SECTIon II. Orthographic Projection. 26. In this projection the eye is supposed at an infinite distance: in which case a great circle BCDF is apparently reduced to a right line equal to its diameter BD, the eye being imagined indefinitely distant in the direction sc. Then, also, every arc CA which has its origin at the apparent centre, has for its projection a right line sa equal to the sine of that arc. The quadrant cB or cd will be pro- jected into its sine, or radius. An arc as AE = CE – CA, will be projected into ae, = sin CE – sin cA = (by equa. U, chap. iv.) 2 sin # (ce — cA) cos (cE + cA) = 2 sin # AE cos (CA + #AE) .... (I.) 27. Every arc, as DA, from the edge of the disc towards the centre, is projected into its versed sine Da. The quadrant pc, therefore, is projected into the radius. Ds the versed sine of 90°; and the semicircle DCB into the diameter DB, the versed sine of 180°. 28. What is here remarked of the circle BCDF applies. equally to all great circles which intersect at s and form. the visible hemisphere: each df these semićircles is reduced to its diameter, and the hemisphere is reduced to. a disc. - - * * 29. In this projection every circle, great or small, whose plane prolonged does not pass through the eye, will be seen obliquely, and under an elliptical form: for. an oblique circle making throughout the same angle. with the plane of projection, its several parallel ordinates are all reduced in a constant ratio; therefore, the projected ordinates are all in a constant ratio to the corresponding ordinates of the circle of equal diameter on the plane of projection, and together constitute an ellipse. (Hutton on Ellipse, prop.3, cor. 1). 30. In general, in the orthographic projection every great circle perpendicular to the plane of projection is represented by its diameter; and every circle perpendicular to that plane is represented by a chord of the primitive circle equal to its diameter. Thus if, by way of showing the use of this projection, we assume the me-- Z! E!" ridian for the plane ... • ** of projection, the horizon will be represented by its diameter Hoo: the prime vertical by its diameter zv, which cuts the former perpendicularly: the six o'clock hour circle will be represented by its diameter which is the axis PP', making with the horizon the angle PCo = the height of the pole = the latitude = L: the equator will be projected into its diameter § making with the horizon an angle hoe = 90° – L: the parallels to the equator will be represented by chords, such as AB parallel to the diameter of the equator: the almucantars are projected into chords, such as Rs, parallel to the horizontal diameter Ho. - -31. AB being the projection of a certain parallel, suppose that the star which in its apparent motion describes that parallel, has its inferior transit of the meridian at B. The point B which is its place on the sphere, is also its place then in the projection. The star being, in the horizon at T, or will be the versed sine of its azimuth, and ct the sine of its amplitude. Also, refer ring the point T to the chord BA, BT will be the versed sine of the arc described from the inferior transit of the star to its rising, or the semi-nocturnal arc; TC will be the sine of the arc which remains to be described before it reaches the 6 o'clock hour circle, or TG will be the sine of the arc which deducted from 90° will leave the semi-nocturnal arc; or TG will be the sine of the arc which added to 90°, equivalent to AG, will give the semi-diurnal arc represented by AT. On AB as a diameter describe the semi-circle ADB ; from any points F and E answering to the position of the star at different instants, erect the perpendiculars. FF', EE"; produce ce to D; then it is evident that if the semi-circle ADB were elevated perpendicularly on the plane of the meridian, F, G, and E, would be the respective projections of the points Fo, D, and E'. 32. To find the value of the arc DE' projected into the rectilinear portion GE, we have Hence, to know the arc which answers to GE, GF, GT, &c. we must divide each of those lines estimated from the middle of the chord by the sine of the chord's distance from its pole. Again, from the triangle cét we have GT = cq tan GCT = sin D tan L.... (3.) D being the declination of the parallel; . 33. Through the point E, a projected place of a star, draw the chord R3 parallel to the horizon, QE will be to the radius QR the cosine of the azimuth, and EL perpendicular to the horizon will be the sine of the alti tude. To determine this from the projection, draw Gá parallel to the horizon, then sin A = EL = La + a F = GM + de cq sin GCM + EG sin EGa sin D sin L + AG sin DE’ cos Gco sin D sin L + cos d cos L cos hour angle = sin D sin L + cos d cos L cos II. . . . (6.) Put A = 90° – D, E = 90° – L, and z = 90° – A ; then will the last equation become - cos z = cos A cos E + sin A sin Ecos H, which accords precisely with the fundamental equation (2) of spherical trigonometry; and the equa. (1 and 4) may readily be deduced from the same diagram. - a SECTION III. Stereographic Projection. 34. In this projection, which appears to have been invented by Hipparchus, all circles, whether great or small, are represented by circles; and a second property, equally general and more curious, is that in the projection all the circles make respectively the same angles as on the sphere. 35. Let Abod be a great circle of the sphere, o the place of the eye: the diameter ocA being drawn, and the diameter BCD perpendicular to it, BCD will be the orthographic projection of a great circle perpendicular to the visual ray oA, or of the circle one of whose poles is o. it is on the plane of this circle that it is proposed to describe all the circles of the sphere, as they would appear at the point o' The circle, then, whose diameter is BD is the plane of projection; the point c its centre; A and o its poles; and the point c is evidently the projection of the point A. . 36. Let P be any point assumed on the circumference oBAD: take PE = PF, and draw the chord EF, it will be the orthographic projection, or the diameter of a small circle whose pole is P. Draw Eo, Fo, to cut BD in s and T, st will be the projection of the chord EF on BD ; and we propose to demonstrate that st is the diameter of a circle which will be the projection of the circle described on EF. Now it is evident that rays from all points of the circumference of the circle whose diameter is EF to meet at o will form the surface of an oblique cone whose vertex will be o, and circle about EF its base; of which all sections parallel to that base will (Hutton’s Geometry, theor. 113) be circles. In order to determine the section of this cone whose orthographic projection is st', we may proceed thus: meas. of Z FEois àfo = }oD + #DF = 45°-- DF; meas. of Z stois $oB + \DF = 45° + \DF: therefore FEo = sto; and consequently, EFo = 180° – FEo — For = 180° — sto – Tos = ost. The triangles EFo, Tso, then, are similar; yet the limes RF, st, are not parallel, but are what is technically denominated anti-parallel, or sub-contrary. Suppose, however, the come Eof to be turned half round upon the axis Po, then (since both slant sides of, of, make equal angles with op) ot would become ot', and os would become os'; in that case T's' would be parallel to the original chord EF, and the section of the cone {. can in no respect of magnitude or shape differ rom the section projected into st) would evidently be a circle. st is, therefore, the diameter of a circular section. . Thus every circle, whether oblique or not to the visual ray directed to its pole, will be represented on the projection by a circle. |