PUB 229 Solution:
Problem 1:
On the 26 February, 1981, your vessel's 1615 ZT DR position is LAT 25°14'S, LONG 57-22'W, you take a azimuth of the sun and the Sun is bearing 266.0° per standard magnetic compass. The chronometer time of the sight is 8h 13m 19s. The chronometer error is 01 m 46s slow, and the variation in the area is 6° E. What is the deviation of the standard compass?
A. 1.7°E
B. 3.4°W
C. 7.7°E
D. 13.7 E
NOTE: SHORT METHOD AZIMUTH WILL USUALLY GET YOU WITH IN 0.5 OF THE ANSWER
ANSWER 1.7 E
STEP 1
Compute the LHA and DEC.
Compute the LHA and DEC.
STEP 2
Round off LHA, LA T, and Dec to nearest whole degree. Since LAT and Dec are "south" we will be on the SAME name page in PUB 229.
Round off LHA, LA T, and Dec to nearest whole degree. Since LAT and Dec are "south" we will be on the SAME name page in PUB 229.
STEP 3
Enter Pub 229 in back half of the book and find the page with LHA of 63°, Same name page. Extract the Z in the LAT column 25 and on the horizontal line with Dec 8 you should get 93.1.
Enter Pub 229 in back half of the book and find the page with LHA of 63°, Same name page. Extract the Z in the LAT column 25 and on the horizontal line with Dec 8 you should get 93.1.
STEP 4
The rule for converting Z to Zn are posted at the top right hand side of the SAME name page for North Latitude; bottom left hand side of the Contrary name page for South Latitude.
LHA greater than 180° Zn = Z
N. Lat LHA less than 180 Zn = 360 - Z
LHA greater than 180° Zn = 180° - Z
The rule for converting Z to Zn are posted at the top right hand side of the SAME name page for North Latitude; bottom left hand side of the Contrary name page for South Latitude.
LHA greater than 180° Zn = Z
N. Lat LHA less than 180 Zn = 360 - Z
LHA greater than 180° Zn = 180° - Z
S. Lat LHA less than 180° Zn = 180 + Z
Zn = 273.1°T
