Hav z = hav (L~d) + cos L cos d hav h
which is derived from the well-known expression:
Cos z = sin L sin d + cos L cos d cos h
where z = zenith distance; L = the latitude; and h = the hour angle.
Solution
| Dec. Arcturus 19° 42´ 29´´. | |||||
| Lat. 55° 00 N. (Assumed). | G. M. T. | 20 d. | 6 h. | 20 m. | 03 s. |
| R. A. M. ⊙︎ | 3 | 51 | 42 | ||
| Acceleration | 1 | 02 | |||
| Lat. 55° 00´ 00´´. | G. S. T. | 10 | 12 | 47 | |
| Dec. 19 42 29 | Long. | 7 | 28 | 32 | |
| —————— | |||||
| L~d 35 17 31 | L. S. T. | 17 | 41 | 19 | |
| R. A. ⁜ | 14 | 11 | 03 | ||
| H. A. ⁜ | 3 | 30 | 16 W. | ||
| (Observer) | 52 | 34 | 00 | ||
| Lat. | 55° | 00´ | 00´´ | = cos. | 9.75859 | ||
| Dec. | 19 | 42 | 29 | = cos. | 9.97378 | ||
| H. A. ⁜ | 52 | 34 | 00 | = hav. | 9.29244 | ||
| 9.02481 = nat. hav. | .10588 | ||||||
| nat. hav. 35° 17´ 31´´ | .09189 | ||||||
| z = 52° 48´ 35´´ | = nat. hav. | .19777 | |||||
| 90 00 00 | |||||||
| Computed altitude | 37 11 25 | ||||||
| Observed altitude | 37 14 50 | ||||||
| Altitude difference = | 3´ 25´´. | ||||||
Fig. 11.
A ship’s position is usually obtained by plotting the lines of azimuth and the position lines much in the manner shown in the chartlet. The azimuth of the body at the moment of observation is readily taken by inspection from the azimuth tables or better still from Weir’s Azimuth Diagram, both published by the U. S. Hydrographic Office.