# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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 --~--~---------~--~----~------------~-------~--~----~ NavList message boards: www.fer3.com/arc Or post by email to: NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---