through a rift in the clouds, bearing to the S'd and E'd, altitude cor.-9'; G. S. T. of 2d observation 8h12m178. As the ship was rolling heavily at the time no accurate bearing could be taken. What star was observed? Watch times 535 43 Interval 2 20 Obs. altitudes 25°33' Diff. 31 31'÷2=13.3' R. From Plate I, ZN. 115° E. = Proceed as directed in paragraph 2 and we find: Log X=9.8011, and X=63°. Also, log Y=8.8901, and Y=4°42′. As L<h, then t<M and t=X − Y = 58° 18′ = 3h53m128. - Note that in the groups for finding X and Y, (a) is the same in both, and (d) and (e) are the complements of (b) and (c) respectively. Enter Plate II with Z=115° on the right margin and the horizontal line through this point cuts the altitude curve for 26° in a vertical line which intersects the horizontal line through hour angle 58° 18′ on declination curve 16° 30′, which is marked S. by Table II. From the star list in the Nautical Almanac we find the star was a Canis Majoris (Sirius), R. A. 6h40m558 and dec. 16°35′ S. To find the hour angle in this case from the Azimuth Tables, enter H. O. Publication No. 120 in Lat. 12° with 115° 740 in the hour angle column, and in the 26° declination column we find 58°04'3h52165 for the hour angle of the star. Example 2.-At sea, February 26, 1901, 6.30 p. m., L. M. T., weather overcast and cloudy; the altitude of an unknown star of about the 2d magnitude, seen through a break in the clouds, was 29° 30' (true), bearing N. 74° W. Lat. by D. R. 35° N., Long. 60° W. What was the name of the star? (c) From Table I, we have: Log X=0.1223, and X=68°. Also, log Y 8.8022, and Y=4° 18'. As L>h, t=H. A.=X+Y; hence, H. A.=72° 18' 4"49m12s. = Enter Plate II with Z=74° on the left margin, the horizontal line through this point cuts the altitude curve for 29° 30' in a vertical line which intersects the horizontal line through the hour angle, 72° 18', on the curve for 28°30′, the declination of the star, which is North by Table II. The star is a Andromedæ, R. A. o103m 198, dec. 28°33′ N. The numbers at the top and bottom of Plate II mark the curves. These are altitude curves when the azimuth is taken on the margin and declination curves when the hour angle is considered on the margin. To find the hour angle from the Azimuth Tables, 74°4h56m. Enter H. O. Publication No. 120 in Lat. 35° with 4h56m as an hour angle, and the altitude, 29°30′, as a declination we find, Z=72°24′ -44936-the required hour angle. Example 3.-At sea, February 6, 1903, Lat. 16° S., Long. 38° 12′ W., observed the following altitudes of a star to the N'd and W'd. Cloudy weather with a heavy sea so that no accurate bearing could be taken by compass. What star was observed? Ans. a Leonis. Obs. alts. 20° 18' W. t. 5h05m078 19 39 5 08 00 C-W-3h08m 138 Chro. fast of G. M. T., o"2013" In 253 change in altitude = 39′ and Rm 13.52', and from Plate IZ S. 110°30′ W. From Table I: Log X-9.8448 and X=66°. Log Y=8.3880 and Y=1°30'. As L<h then t=X-Y=64°30', or hour angle=4h18m. From Plate II with Z=110°30′, h=20° and hour angle 64°30′, we find the declination, 12°30', marked North by Table II. To find the hour angle from the Azimuth Tables, enter H. O. Publication No. 71, Lat 16°, dec. 20°, contrary name. We find |