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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Q: how to calculate refraction at higher altitudes on land?
From: Dov Kruger
Date: 2002 Feb 28, 16:16 -0500
From: Dov Kruger
Date: 2002 Feb 28, 16:16 -0500
Craig wrote: >I was thinking, in using trigonometry to solve a right triangle, you know >the adjacent side from your GPS (Difference between your lat/lon position >and lat/lon of mountain peak. You measure the angle, can get the tangent, >and solve for the opposite side, which is the difference between height of >mountain peak and your height. Subtract length of oppposite side from >mountain peak height, doesn't that give you your height? >Tangent=opposite/adjacent, therefore opposite=tangent*adjacent. > My mistake, you are right. Dov
