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    Re: Sun sights during an eclipse: "bad limb" calculation
    From: Antoine Couëtte
    Date: 2023 Oct 10, 22:33 -0700

    Lars, you brought the subject of the Contraction of the SD's, to which I indicated that I have not investigated this subject any further.

    Let's then investigate here, with a simple numerical example.

    Start from Sun center unrefracted height = 50° with Sun SD = 16.0000' as measured alongside a Great Circle or through a sextant.

    My Éphémérides Nautiques indicate: Refraction correction (from unrefracted to refracted) = + 0.8' [ = + 23 * E-5 rad.]

    The Sun vertical limbs cover an Azimuth difference ΔAz equal to = SD / cosh , hence SD = ΔAz cosh. Numerical application : ΔAz = 24.8916'

    dSD/SD = dΔAz/ΔAz + dcosh/cosh . Refraction elevates both vertical limbs with no change in Azimuth. This implies  dΔAz = 0, hence :

    dSD/SD = dcosh/cosh = -dh sinh/cosh = -dh tgh

    Numerical application with values hereabove (dh = 23*E-5 rad. and tgh = 1.19175) gives dSD/SD = - 28 * E-5. Since SD = 16.0000', the SD contraction dSD is equal to -0.0044', a totally negligible value here.

    With ΔAz = 24.8916' we can also recover this value through brute force computation :

    For h=50°00.0' , SD = ΔAz cosh = 16.0000'

    For h = 50°00.8', SD = SD = ΔAz cosh = 15.9556' . Difference : -0.0044'

    In our case and again the Contractions of the SD's are totally negligible.

    Kermit

       
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