Shootingwind flag

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The flags are raised manually by UN Security officers every weekday morning at approximately 8am and lowered every weekday at around 4pm, except in the ...

Flagin thewindmeaning

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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.

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.

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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).

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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).

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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.

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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 $$v = \sqrt{\frac{2mg}{A \rho}}$$

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Flagwaving in thewindvideo

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 $0 \rightarrow 20 \;[\frac {m}{s}]$.

Benchrestwindflags

The blue triangle on flag of the Czech Republic, with which Slovakia formed Czechoslovakia up to the dissolution of Czechoslovakia, was taken over from the blue strip of Slovakia's flag in 1920 into the flag of Czechoslovakia. Since the Slovak flag without the coat of arms is identical to that of the modern flag of the Russian Federation and it can also be compared to the modern flag of Slovenia.

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Flagwaving in thewindGIF

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Given the flags dimensions, $\ell = 300 \;\text{[ft]}$ and $D = 150 \;\text{[ft]}$: $$\therefore C_{D \;@\frac {\ell}{D}=2} \approx 0.12$$

Flagblowing in thewinddrawing

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.

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 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.

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The current Slovakian flag can be dated back to the revolutionary year 1848. It was also used, but not officially, in Czechoslovakia before World War II, by the Slovak Republic during WWII, and finally adopted (without the coat of arms) on 1 March 1990 as the flag of the Slovak Republic within Czechoslovakia. The coat of arms was then added to the flag on 3 September 1992 and a law describing the details of the flag was writen just after in February 1993.