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Re: Moon Parallax, Math Trivia (was Re: venus)
From: George Huxtable
Date: 2004 Oct 21, 00:42 +0100

I think we need to take this question of the fine-print of the parallax
correction somewhat further, if there isn't too much objection from
non-purists. We are discussing tiny errors here, but there's all the
difference between a tiny error and an exact solution.

There are two problems to solve.

1. deducing what the true altitude TA would be, knowing the observed
altitude OA and calculating the parallax. This is the normal procedure when
clearing a lunar distance, or any other Moon observation.

The exact expression for this can be easily deduced: see Smart, "Textbook
of spherical astronomy" chapter IX (parallax), eq 18, to be-

TA =  OA + arc-sine ( (a/r) cos OA)    (eq. 1a)

A CLOSE APPROXIMATION to this, for small angles (and we're talking about 1
degree or less)

is

TA = OA + (a/r) cos OA, where the angle [(a/r) cos OA] has to be expressed

This can be written more conveniently as

TA = OA + HP cos OA     (eq. 1b)

where the constant terms are gathered into the quantity HP, horizontal
parallax, the parallax at zero observed altitude.

But remember, this is only an approximation; the exact answer calls for the
use of arc-sine above.

2. Deducing what the observed altitude OA would have been if it had been
measured, knowing the true altitude TA and calculating the parallax. This
is the reverse problem, which is called in when clearing a lunar and the
Moon's altitude has not been measured.

In this case, the exact solution appears to me to be as I gave it in an
earlier posting-

OA =  TA - atn ( cos TA / ( (r/a) - sin TA))       (eq. 2)

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

Frank wrote-

>The original direct parallax equation is
>  OA = TA - HP cos OA .
>Can you invert this and solve for OA in some "closed-form" equation? ....No.

Where has Frank taken this equation from? Can he justify it as exact?

It seems to me that the inexactitude is not in the arctan solution to the
inverse problem 2, but in Frank's chosen expression for the direct problem
1.

Try putting in some numbers.

Assume HP = 55 minutes. or .91666666667 degrees

then a/r = .01599816897

If we take OA as 45 deg then using eq. 1a, TA is 45.64816739

Putting those values of TA and (a/r) into eq. 2 brings us back exactly to
OA = 45.00000000, just as it should.

But if we use 1b, starting at OA = 45, we calculate TA = 45.64815357

and putting that into eq.2 takes is back to
OA = 44.99998602, marginally different from the starting point.

So the question is whether eq. 1a or eq. 1b is an exact solution to problem
1. If it's 1a, then eq.2 is an exact solution to the reverse problem.

Frank's argument holds only if eq. 1b is the exact solution. Can he justify
that?

George.

================================================================
contact George Huxtable by email at george---.u-net.com, 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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