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Lat/Long and timekeeping system for Mars
On Thu, 06 Dec 2018 08:19:22 +0100, Paul Schlyter
wrote: The Hour Angle of a celestial object, which is needed to compute its local altitude and azimuth, is most easily computed by subtracting the object's RA from your local sidereal time. Of course you can compute your local sidereal time without labeling it as such or even without even being aware of what it is, but I don't see any convenient way around computing it if you compute the local altitude and azimuth for some celestial object. Sure, but that's exactly my point. We can understand the concept, but are still going to have a "huh?" moment if somebody talks about a "sidereal hour", simply because that's not a conventional term that anybody really uses. |
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