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From: George Huxtable (no email)
Date: Sun Oct 03 2004 - 10:05:15 EDT
Chuck Taylor has taken some of the mystery out of haversines and the
cosine-haversine method, and shown how such calculations were made.
I think there's a bit more to be add, if Chuck doesn't mind.
====================
He wrote-
>Haversines are merely a vehicle for simplifying the
>computations. While sines and cosines range from -1
>to + 1, haversines range only from 0 to + 1, and the
>haversine of a negative angle is the same as the
>haversine of the absolute value of that angle.
That's all correct, but Chuck has omitted the important reason WHY it was
necessary to avoid negative values. The reason was the USE OF LOGS.
Unless a navigator was keen to do 4- or 5-figure long-multiplication and
long-division (and who would be?), before the days of calculators and
computers such calculations had to be done by logs. Logs were necessary and
useful, but had a great drawback; that for negative numbers, the logarithm
is meaningless. There was a trick to make get such log calculations to work
with negative numbers, but it never found favour with navigators. Instead,
their calculation formulae were rewritten in a form so that when logs were
taken, the numbers were always positive.
First, a new function, the "versine", (1 - cos), was introduced, in
addition to the usual sin and cos; clearly, this was always positive,
varying between 0 and +2. Later, it became clear that a function which
never exceeded 1 would be more useful still, so the versine was simply
halved, becoming the haversine (hence its name). Then the altitude formula
was reconstructed to use the haversine and avoid negative quantities.
Things have changed. In these days of pocket-calculators and computers,
there's no need to involve logs or haversines, and everything becomes much
simpler. But Chuck's description is useful, and interesting, in showing how
the old navigators had to do it, and you can do it their way if you wish.
To compute an intercept, you need lat and dec, and the separation in
longitude, (or HA, which Chuck abbreviates as t) and then require the alt,
to compare with the corrected alt from your sextant.
===================
Direct method (can't be used with logs).
From the basic cosine formula for a spherical triangle, then
sin alt = sin lat sin dec + cos lat cos dec cos HA
With a calculator, that's easy to work out. For this method, you have to
remember to give lat and dec their appropriate signs, + for North, - for
South. The sign of HA (i.e. whether it's East or West) doesn't matter.
==================
Cosine-haversine method (using logs).
If you're interested in how the haversine calculation works out in the
method Chuck describes, the expression above can be manipulated into-
hav ZA = hav (lat ~ dec) + sin lat sin dec hav HA
ZA is the zenith angle, or (90 - alt)
Lat and dec are now always positive quantities, 0 to 90 degrees, whatever
the hemisphere.
(lat ~ dec) is the amount of North-South angular difference between them,
taking account of which hemisphere each is in, therefore in the range 0 to
+ 180 degrees.
Instead of sin lat cos dec hav HA in the expression above, we can
substitute hav X; we don't need to bother what X actually is, as will be
seen.
So hav ZA = hav (lat ~ dec) + hav X
and hav X = sin lat sin dec hav HA
Then taking logs,
log hav X = log sin lat + log sin dec + log hav HA These logs always
turn out now to be of positive quantities, so that's OK.
Chuck's procedure adds those logs to get log hav X.
Most haversine tables (but not all) list the angle (0 to 180 deg), then its
log hav, and then its hav (as "nat", or natural, haversine). In my modern
edition of Norie's (1970), this table, to 5 decimal places, occupies over
100 pages.
So, knowing the log hav, you search down that column for the value of log
hav X that you have just calculated. Then alongside it in the adjacent
column, is the value of the haversine, hav X. You don't need to bother to
find what the corresponding value of angle X actually is.
So now we have hav X, to which we must add hav (lat ~ dec), and the result
is hav ZA.
Then take out the angle ZA, corresponding to this value of hav, from the
table, subtract it from 90 degrees, and you have the required calculated
alt, to compare with the sextant and obtain the intercept.
=====================
The method for deriving azimuth, from its cosine, quoted by Chuck, has two
problems.
One was the difficulty that for azimuths near due East, say, angles the
same amount North and South of East, such as 80degrees and 100 degrees,
have the same cosine, so the expression for cos az can't distinguish
between them. Similarly for azimuths near West. A special procedure was
quoted to distinguish which was which, which added quite a bit of
complexity.
By the way, if observations of the Sun were made between the Autumn and
Spring equinox, there's no need to make such a test, because during the
Winter months the Sun must always be in the Southern quadrants from the
Northern hemisphere (and vice versa). Similarly for stars with a South
declination observed from the Northern hemisphere.
The other difficulty is that cos az changes hardly at all for angles near
to East and West, so in that situation, the azimuth is predicted only
approximately and is very sensitive to small errors in calculation.
Unless one is forced to use logs, both these difficulties are avoided by
using a different expression which derives az from tan az and is better in
every way. This alternative has been referred to earlier on Nav-L and I
won't discuss it further unless anyone asks.
George.
================================================================
contact George Huxtable by email at , by phone at
01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
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