In the spherical triangle APV, P is the pole of the earth, V the pole of the vortex, A the point of the earth's surface pierced by the radius vector of the moon, AQ is the corrected arc, and PV is the obliquity of the vortex. Now, as the axis of the vortex is parallel to the pole V, and the earth's centre, and the line MA also passes through the earth's centre, consequently AQV will all lie in the same great circle, and as PV is known, and PA is equal to the complement of the moon's declination at the time, and the right, ascensions of A and V give the angle P, we have two sides and the included angle to find the rest, PQ being the complement of the latitude sought. We will now give an example of the application of these principles. _Example._[10] Required the latitude of the central vortex at the time of its meridian passage in longitude 88° 50′ west, July 2d, 1853. CENTRAL VORTEX ASCENDING. Greenwich time of passage 2d. 3h. 1m. Mean longitude of moon's node 78° 29′ True " " 79 32 Mean inclination of lunar orbit 5 9 True " " 5 13 Obliquity of ecliptic 23 27 32″ Mean inclination of vortex 2 45 0 Then in the spherical triangle PEV, PE is equal 23° 27′ 32″ EV " 7 58 0 E " 100 28 0 P " 18 5 7 PV " 26 2 32 Calling P the polar angle and PV the obliquity of vortex. [Illustration: Fig. 13] To find the arc AR. By combining the two proportions already given, we have by logarithms: M.R.V. minor = 3256 Log. 3.512683 M.S.D. of moon = 940″ " 2.973128 P.S.D. of earth = 3950 A. C. 6.403403 Radius 10.000000 T.S.D. of moon 885″.5 A. C. 7.052811 Log. Cosine arc AR = 28° 57′ 3″ 9.942025 --------- As the only variable quantity in the above formula is the "True" semi-diameter of the moon at the time, we may add the Constant logarithm 2.889214 to the arithmetical complement of the logarithm of the true semi-diameter, and we have in two lines the log. cosine of the arc AR. We must now find the arc RK equal at a maximum to 2° 45′. The true longitude of the moon's node being 79° 32′, and the moon's longitude, per Nautical Almanac, being 58° 30′, the distance from the node is 21° 2′, therefore, the correction is -2° 45′ × sin 21° 2′ -arc RK = --------------------- = -59′ 13″ R To find the correction for displacement. True longitude of sun at date 100° 30′ " of moon " 58 30 Moon's distance from quadrature 48 0 As the moon is less than 90° from the sun this correction is also negative, or -90′ × sin 48° Arc Kq = --------------- = -1° 6′ 46″. R