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    Re: Distance by Vertical Angle Adaptation
    From: Greg Rudzinski
    Date: 2009 Jul 27, 23:30 -0700

    Frank,
    
    Yes it does appear that small angles are not covered by this
    compromise formula (below 8' for 250ft , 10' for 500ft, 12' for
    1000ft).
    
    Greg
    
    On Jul 27, 9:25�pm,  wrote:
    > Greg,
    >
    > Does this really work? I've tried a few cases, and it seems to converge 
    rather slowly. Let's suppose I measure the angular height of some distant 
    object with known height H=250 feet and I find the angle a=10 minutes of arc. 
    Assume my height is h=16 feet. For my first distance estimate, dist0, I 
    assume that the horizon is not obstructing my view and calculate
    > � dist0(n.m.) = (3438/6076)*H(ft)/a(m.o.a.) � ;your first formula
    > or
    > � dist0 = 14.1 n.m.
    > Next using the formula for the visibility range between two objects,
    > � range(n.m.) = 1.17(sqrt(h1) + sqrt(h2)),
    > I calculate the maximum height, dH that would be visible at a range of 14.1 
    n.m. by re-arranging, solving for h2 and calling that dH:
    > � dH = [dist/1.17 - sqrt(h)]^2 � ;your second formula
    > This is the number of feet in some distant object that would be obscured by 
    the horizon at that distance (with dist in n.m. and h in feet). Note that the 
    factor 1.17 depends somewhat on the terrestrial refraction so you could use 
    1.14 or 1.20 with no worries. Ok, so for this case, I get dH=65 feet. That 
    means that the actual height sticking up above the horizon is really only 185 
    feet. So I have to go back and calculate a new distance, call it dist1, from 
    the simple angle formula:
    > � dist1 = 10.5 n.m.
    > But now I have to calculate a new estimate of the height obscured by the 
    horizon and now I get dH=25 feet. So the height sticking up above the horizon 
    is better estimated at 225 feet. Now I need to calculate dist2 and so on... 
    There is, of course, a direct equation for this (which is rather long for 
    hand calculation and not easy to remember), but I like this indirect approach 
    on general principles and I am just trying to see if there's some way to make 
    it work better. Maybe the distances at each step should be averaged?
    >
    > Another case to try: h=16 ft, H=250 ft, a=1' (just peeking above the horizon). Determine the distance.
    >
    > -FER
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