On 31 July 198'1, at 0604 ZT, in DR position LAT 16° 42.3' S, LONG 28° 19.3' W, you observed an amplitude of the Sun. The lower limb was a little above the horizon, and the Sun was bearing 069.8° pgc. At the time of the observation, the helmsman reported that he was heading 143° pgc and167 per magnetic compass. The variation in the area was 23° W. What were the gyro error and deviation for that heading?
A. 1° W GE 2 W DEV
B. 1° E GE 1 E DEV
C. 2° W GE 1 E DEV
D. 1° E GE 0 DEV
ANSWER D
BODY SUN CELESTIAL HORIZON
STEP 1
You can see that you must enter the left-hand column with your ship's DR latitude: You can also see that the sun's declination is listed across the top of the table. Since latitude 16° 42.3'S, and declination N 18- 15.8' are not right on the tabulated values, we determine the amplitude BY INTERPOLA TION. Convert the minutes of the latitude and declination into tenths of a degree.
STEP 2
The amplitude of the sun when it is on the celestial horizon is 19.2°. Now that we have the amplitude, what do we do with it? First of all, there are some basic rules that must be applied to convert 19.2° to a true bearing. The explanations are given to Table 27 on page 12 of Bowditch, states.
"The amplitude of a body is given the prefix E (east) if the body is rising and the prefix W (west) if the body is setting. Additionally, the amplitude of a body is given the suffix N (north) if the body has northerly declination and the suffix S (south) if it has southerly declination." The table was computed by the following formula: "sin Amp = see Lat x sin dec", If you know your
trigonometry-the formula sin Amp = sin dec divided cos Lat.
SS SR
270 090
-19.2 N (dec)
070.8
Our amplitude was taken when the sun was rising, and its declination name is north. The amplitude would be labeled as follows:
E 19.2°N, and we would have to subtract it from 090, thus 070.8° T is the true bearing.
STEP 3
Gyro error and deviation can be determined as
