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    Re: Combine DR and celestial LOP
    From: Michael Bradley
    Date: 2013 Mar 29, 18:14 -0700

    Nice easy one Frank

    The two sds are equal, so this is a simplified problem in probabilistic pattern recognition.
    The pattern recogniser guys would be dancing in the street because all they have to do use is a 'minimum distance classifier', not a 'maximum probability classifier'. All of which is a fancy way of saying they would put the MPP at the half way point along the perpendicular line between the DR and the EP, where the probabilities will be both equal and the largest avaiable.

    I've tried to find a free download of Duda and Hart's 'Pattern Recognition and Scene Classification' which is/was in the 80s the classical text - no luck. Plenty of second hand copies on good old abebooks etc. Gave my copy away decades ago, too mean to buy another one now. It's all in there, if you want to look it up.

    Not such an easy one from your text Gary.

    In the first place the formula is not dimensionally sound, mixing time and distance in the t+p term.
    I tried a little wrestling with the formula to get a feel for the underlying simplifying assumptions in the hope of getting inside the head of the guy or gal who derived it. Not much luck there, gave up fairly quickly, but did get out of it yet another interesting case of where, apparently, the two probabilities are equal:

    For d = p/2, ie. for the MPP to be half way along the perpendicular between the DR and the LOP, then t=p. That is, the time duration since the DR in minutes = the perpendicular distance between the DR and the LOP in miles. If you look up values that suit that case in the table, the t=p holds. Very mysterious. Aviators magic, 1 mile exactly equivalent to 1 minute. Is there some background assumption / rule of thumb in aviation DR that would explain that assumed equivalence?


    Good guessing ...

    Michael B
    55 North
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