NavList:

Dear Greg,

Pending further insight and expected explanations form other people - including from Frank who has not published his solution yet - the 90° case is "very special" in the sense that it can be readily solved for example through extensively using Pythagoras's Theorem (see : [NavList 14680] Re: A simple three-body fix puzzle) which holds true whatever the Star Azimuts, JUST AND ONLY because the 2 SUN LOP's are perpendicular to each other.

When you have have 120° between the TWO SUN LOP's, it becomes a quite different story, and there seems to be absolutely no reason why the Least Quare distances Point should remain at any specific fixed distance - whatever it might be - from point P.

And actually the fact that you mention that such Least Square Distance "oscillates" between .57d and .66d seems quite reasonable when the third body - i.e. the Evening Star - azimut is allowed to take all directions around point P. I would personnaly feel : "no need to really worry here". Your results seem excellent for the case you have studied, and by the way thank for having taken taking the time to draw and publish your detailed drawings. But your such "120° case" is - again - significantly different from the "90° case" submitted by Frank E. Reed just recently. So we should should except that both cases being different, we are to get significantly different end results.

Time permetting, it would certainly be interesting to study your "120° case" in further depth. Or even better, treating the general case with the angle between both SUN LOP's being considered as a variable. Then after one has derived the general solution, checking how its "big resulting formula" can be simplified into "d/2" for the 90° case - which is an alternative to Frank's suggestion about using the least square formulae published in the NAL - and also cheking how this same "big formula" can be transformed into some (simplified?) oscillating solution for your specific value of 120°.

This typically would be a standard mathematic exercise for any young man trying to join his home country Naval Academy whether in Annapolis (Hello to all of you on the Western side of the North Atlantic Ocean !), in Dartmouth (a "royal" Hello to both of you George and Douglas !) or even Brest (a "republican" Hello to Kermit !).

I am just hoping that I correctly understood your question, and that I am hereby bringing you a reasonable - although very limited - reply.

Thank you for your Kind Attention and

Very Best Regards

Antoine

Antoine M. "Kermit" Couëtte

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[NavList] Re: A simple three-body fix puzzle
From: gregrudzinski---com
Date: 8 Dec 2010 08:14

The linked diagram shows a variation of the three-body fix puzzle for an ideal triangle formed with azimuths differing by 120°. The plotted fix shows a variation of .57d to .66d when rotating the third LOP around point P. How does this case differ from the original 90° P three-body puzzle ?

Greg Rudzinski
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