Wind velocity required to get a flag flying/waving - is there a formula to estimate this? For example, the giant flags here are 300' x 150 ' with 1,100 lb weight. A question was asked about how much wind would be required to "fly" such a flag given a tall enough flag pole. A Google search marathon yielded nothing, as did a cursory search of some fluid mechanics books. Is there a formula to estimate this (an an estimate is fine +/- 10% is not a problem).
 A: The following (approximate) solution relates the flag's angle to wind speed.  Three forces act on the flag: Weight, Drag, and the Reaction Forces (from the flag pole).

Because the flag is flexible, it is modeled as a parallelogram (ie. the corner angles are not fixed) to capture the general shape and center of mass of the flag.  This is illustrated below.  Note that only the center of mass and corner angles change- force vectors and area (length and width) do not change.


After hours spent trying to calculate drag force, $F_{D}(Shape, Re_{x}, Roughness)$, I stumbled across the "drag coefficient of a fluttering flag" in Figure 9.30 of '6th Ed. - Fundamentals of Fluid Mechanics by Munson, Young, Okiishi, and Huebsch', shown below.

Given the flags dimensions, $\ell = 300 \;\text{[ft]}$ and $D = 150 \;\text{[ft]}$: $$\therefore C_{D \;@\frac {\ell}{D}=2} \approx 0.12$$
The overall drag coefficient combines the effects of friction (shear) drag and form (pressure) drag.  Assuming constant density, $\rho_{stp} = 1.2 \;[\frac {kg}{m^3}]$, the drag force is a function of velocity.
$$F_D = C_{D}(\frac {1}{2}\rho v^2) \qquad \Rightarrow \qquad F_{D} \approx 0.072v^2 \;\text{[N]}$$
The angle of the flying flag is calculated by summation of moments about Point O at a specific wind velocity.  I am less error prone in SI units, where $W = 4893 \;\text{[N]}$, $\ell = 45.72 \;\text{[m]}$, and $D = 91.44 \;\text{[m]}$. The plot is created from this equation, where $\theta$ is solved at wind speeds from $0 \rightarrow 20 \;[\frac {m}{s}]$.
$$\sum M_{O} = 0 = W(\frac {\ell}{2} \cos(\theta)) - F_{D}(\frac {D}{2} \sin(\theta))$$

$$\boxed{\therefore \theta = \tan^{-1} (\frac {2W\ell}{C_{D}D\rho v^2})}$$
The plotted equation is valid for all flags of with an aspect ratio of 2.  As expected, the flag hangs completely limp when there is no wind ($\theta = 90 \;\text{[deg]}$ at $v = 0 \;[\frac {m}{s}]$) and approaches $\theta = 0 \;\text{[deg]}$, where the flag stands straight out at high wind speeds.

For more detailed analysis, see the following papers:


*

*Fluttering flags: an experimental study of unsteady forces

*EXPERIMENTAL STUDY OF DRAG FROM A FLUTTERING FLAG
A: The following is a back of the envelop estimate. I had a minor itch with the numbers though, non-metric you see.
Assume the flag is tied on a horizontal pole, so we're trying to lift/fly it so as to make it parallel to the ground. The force needed to do that must equal the pressure difference created due to flowing air. Hence, 
$$F = mg  = PA$$
$$P = \frac{mg}{A}$$
This pressure is of the order of 
$$P = \frac{1}{2}\rho v^2$$
Equating the two equations for pressure, you have
$$v = \sqrt{\frac{2mg}{A \rho}}$$
Putting in your numbers, we have
$$v = 1.38 \ \mathrm{ms^{-1}}$$
Well, you can see that this is highly unrealistic, which apparently stems out of the assumption that the flag is tied to a horizontal pole.
If you guess that when the flag is tied vertically, about $1\%$ of its area is exposed, you will get a velocity $10$ times higher than above, which is closer to intuition.
Comments are welcome.
