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Sideral question

  1. Nov 21, 2009 #1


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    I'm re-reading the calculations for Sideral Time and want to make sure that I've got the basics down. I want to make sure that I have grasped a few things though.

    Sideral Time = local time on the autumnal equinox (Sept. 22)
    Sideral Time = local time + (2hours * months since Sept. 22)
    Sidereal Time = Right Ascension + Hour Angle

    So, if I go outside at 6pm local time on Nov. 22, ST will be 1800 + 4 = 2200 hours. And anything directly overhead on my meridan at that time will have an RA of 2200 hours.

    Did I get this right?
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  3. Nov 21, 2009 #2


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    Not quite. Sidereal time doesn't equal local time on the autumnal equinox because time zones differ by an integer number of hours, while the Sun's path across the sky is continuous. All you can say is that the apparent solar time on the autumnal equinox is roughly equal to sidereal time. To convert from apparent solar time to clock time, you'd have to first convert to mean solar time using the equation of time, then convert to clock time by accounting for time zone differences & location within a time zone.
  4. Nov 22, 2009 #3


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    Re: Sidereal question

    OK, thanks. I should have been more liberal with the word approximate. The article that I was reading was discussing how to estimate the sidereal time to within a half-hour and didn't discuss time zones other than to mention DST. My first goal is to be able to stand outside without any equipment and approximately know where I'm looking on the celestial sphere.

    So if we're only interested in a rough half-hour estimate, it sounds like I understand the basics and I can move on to the next level of reading with respect to the equation of time, position within a time zone, DST, etc.

    Now if I could just spell sidereal correctly... :blushing:
    Last edited: Nov 22, 2009
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