Question 40Navigation Problems70% to pass
You observe the lower limb of the Sun at a sextant altitude (hs) of 62°22.2' on 6 June. The index error is 1.2' on the arc. The height of eye is 28 feet (8.5 meters). What is the observed altitude (Ho)?
AI Explanation
The correct answer is B) 62°31.4'.
To calculate the observed altitude (Ho), we need to apply the index error to the sextant altitude (hs) and then add the correction for the height of eye.
The sextant altitude (hs) is 62°22.2', and the index error is 1.2' on the arc. Therefore, the true altitude (ht) is hs + index error, which equals 62°22.2' + 1.2' = 62°23.4'.
The correction for the height of eye (28 feet or 8.5 meters) is approximately 0.8'. Adding this to the true altitude (ht) gives the observed altitude (Ho) of 62°23.4' + 0.8' = 62°31.4'.
The other options are incorrect because they do not properly account for the index error and the height of eye correction.
Related Questions
#38 On 9 February your 0830 zone time DR position is LAT 22°19.0'N, LONG 64°37.0'E. Your vessel is on course 128°T at a speed of 19.0 knots. What is the zone time of local apparent noon?#39 On 12 February your 0930 zone time DR position is LAT 25°20.0'N, LONG 30°40.0'W. Your vessel is on course 135°T at a speed of 11.2 knots. What is the zone time of local apparent noon (LAN)?#41 On 3 October your 0830 ZT position is LAT 26°15.0'S, LONG 73°16.0'E. Your vessel is on course 280°T at a speed of 19.0 knots. What is the ZT of local apparent noon (LAN)?#43 On 22 February your 1857 DR position is LAT 23°46.0'S, LONG 93°16.5'E. You observe an unidentified star bearing 108°T at an observed altitude (Ho) of 67°53.9'. The chronometer reads 01h 00m 35s, and is 03m 25s fast. What star did you observe?#44 The true course between two points is 078°. Your gyrocompass has an error of 2°E. You make an allowance of 3° leeway for a north wind. What gyro course should be steered to make the true course good?