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On 2014-09-29 17:41, Brad Morris wrote:
> The USNO website provides the time of transit as 12:51 PM.
>
> My longitude is 72°47'.6.  Therefore my delta longitude is 2°12'.4 or 2.21
> degrees.  At 4 minutes of time per degree, the offset is 8.82 minutes or 8
> minutes 49.6 seconds.
>
> 12h51m - 8m49.6s  is 12h42m10.4s

Brad, the .1s precision of your result is an illusion. Since the transit
time at Greenwich is rounded to one minute, local transit time computed
from that value cannot be any better. The USNO celestial navigation
calculator says GHA is off by 5.8′ at that UT1, which is equivalent to
about 23s error.

Time of local apparent noon can be calculated from the equation of time.
It's a good deal easier to use GHA, but that's been explained already.
Let's look at a different method.

Sep 28  +9m19.5s equation of time
Sep 29  +9m39.5s

Those values are valid at noon UT. But transit at your meridian occurs
about 5 hours after noon UT. At that time, EoT is somewhere between the
tabular values. Specifically, it's (72°47.6′ / 360°) of the way from the
first to second value. That fraction is .2022.

EoT increases 20.0s between table entries. Therefore, at your transit
it's .2022 * 20.0 = 4.0s more than the tabular value for Sep 28. So at
12h local mean time, local apparent time is 12h09m23.5s. Or, noon LAT =
11h50m36.5s LMT.

Zone time would be more convenient. Your computation of the offset from
zone center, 8m49.6s, is correct. Subtract that from 11h50m36.5s LMT to
obtain 11h41m47s standard time.

Though not rigorous, the computation gave an excellent result, as
confirmed by the USNO calculator. Sun GHA at that time exactly equals
your longitude.

   

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