Слике страница
PDF
ePub

through a rift in the clouds, bearing to the S'd and Ed, altitude cor.-9'; G. S. T. of 2d observation 8"12" 17".

As the ship was rolling heavily at the time no accurate bearing could be taken. What star was observed ?

26 04

Watch times

Obs. altitudes
5"33" 235

25° 33
5 35 43
Interval 2 20

Diff.

31
31' +21=13.3'=R... From Plate I, Z=X. 115o E.
r(a)

Z= 57°30'.
X (b)

}(h-L)= 7°09'.
90° — }(L+h)=71°14'.

(c)

[(a)

#2 = 57° 30'. Y (d) o° -}() - L)=82°5.

(e) I (L+h) = 18° 46'.

Proceed as directed in paragraph 2 and we find: Log X =9.8011, and X=63o. Also, log Y=8.8901, and Y=4°42'.

As L<h, then < M and t= X - Y = 58°18'=3"53"129.

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.

G. S. T. of 2d Obs. 8"12" 174
Long., West

5 24 36
L. S. T.

2 47 41 Star's H. A.-t

3 53 12, add. Star's R. A.

6 40 53

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. 6"40M558 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 115o = 7" 40in

[graphic][merged small][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][ocr errors][subsumed][subsumed][subsumed][subsumed][merged small][ocr errors][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][merged small][merged small][merged small][merged small][merged small][merged small]

AND GREAT CIRCLE DIAGRAM

PLATE II. 0°

55° 50° 45° 40° 35° 30° 25° 20° 15° 10°0°

[graphic]
[ocr errors]

ALTITUDES AND DECLINATIONS ON THE CURVES

[ocr errors]

150°

160°

65°

[ocr errors]
[ocr errors]

the hour angle column, and in the 26° declination column we find 58°04' = 3"52"165 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?

(a)

12=37°
X (b) }(L-1) = 2°45'.

(c) 90° – }(L+h) = 57°45'.

[ocr errors]

4 54 16

(a)

Z=37°
y (d) 90° – }(L-1) =87°15'.

l (e) 3(L+h) = 32°15.
From Table I, we have: Log X=0.1223, and X=68o. Also,
log Y=8.8022, and Y=4°18'.
As L>h, t=H. A.=X+Y; hence, H. A.=72°18'=4"49"125.

L. M.T. 6h3000
R. A. M.S. 22 22 33
Cor, G. M. T.

I 43
L. S. T.
Star's H. A. 4 49 12 (West)
Star's R. A.

0 05 04
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 verti-
cal 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 Andromeda, R. A. 0"0319",
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° = 4"56m. Enter H. O. Publication No. 120 in Lat. 35° with 4"56m as an hour angle, and the altitude, 29°30', as a declination we find, 2=72°24' -4"49m 365—the required hour angle.

E.rample 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, W. t. 20°18' 5"05"078 C-W=3"08m 138 19 39 5 08 oo Chro, fast of G. M. T., o"20" 138

Altitude correction, -7'. In 2534 change in altitude= 39' and Rm=13.52', and from Plate IZ=S. 110°30' W.

(a)

Z= 55°15'.
X (b) }(h-L)=

(c) 90° — }(L+h) =72o.

ylla
(a)

{Z=55°15'.
(d) 90°-3(h-L)=88.

|(e) (L+h) = 18°. From Table I: Log X=9.8448 and X=66o. Log Y=8.3880 and Y=1°30'.

As L<h then t=X – Y=64°30', or hour angle=4"18m.

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.

w. t. (mean) 5"06m 338
C-W

308 13
Chro.

8 14 46
Chro. cor.
G. M. T., Feb. 5 19 54 37
R. A. M. S.

20 57 50
Cor, G. M. T.
G. S. T.

16 55 43
Long., West

2 32 48 L. S. T.

14 22 55 Star's H. A. 4 18 00 Star's R. A. IO 04 55

20 09

3 16

To find the hour angle from the Azimuth Tables, enter H. O. Publication No. 71, Lat 16°, dec. 20°, contrary name. We find

« ПретходнаНастави »