The solution for this apparently is $\frac{1}{2}C \rho A V^{2}$ where $C$ is the drag coefficient, $A$ is the area and $\rho$ and $V$ are the wind density and velocity. Let's do it without the drag coefficient.

Assume that the wind hits the panel normally with speed $V$ and is then stopped (or escapes along the panel). The change in momentum of the wind per unit time is \begin{align} F&=\frac{d}{dt}M_{wind}V\\ &=\frac{d}{dt}(\rho A h) V\\ &=\rho A V^{2}, \end{align}

where $h$ is some height above the panel. Thus, the force on the panel is $\rho A V^{2}$. What happened to the factor of $1/2$?

  • 1
    $\begingroup$ What is $C$ for your case? $\endgroup$
    – Bernhard
    Commented Oct 1, 2014 at 12:10
  • $\begingroup$ So your drag coefficient is about 2. (your wiki link shows that for a sphere Cd is >1 for low Re.) $\endgroup$ Commented Oct 1, 2014 at 14:02
  • $\begingroup$ I was going to leave C out of it, but for a flat plate, it is apparently 1.28. The point was to get the factor of 1/2 (and its existence doesn't have anything to do with C). $\endgroup$ Commented Oct 3, 2014 at 3:43
  • $\begingroup$ The problem is that your derivation is a bit suspect. $\endgroup$ Commented Oct 4, 2014 at 15:47

3 Answers 3


Assume that the wind hits the panel normally with speed V and is then stopped (or escapes along the panel).

That's not the best of assumptions. The air will flow around the plate. Only in the center of the plate will it hit the panel normally and stop. Your analysis does however capture some of the key dynamics in that the drag force is proportional to $\rho A V^2$.

Some shapes will do a much better job directing the flow around the object than others. Consider the drag on a flat plate, a sphere, and a nicely shaped airfoil, all with the same cross sectional area. The flat plate is going to do a rather lousy job redirecting the air flow and will experience a large drag. The sphere will do a good deal better than a flat plate and suffer less drag. The airfoil will do an even better job and suffer even less drag that the sphere. That the shape of an object has a marked impact on the drag force is what the coefficient of drag tries to capture.

Update: Why a factor of 1/2?
A few simplifying assumptions lead directly to that factor of 1/2. I'll assume a somewhat slowly moving fluid. Not so slowly as to result in Stokes drag, but not so quickly as to result in turbulence. That means a Reynolds number between 10 and 2000 or so. With these assumptions, the flow will be dominated by convective acceleration but will still be steady.

Consider a packet of fluid moving horizontally with velocity $v$, density $\rho$, and pressure $p$. Given the assumptions, the conservation laws lead to a simplified form of Bernoulli's equation, $\frac {d}{dx} \left( \rho \frac {v^2} 2 + p \right) = 0$. You can find many derivations of Bernoulli's equation on the internet. The term $\rho \frac {v^2} 2$ is the dynamic pressure that results from the flow. Naively, this leads to the force on an object being the product of the dynamic pressure and the cross section area: $F = \rho A \frac {v^2} 2$.

This is not quite the case. As noted earlier, different objects with the same cross section have markedly different drag forces. The coefficient of drag is a unitless parameter that accounts for these variations. Also note that the coefficient of drag is not a constant. It is a function of both shape and velocity.

  • $\begingroup$ Yes, fair enough the assumption that it hits normally and stops is pretty bad. But the shape of the object is not relevant here (that information is in C). I want to know why there is a 1/2 in the formula. Regardless of what shape is chosen, there is always a 1/2 but why? $\endgroup$ Commented Oct 3, 2014 at 3:52

The expression for the drag coefficient ($F_D = C_D \tfrac{\rho}{2} U_\infty^2 A$) appears from dimensional analysis, not by integrating the momentum equations over a control volume. Applying the Buckingham's $\Pi$ theorem you get different ways to express your dimensionless parameter $C_D$, but the dynamic pressure ($\tfrac{\rho}{2} U^2$) is more suited. You could easily change to something like $F_D = C \, P_\infty \, A$, but if you consider a flat plate parallel to the flow with null pressure gradient, it would be a lousy description of $F_D$ as the flow is not pressure driven.

You can relate force $F_D$ with momentum losses with a control volume approach if you happen to know the distribution of pressure and velocity over the respective control surfaces (difficult unless you have CFD simulation results). Instead, it is easier to look to the body (flat plate in this case) instead of a control volume of fluid, and establish the free-body diagram. For example, to know the drag of a model in a wind tunnel, you attempt to measure the force required to keep the model static, which will equal $F_D$. You could measure the momentum deficit due to the wake of the model, but that would be difficult and prone to error.

As such, most of external flow problems focus on the body, not a control volume of fluid.


V1 : Velocity in front of panel

V2 : Velocity behind the panel

Force acting on the plate (Newton second) : F=m*a The accelleration is found by assuming a decrease of velocity from V1 to V2 over a distance s. The acceleration is thus found as: 2*a*s=v1^2-v2^2 .

The mass transport over the distance s is m=rho*A *s

By insertion: F=rho* A * s* (v1^2-v2^2)/ (2*s)

The distance s will vanish and thus:

F=1/2* rho* A* (v1^2-v2^2)

Due to conservation of mass the velocity behind the plate will never reach zero. According to Betz limit the velocity v2>1/3*v1.


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