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The previous post answers were based on a Monte-Carlo result.  There's
no doubt, that Monte-Carlo methods are a powerful tool.  But I feel
sometimes they're "too powerful."  They make is so easy to get to a
numerical result, one can "stop thinking" about the problem too soon.

After producing the result above, I felt sure the results were "in the
equations" somewhere, if I just looked hard enough.  Indeed, as
expected, thats true. The below compact FORTRAN program reproduces the
average distance results for various numbers of equally spaced lines of
position based on analytics.  (Well, some numerics, since the
exponetial integrals involved have no closed form solutions.)  No
comments/clues are provided so you can enjoy the chase yourself.

Details if there's interest.

    pi=4.*atan(1.)
    print*,"number of                   average distance error"
    print*,"equally spaced LOP's   (SD obs error 0.5)"
    do 10 ilop=3,8
    dx=0.
    do 9 jj=1,ilop
    dx=dx+((ilop/2.)/(ilop/2.)**2*sin((-75.+jj*360./ilop)*pi/180.))**2
9   continue
    sd=sqrt(dx*.5*.5)
    sum=0.
    do 1 i=-300,300,5
    do 1 j=-300,300,5
    x=i/100.
    y=j/100.
    sum=sum+1./sqrt(2.)*sqrt(x**2+y**2)*1./(sqrt(2.*PI)*SD)*EXP(-X**2/(2*SD**2))
     .     *1./(SQRT(2.*PI)*SD)*EXP(-Y**2/(2.*SD**2))
1   continue
    print*,'      ',ilop,'                 ',sum*.05**2*sqrt(2.)
10  continue
    end


knoppix@ttyp0[knoppix]$ g77 big.f -obig.exe
knoppix@ttyp0[knoppix]$ ./big.exe
 nunmber of                   average distance error
 equally spaced LOP's   (SD obs error 0.5)
        3                   0.511632323
        4                   0.443074971
        5                   0.396285892
        6                   0.361745924
        7                   0.334897667
        8                   0.313253939
knoppix@ttyp0[knoppix]$
knoppix@ttyp0[knoppix]$


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