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(no email)
Date: Wed Mar 15 2006 - 10:38:17 EST
Frank in a archived post I found this working of a time sight:
Almanac data:
GHA Sun: 171d 27.4'
Dec Sun: 1d 17.6' S
EqT: + 8m 38s
GMT: 23-17-12
GAT: 23-25-50
Method 1: Direct Computation
Lat = L = 48.50167
Dec = d = -1.27833
Ho = h = 24.90500
cos t = (sin h - sin L sin d) / (cos L cos d)
cos t = 0.66093
t = 48.62894
t = 48d 37.7'
For an afternoon sight,
Lon = GHA - t
= 171d 24.4' - 48d 37.7'
= 122d 49.7'
OK, a couple of questions:
(1) I thought you had to use co-latitude , co-declination, & co-Ho? (If
this was already done I must have missed it).
(2) This looks suspiciously like every other spherical / navigational
triangle so why do you say stars were not used except for lunars?
(3) Are you saying you can't get a good longitude from just the noon
site?
My reason for asking is that I do a lot of rev-war & pre-1840 re-enacting
and I'm trying to understand exactly what was done
on land & sea. I know for example that on land both David Thompson & Lewis
& Clark took Lat via Polaris.
-Greg
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