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

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Re: Sextant double reflection: angles halved
From: David Pike
Date: 2021 Feb 16, 14:32 -0800
Richard Toohey you said: Not sure I understand Dave. By “goes wonky” are you saying my proof is only valid for a certain range of altitudes?

No. First, I'm saying 2v isn't 45°; it's 135°.  2u does happen to be 90°, but that's only because I've drawn my diagram with the ship on an even keel and the horizon mirror at 45° to the horizon so 180-2u =  2u.  As I said, it was a poor choice of Hs and sextant tilt on my part. It would interesting to draw the diagram with the sextant tilted a few degrees in the line of sight.  Theta has to remain = 2Phi, but the angles of incidence v and u at the mirrors would be different.   E.g. if a big wave comes along and tilts me forward 2.5°, v becomes 70° and 2v = 140°.  At the same time u reduces to 42.5° and 2u = 85°.  Now 2u no longer equals 180-2u.

By wonkey, I simply meant You can't have u-v in the first half of the proof if you've got v-u in the second half of the proof.  If you state correctly that the angle reflected though is 180 minus twice the angle of incidence, you have to bring in clockwise and anticlockwise to be able to say 2(u-v) is effectively = minus theta.  Then everything fits.  DaveP

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