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Subject: Re: Towards a basis for Bruce Stark's Tables
From: Bruce Stark (Stark4677@XXX.XXX)
Date: Sun Jan 05 2003 - 17:35:43 EST
Fred,
If I understand what you mean by "intercelestial distances" you're right: the
calculation that gets comparing distances from the Almanac is the
cosine-haversine formula. But you're not out of the woods yet with the
formula for clearing the distance.
But first, let's talk about the need to check for errors of arc and centering
in your sextant. You can use my tables as long as one of the objects is the
moon. Otherwise use another method of clearing. If you use one such as
Duntorne or Borda, don't use the "logarithmic difference" short cut. Nothing
wrong with the log-dif tables as long as one of the bodies is the moon, but
if you use them for two stars you run into the same problem you would in
using my method with its "Q." You could get around the problem posed by
log-dif or "Q" but it would be tricky. So if you're checking your sextant
with a distance from the moon use my tables; if between two other bodies,
Borda, without the log-dif table, is a good choice.
If you're woking Borda's with logarithms you'll want to work to six or seven
decimal places. So far as I know, mine is the only rigorous method of
clearing that doesn't require more than five places of decimals.
For star-to-star distances above 25° you might want to use one of the
approximate methods. These need only four places of decimals. Unfortunately
they aren't reliable with short distances no matter how many decimals you use.
I wouldn't be surprised if you can develop my formula from Borda's, or vice
versa. Would be an interesting exercise.
Or you can develop the formula the way I did, from the cosine-haversine and
the old time-sight formula.
That time-sight formula probably was used by mathematicians centuries before
it was used at sea. Here it is, spelled out in a way I hope will come through
un-garbled:
Call the sides of spherical triangle "a," "b," and "c."
Call half the sum of those three sides "s."
Opposite side "a" is angle "A."
Then the sine of one-half angle A is equal to the square root of:
[sin(s-b)sin(s-c)]/(sin b sin c).
I combined that formula with the cosine-haversine formula to get the one for
the Tables.
Bruce
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