On 27 March in DR position LAT 32° 31.0' N, LONG 76° 25.0' W, you take an ex-meridian observation of the Sun's lower limb. The chronometer time of the sight is 05h 23m 32s, and the chronometer error is 01m 30s fast. The sextant altitude (hs) is 59° 59.0'. The index error is 1.8' off the arc, and your height of eye is 52 feet. What is the latitude at meridian transit?
• Ex-meridian sights and how they differ from true meridian passage (LAN) • Converting hs → ho (applying index correction and dip) and using the Sun’s declination on 27 March • Relationship between latitude, declination, and meridian altitude: Lat ≈ Dec ± (90° − ho)
• First, carefully reduce the sextant altitude to get the observed altitude (ho). Is your correction for index error in the right direction for ‘off the arc’? Is the dip sign correct? • Next, think about what an ex-meridian sight is doing: you’re slightly before or after LAN, and you use a small correction to estimate what the meridian altitude would have been. Are you using the correct sign so that your corrected altitude represents the maximum noon altitude? • Finally, once you have the meridian altitude, compare it with the Sun’s declination (on 27 March) using the classic noon-sight relationship. Given that both latitude and declination are north, should you add or subtract the zenith distance to get latitude?
• Be sure the index error is applied with the correct sign for ‘off the arc’ and that dip is subtracted from apparent altitude. • Confirm that you are using the Sun’s declination for 27 March at the correct GMT, based on chronometer time and error. • Check that when both latitude and declination are in the same hemisphere, the correct noon formula is: Lat = Dec ± (90° − ho), with the sign chosen based on whether the body is due south or due north at meridian passage.
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