PROBLEM II. To project a right circle, or one, that is perpendicular to the plane of projection. Through the centre C of the primitive draw the diameter AB, and take the distance from its parallel great circle, and set from A to E, and from B to D, and draw ED, the right circle required. E D B PROBLEM 5. Hence it is evident, that a line parallel to the plane of the projection is projected into a right line equal to itself; but a line, that is oblique to the plane of projection, is projected into one, that is less than itself. 6. A plane surface, as ACBD, perpendicular to the plane of the projection is projected into the right line, as AB, in which it cuts that plane. A Hence it is evident, that the circle ACBD perpendicular to the plane of projection, passing through its centre, C E is projected into that diameter AB, in which it cuts the plane of the projection. Also any arc, as Cc, is projected into Oo, equal to ca, the right sine of that arc; and the complemental arc cB is projected into oB, the versed sine of the same arc cB. 7. A circle parallel to the plane of the projection is projected into a circle equal to itself, having its centre the same with the centre of the projection, and its radius equal to the cosine of its distance from the plane. And a circle oblique to the plane of the ED, and let fall the perpendiculars EF, DG; bisect FG in H, and erect the perpendicular KHI, making KH= HI= half ED; then describe an ellipse, whose transverse is IK, and conjugate FG; and that will represent the given circle. PROBLEM IV, To find the pole of a given ellipse. Through the centre C of the primitive draw the conjugate of the ellipse; on the extreme points F, G, erect the perpendiculars FE, GD, or set the transverse IK from E to D, bisect ED in R, and let fall RP perpendicular to AB; then is P the pole. E B CHPG K PROBLEM the projection is projected into an ellipse, whose greater axis is equal to the diameter of the circle, and its less axis equal to double the cosine of the obliquity of the circle, to a radius equal to half the greater axis. PROBLEM V. To measure an arc of a parallel circle; or to set any number of degrees on it. With the radius of the parallel, and centre C, describe a circle Gg, draw CGB and Cgb; then Bb will measure the given arc Gg; or Gg will contain the given number of degrees, set from B to b. PROBLEM VI. To measure any part of a right circle In the right circle ED, let EAD; and let AB be the part to be measured. On the diameter ED, describe the semicircle END, draw AN, BO, LP, perpendicular to ED. Then ON is the measure of BA, and NP of AL; and ON, or NP, may be measured as in Prob. V. N COR. If the right circle pass through the centre, it is only necessary to raise perpendiculars on it, which will cut the primitive, as required. PROBLEM PROBLEM VII. To set any number of degrees on a right circle. [See Figure under last Problem.] On ED, the given right circle, describe the semicircle. END; then, by Prob. V. set off NP the given de grees, and draw PL perpendicular to ED; then AL contains the degrees required. PROBLEM VIII. To measure an arc of an ellipse; or to set any number of degrees on it. About AR, the transverse axis of the ellipse, describe a circle ABR; erect the perpendiculars BED, KFI, on AR; then BK is the measure of EF, or EF is the representation of the arc BK. And BK is to be measured, or any degrees set on it, as in Prob. V. CD I STEREOGRAPHIC STEREOGRAPHIC PROJECTION. PROBLEM I. To find the poles of any projected great circle. 1. The poles of the primitive circle. They are in the cen tre C.* C 2. The The following are laws of the stereographic projection. 1. In this projection a right circle, or one perpendicular to the plane of projection, and passing through the eye, is projected into a line of half tangents. 2. The projection of all other circles, not passing through the projecting point, whether parallel or oblique, is into circles. |