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).

• This could be tricky, because I think flags fly/wave partly because of the effects of turbulence. – innisfree Nov 3 '15 at 3:52
• Neglecting turbulence and assuming a horizontal wind, it then becomes a case of what you mean by "flying" - a flag flying perfectly horizontally would require an infinite velocity wind. – innisfree Nov 3 '15 at 3:53
• @innisfree: I though so myself - the fruitless search (there was one paper, quite interesting, about some experimental results about why it flaps, but otherwise zilch) hinted this might be something not simple. – rasher Nov 3 '15 at 6:18

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:

• +1, that looks promising - going through details now. The "Fluttering flags..." paper is the one I alluded to in an earlier comment. Might I inquire as to the author of the "Fundamentals of Fluid Dynamics" text? – rasher Nov 4 '15 at 21:46
• No, have not been on in a bit - done, with the proviso that I'm not a physicist: the mathematics looks fine, I'm not qualified to determine if the physics is, but I'm sure if there's an issue someone will chime in. Sorrt for the delay. – rasher Nov 6 '15 at 23:42
• NVM - I see you updated post with full title of book - thanks . – rasher Nov 6 '15 at 23:42
• @rasher Hey, an update. I have been testing this and it is surprisingly accurate. I estimate the angle of a flag, use the plot, then check weather conditions online- I am always within a few [mph] difference. – OnStrike Nov 21 '15 at 0:04
• Appreciate the testing and update - I'd vote again if I could. Thanks again for the effort! – rasher Nov 21 '15 at 0:14

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.

• Which area $A$ is it in $PA$? the bottom edge of the flag? If the bottom, why isn't there a equal force pushing down on the top of the flag? If the side of the flag, how does that give a vertical force to counter gravity? – innisfree Nov 3 '15 at 3:55
• @innisfree In my estimate, the total area of the flag as I assume the flag to be tied to a horizontal pole. But yes, there will be an equivalent force on the other side. My $v^2$ is ideally the difference $v_1^2-v_2^2$ but I just wanted to give a rough calculation, to get an order of magnitude, which apparently doesn't work so well. – Cheeku Nov 3 '15 at 6:37