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Re: Sextant double reflection: angles halved
From: Richard Toohey
Date: 2021 Feb 16, 06:51 -0500
From: Richard Toohey
Date: 2021 Feb 16, 06:51 -0500
Hmmm. Your first proof made perfect sense to me. I considered the starlight’s reflections as course changes. The starlight was initially on course 140 let’s say. Then the light changed course to 210 (a positive course change of 2N) before coming back to its final heading of 090 ( a turn back to the left or negative 2U).
So, in my mind, both of your proofs work.
And I guess my approach removes the positive/negative labels all together!
On Mon, Feb 15, 2021 at 10:32 PM David Pike <NoReply_DavidPike@fer3.com> wrote:
David - I sent my proof a bit ago, but damn, yours is about 10 times simpler! Love it!
Except I was too slick. You have to consider direction of bending. A better argument is:
Letting clockwise be positive, starlight is bent (180-2v) - (180-2u) = Minus Theta, so Theta = 2(v-u)
For triangle ABC, (90-v) + (90+u) + Phi = 180, or Phi = v-u, so Theta = 2 Phi. Got there in the end. Sorry. DaveP
