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scribed, the vertical projection A' of the point A will assume the position A". In the same manner, the vertical projection B' of the point B will assume the position B, and the line joining A" and B" will be the vertical projection of a line in space, whose horizontal projection is AB.

14. Again, in the subjoined figure, let xyu and xyu' represent the two planes of projection, xyu being the vertical, and xyu' the horizontal.

Let the vertical plane xyu revolve about xy until it takes the position syu', when xyz the horizontal plane and xyu' form one plane. Let A be a point in space; a line drawn at right angles to the vertical plane will intersect it in A', which is the vertical projection of the point A. Next, drop a perpendicular from A to meet the horizontal plane xyz in the point Ā, which will be the horizontal projection of the point A. On revolving the planes as stated, the vertical projection A' will assume the position A".

After the vertical plane has been turned down to coincide with the horizontal plane ; wyr, i.e., that portion of it which was below the line of intersection, will take the position syr'. Consequently, any elevation on it will be in front of the line of intersection.

15. It will be seen from the foregoing figures, that the distance of the elevation of a point from the ground line (xy) shows the distance of the point from the horizontal plane ; also that the distance of the plan of a point from the ground line (wy) shows the distance of the point from the vertical plane.

NOTE.- In the course of solving problems in solid geometry it is fre. quently necessary to revolve a plane, until it coincides with the plane of projection. This is termed rabatting, or constructing the plane.

Section 11. PROJECTION OF POINTS, LINES, &c.

Problem 1.
To find the plan of a point, its elevation being given.

Let A' be the elevation of the given point. It is required to find its plan.

From A' draw a line perpendicular to xy. The plan of A' will be in this line. Let A be its plan. Then the points A' A are the ele

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vation and plan of a point in space, the height of which above the horizontal plane is equal to A'a, and the distance of which from the vertical plane is equal to Aa.

Problem 2. To draw the plan of a line, its elevation being given at right angles to the vertical plane.

As the line is at right angles to the vertical plane, it will be projected on that plane in a point, just as (see figure, page 191) the line BC is projected on abxy in the point C'.

Let A' then be its elevation.

The plan of the line is found by dropping a perpendicular from A', and making Aa equal in length to the given line. The line Aa

is parallel to the horizontal plane, and is projected on that plane in length equal to the original line.

NOTE.-The line Aa, viewed in the direction of its length, will be seen as the point A' in the vertical plane; and A’, viewed from above, i.e., at right angles to xy, will be seen as Aa. Thus, Aa and A' are the plan and elevation of a line at right angles to the vertical plane, parallel to, and situated above the horizontal plane at a distance equal to A'a.

Problem 3. To find the plan of a line, its elevation being given parallel to the two planes of projection.

As the line is parallel to the two planes, its projections will be parallel to xy. Let A'B' then be its elevation.

Its plan AB is parallel to cy, and equal in length to the original line. The lines A' B' and AB are the projections of a line in space, elevated above the horizontal plane a distance equal to A'a or Bob, and removed from the vertical plane a distance equal to Aa or Bb.

Problem 4. To find the plan of a line, its elevation being given parallel to the horizontal plane, but inclined to the vertical plane of projection.

Here the elevation of the line is parallel to sy, as in Pr. 3.
Let A'B' be its elevation.

The plan AB shows that the line meets the vertical plane in A, also that it is inclined to that plane at an angle BAb.

The plan AB is the real length of the line.

Problem 5. To find the plan of a line, inclined to both planes of projection, its elevation being given.

Let A'B' be the elevation. Taking A and B as the distances of

points A and B respectively, from the vertical plane, the line joining those points will be the plan of A'B.

NOTE. -Neither the plan nor the elevation expresses the real length of the line, nor its inclination to the two planes of projection. (See Pr. 4.)

Problem 6. To find the plan of the end elevation of a rectangular surface given parallel to the horizontal plane.

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Let A'B' be its elevation. Now, the points A and B' represent

lines perpendicular to the vertical plane, the projections of which will be found as in Pr. 2.

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