Question 2 of 6: A2 — Drag on a dimpled vs. smooth golf ball

Question 2: A2 — Drag on a dimpled vs. smooth golf ball

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Air $\nu=1.69\times10^{-4}\ \text{ft}^2/\text{s}$, density $\rho=0.00237\ \text{slug/ft}^3$; sphere diameter $D=1.65\ \text{in}=0.1375\ \text{ft}$, frontal area $A=\pi D^2/4$. For the dimpled ball, $C_D$ is read from the table; the smooth sphere over this Reynolds range is subcritical (below the drag-crisis $Re\approx3\times10^5$), so $C_D\approx0.5$ is constant.

Find. (a) drag $F_D(v)$ for the dimpled ball; (b) drag $F_D(v)$ for a smooth sphere, and the comparison.

Approach. Convert each tabulated $Re$ to a velocity through $v=Re\,\nu/D$, then evaluate $F_D=C_D\cdot\tfrac12\rho v^2 A$ using the table $C_D$ (dimpled) or the constant $C_D=0.5$ (smooth).

1. Velocity at each Reynolds number. $v=\dfrac{Re\,\nu}{D}$; e.g. at $Re=7.5\times10^4$, $v=\dfrac{(7.5\times10^4)(1.69\times10^{-4})}{0.1375}=92.2\ \text{ft/s}$. The full mapping is $v=\{92.2,\,122.9,\,184.4,\,245.8,\,307.3\}\ \text{ft/s}$.
2. Drag law. With $\tfrac12\rho A=\tfrac12(0.00237)(0.01485)=1.760\times10^{-5}$, $$F_D=C_D\left(\tfrac12\rho A\right)v^2=1.760\times10^{-5}\,C_D\,v^2\ \text{(lbf, }v\text{ in ft/s)} .$$
3. Dimpled ball (part a). Multiplying by the table $C_D$ at each speed gives $$F_D=\{0.072,\,0.101,\,0.132,\,0.128,\,0.166\}\ \text{lbf}.$$ The drag actually falls between $Re=15$ and $20\times10^4$ because $C_D$ drops faster than $v^2$ rises — this is the dimple-induced drag crisis.
4. Smooth sphere (part b). With $C_D=0.5$ fixed, $F_D=8.80\times10^{-6}\,v^2$, giving $$F_D=\{0.075,\,0.133,\,0.299,\,0.532,\,0.831\}\ \text{lbf}.$$
5. Compare. At the top speed ($Re=2.5\times10^5$, $v=307\ \text{ft/s}$) the smooth sphere suffers $$\boxed{\dfrac{F_{smooth}}{F_{dimpled}}=\dfrac{0.831}{0.166}\approx5\times}$$ the drag of the dimpled ball. The dimples force an early turbulent boundary layer that clings farther around the sphere, shrinking the wake and slashing pressure drag — which is why a driven golf ball carries so much farther than a smooth one.