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Coriolis force effect on skydivers? |
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| Apr19-07, 02:36 AM | #1 |
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Coriolis force effect on skydivers?
Galileo observed that objects dropped from height hit ground to the east of a plumb bob. I guess this is due to the Coriolis force. How would this distance be calculated for something like a baseball off the Empire State Building or a skydiver?
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| Apr19-07, 08:42 AM | #2 |
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Please, leave Coriolis alone! |
| Apr19-07, 11:58 AM | #3 |
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Curiosity is a powerful thing :)
You may be right about Galileo's observations being myth, Lpfr. Whether he did or didn't Galileo certainly isn't credited for discovering the Coriolis effect. Can anyone post the equations used to determine the horizontal movement? To many people it would be a concrete example of an abstract phenomenon. |
| Apr22-07, 07:43 PM | #4 |
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Coriolis force effect on skydivers?
Fcor = 2mr' x Omega = 2mv x Omega
Omega depends upon the location of the object since the Earth isn't a perfect sphere. But it's generally seen as about 7.3 x 10^-5 s^-1. When Coriolis is added to free fall, the equation becomes r''= g + 2r' x Omega. r'= (x',y',x') and Omega (0,Omega sin(theta), Omega cos(theta)) The equation of motion for the x-dimension becomes 2Omega(y'cos(theta) - z'sin(theta)). Y-dimension= -2Omega x' cos(theta) and z-dimension= -g + 2Omega x' sin (theta). Working through approximations, working those approximations back into the original equations and so forth, you get a rough approximation that x= 1/3 Omega(gt^3)sin(theta). Let's say you put theta at 90 degrees and let an object fall for about 20 seconds with g at 10 m/s/s. The x displacement is only about 2.2 cm, while it fell 100 meters. |
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