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Attn: williamashbless Re: terrestrial tides. | by AndyA | 2006-11-19 12:55:59 |
| Yyyyeah, those are a little large. :D |
by williamashbless |
2006-07-26 16:36:17 |
You do have to be careful about the periods you're averaging over. You need to get the expected tidal period right, otherwise you'll average out the very signal you're looking for.
Lesse if I can reproduce your 12 hours and 24 minutes.
The moon orbits the earth. We're ignoring the effects of the tides caused by the sun for the sake of this argument. What we want is to find out how long it takes for the moon to be over the same spot of the earth (in principle, assuming the Earth's tilt and the moon orbital plane were the same).
The sidereal period of the moon is 27.32166155 days. This equates to 13.17635822 degrees per day. The earth is of course rotating once every day, for 360 degrees per day.
After one revolution, the moon will have moved on a bit, in fact, 13.17635822 degrees. We have to wait a little longer for the moon to be over the same spot of the Earth.
So, we have a relationship here... For t being the number of days past one, we have:
360t = 13.17635822 + 27.32166155t
332.67833845 t = 13.17635822
t = 0.03960690 days past one day.
That's 1.03960690 days to get the moon over the correct repeatable spot on the globe.
24h 57 m 2.0363s
We want half this time, because the tides actually occur twice in an orbital period. That's
12h 28m 31.0181s
So I disagree slightly with your period.
Please check my math? I'm sick.
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[ Reply ] |
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I just ran through the calculations from scratch | by AndyA | 2006-07-26 16:57:52 |
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