Question 1479Navigation General

70% 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?

A Ho 4°18.6'
B Ho 4°29.1'
C Ho 4°36.3'
D Ho 4°42.2'
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.

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