A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Robert VanderPol II
Date: 2020 Oct 30, 17:39 -0700
I do not have answers to your questions but I do have an observation:
The length of a arc-min angle across the surface of the earth varies by about 0.5% (10m/1852m = 0.0054).
If CelNav was using the distances rather than angles the accumulating error could become significant. For example shooting the sun at the horizon the error would be 5400nm x 0.0054 = 29.2nm. That's significantly larger than the expected 5-10nmm error for small navigation.
But the intercept method generally used sticks with angles almost to the end and only uses a relatively very small distance. This method uses angles for both the calculated and apparent altitudes to they are both using the same basis which is a perfect sphere. Distance in nautical miles only appears when you determine the altitude intercept distance. If you have a big intercept it would be a degree. The distance error from 1 degree on intercept distance would be 60nm X 0.0054 = 0.324nm. That is significantly better than the normally expected error.
So the varying length of a nautical mile only creates a relatively small error for celestial navigation as it is normally practiced.
That said I do not wish to discourage your from pursuing the answers to your questions, they are of interest in and of themselves and look forward to your discoveries.
I have found an online calculator that claims up to 130 place precision. Precision is user defined. I have not tested it and have no opinion about it's accuracy.
From: Mike Freeman
Date: 2020 Oct 30, 00:59 -0700