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    Re: Moon Venus Lunar - Interpretation of results
    From: Frank Reed
    Date: 2020 Feb 3, 09:58 -0800

    Jim, you wrote:
    "This is all very interesting.  What do you (Paul and/or Frank) use to determine the relative angular velocities of, say, the Moon and Venus?  Are you using the prediction software on ReedNavigation and finding the rates in a one or three hour period?"

    Yes, it's quite simple. You can just look at tabulated lunar distances. Take the difference in the tabulated distances (convert to minutes of arc if necessary). Determine the number of elapsed seconds of time between the entries (3600 for one hour tables). Then get the rate of seconds of time to minutes of arc:
      Rate = 3600sec / Diff.
    The rate that you find will be about 120 seconds per minute of arc of distance. Does this sound familiar? It should. It's prominently spelled out on page 2 of the notes from the Lunars class you attended.

    You asked:
    "Can Stellarium be used for this purpose?"

    Only for very rough estimates, not really relevant for lunars work. There is an angle measuring tool in Stellarium, but it's difficult to use over large angles, and it has no "snap-to" capability that would allow it to measure exact angles from one celestial object to another. It's only as good as your point-and-click skills.

    You wrote:
    "I have heard several times ove the past weeks that the effects of small errors in observing a Lunar are magnified by the time you calculate GMT.  I had no idea that it was a 30x factor, though I'm sure I had that fact in my notes somewhere.  "

    Of course it's in your notes from class and ten times over! One problem with celestial navigation is that many people come to it with an assumption that they don't need to pay attention to the "easy stuff". But you do.

    Let's do the numbers again. In typical cases, the Moon moves about 30 minutes of arc per hour. That is approximately its own diameter in one hour. So if we can measure the Moon's position to the near minute of arc, we can determine the time to two minutes. Then dividing by ten, if we can measure the Moon's position to the nearest 0.1' (certainly not guaranteed, but quite possible with averaging, as you have already proven with your own sights), then we determine time to the nearest 12 seconds which is equivalent to 3 nautical miles at the equator.

    Frank Reed
    Clockwork Mapping / ReedNavigation.com
    Conanicut Island USA

    PS: Finally, I must emphasize that Paul Hirose's numbers were simply wrong. He sometimes buries himself so deeply in the computations that he loses track of the goal.

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