What is the math on how much wind speed is required to lift a given mass in a vertical wind tunnel, assuming a spherical object?

  • $\begingroup$ -1. No research effort. $\endgroup$ – sammy gerbil Aug 17 '17 at 22:04

The formula for drag force is $$ F = \frac{1}{2}*p*v^2*c_d*A $$

  • $C_d$ is your coefficient of drag, which is about 0.5 for a sphere. (Independent of size or material)
  • $p$ is the density of the air. Approximately $1.225 \frac{kg}{m^3}$ at sea level.
  • $v^2$ is your wind velocity, what you need to solve for
  • $A$ is the 2-D cross sectional area of the sphere Use $A = \pi * r^2$

Assuming you want your sphere to levitate you want to set $F = m*g$ of the sphere. So in your case you want to solve for $v$.

$$ v = \sqrt{\frac{2*mg}{p*c_d*A}} $$


There are three comments that need to be made on this.

  1. The drag coefficient of a sphere depends quite strongly on several things, including the roughness of the surface. This is an important aspect of many ball games, as you will learn if you Google "dimpled golf balls"

  2. It will also depend on the ratio between the diameters of the sphere and the tunnel.

  3. The airflow over a sphere is seldom symmetrical, even though the sphere itself is symmetrical. There is usually a significant side force, which means that the sphere will move unpredictably off center in a wind tunnel. However, a sphere will often sit stably inside a free jet. This time Google "Coanda effect"

The aerodynamics of a sphere is surprisingly complex.

  • $\begingroup$ -1. No research effort. $\endgroup$ – sammy gerbil Aug 17 '17 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.