be the projection of the given line. Whence follows a first con. struction.

Make I B = 6 B, and I a = a A, the angle at I being the inclination as in the former problem; and find b, a, the projections of B, A; and the line a, b, is the projection sought.

To find the inclination of the line AB to the plane of projection, if we draw in the projecting plane of AB the line BC parallel to the plane of projection, this will form with AB the same angle that ab, forms with it, and consequently that AB forms with the plane of projection. Now, in this rightangled triangle we have AB the given line, BC the equal of ba, which is already found, and CA the difference of the orthographic projectors of A and B which are exhibited as BB,, ar, in the figure. Whence, if we make a,C' perpendicular to ab, and either make a equal to aD the difference of as, and BB,, or make 6,C' equal to AB, we shall have a triangle equal in all respects to that just verbally described, and ab,C' is the inclination of AB to the plane of projection,

This method, though capable of application in all possible cases, is often superseded in practice by more facile processes, of which two are given here.

(6.) When the figure line can be produced to meet the trace of the figure plane within the limits of the drawing, the following method is that most frequently employed in practice.

Let AB be the figure-line, produced to meet the trace in B: find, as before, the projection a,, of one point A, and join a B. This is the projection sought.

For a, is one point by construction ; and B is a point also in the projection by Art. 16.

Again, make a, C'at right angles to a,B, and equal to a, A', and join BC': the angle a,BC' is the inclination of AB to the plane Á of projection.

For it is obvious that a,C'B is but the triangle a,AB turned about a,B on the plane of projection, and hence that the angles a,BC' and a BA are equal, whilst the latter of these is the inclination itself.

(c.) The following method also possesses advantages in practice which will often be recognised.

Let A, B be points in the line, and take any other point C in the figure plane that may_appear convenient, and draw CA, CB to meet the trace in D, E: find the projection c, of C; draw De,,



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Ec, which will be the projections of CD, CE; and draw Aaa,, Bbb, parallel to Cc, meeting De,, Ec, in a, and b, Then ab, is the orthographic projection of AB.

For, always after the figure plane is turned round on the plane of projection, the point and its projection are in one line perpendicular to the trace; as Aa, is perpendicular to DE. Also, the projection of A is in Dc,, and hence a, the intersection of Aa and Dc,, is the projection of A. In like manner for b,.

The inclination is found as in the first construction, the altitudes or projectors being an', BB' found by drawing parallels from a, and b, to the trace.

(d.) When the given line is parallel to the trace, it will be parallel to the plane of projection. It will in that case be sufficient to construct the projection of one point A, and draw through a,, a parallel ab, to the trace aX.

The inclination is zero; and since for all the points the projectors are of the same length, the same result is obtained by construction as by our previous reasoning we were led to deduce.

(e.) If the given line AB be perpendicular to the trace, only one point will be necessary to be found.

For find the projector a, of A in it as before ; and then a itself will be the projection of the trace of the line.

The inclination of the line AB is in this case the same as the inclination of the figure-plane itself; and is therefore given.

(f.) If the figure-plane be perpendicular to the plane of projection, the projections of every line in it will be upon the trace.

The constructions will terminate in the trace, and the inclinations of the lines to the plane will be the same as their inclinations to the trace.

SCHOLIUM. The converse case often occurs in a series of consecutive constructions, viz. :

Given the projection of a line with the trace and inclination of the plane, to find the original line.

The requisite operations follow so readily from the construction of the direct problem, as to scarcely call for special remark : yet we shall give one example on account of its frequency of use.

Let aB be the trace of the given plane, and AaA' its inclination ; and let a'B be the projection of the line in the given plane. On AB



describe a semicircle and join Aa; it will pass through a' the projection of A, and a'B will be the length of the projection of AB. Make, therefore, Bæ = Ba' and draw Aa. Then AaB is the projecting triangle, and aBA the angle of inclination of the line.


Given any rectilineal figure in a given figure plane to construct its projection.

(a.) We may project the angular points (as in Prop. I.) and draw lines to join them. Thus in the triangle ABC we should have the figure as follows, according as we keep the figure after revolution, and the projection in the same or different sides of the trace of the figure plane.


(6.) We may project the sides of the given figure, viewed as indefinitely extended lines (as in Prop. 11.): these projections by their mutual intersections give the limited projection of the figure. To render this method useful, and avoid a great deal of unnecessary labour, and the consequent complexity of the figure, some degree of discretion is necessary. Take, for example, a triangle ABC, and let the sides AB, BC, ČA be projected by taking points D, E, F arbitrarily in these respective sides, as those which with the traces are to define the positions of the sides of ABC. Then the figure would resemble the following:

The complexity of such a figure is sufficiently discouraging in practical operations; and evidently, this complexity becomes greater with every new line of the figure to be projected. This complexity is, however, in a great degree evaded by taking, instead of D, E, F, some specific points either in the figure, or having known relations of position with respect to it. It is impossible to give any general rule for this choice: but the nearest approach to generality is to select an intersection of a line with an adjacent one-in fact an angular point of the figure—as the constructive one. The same triangle so constructed in projection would have its delineation as follows: but, on the whole, the method itself is not the most facile, and is only dwelt upon here for the sake of completeness of discussion.

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(c.) In a practical point of view, the greatest facility is afforded in construction by the following process. As the method is precisely the same how numerous soever the lines may be, we shall give it in detail for the irregular hexagon ABCDEF.

Produce some two adjacent edges AB, AF to meet the trace a X of


the figure plane in B, 4, and project the point A in a, by Prop. I. ; draw AC, AD, AE to meet the trace in y, d, e, and then draw Ba,, γα, δα, εα, φα. These will be the projections of , Αγ, ... A0; and hence the projections of B, C, ... F in them will be in Ba,, ya,, ... pa,. But these projections are also in lines Bb, Cc, ... Ff perpendicular to the trace; and hence at their respective intersections 6,, c, d, e, f, We are, therefore, only required to draw these two sets of lines Ba, and Bb, ... to obtain the angular points of the projection; and these being joined in the same order as the original points, the projection is complete.

(d.) It will sometimes happen that no two adjacent sides are well adapted for the application of this method. In such case any arbitrary but more convenient point may be assumed as that to be projected; and the other points found by drawing lines through the given angular points to the trace, as was done in the preceding construction. Very often such a point may be conveniently found by producing two alternate sides, or indeed any two sides of the figure to meet; but in all cases the selected point should be such as to throw all the lines from it to the given angular points in such positions as to cut the trace within the limits of the drawing.

For example, if in the preceding figure the pentagon BCDEF had been given, no two adjacent sides of it would have been well fitted for the operation; but by taking the point A arbitrarily as in the figure, we should have found b, c, d, e, f,, as already shown with much facility.

(e.) It will sometimes happen that where one or more pairs or sets of parallels enter into the composition of the figure, the work may be still further simplified by the application of the principle that the projections of parallel lines are parallel.

Let there be given, for instance, the parallelogram ABCD, having a line drawn through E parallel to AD or BC, and lines from F and G parallel to AB or CD; then of two adjacent sides AB, AD, as before, find the projections of A, E, B, F, G, D; and draw parallels from e, b, fi, 9,,d,: these will constitute the projection required.

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