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On 11/20/05 2:22 AM, "Frank Reed"  wrote:

> Here's a little trick that might entertain some of you...
>
> Suppose you want to know how rapidly the Moon's position in the sky is
> changing. The average rate is 13.2 degrees per day or 33.0 minutes of arc per
> hour. But the range is large, running from 29.5 to 38.5 minutes of arc per
> hour.
> It turns out that you can get a very good estimate of the actual angular speed
> (geocentric!) at any moment by looking at the Moon's HP.
>
> Since the Moon  slows down when it is further from the Earth in accordance
> with Kepler's Second  Law, it might at first seem that the change in angular
> speed would be directly  proportional to the change in horizontal parallax
> relatve to its mean value of  57.0 minutes of arc, but in fact it's double
> that.
> When the Moon is, for  example, 1% farther from the Earth, its linear speed is
> reduced by 1% and in  addition, since it is visually further away, the angular
> rate is reduced by 1%  more. So if I look up the Moon's HP in an almanac and
> find that it's 60.0',  that's 3 minutes of arc or 5.3% above average. I double
> that percentage to get  10.6% and apply that to the Moon's mean angular rate
> of
> 33 minutes per hour to  get 36.5 minutes per hour. This calculation is
> accurate to +/-1% at worst and  typically accurate to +/-0.5%. Of course, it's
> not
> complicated to do a really  accurate calculation if you have a computing
> device
> handy (you can calculate the  actual great circle distance, or you could use
> a formula from Meeus), but  this little method is something that you can do in
> your head if you have  access to the Moon's actual HP for the time in
> question.
>
> Note: I think I've got this all right, but as always, check it out for
> yourself.
>
> -FER
> 42.0N 87.7W, or 41.4N  72.1W.
> www.HistoricalAtlas.com/lunars
>
Frank,

Don't you mean 13.2? per HOUR instead of per day?

Ken Gebhart

   

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