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The sum of the acute angles which a line makes with a plane varies between 0° and 90°. In fig. 113m, where the line is parallel to both planes, it is 0°; and in figs. 113p, 113q, likewise in the case of CD, fig. 118, the sum is 90° in each case.

Let the student cut out a square piece of cardboard, fig. 119c, of 3" side, cut it half through and fold it half over along the line AB, and fit it in the first angle of his planes as indicated in the view d, so that if AB represent a line in space, ab will be the plan, and a'b' the elevation. The model used in conjunction with the following explanations will make this part of the subject clear. Imagine a line AB in space as in fig. 120.

If the plan ab, and the elevation a'b', be assumed anywhere in the planes of projections as indicated, it will be observed that by drawing imaginary projectors through the extremities of these projections, they will meet in AB, the actual line in space; and by producing a'b' and ab to meet xy, the elevation marked, h'.t'., of the horizontal trace, and the plan of the vertical trace are located. h.t. and v'.t'. are then at once found by drawing perpendiculars to xy, which, as we have already seen, adequately represent the traces of the line. If AB be produced both ways it will, of course, pass through both traces, as shown.

Referring to the projection on the H.P. (ie., the plan), the form of the plane figure ABba is quadrilateral, with right angles at a and b. If, therefore, the true shape of the quadrilateral be constructed on plan ab as a base, and folded over or rabatted, as it is called, into the H.P. about ab as an axis or hinge-line, we get the figure A,B,ba, having on the H.P. the actual size of the quadrilateral, and therefore the true length A,B, of the line AB. a B1, b B, are the heights of the projectors through A and B respectively, and are evident in the model.

Further, if A,B be produced to h.t., the angle so formed with the foreshortened plan ab is the angle of inclination or slope of the line AB to the H.P., sometimes indicated by the Greek letter (theta) when the measure of the angle in degrees is unknown.

In a similar manner, referring to the V.P., ABb'a' is a second quadrilateral, with right angles at a' and b'. If, therefore, the true shape be constructed on elevation a'b' as a base, and

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Fig. 120.-Showing the perspective of a line A B in space, together with its projections, traces, inclinations, and true length.

rabatted into the V.P. about a'b' as a hinge-line, we obtain the figure A,B,b'a', giving on the V.P. the actual size of the quadrilateral, and therefore A,B,, the true length again of the line AB. a'A, b'B, are the distances of the projectors A and B respectively from the V.P.

Further, if A,B, be produced to v'.t., the angle so formed with the elevation a'b' is the angle of inclination of the line AB to the V.P., sometimes indicated by the letter (phi).

These principles will be employed in the following problem.

PROBLEM 85.

Having given the projections of an oblique line, AB, to determine (1) its traces; (2) its actual length; (3) its inclinations to the planes of projection.

For the purpose of comparing this figure with the preceding one, practically the same lettering is adopted, and the sizes used are proportionate to those given for the model in fig. 119.

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ab, a'b' would first be drawn in any desired position. The horizontal trace being the intersection of the line (or the produced line) and the H.P.; and xy being the elevation of the latter, the elevation of the horizontal trace will be determined by producing a'b' to intersect xy. Project the intersection down to cut the produced plan ab in h.t., the required horizontal trace.

Similarly, for the vertical trace, produce the plan ba to intersect xy. Project the intersection up to cut the produced elevation b'a' in v.t., the required vertical trace.

It must not be forgotten, though h.t. and v.t. are said to be the traces of the line, they are virtually only the plan and elevation of the horizontal and vertical traces respectively.

At the points a and b erect a▲,, bв1, perpendiculars to ab, equal in lengths to o'a', p'b', the heights of A and B respec

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tively. Join A,B,, and produce to h.t. 0 is the inclination of AB to the H.P., and AB is the actual length of the line.

At the points a', b', erect a’A2, b'в, perpendiculars to a'b', equal in lengths to o'a, p'b, the respective distances of A and B from the V.P. Join B2A2, and produce to v.t. is the inclination of BA to the V.P.

The determination of the vertical and horizontal traces by producing the true length to meet the respective projections produced, as shown by the latter construction, would, of course, do instead of the first method. It will be found on measurement that the angle is not equal to the angle which the plan ab makes with ay. Similarly, is not equal to the angle which a'b' makes with xy. As a fact, when a line is oblique to both planes, and are always less than the corresponding apparent angles. Should the traces be inaccessible, the angles are found by drawing the dotted lines B1s and At′ parallel to the projections of AB as shown.

To make the student's solution still clearer, cut through the three border lines above ry, as shown in strong lines, and rotate about xy the piece cut, till the V.P. is represented perpendicular to the H.P. Cut also along the lines Bv.t. and B'; likewise cut Ah.t. and A,a. Now, folding the triangles about their bases b'v.t. and ah.t. respectively, till they are at right angles to their planes, it will be found that as the whole arrangement is in its natural position, A,B, and A,B, will coincide.

PROBLEM 86.

Given the projections of a line EF, to find its inclinations to the planes of projection; and (1) to extend its length to a point 9, g', so that EG 2", (2) to cut off a segment, FH, of its length,

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The projections of the oblique line ef, ef' are given perpendicular to xy, and represent an oblique line in a plane at right

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angles to both planes of projection and intersecting both planes in the ground line. As they are only the apparent length of the line, it is not possible to measure the true lengths on them. E is in both planes, and F is in space.

From ƒ draw a line fr parallel to xy, equal to eƒ', and join er.

The angle is the inclination to the H.P., and its complement is the inclination of the line to the V.P.

Produce ef, ef' and er. Make eG = 2". Draw Gg (which the real height of G) parallel to Ff. Set up e'g' equal to go. eg, é'g' is the solution to (1). Next, cut off FH = '', in Ge

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