Wind velocity

1. Apr 14, 2008

Ry122

If I know the force that a wind is exerting on an object, is it be possible to determine what velocity the wind is moving at? I also know the air density.

2. Apr 14, 2008

Danger

That's pretty much how anemometers work, so it's definitely possible.
I'm afraid that I can't supply any formulae, though.

3. Apr 14, 2008

Daiquiri

The aerodynamic force acting on a solid body can be expressed by a general equation:

F = $$1/2 \rho V^2 S C$$

Where:
C is a force coefficient (either lift Cl, drag Cd or Cx, or whatever you're looking for)
S is a reference area (either frontal area, wing area for airplanes etc.)

The problem lies in the coefficient C. It depends on some adimensional numbers (Mach number and Reynolds number in most cases) and is usualy measured in a wind tunnel on a scaled model of the object.

If you were able to measure the force acting on the solid body, and if you knew the value of the aerodynamic coefficient C, then you wuld be also able to calculate the velocity from the formula above.

If you only knew the values of C vs. V (through Reynolds number) then it would necessarily be an iterative process, since you would have to estimate an initial V, then you would calculate C for that V, then you would recalculate V with that value of C, and then recalculate C with the new value of V, and so on and so on and so on... till the convergence of the result. :zzz:

On the other hand, if you had data (from real-scale or wind-tunnel measurements) which relate directly F to V (usually for incompressible flows and near-standard temperatures) then the process is straightforward: measure F --> read V from the F-V curve.

Last edited: Apr 14, 2008
4. Apr 14, 2008

Staff: Mentor

In practice, that method is difficult, but what you could do is use a pito-static tube.

5. Jun 7, 2008

Can you add units I shall use to calculate force?

I found that Cx for flat wall is 1.5

R

6. Jun 7, 2008

Redbelly98

Staff Emeritus
C has no units. The other quantities on the right-hand-side of the equation combine to give force units.

As long as you use a consistent set of units for the different quantities, the result will be a force. For example:

$$\rho$$ in kg/m^3
V in m/s
S in m^2

Resulting units are
(kg / m^3) * (m/s)^2 * m^2
= (kg / m^3) * (m^2 / s^2) * m^2
= (kg / m^3) * m^4 / s^2
= kg * m / s^2
= Newtons

7. Jun 7, 2008