These orthographs are representative of the elevation and section, and of the plan and section respectively.

SCHOLIUM. 1. When the line is given under some particular circumstances, the construction is simplified. Three cases deserve notice from their frequent occurrence in practice.

1. When the traces of the line are given.

A

Let a,, B, be the traces: draw as in the figure: then the dark line is evidently the projection on the third plane.

2. When the line itself is the section of any plane whatever with a profile plane; and that profile plane can be taken as the coordinate profile plane.

This is obviously only the particular case of the preceding, in which a coincides with a and B, with ß.

3. When the line is parallel to one of the coordinate planes.

SCHOLIUM 2. It will be seen hereafter that the same process can be opplied to any figures whatever. In fact it will be obvious that we can always construct by points as we have here done the line aß.

PROPOSITION III.

Given the traces of a line to find its projections; and, conversely, given the projections of a line to find its traces.

(1) Let α, ß, be the traces of the line, and through a18, draw planes a1BB, a1aß, perpendicular to the planes XY, XZ, cutting the two planes as in the figure of the eidograph.

Then, since XZ and a‚ß‚ß, are perpendicular to XY, the line ẞ,8, is perpendicular to XY, and to the line OX in it. (Pls. II. 18). In like manner is perpendicular to OX. Also, the projection on the horizontal plane is aß, and that on the vertical is B. Wherefore these projections can be found without actually describing the projecting planes; since from the positions of a,, B, and the axis OX, we can find B, and α.

Again, had the vertical plane been turned to the horizontal as in the orthograph, and the points a1, B been given, the same processes might be performed upon the united plane that were separately performed upon the two planes in eidograph. Whence it follows, that if

we:

Draw BB, and aα, perpendicular to OX, and join a11, aß, these will respectively be the horizontal and vertical projections which were required.

(2) The converse follows from the same considerations, viz. :—

Draw the projections ab1, ab, to meet the axis in B1, a; and then perpendicular to the axis, draw aα, to meet a,b, in a, and BB, to meet ab in B: then a1, B are the traces of the line whose projections are ab, abջ.

SCHOLIUM. In the preceding figure, the traces are given for the line lying in the first region; but the process is the same for all the others. COR. 1. If the projections of two points in the line be given, the same things may be found, viz., the traces and the projections of the line passing through them: for the projections of the line will pass through the projections of the points.

The projections being found by drawing these lines on the coordinate planes; then the traces of the lines will be found as in the problem itself.

COR. 2. Hence, also, if a point and a line be given by means of their projections, the projections and traces of a line through the given point parallel to the given line can be found.

Thus, let a (or a,a,) be the given point, and be (or b,c,, b,c,) the

given line; then lines through a, and a, parallel to be, and be, will be the projections of a line through a parallel to be.

For, when lines are parallel their projections on any plane will be parallel, and the converse: but in the present" case, the projections are, by construction, parallel; and hence they are the projections of parallel lines.

PROPOSITION IV.

Given the projections of two points in a line, to construct a line equal to the distance of those points.

In the eidograph, let AB be the segment of the line a, a line equal to which is to be constructed in orthograph, having given the projections a,a,, b1b2.

Let a BB, be the plane which projects the line horizontally, and Bb, Aa, the lines which project the points A, B. Draw AC parallel to a, B, meeting Bb, in C.

Then snce a BB, is a right angle, and AC is parallel to a11, ACB is a right angle, and therefore given. Also, since Aa, is parallel to

=

Bb, and AC to ab, we have AC ab1, and therefore given. And, finally Aa1, Bb, are the heights of the projections on the vertical plane of A and B above the axis OX; and, therefore, likewise given, as well. as the difference BC.

Wherefore, we have given the two sides AC, CB and the included right angle C; the hypothenuse AB of which triangle is the length required.

Whence, in the first orthograph, if we draw aa parallel to the axis, and take in it ac ab, and join cb, this line is equal to AB: for its sides about the right angle are equal respectively to those of the triangle ABC about its right angle.

COR. 1. Had we employed the plane which projects the line upon the vertical coordinate plane, the same reasoning would have led to the construction indicated by the second orthograph.

COR. 2. When the traces of the line itself are given, as well as the projections of the points upon one of the planes are given, the following method will be convenient.

Conceive the projecting plane for the horizontal projection to revolve about BB till it coincides with the vertical coordinate plane XZ: then the points a, a, b, will describe circular arcs aa, a,a, bb; and the lines a,B, Aa,, Bb, will take the positions aß, A'a, B'b; and AB will coincide with A'B'. Whence, in the orthograph we have the rule:

With centre B, describe (the figure is easily conceived) the arcs bb, ɑɑ, a1a; join aß,; and draw aA', bB' parallel to BB: then A'B' is the length required.

COR. 3. The following method, adapted to the case of Cor. 2, is rather simpler as to actual work: the deduction of it from the eidograph is left as an exercise for the reader.

From centre a with radius aß, describe the arc BB to cut the axis in ß; join ∞‚ß; and draw b1B', a,A' parallel to the axis: then A'B' is equal to the segment of the line required.

SCHOLIUM. The student should take into consideration the circumstances of this problem

3.

A

under special relations of the data, as several such occur in subsequent problems.

PROPOSITION V.

Given one of the projections of a point situated in a given plane to find the other.

In the eidograph, let the horizontal projection a, of the point A situated in the plane P be given. Through A let a line Ab, be drawn in the plane P parallel to its horizontal trace PP1, and let the projecting planes of this line be drawn.

Then since Ab, is parallel to PP, the section of the planes P and XY, it is parallel to XY, (Pls. 1. 3): wherefore since the point a, is given, the line a,b, through it parallel to the given line PP, is also given.

Again, the line Ab, being in the plane P, has its trace b2 in the trace PP, of the plane; and the line bb, is perpendicular to the axis. Whence, having the point b, and the trace PP, given, the point b, is given.

Finally, the projection of Ab, on the vertical plane is parallel to the axis, since Ab, is parallel to XY, and its projecting plane is evidently parallel to the axis, and the point be in it is given; whence the line ba,, the projection of Ab, is given; and its intersection with aa, is given. That is, the projection on the vertical plane is given.

Wherefore the orthographic construction.

Through the given horizontal projection a, draw a line parallel to the horizontal trace PP, meeting the axis in b,; draw bb, perpendicular to the axis, meeting the vertical trace in b,; draw ba, parallel to the axis; and finally a,a perpendicular to the axis meeting ba, in a,. This is the vertical projection sought.

We might have operated in a similar manner if the vertical projection a, had been given instead of the horizontal one a1.

COR. If the projections of a point and the traces of a plane be given, either of the constructions (the horizontal or the vertical) will furnish a criterion, whether a given point be in a given plane or not—a criterion that in practice is often required. One alone of them is sufficient; for if one be fulfilled, the other will necessarily be so too.

PROPOSITION VI.

Through a given point to trace a plane parallel to a given plane. Since the traces of parallel planes upon each coordinate plane are parallel, it is manifestly sufficient that we find one point in either of