# Does the terminal velocity of a sphere in free drop increase with increasing diameter?

I was trying today to calculate the terminal velocity of a hailstones with increasing diameter and mass. I was trying to figure out if larger hailstones will have a higher impact velocity or lower.

I performed the calculation (I was a bit surprised by the end result), which I am presenting as a potential answer. I would be very happy to see if there are any improvements to this answer (if there are additional factors that I need to consider)

The terminal velocity $$V_t$$ will be reached when the drag coefficient is equal to the force of gravity:

$$F_{drag} = mg$$ $$C_D\frac{1}{2}\cdot \rho_{air} \cdot A \cdot V_{t}^2= m\cdot g$$

where:

• $$C_D$$: is the drag coefficient (for a sphere is 0.5)
• $$\rho_{air}$$ is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)
• $$A$$: cross-sectional area of the hailstone assuming its a sphere $$\frac{\pi d^2}{4} =\pi r^2$$
• $$V_{t}$$: the terminal velocity of the sphere
• $$m$$: the mass of the sphere (assuming its a sphere the volume is $$\frac{4}{3}\pi r^3$$, and the density of the sphere is $$\rho_{sphere}$$)
• $$g$$: the acceleration of gravity

Therefore: $$C_D\frac{1}{2} \rho_{air} \cdot \left(\pi r^2\right)\cdot V_{t}^2= \frac{4}{3}\pi r^3 \rho_{sphere}\cdot g$$

$$V_{t}^2= \frac{4\cdot 2 \cdot\pi r^3 \rho_{sphere}\cdot g}{3C_D \rho_{air} \cdot \left(\pi r^2\right)}$$ $$V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere}}{\rho_{air} }\cdot r }$$

which indicates that the terminal velocity of a sphere increases proportionally to the square root of the diameter (radius).

if the buoyancy is considered then:

$$V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere} - \rho_{air }}{\rho_{air} }\cdot r }$$

• Which could probably be qualitatively demonstrated by timing the fall of a rock and a grain of sand. Have to go back and generate the acceleration formula so as to drop from sufficient height to have at least the sand grain reach $V_t$ Aug 17, 2021 at 13:22
• I'm actually working on that right now. I will probably post that as a separate question :-)
– NMech
Aug 17, 2021 at 13:26