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I am trying to derive the wind load formula:

$F_d = \frac{1}{2} \rho v^2 A C_d$

in which I assume the drag coefficient to be equal to 1.

My starting point is that the momentum of a rectangular wall gained due to the elastic collision with air molecules is equal to twice the original momentum of a single air particle multiplied by the number of particles:

$p_w=2mvN(t)$

$N(t)=V\rho_N=Aa\rho_N=Avt\rho_N$

$p_w=2 m v^2 A \rho_N t=2 \rho v^2 A t$

where $v$ is the wind wpeed$\rho_N$ is the particle density, $\rho$ is the mass density, $A$ is the wall area, and $V$ is the volume containing air particles that will collide with the wall within time period $t$. Air is treated as an ideal gas, so the air-air particle collisions are ignored.

The coefficient of 2 arises in the equation for $p_w$ from the assumption that the wall has a very large mass and is stationary at the initial point of time.

The corresponding force on the wall:

$F=\dot p_w=2 \rho v^2 A$

is close to the correct answer but has a factor of 2 instead of 1/2. Apparently, I made a mistake somewhere but cannot find it. Could you help me with that?

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  • $\begingroup$ It's unclear where your factor 2 comes from. Even if the air molecules are slowed down to a stand-still, then momentum change is still only $mvN(t)$, without the factor 2. In reality, depending on the drag coefficient, they will keep some part of their momentum. Apparently, for a drag coefficient of 1, that part is 1/2 on average. $\endgroup$ – fishinear Apr 11 at 16:36
  • $\begingroup$ The coefficient of 2 comes from the momentum and energy conservation for one-dimensional elastic collision with $u_2=0$ and $m_2 >> m_1$ (en.wikipedia.org/wiki/…). Inelasticity of collisions would surely decrease the wind load but I don't see how it would convert the leading factor from 2 to perfect 1/2. $\endgroup$ – Marat Talipov Apr 11 at 16:55
  • $\begingroup$ But that is in a static gas. That is the force that results with a static gas on one side and a vacuum on the other. In the case of an object in no wind, that static force is there on both sides of the object, and cancel each other out. The extra force because of the wind, is the difference between the (average) speed of the gas molecules on one side, and that on the other side. If the molecules behind the objects are standing still, then that difference is only once the wind speed. $\endgroup$ – fishinear Apr 11 at 16:58
  • $\begingroup$ @fishinear, thank you for contributing. So, we have atmospheric pressure from one side, and atmospheric pressure + wind pressure on the other side. The atmospheric pressure cancels out thus leaving only the wind pressure, i.e. $P_w=2\rho v^2$. I don't see why the wind pressure or force should be reduced here. $\endgroup$ – Marat Talipov Apr 11 at 17:07
  • $\begingroup$ I think your formula is correct in approximation if you remove the factor 2. But note that is the case when the air behind the object is standing still and the air on the wind side looses all its momentum. That is, it is an extreme case with a drag coefficient of 2, which is almost never reached. In reality, the drag coefficient is simply defined in such a way that a coefficient of 1 corresponds to a factor 1/2 in the formula. They could just as well have taken another definition. $\endgroup$ – fishinear Apr 11 at 17:54

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