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    Re: Distance by Vertical Angle Adaptation
    From: Greg Rudzinski
    Date: 2009 Jul 28, 10:45 -0700

    The visible range formula 1.144(sqrt.HE + sqrt Object Height) can be
    used for determining low angle vertical distance by multiplying the
    minutes of arc by 1.2 then subtracting from the visible range.
    16ft. HE 125ft. Object Height try 1' to 7'
    16ft. HE 250ft. Object Height try 1' to 8'
    16ft. HE 500ft. Object Height try 1' to 9'
    16ft. HE1000ft. Object Height try 1' to 10'
    Not sure why this works but it does seem to cover low angles to a
    reasonable accuracy. There may also be a useful distance approximation
    application for the other formulae. A full table would have to be
    created to see where other weaknesses might be.
    On Jul 27, 11:30�pm, Greg Rudzinski  wrote:
    > 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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