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    Re: A simple three-body fix puzzle
    From: Antoine Couëtte
    Date: 2010 Dec 8, 01:37 -0800

    [NavList] Re: A simple three-body fix puzzle
    From: antoine.m.couette---fr
    Date: 7 Dec 2010 23:49


    Frank, you wrote :


    So WHY is this solution rotationally symmetrical? Why is the three-LOP fix located at a distance d/2 from the two-LOP fix (when the first two LOPs are perpendicular --not otherwise) for any azimuth despite the fact that the "cocked hat" diagrams are so very different for different azimuths? I have one answer that leads to some interesting tricks (and has some bearing on John Karl's "running fix" technique), but before I describe that, does anyone else have an easy way of explaining why this should be rotationally symmetrical?




    Instead of setting course to a (funny) two variable partial derivatives calculus exercise, let's do it differently, just through keeping our eyes "wide open".

    I am encouraging "bold readers" to simply take a piece of paper and a pencil and draw just a few lines as they read this post through.

    I have done my best to use the definitions previously addressed by Frank, and namely with Point "P" being the intersection of the 2 SUN LOP's and with the quantity "d" being the distance between the STAR LOP and point P.

    Also, I have used the following conventions :

    - a POINT is designated by a Capital letter with/(without) a number, e.g. "P", "Q", "T", "T1","U" ..., and

    - a SET OF POINTS (such as a line) is designated by (a) Capital Letter(s) between brackets, e.g. (LQ), (R) ... and

    - a NUMBER is designated by (a) small single/(dual) letter/(s) with/(without) a number, e.g. d, d1, d2, d3, t, up, us ....


    1 - Let's pick up ANY point INSIDE the cocked hat, i.e. the triangle defined by the three LOP's. Its distances from the 2 SUN LOP's are "d1" and "d2". Its distance from the Star LOP is "d3".

    From any such arbitrarily chosen point "Q", let us draw a Line "(LQ)" parallel to the STAR LOP. For every point "forced" to stay on (LQ), the quantity d3**2 is constant and therefore minimum since it can not decrease.

    NOTE : We will also notice that EVERY and ANY point inside the cocked hat can be uniquely addressed/reached through this procedure, i.e. from EVERY and ANY point inside the cocked hat we can draw ONE AND ONLY ONE SINGLE line parallel to the STAR LOP (EUCLIDES's AXIOM).


    2 - Let us now look for the point on this Line (LQ) which also minimizes the quantity " d1**2 + d2**2 ". For any variable point "T" on Line (LQ), let us call "t" its distance to point P. Since both SUN LOP's are perpendicular, we always have :

    t**2 = d1**2 + d2**2.

    We see that the quantity " d1**2 + d2**2 " and accordingly the quantity " t**2 " becomes mimimum when point T becomes closest from point P, which means that such point "T1" lies at the intersection of Line (LQ) with the Radius "(R)" which is perpendicular to (LQ).

    We therefore know that our final (and unique) solution NECESSARILY (i.e. it can NOT be differently) lies somewhere on Radius "(R)" which is perpendicular to both (LQ) and the STAR LOP, both latter Lines being parallel by definition.


    3 - Let's pick up any variable point "U" on Radius (R), and let us call :

    - "up" the distance between U and point P (the intersection of the 2 SUN LOP's), and
    - "us" the distance between U and the STAR LOP.

    With d being the distance between Point P and the Star LOP, we see that :

    up + us = d

    We also see that, BECAUSE the angle beween the 2 SUN LOP's IS A SQUARE ANGLE, we always have the equality " up**2 = d1**2 + d2**2 " (Pythagoras's theorem).

    So the quantity " d1**2 + d2**2 + d3**2 " = up**2 + us**2.

    Since the sum " up + us " is constant (and equal to d), the quantity " up**2 + us**2 " is minimum when up = us.

    Therefore point "S" which is our "3 LOP least square Solution" is always at a distance d/2 from point P.




    Simply because - and again - with the 2 SUN LOP's being perpendicular, WHATHEVER THE AZIMUT "z" of the STAR, for any point T belonging to radius (R), we can always write the following equality " up**2 = d1**2 + d2**2 " (Pythagoras's theorem).


    Congratulations to you my dear Friend "Bold Reader" ... since you went through !!!

    Now, there are certainly ways of "explaining faster" and with "more immediate light shed on" this topic.

    Standing by for easier and more immediate explanations ... :-))

    Best Regards to U ALL


    Antoine M. "Kermit" Couëtte



    LAST NOTE 1 : More accurately, we can see that there is a SYMMETRICAL ROTATION because - thanks to the orthogonolity between the 2 SUN LOP's, and WHATEVER THE STAR AZIMUT - we did use the Pythagoreas's Theorem TWICE, and not once as incompletely summarized hereabove :

    - First in Part 2 hereabove, with the equation " t**2 = d1**2 + d2**2. " , and

    - Second in Part 3 hereabove, with the equation " up**2 = d1**2 + d2**2 " (again thanks to the Pythagoras's theorem which holds for ANY VALUE OF THE STAR AZIMUT). And

    LAST NOTE 2 ( Ref Part 2 hereabove ) : the Radius (R) drawn from point P in a direction perpendicular to (LQ) exists and it is unique (from a theorem derived from Euclides's Axiom). And

    LAST NOTE 3 : Ref previous post [NavList 14677] Re: A simple three-body fix puzzle by Peter Hakel

    Just a reminder here :

    Once two LOP's cross at a non zero angle - and here we have an "optimum case" since the 2 SUN LOP's cross at right angles - whatever the 3rd LOP Azimut (here the STAR LOP Azimut), and even if it were strictly parallel to one of the previous SUN LOP's (in which case there would be one of the three LOP's intersections at an infinite distance as Peter rightly points out), the "least square distances point" between such 3 LOP's would not be rejected away to an infinite distance. In our specific case with the 2 SUN LOP's being perpendicular, the "least square distances point" would simply stay on the SUN LOP which is perpendicular to the other 2 LOP's and halfway between them.

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