How to estimate wind speed from a pressure difference?

Does a formula (or a rule of thumb) exist to estimate the wind speed between a high pressure area and a low pressure area given the pressure difference between the two areas? Only the wind resulting from a pressure difference is of interest, additional influences like Coriolis force or centrifugal force can be neglected.

• You might want to look up "Pitot Tube" – Carl Witthoft Jan 14 '16 at 12:39
• Coriolis force is the controlling force of atmospheric dynamics outside of a short band around the equator. Outside of the tropics, it makes no sense to neglect the Coriolis force between low and high pressure areas in the atmosphere. – Vladimir F Jan 26 '16 at 23:08

3 Answers

I have no direct experience with meteorology, but if you want the "rule of thumb", study the Euler equations. Specifically:

$$\nabla p = - \rho\frac{\mathrm{D}\vec{v}}{\mathrm{D}t}$$

where D denotes the material derivative. That's the root of all other derivations.

A rule of thumb exists if coriollis force is the dominant force balancing the pressure gradient. This is known as the geostrophic balance :

$$\overrightarrow{V_g} = {\hat{k} \over f} \times \nabla_p \Phi$$

However if only a pressure gradient is being maintained by some source then the velocity will keep increasing as the pressure gradient results in accelerations as the previous user (@Victor noted). However in the real world the balance will eventually be between pressure and some parameterized viscosity :

$$\nabla p = \nu_{eddy} \nabla ^2 v$$

Calculating $\nu_{eddy}$ is non trivial and really depends on case to case basis.

In meteorology the Coriolis force can very seldom be ignored. It gives rise to the geostrophic wind. This is a flow parallel to the isobars (lines of constant pressure) in which the force due to the pressure gradient is exactly balanced by the Coriolis force. This gives $$(2\Omega \sin \phi) \rho v_y = \frac {\partial P}{\partial x}\\ (2\Omega \sin \phi) \rho v_x = -\frac {\partial P}{\partial y}$$ Here $$\phi$$ is the latitude, $$\Omega$$ is the angular velocity of the Earth ($$2\pi$$ radians in 24 hrs). The geostrophic approximation is never exact, but is very good for most meteorological conditions with exception of tornados and hurricanes.

There is wikipedia page:https://en.wikipedia.org/wiki/Geostrophic_wind