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astronomical tidal calculation

Is it possible to reliably define the times (for a specific location on earth) when the tidal gravity from astronomical objects (moon, sun etc.) are at their weakest & strongest?

I have considered using sea & ocean tide times, but these factor in unwanted variables, such as ocean currents and coastal water flows, which would invalidate the results somewhat.

I am attempting to define whether any noticeable effects occur from the moon on athletic performance, but need specific times with which to test and compare (hence the strongest & weakest times are my targets).

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  • Asked by 25107945
  • on 2010-03-10 11:09:14
  • Member status
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Categories: Our universe.

Tags: humanbody, gravity, moon, tide.

 

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tbrucenyc says:

We don't step onto a scale to measure our weight and think, "oh, where's the moon, so I can compensate for its effects".

No... the tidal forces won't change your performance.

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Tags: humanbody, gravity, moon, tide.

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posted on 2010-03-15 13:01:00 | Report abuse


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HughC says:

It depends whether you want the vertical or horizontal component of the force.  Common sense tells me the vertical component should be greatest when the attracting body is on the meridian, the horizontal when it's rising or setting.

The sun crosses the meridian at local noon.  The mean period between successive lunar transits is 1.0350501 days, but it varies quite a bit (la luna e mobile).  See the book Astronomical Algorithms by Jean Meeus.  His English is sometimes a bit quaint but his maths are generally sound (I did find a confusion between sin and cos in one formula in an early edition).  The same book will tell you how to calculate the times of sunrise & sunset, moonrise & moonset.  I suspect that for your purpose you could ignore the effects of refraction and parallax, and so simplify the formulae.

 

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Tags: humanbody, gravity, moon, tide.

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posted on 2010-03-15 20:24:17 | Report abuse


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