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Jeremy wrote-

| Here is an example of one of those archaic amplitudes.
|
| A ship is at sea and its DR Latitude is 14 degrees North.  At about
| 0830 UTC on 6 May 2008, an amplitude of the sun is observed while the
| body is on the celestial horizon giving a bearing of 288.0 deg per
| gyro compass.  Local variation is 2.0 deg East.  The ship is heading
| 208 deg per gyro compass and the magnetic compass heading is 207
| degrees.  Determine Magnetic Compass Error, Gyro error, and Deviation
| at that heading.    Please let us know if you used tables or formula
| to find the amplitude value.

====================

Here's my shot at it.

First thing needed is the Sun's declination at the given moment. I no longer
take an annual almanac, relying for astro. predictions on my home-brewed
program based on Meeus, which runs on a 20-year-old programmable pocket
calculator. At the moment in question that tells me that the Sun dec. is
+16deg 40.9'. (North positive)

But dec., for this job, isn't needed to very high accuracy, and a
declination from the almanac for a different year, at the same date, should
come close. A 2005 almanac gives me 16deg 36', which provides a good
cross-check. A 2004 almanac would have come closer still.

So we can take Sun dec. to be +16.7deg.

Burton's (or any other) amplitude table , for a latitude of +14 deg, giveas
at arc of 16.5deg at dec=16, and 17.5 at dec=17, so it's easy interpolation
to get an amplitude arc of 17.2 deg for the moment in question.

What does that mean? Well, we know, from the azimuth given of 288 deg, that
we're looking at a sunset. And it's from a Northern latitude, in Northern
Summer, so the Sun must be setting somewhat North of due West (270). So we
must add the amplitude, 17.2 deg, to 270, to predict that celestial sunset
with the Sun a bit above the horizon, would be at an azimuth of 287.2.

So, because the gyro bearing of the Sun was then 288.0, we can say
straightaway that the gyro error must be 0.8 degrees. Which way? Well, the
gyro reads a bit high, so that corresponds to a clockwise, or Easterly,
error.

And we know, then, that at a heading by gyro of 208, the true heading should
be 0.8 deg less, at 207.2. So if the magnetic compass then reads 207, at
that heading, it has a tiny error of only 0.2 deg, anticlockwise, or
Westerly.

And as we know that the local variation is 2.0 deg East, then for the total
error to add up to 0.2 deg West, the deviation at that heading must be 2.2
deg W.

(So the net sum of those three quantities to be found, which I offered as a
sort of hostage, comes out as 1.6deg, Westerly.)

Have I got it right, I wonder? I hate these mnemonic rhymes for getting the
signs right, preferring to sketch little pictures instead, as now litter the
margins of all my charts. But I can still get things wrong.

Of course, quoting angles to a tenth of a degree doesn't imply that you can
measure or use angles to that precision.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.


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