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