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I'm doing correlation analysis between atmospheric pressure and wind speed during hurricanes. The problem is, I only have data for atmospheric pressure. Is there an equation where I can plug in the pressure so I can get the equivalent wind speed? Thanks for your help!

2006-11-21 22:33:50 · 7 answers · asked by Chris 2 in Science & Mathematics Weather

7 answers

Say---you could hurt someones brain with a question like that. ROTF-----Good luck hope you can find the real answer. seems like an unknown value without some kind of meter.

2006-11-21 22:44:25 · answer #1 · answered by EZMZ 7 · 1 0

Hmmm... I don't think so, I know that during recent hurricanes on TV they would talk about how this or that hurricane had very low pressure, and this is related to the damaging effect. But you can imagine that pressure is only one part of the equation.

Difference in atmospheric pressure is what causes wind; air moving from high pressure to low pressure - So the higher the pressure gradient the higher the wind - But that is exactly it, the GRADIENT - this will be effected by the size of the eye, and other atmospheric features nearby.

2006-11-22 04:38:51 · answer #2 · answered by Leonardo D 3 · 0 0

There's a relationship between horizontal pressure gradient (not absolute pressure) and windspeed, but it depends on your latitude. It gives you a slight overestimate of surface windspeed because of friction with the ground. It's called the geostrophic windspeed and above an altitude of about 3000 feet it's quite accurate. For a given pressure gradient (hectopascals/km) the windspeed in metres/sec. goes up as the sine of your latitude, so it's a maximum at the poles, in opposite directions in the northern and southern hemispheres (hurricanes rotate in opposite directions) and zero at the equator. If you can find a weather map which gives you an accurate windspeed and isobar map for a given location, you can work it out yourself in whichever units you're used to. It's a linear relationship; for a given latitude if you halve the spacing between isobars you double the windspeed. And for a given isobar spacing the windspeed at the pole would be twice that at 30 degrees latitude because sin 90 is twice sin 30.

2006-11-22 10:40:32 · answer #3 · answered by zee_prime 6 · 0 0

To a first approximation (neglecting friction, mixing, curvature, etc.), you can use the finite difference form of the geostrophic approximation, given information about how pressure changes with horizontal distance (the pressure gradient) and the latitude (for f). You can use the gradient wind approximation if you want to include the curvature effects. See the links for more information.

2006-11-24 12:19:35 · answer #4 · answered by stormfront105 2 · 0 0

It's not so much the actual pressure as the pressure difference over a given distance.

Tightly squashed isobars are usually associated with high wind speeds (cyclone). High pressure (anti cyclone is associated with very gentle winds or windless conditions.

Reminder - isobars join points of equal pressure.

2006-11-22 05:52:01 · answer #5 · answered by rosie recipe 7 · 0 0

On the scale of dbV/m3 would be normaly, would've been measurable on a Normale Instrumentary, be it whats yoyu can be, ion the armfull. FoolStop bit myself, Dodge @ Chev puck it and scent it for MayFlavar,pint.

2006-11-22 00:01:59 · answer #6 · answered by Anonymous · 0 2

no ;:: no relation between

2006-11-21 22:37:08 · answer #7 · answered by alex a 6 · 0 0

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