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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.

Let QB, 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 O, coincides with a and B, with B.

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 .

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 , B, be the traces of the line, and through a,ß, draw planes QBLB., Q, Usb, perpendicular to the planes XY, XZ, cutting the two planes as in the figure of the eidograph.

Then, since XZ and Q, BB, are perpendicular to XY, the line B.B, is perpendicular to XY, and to the line OX in it. (Pls. 11. 18). In like manner a, is perpendicular to OX. Also, the projection on the horizontal plane is , and that on the vertical is Bodą. Wherefore these projections can be found without actually describing the projecting planes; since from the positions of an, B, and the axis OX, we can find B, and legs

Again, had the vertical plane been turned to the horizontal as in the orthograph, and the points au Bg 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 Beßi, and 0, 0, perpendicular to OX, and join abi, apy, these will respectively be the horizontal and vertical projections which were required.

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

Draw the projections abı, abto meet the axis in B., Qg; and then perpendicular to the axis, draw and, to meet a,b, in a, and BB, to meet a,bin ßg: then ay, B, are the traces of the line whose projections are

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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 aja,) be the given point, and bc (or b,, b,c,) the given line; then lines through a, and a, parallel to b,c, and bec, 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 aßg, a line equal to which is to be constructed in orthograph, having given the projections a,Az, b,bz.

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

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

Bb, and AC to abı, we have AC = a,bi, and therefore given. And, finally Aa, 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 aga parallel to the axis, and take in it ac = a,bi, and join cbą, 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 B.B, till it coincides with the vertical coordinate plane XZ: then the points ang an, b, will describe circular arcs a, a, a,a, 6,6; and the lines Ba, Aa, Bb, will take the positions aßa, 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 6,6, aja, a,Q; join ,; and draw aA', 6B' parallel to B.B, 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 dg with radius Qaße describe the arc B2B to cut the axis in ß; join ; and draw b,B', a,A' parallel to the axis : then A'B' is equal to the segment of the line required.

SCHOLIUM. The student should take into con- lost sideration the circumstances of this problem 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 PP, 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 ab, through it parallel to the given line PP, is also given.

Again, the line Ab, being in the plane P, has its trace be 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 b, in it is given; whence the line b,a,, the projection of Abg, 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 b, b, perpendicular to the axis, meeting the vertical trace in ; draw b,a, parallel to the axis ; and finally a,a perpendicular to the axis meeting b,a, in ag. 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 a,.

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

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