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From: Bill Noyce (no email)
Date: Sat Jun 03 2006 - 17:24:00 EDT
Good questions!
Although the formulas we use with calculators are relatively straightforward,
they're awkward iv you have to compute them with log tables. For example,
to evaluate
LHA = arccos{ [ sin(Ho) - ( sin(Lat.) * sin(Dec) ) ] / [ cos(lat) *
cos(Dec) ] }
you would start by looking up log(sin(Lat)) & log(sin(Dec)) and adding them.
Next you would need to find the antilog of the result, add sin(Ho), and take
the log of the result -- accumulating roundoff error along the way too.
There were numerous formulas devised to avoid coming "out of logs" -- or
equivalently, expressing the formula as a product. I assume the formula
Sumner used was the standard one in his day for reducing a time sight.
Let's see how it works.
As you point out, log(sec(Lat))+log(sec(Dec)) forms the denominator of the
formula -- no trouble there.
From the angle-addition formulas we memorized in high school,
cos(Lat+Dec) = cos(Lat)*cos(Dec) - sin(Lat)*sin(Dec)
so your [value 3] - [value 4] comes out to
100000*{ cos(Lat)*cos(Dec) - sin(Lat)*sin(Dec) - sin(Ho) }
Aside from the extra factor of 100000, and the presense of cos(Lat)*cos(Dec),
this looks like the numerator of our present-day formula. (Recall that Lat
and Dec are in opposite hemispheres, so our modern approach would
make one of them negative, so the product of their sines would also be
negative.)
Finally we use logs to multiply this by the product of the secants, ending
up with
5 + log( [cos(Lat)cos(Dec) - sin(Lat)sin(Dec) - sin(Ho)]/[cos(Lat)cos(Dec)] )
which equals
5 + log( 1- [sin(Lat)sin(Dec) + sin(Ho)]/[cos(Lat)cos(Dec)] )
This can now be looked up in a "log rising" table that lists
LHA against 5 + log(1+cos(LHA)).
I apologize in advance for any blunders above, but I hope this gives the
flavor of the approach. I'm sure the derivation of the formula is discussed
in Bowditch, for example.
-- Bill
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