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    Re: On LOPs
    From: Bill Murdoch
    Date: 2002 Apr 19, 17:48 EDT
    In a message dated 4/19/02 3:23:23 PM Eastern Daylight Time, Gadus{at}ISTAR.CA writes:


    I follow that in the case of two dimensions of error, i.e. two LOPs.
    Indeed, I suspect that the two LOPs would lead to normally-distributed
    errors in two perpendicular  dimensions even if the LOPs were not
    themselves perpendicular. What I am not sure about is whether three or
    more LOPs, each with normally-distributed errors, produce
    normally-distributed errors in each of two dimensions.


    Here is a plan.  Remember, I am an engineer so this is pre-Newton and rather inelegant.  I am also American and nm is nautical miles.

    Open four worksheets in one Excel workbook.  Pick a space 100 rows by 100 columns in each.  Assume each space is a map 10 nm x 10 nm and that each cell is a spot 0.1 nm x 0.1 nm.  Consider each to be a first quadrant graph with (0,0) at the bottom left and (10,10) at the top right.  Number the cells in the column to the left of the graph and in the row below bottom of the graph.  These can be used to locate any cell on the graph.  Off to the top of each of the last three maps fill two cells with the two constants (slope and intercept) that describe a line crossing that map.  Chose the three sets of constants to make lines reasonably represent three LOPs, one on each graph or map.  In each cell of each of these last three maps calculate the distance from the center of each cell to the LOP on that graph.  When you have that working, modify the formulas to give the distance to the LOP in units of standard deviation picking 1.0 nm! or so as your standard deviation.  Next, modify the formula again to give the differences in the cumulative standard normal distribution from one side of the cell to the other along a line through its middle.  (My guess is that you should consider the probability of being in a circle centered on the middle of the cell and maybe avoid a sine function if you did the square.)  You should then be looking at a ridgeline of probability centered on the LOP running across each of the three maps.  Now, in the first map find the product of the corresponding cells in the last three maps.  That would be your total probability of being in each of the 10,000 squares.  Play around with the orientation of the LOPs, the number of the LOPs, and maybe the standard deviation of the errors of each line (who says they all have to be the same) and see if you can make funny shaped hills or hills with more than one peak.

    This is a plan because the weather here is too good to do it right now.

    Bill Murdoch
       
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