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    Re: Sextant double reflection: angles halved
    From: Richard Toohey
    Date: 2021 Feb 16, 15:31 -0500
    Not sure I understand Dave. By “goes wonky” are you saying my proof is only valid for a certain range of altitudes?

    On Tue, Feb 16, 2021 at 1:16 PM David Pike <NoReply_DavidPike@fer3.com> wrote:

    Richard Toohey you wrote:  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!

    If it works for you, that’s fine, but I thought about this long into the night on Sunday.  The beam is reflected clockwise through the supplement of 2v and back anticlockwise though the supplement of 2u, so the total angle turned though is (180°-2v) – (180°-2u) = 2u-2v clockwise. The overall turn theta is anticlockwise, so theta=2(v-u).

    In my diagram Hs = 45°, so the light ray changes from 135° to 180°. I.e. 45° clockwise.  By inspection v is 67.5°, and 2v is 135°.   Therefore, to move from 135° to 180°, the beam is reflected through 180° – 135° = 180° – 2v = 45°.  The pedantic approach has to include clockwise and anticlockwise to end up with v-u as opposed to u-v.  u-v hits a nerve when you compare it to triangle ABC, and your proof goes wonkey.  A better teacher wouldn't have picked u and v; as letters, they're too easily to confused, and they wouldn't have picked Hs = 45°.  45° is half 90°, and that can lead you into too many slick first impressions as it did me. DaveP

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