the horizontal trace of the plane. A and P are the traces, and consequently Ap, a'P the projections, of a line lying in the plane Aa P and having an inclination of 36°.

1'

Now let the plane Aa'P be rabatted on the horizontal plane; the line AP will obviously come into the position AP, Make AB,=1" and on AB, construct the square ABCD, which will be the rabatment of the square required. When the plane containing the square is raised to its former position the points b, c, d come to b, c, d, which are therefore the vertical projections of A, B, C, three angular points of the square, and Abcd is the plan required.

It will be observed in this Problem how much the constructions are simplified by taking the plane containing the figure perpendicular to the vertical plane of projection.

X

Fig. 57.

PROBLEM II. Fig. 58.

The side of a regular hexagon is 1"; its plane is inclined at 50°; and two adjacent angular points A and B are 0.4" and 1' respectively above the horizontal plane of projection. Draw its plan.

As in Problem I. draw Ss'a', the traces of a plane inclined at 50° and perpendicular to the vertical plane of projection. Find the two points a' and b' on the vertical trace of the plane at a distance 0.4" and 1′′ respectively from xy: a', b' are the traces of two horizontal lines in the plane Ss'a'. Now let the plane be rabatted on the horizontal plane and draw the two lines am and b''n parallel to Ss'; these are the rabatments of the two horizontal lines through a' and b'.

1 1

Next draw A,B, an inch long and having one extremity on each of the lines am and b'n; on AB ̧ construct the regular hexagon  ̧‚ǂDEF ̧.

1 1 1

Now let the plane be raised to its former position as in the last problem; abcdef is the plan of the hexagon.

As a check on the accuracy of the construction it may be noticed that if any line of the hexagon A,B,C,... be produced to meet Ss', the horizontal trace of the plane, as ÂB, and EF, the projections ab and ef of these lines must meet Ss' in the same points S and T respectively, for as the plane containing the regular hexagon turns about Ss' as an axis every point in that line remains fixed.

T

Fig. 58.

m

PROBLEM III. Fig. 59.

Two adjacent sides, AB and AD, of a parallelogram are 1′′ and 11′′ respectively and the contained angle 45°. Find the plan of the parallelogram when AB is inclined at 30° and AD inclined at 50°. Draw also an elevation on a vertical plane containing AB, and another on a plane making an angle of 80° with ab.

Taking any ground line xy and any point A find by Problem IX. Chap. II. the projections of two lines intersecting at A, containing an angle of 45°, and having inclinations of 30° and 50° respectively. AF is one of these lines, in the vertical plane of projection, and aE, Ae,' the projections of the other. Set off on AF the distance AB=1" and on AE, the first position of the second line, the distance AD1=11". ab and ad (=ad,) are the horizontal projections of the two sides of the given parallelogram in the position required. The figure is completed by drawing bc and cd parallel to ad and ab respectively. ABc'd' is the elevation on the plane containing AB.

To find an elevation on a plane making an angle of 80° with ab. Draw x,y, making an angle of 80° with ab for a new ground line, and draw aa", bb", cc", dd" at right angles to xy, making the distances of a", b", c", d" from x1y, equal respectively to the distances of A, B, c', d' from xy; for these distances are the heights of A, B, C and D respectively above the horizontal plane of projection. a"b"c"'d" is the elevation required.