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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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