# 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 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 Rudzinskiwrote: > 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 -~----------~----~----~----~------~----~------~--~---