Question 1479Navigation General70% to pass
At about GMT 1436, on 3 December, the lower limb of the Moon is observed with a sextant having an index error of 2.5' on the arc. The height of eye is 32 feet. The sextant altitude (hs) is 3°38.8'. What is the observed altitude?
AI Explanation
The correct answer is B) Ho 4°29.1'.
To calculate the observed altitude (Ho), we need to apply the index error to the sextant altitude (hs) and then correct for the height of eye.
The sextant altitude (hs) is given as 3°38.8', and the index error is -2.5'. Therefore, the true altitude (ht) is:
ht = hs + index error = 3°38.8' + (-2.5') = 3°36.3'
To correct for the height of eye, we use the formula:
Ho = ht + Dip
Where Dip = 0.983 * √(height of eye in feet)
Dip = 0.983 * √32 = 0.983 * 5.66 = 5.55'
Therefore, the observed altitude (Ho) is:
Ho = ht + Dip = 3°36.3' + 5.55' = 4°29.1'
The other options are incorrect because they do not correctly apply the index error and height of eye correction to the sextant altitude.
Related Questions
#1477 At upper transit, if the zenith distance is 34°, the geographical distance from the observer to a body's GP is _______________.#1478 You determine your vessel's position by taking a range and bearing to a buoy. Your position will be plotted as a(n) _______________.#1480 What sextant correction corrects the apparent altitude to the equivalent reading at the center of the Earth? #1481 The magnitude of three stars is indicated. Which star is the brightest? #1482 Which towing vessel(s) is/are exempt from carrying radar?