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Re: Q: how to calculate refraction at higher altitudes on land?
From: Dan Allen
Date: 2002 Feb 28, 15:23 -0800
From: Dan Allen
Date: 2002 Feb 28, 15:23 -0800
Great ideas all! I'll have to do the barometer experiment (the ocean is less than an hour away), and I'll also mess around with measuring the angle to the top of Rattlesnake Ridge, the mountain south of my house. Dan -----Original Message----- From: Navigation Mailing List [mailto:NAVIGATION-L@LISTSERV.WEBKAHUNA.COM]On Behalf Of Craig Sent: Thursday, February 28, 2002 12:41 PM To: NAVIGATION-L@LISTSERV.WEBKAHUNA.COM Subject: Re: Q: how to calculate refraction at higher altitudes on land? 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.