tude. To determine this from the projection, draw ca parallel to the horizon, then sin A Put A EL = LA + αE = GM + аE = CG 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 H.... .... (6.) 90° — D, E = 90° - L, and z = 90° then will the last equation become A; cos z = cos ▲ cos E + sin ▲ sin E cos 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. 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. E' A E 35. Let ABOD be a great circle of the sphere, o the place of the eye: the dia meter 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 B F T m T SK D 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 o 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 orthographie 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 FEO is FOOD + DF = 45° + {DF; meas. of STO is OB + DF 45° + $DF: therefore FEO ➡ STO; and consequently, FEO FOE 180° STO TOS OST. EFO 180°. The triangles EFO, TSO, then, are similar; yet the lines EF, ST, are not parallel, but are what is technically denominated anti-parallel, or sub-contrary. Suppose, however, the cone EOF to be turned half round upon the axis po, then (since both slant sides OE, OF, make equal angles with or) or would become or', and os would become os'; in that case T's' would be parallel to the original chord EF, and the section of the cone (which can in no respect of magnitude or shape differ from 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. 37. Draw from the centre to the pole P, the radius CP, and through the point r the tangent E'PFG, to mees in G the plane of projection BD prolonged. Then, meas. of GPK = PDO = 45° + PD; meas. of ▲ PKG = {BO + }PD ≈ 45° + †PD; that is, the targent PG is projected into a line Ko equal: to it. 38. Since POE and POF are equal, the line or will not bisect ST; but KS < KT. Bisect ST in m, then ms mT = radius of circle of projection; and cs, cm, CT, will be in arithmetical progression. Hence, cm = {CT + {cs tan AF+tan AE= = sin AP 2 cos &AF COS AE 2 cos § (AP. + PE) COS § (AP — PE) Let dcm distance of the centre m from the centre c of the plane of projection, ▲ = PE = polar distance of the circle EF, DAP distance of the two poles, r = 1ST = } (CTCs) (tan AF-tan AE) sin (AF - AE) 2 COSAF COSTAE = [See formulæ (u), &c. chap. iv.] Consequently, d:r:: sin D: sin A.... (9.) From these three theorems the whole doctrine of stereographic projection may be deduced, by tracing the mutations of D and 4. The chief maxims and principles of construction may also be developed geometrically, thus: 39. Beginning with great circles, let PE 90°, and PFPE; then will EF be a diameter. Draw the right lines OTF, OES, bisect Ts in r; then rs =rT, will be the radius of the circle into which the great circle whose diameter is Er will be projected. Through o and r, draw ORTI. The circle described on ST will pass through o, because FOE is a right angle. In like manner, it will pass through A, because cs is perpendicular to the middle of A great circle to the plane of projection. Hence, or is manifestly sec, inclination, and crtan inclination, to rad co. Thus, with the radius sec D, and at the distance from the centre tan », it will be easy to describe the circle. Again, take AI = 2be2ap = 2D, and draw oI, the point of intersection r with DS will be the centre, and ro the radius. = 90° Also, since meas. of cor= BE, meas. of OCR = OE - BE; we have COR + OCR = 90°, and conseq. CRO 90°. Hence, by drawing oRr perpendicular to FE, we find the centre r and the radius ro: also CR = sin BE sin D, OR = cos D. 40. or, therefore, will always make with oc an equal equal to the inclination or distance of the two poles. Let there be, then, a second circle whose inclination shall be cor'; the radius of its projection will be r'o, and the radii ro, ro, will make at their points of intersection an angle ror' which will be their difference of inclination, or the mutual inclination of the two circles, This is a particular case of the general theorem. 41. Suppose now that AP = 90°, P will coincide with B, and the pole of the circle will be upon the limit of the projection. Let BH = BE = A polar distance of the circle to be projected, the chord HE will be perpendicular to DB. Draw OH and OES, GS will be the diameter sought. Bisect cs in n, n will be the centre, ne = nH = ns = nG, will be the radius of the projected circle. Meas. of EnG= 2nse od — BE 90°. '— BE 90° — BCZ CNE + nce = 90° .•. CEn = 90°.. En tan A, and cn sec EE sec A. tan BE These values serve for all circles which have their pole on the circumference of the circle of projection. If these are great circles, then A = 90°, and tan ▲, sec ▲ are infinite: consequently, the centres of the projections falling at an infinite distance from c, the projections themselves will be right lines passing through c, and intersecting under the angles which such circles, taken two and two, form on the sphere. A t 42. Let o be the point of observation, or place of the eye, a the pole of the projection, BDECB the plane of projection, PD an arc of a great circle which has its origin at any point whatever M" of the circle OBPE, Pt the B tangent of PD; ct will be the secant, and s the projection of P. Draw st: then from the rectilineal triangle sct we shall have, M' M m N st2 cs2 + ct2 2cs. ct. cos sct tan2 AP + sec2 PD 2 tan AP sec PD COS DE tan2 AP + tan2 PD + sec2 AP+tan2 PD 1-2 tan AP SEC PD COS DE 2 tan AP Sec PD COS DE. But the spherical triangle PDE right angled in gives, (chap. vi. equa. 6), Substituting this in the last value of st2, we have |