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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; to 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 cp 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 F', D, and E'. 32. To find the value of the arc DE’ projected into the rectilinear portion GE, we have * rad: sin DE’:: AG: GE; whence GE x rad GE GE * ge GE 2 AG GA sin PAT cos Ae " cos D ' ' .(2.)

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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 cet we have GT = cq tan GCT = sin D tan L.... (3.) D being the declination of the parallel; GT sin D

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From the same triangle we have, also,
Co- sin D
cos gct T cos L

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... (5.)

33. Through the point E, a projected place of a star, draw the chord Rs 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

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tude. To determine this from the projection, draw ga parallel to the horizon, then sin A = EL = La + a E = GM + a B = 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.

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 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 of 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.

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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 sand 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 Er. 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 geois fo = }od + 4DF = 45° + #DF;

meas. of Z sto is $oB + 3LP = 45° + DF: therefore Foo = sto; and consequently, EFo = 180° – Feo — For = 180° - sto - Tos = ost.

The triangles Ero, Tso, then, are similar; yet the lines E.F, 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) or 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 which 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.

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... [See formulae (u), &c. chap. iv.] Censequently, d: ri; 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 A. The chief maxims and principles of construction may also be developed geometrically, thus: * 39. Beginning with great circles, let re = 90°, and rf = Pe; then will EF be a diameter. Draw the right lines otr, ors, bisect Ts in r: then rs = rT, will be the radius of the circle into which the great circle whose diameter is EF will be projected. Through o and r, draw ortri. 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 Ao. Therefore ro = rs = rT ; conseq. sor = osr, and cro = 2 osr. —cor = 90° – cro = 90° – 2 osr = 90° — (po — BE) = 90°-90° 4- BE = BE. That is, cor = inclinatian of the plane of the O great circle to the plane of projection. Hence, or is manifestly = sec. inclination, and cr = tan inclination, to rad co. Thus, with the radius = sec D, and at the distance from the centre = tan D, it will be easy to describe the circle. Again, take A1 = 2BE = 2AP = 2n, and draw or, the point of intersection r with Ds will be the centre, and ro the radius. Also, since meas, of cor = BE, meas. of ocR = or = 90°– 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, ro, and the radii ro, roo, 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 s, and the pole of the circle will be upon the limit of

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