In this case we proceed as if to find the inclination of A'BC, and at a perpendicular distance equal to ab, draw DE' parallel to FE".
Then E' is one point in the vertical trace of the required plane, and as parallel lines have parallel traces, G'H and HK drawn parallel to A'B, BC, will be those of the plane required.
Let ABCD and ABEF be the two given planes, of which BC and BE are the traces; and GH, GK; AD, AF, horizontals.
First, find the elevation of AB, the intersection of the planes ; that is to say, make AA' drawn perpendicular to AB equal to the height of A above the plane of projection, and join A'B. Draw mn at right angles to AB, and consider it as the trace of a plane cutting the traces of the given planes in m and n.
Next, if we consider mn as the trace of a plane at right angles to the horizontal plane, it would cut the given planes in a triangular section. This triangle will be found thus—make Gb equal to Ga (because G is elevated above the plane of projection a distance equal to Ga) and join mb, nb.
The solution of the problem then consists in finding the sections of the planes when cut by a third plane at right angles, not to the horizontal plane, but to AB, the intersection of the planes. Hence, draw Gc at right angles to A'B; Gc will be the elevation of the plane drawn perpendicular to A'B; and as this plane contains the lines which measure the angle between the given planes, we have only to construct Go to find this angle. Make Gd equal to Gc and join md, nd; then mdn is the angle between the planes. The angle mdn is called the dihedral angle and profile angle of the planes.
NOTE 1.-A dihedral angle is the angle contained by two intersecting planes.
NOTE 2.-The profile angle of two planes is the angle contained by the two straight lines in which these planes are cut by a third plane, at right angles to both of them. This third plane is called a profile plane. Since these lines are perpendicular to the intersection of the two given planes (Eac. XI., Def. 3), the profile angle will be the measure of the dihedral angle (Euc. XI., Def. 6).
To draw a plane, so that it makes a given angle with a given plane and passes through a line in the first.
Let ABEF be the given plane, the line in which the planes are to intersect each other being AB.
Find A'B, the elevation of AB. Then draw mn perpendicular to AB, intersecting it in G, and from G draw Gc perpendicular to A'B.
Make Gd equal to Gc, and join dn. At the point d in dn, make the angle ndm equal to the angle which the required plane is to make with the given plane ABEF.
Then the point m, where dm meets min, will be a point in the horizontal trace of the plane sought; and as B is another point in this trace, join Bm, and produce it to C, then BC is the horizontal trace of the required plane, which will be completed by drawing AD parallel to BC.
NOTE.—This problem is the converse of the preceding.
To determine the plan and elevation of any line inclined at 60° to the horizontal plane and 20° to the vertical plane.
If a number of lines lie upon the surface of a cone standing with its base upon the paper, and each line having one of its extremities
in the apex, these lines will all be equally inclined to the horizontal. They will also make with the paper the same angle which the sides of the cone makes with its base. The surface of a cone whose base angle is 60° is the locus of all straight lines which pass through the apex, and are inclined at that angle to the horizontal plane.
In sy take any point A', draw a line A'B' making the angle of 60° with it. Draw BB at right angles, and consider A'B'B' as half elevation of a cone. Then an arc, having B' for its centre, and B'A' radius, will represent part of its plan.
Now, if lines through B' be conceived to lie upon the surface of the cone, only two of them-2.., those on the extreme right and left
–will be shown in their full length in elevation. As the line travels round the solid, its elevation alters its length ; when it is in such a position as this, it makes an angle with the vertical plane. In the given case, therefore, we have to determine the exact position upon the cone, when the line is inclined 20° to the vertical plane.
There will be four solutions—two when the line is in front of the cone, and two when it is behind.
At A', set out a line A'B equal to the side of the cone, and making an angle of 20° with y. Draw BC at right angles to the base line, and the length A'C is that of the elevation of the line when it makes an angle of 20° with the vertical plane. With B as centre, and radius equal to AC, describe an arc intersecting xy in A". Join A"B', which is the required elevation.
A projector from A" meeting the arc first drawn in A, gives the plan of A", one of the extremities of the line. Join A B' and the required problem is solved, i.e., AB' is the plan of the line.
Given an equilateral triangle with two of its sides inclined, at 60° and 30° to the horizon, to draw its plan, and determine the inclination of the plane in which it is situated.
Let ABC be the given equilateral triangle.
First, draw BD, CE making with AB and AC angles of 60° and 30°.
From A, draw AF perpendicular to EC, and with centre A, and radius AF, describe a circle. Draw Gc parallel to DB, tangential to this circle, and cutting AB in c. Then as AH is equal to AF, C and c will have the same altitude above the horizontal plane. Let Cc be joined, and we have one of the horizontals of the plane sought.
Draw xy perpendicular to Cc produced, and cutting it in C'.
Project the point A to A, and with centre C', and radius C'A', describe an arc A'é.