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    Re: Sun sights during an eclipse: "bad limb" calculation
    From: Frank Reed
    Date: 2023 Oct 4, 05:39 -0700

    Sorry for answering in reverse order (LIFO!), but just a quick thought here...

    Geoff Hitchcox, you suggested a "Pythagorean" calculation of the distance between the center of the Sun and Moon. There's nothing wrong with that since they're so close together. But there is one little catch: just as lines of longitudes on the globe converge towards the poles, lines of azimuth converge towards the zenith. On the globe, to put differences of longitude (over small distances) on the same scale as differences of latitude, we know we have to multiply by cos(Lat). Looks like this for lat/lon:
       dy = dLat,
       dx = cos(Lat) × dLon.
    And then we can apply Pythagoras:
       dist = √(dx² + dy²).

    Similarly with Alt/Azm (altitude and azimuth or "horizontal") coordinates on the sky, for small enough differences in coordinates:
       dy = dAlt,
       dx = cos(Alt) × dAzm.  [because azimuth lines converge at higher altitudes]
    And then Pythagoras:
       dist = √(dx² + dy²).

    Try that. It should be a quick modification of your code, Geoff. Does that match up with results from Lars Bergman and others? Does it match up with other simulations [try it in Stellarium]? 

    Frank Reed

       
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