# 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 31, 12:12 -0700

Frank's modified Bowditch formula for distance by vertical angle is now inked into my piloting note book and useful for generating a table of distances for frequently used object heights. Table for Anacapa Light at 250 ft. from a height of eye of 6 ft. MOA Distance (NM) 1' 19.8 2' 18.6 3' 17.4 4' 16.3 5' 15.3 6' 14.3 7' 13.4 8' 12.6 9' 11.9 10' 11.2 15' 8.5 20' 6.7 25' 5.5 30' 4.7 40' 3.5 142/MOA (NM for MOA above 40') On Jul 29, 6:51�pm, Greg Rudzinskiwrote: > Frank, > > I've practiced a bit with your formula and find it satisfactory if I > use an index card to keep track of things. The head calculation is > quite a test and I have to admit that I couldn't get the same result > two times in a row ;-) > > Greg > > On Jul 29, 6:19�pm, wrote: > > > I wrote: > > > "D = 1.21*[sqrt[a1^2+0.94*(H0-h)] - a1] > > � where a1=a-0.97*sqrt(h)) " > > > And Greg, you wrote: > > > "The over the horizon distance by vertical angle problem remains begging for a calculator/slide rule friendly formula." > > > Well, it is what it is, right? This certainly isn't hard to work on a calculator, and in fact it's not that hard to do in your head. Compare it against the equation in Bowditch which supposedly "explains" the origin of Table 15. In what school of obfuscation do they teach people to put 0.000246 in the denominator of an equation?? And the difference between tan(x) and x/3438 is much less than 1% for any angles that might ever be used for this calculation, so why do they bother with explicit use of the tangent in the equation? To make it look more "serious"? > > > What should be said, and isn't said enough, is that these equations (and tables) only provide an estimate of the distance since they depend on a mean state of the terrestrial refraction. Terrestrial refraction (meaning refraction of light traveling nearly horizontally close to the Earth's surface) on average bends light rays downward by about one minute of arc for every 6 nautical miles, but a range of 0.75 to 1.25 minutes in the same distance should be expected as normal variation (on the order of one s.d.). This matters for angles over the horizon because the we're measuring an angle between things which may be at very differnt distances. If the horizon is five miles away, and the vessel beyond the horizon is twenty miles away, then a small increase in refraction lifts the image of the vessel much more than the horizon (the distance to the horizon itself changes so the comparison is not one-to-one). > > > And it's possible to have even larger variations in terrestrial variation than this. If the air gets cooler with altitude more rapidly than normal, the terrestrial refraction can be zero (never less than zero for more than a few seconds so that really is the lower limit). If, rather than cooling with altitude, the atmosphere gets warmer with altitude at just the right rate, the terrestrial refraction can be 6 minutes of arc in 6 nautical miles which means that horizontal light rays remain parallel to the ground from all distances --and that means that the Earth looks flat and objects at all distances are in front of the sea horizon (the sea horizon would not actually be visible in this case, and note that this is not an upper limit). > > > -FER --~--~---------~--~----~------------~-------~--~----~ NavList message boards: www.fer3.com/arc Or post by email to: NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---