A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Samuel L
Date: 2015 May 11, 05:29 -0700
How do you determine local time using the method you described below that in effect is using the sextant as a sundial. Here's what I did;
I'm at W075°
Sun's Declination- 16° 50.1'
Ho= 59° 19.5'
Calculations- (90°- 59° 19.5') / 15= 2 hours 2 min 42 seconds
Including Declination in the problem;
Calculations- (90°- 59° 19.5') + 16° 50.1' / 15= 3 hours 10 min 2.4 seconds
The results of my calculations don't provide anywhere near correct time before before noon.
Can you explain how to get the right result?
Frank- you wrote-
"A time sight turns a sextant in a very accurate sundial. It may be useful to consider a geometrically simple case that requires no logarithmic math to work things out. Suppose you're nearly on the equator on an equinox (for example, off the Galapagos Islands on March 21). The latitude is near zero, and the Sun's declination is near zero. You watch the Sun rise from the eastern horizon in the morning. Then you measure its altitude with your sextant, correct it as usual, and then subtract that from 90° to convert it to a zenith distance. If we then convert that angle to hours at the usual rate of 15° per hour, we immediately get the number of hours before noon. That's local time which we can compare with Greenwich local time yielding the longitude. So if we the corrected altitude is 30°, yielding a zenith distance of 60° or equivalently 4:00 hours, then the local apparent time is 4 hours before noon or 8:00am. And that's just what a sundial on land would read at that same location and time. The sextant gives sundial time."