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.