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be of a different scale, either greater or less than the horizontal one, then these divisions 01, 12, etc., will represent only parts of the projection corresponding to that relation. If aa, be a multiple of the vertical unit, 01 will be the same multiple of the projective unit ; and similarly, if aa, be only a part of the vertical unit, oi will be the same part of the projective unit.
Moreover, it is clear that o'a is also one of the lines which fulfils the conditions of the problem.
PROPOSITION II. Having given two figured lines, and a point in one of them, to draw
the projection of a horizontal line through the given point to meet the other line.
Let ab be a given line, and p a point in it; to draw the projection of a horizontal line through p to meet another line cd.
Figure the line cd as before directed, and let p' be the point in it which is figured with the number at p. Then draw pp'; it will be the projection required.
For since p and p' are at the same distance from the plane of comparison, they are points in a line parallel to that plane.
In a manner precisely similar may any number of lines be drawn through as many given points; in whichever of the given lines the points be situated : or again, whether the points be in a specific line or not.
PROPOSITION III. Through a given point to draw a line parallel to a given line. Let ab be the projection of the given line figured at a and b, and let a' be the projection of the given point, also figured.
Now, since the lines are to be parallel, their projections will be parallel, and hence the position of the projection a' b' through a' is given.
Make a A, 6B, a' A' perpendicular to ab, a'b', and equal to the given B figures. Draw AB to meet ab in 0: then 0 is the zero point of the given line. Through Al draw A'O' parallel to A0: then O' is the zero point of the line sought. Draw 11', 22'... parallel to 00': then the projection a' b' is graduated.
PROPOSITION IV. Two lines are given, to find whether they intersect, and if so, at what
figure. (1.) If their projections be parallel, they can never meet, whatever their scales may be, for they are then in parallel planes.
(2.) Let their projections intersect (viz. ab and a' b') at c. Gra
duate them; then if c represent the same number on both lines, each in reference to its own scale, the lines meet in the point which is projected in c; and the figure of that point is the altitude of their point of intersection.
If c upon the two lines do not represent the same altitude, the lines do not meet; and the difference of these numbers is the difference of the altitudes of the poi.its in the two lines which have a common projection.
Given the figured projections of two lines which meet, to find the eido
graphic inclination of these lines. Let c be the intersection of the projections, and 0, 0 the zero points of the given lines. Make cC, C' perpendicular to Oc and O'c respectively, and each equal to the figure of the point c. From centres 0, 0' with radii OC, O'C', describe arcs intersecting in D; and join OD, O'D: then ODO' is the eidographic angle of the given lines.
For it is evident that ODO' is the angle made upon the plane of projection by the two given lines after revolution.
ScHolium. If the lines should have but slight declivity, the zero points may fall at an inconvenient distance. In this case any two points having the same figure may be taken instead of them, only remarking that the perpendiculars at c in the preceding construction must be diminished also in the same ratio.
For we are thus in reality only changing the plane of comparison and all the work remains, like the reasoning, precisely similar to that already employed.
PLANES, 1. In this kind of projection a plane is defined by its trace on the plane of comparison, and its inclination to that plane.
If we cut the trace of a given plane by a profile plane, that plane is perpendicular both to the given plane and to the plane of comparison ; and the common sections of the profile plane with those which it cuts are two lines, one in each of the planes, that contain an angle equal to the inclination of the given plane to the plane of comparison (Pls. 11., 18, and Def. 4). One of these lines is also the projection of the other; and by being properly figured, may represent the angle of inclination of the given plane to the plane of comparison. This is the mode of representation adopted for planes in this system; and this angle we shall express by declivity, or the declivity of the plane. One form of expression and construction founded on it will be-a plane is defined by its trace and declivity.
2. If lines be drawn in the given plane parallel to its trace, and at distances from that trace equal to the unit of the line of declivity (or at distances from the plane of comparison equal to the unit of altitude), their projections will be parallel to the trace, and at distances on the projection of the line of declivity equal to the unit of the general scale of projection in the system employed.
PROPOSITION I. To form the scale for figuring a given plane. Let X'X be the trace of the plane, and conceive a plane drawn perpendicular to it, cutting the given plane in a line and the plane of comparison in 04: about 04 let the profile plane be turned to coincide with the horizontal plane, so that the line of declivity shall take the position 0A,, draw the perpendiculars A, Q., A, 22, ... to OX : then 0a,, Oan, Odz ... are equal to the altitudes of the points in the line of declivity above the plane of comparison.
Wherefore we have this construction :
Draw (first fig.) a line 04. .. perpendicular to the trace X'X of the given plane: make the angle 40 A, equal to the given angle of declivity: make a, a,, ... a scale of altitudes, and draw a, A1, A, Ag, parallel to 04: then, lastly, draw A,1, A,2, A33, ... parallel to the trace. The scale of the plane is then graduated in 0, 1, 2,...
In practice it will be easier to find merely a, A., and step it off from 0 along 04.
Cor. We may readily find whether a given figured point be in a given figured plane or not. Let c be the projection of the point (second fig.): draw cc' parallel to the trace XX, and the point in question will be in the plane if Oc', be the figure in the given point: otherwise not.
SCHOLIUM 1. Should the trace be inconveniently situated for use, any line parallel to it may be used instead, if its altitude be given ; always recollecting that the numbering must proceed both ways from that point in the scale.
SCHOLIUM 2. For the sake of distinguishing to the eye the graduations of a line and a plane, it is usual to mark the scale of a plane with a double line ; and the one is drawn a little thicker than the other, for a reason which will presently appear.
To draw a plane through three given points. Let a, b, c, be the projections of the three given points A, B, C. Construct the lines through CA, CB as before (Sec. 11., 2), and let 0, 0, be their traces: join 0,02 and draw the perpendicular c0 to 0,0.. From the graduations 12, 22, 3, ... of one of the lines, draw parallels to 0,02, cutting co in 1, 2, 3, ... we shall then have c0 graduated for the plane through the three points.
Cor. 1. The construction is evidently the same as that for a on plane to pass through two given lines.
Cor. 2. If we wish to exhibit the inclination of this plane to the plane of comparison, we must draw cC parallel to 0, 0,, and make it equal to the figure of the point c, and join Co. The angle cOC is that required.
For it is in fact the angle of the profile plane placed on the horizontal plane, and OC is the line of declivity.
To draw a plane through a given point parallel to a given plane.
Let Oa be the graduation of the given plane, and a, the projection of the given point : draw the scale 0,a, of the plane through a, parallel to Oa: make Oo equal to the figure of the given point : draw OA to make the declivity of the given plane with Oa, meeting the perpendicular Aa in A, and draw Aa parallel to Oa. Make a,,, = a0 : then 0, is the zero point of the line of declivity of the plane sought.
Also, since the planes are parallel their traces are parallel, and the point 0, is given; therefore the trace and scale of the plane are found.
PROPOSITION IV. Through a given point in a given plane to draw a line that will make
a given angle with the plane of comparison. Let a' be the projection of the given point ; draw a'B perpendicular
to X'X the trace of the given plane, and a'A parallel to it and equal to the figure of the given point: make a'a the unit of the vertical scale to which the given plane is figured, and draw aß to make the angle aßa' equal to the given angle with the plane of comparison, and draw AB parallel to aß. With centre a' and radius a'B'describe the arc BO', cutting the axis in O'. Then ('a' is the projection of the line ; and its figures are found by parallels l'... to XX.
For, if we consider that AB of this figure is a line which fulfils the condition ; and that O'a' is the projection of same line revolved about the projection a' till it comes to the given plane, the truth of the construction will be apparent.