# Computing Those Lift and Drag Coefficients

In the Fluid Mechanics Laboratory course, when we get to the Wind Tunnel Testing, we have to compute two things which students find tricky: Reynolds Numbers and lift and drag coefficients.  We’ve already covered Reynolds Numbers, but lift and drag coefficients are equally problematic, mostly because of the units involved.  Here are some sample calculations to help avoid errors.

The equation for drag force is

$F_D = \frac {C_D A \rho u^2}{2}$

and, for objects with lift, the lift force is

$F_L = \frac {C_L A \rho u^2}{2}$

The variables are as follows:

• $F =$ lift $(L)$ or drag $(D)$ force, lbs.
• $C =$ lift $(L)$ or drag $(D)$ coefficient, dimensionless
• $A =$ cross-sectional area of object, $ft^2$
• $\rho =$ density of fluid, $\frac{slugs}{ft^3}$
• $u =$ free-stream velocity of fluid, $\frac{ft}{sec}$

I’ve been very specific about the units because this is the key to solving this problem.

Let’s say that we want to determine the drag coefficient of a golf ball such as shown below.

Golf Ball. A USGA standard golf ball is 1.68″ in diameter.

Let’s say that, in our wind tunnel, the drag force on this golf ball is 0.11 lbs. given the conditions described in Getting the Reynolds Number Right.  This means that the wind speed is 50 mph at an altitude of 600′.  What is the drag coefficient?

We start by solving for the drag coefficient, thus

$C_D = \frac {2 F_D}{A \rho u^2}$

The area for the golf ball is simply the cross-sectional area of the sphere that “faces” the wind, or the area of a circle.  We start by converting the diameter of the golf ball to feet, thus $D = 0.14'$.  From there $A = \frac{\pi D^2}{4} = \frac{\pi 0.14^2}{4} = 0.0154\,ft^2$.

The density must be expressed in the proper units; this is what messes up people the most.  Based on this calculator, $\rho = 0.00233\,\frac{slug}{ft^3}$.  The free stream velocity must be converted to the proper units, thus $u = 73.333\,\frac{ft}{sec}$.

Now we substitute, with all the units and constants stated, as follows:

$C_D = \frac {2 \times 0.11\,lb_f \times 1\,\frac{slug-ft}{lb_f \times sec^2}}{0.0154\,ft^2 \times 0.00233\,\frac{slug}{ft^3} \times 73.33^2\,\frac{ft^2}{sec^2}} = 1.14$

Comparison with published data will reveal that this figure is high.  In this case, the interference from the stand (described in Wind Tunnel Testing) is the reason for the discrepancy.

Lift and drag coefficients for airfoils are computed in the same way, except that the area is customarily the planform area of the airfoil.  This means that, if you’re looking straight down on the “flat” surface of the airfoil, the area you see is the planform area you use.

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