# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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 in radians 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@huxtable.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. ================================================================