I just thought of a problem that seems quite fundamental but I couldn't find a clear answer or explanation.

Consider a solid sphere falling in a Newtonian fluid due to gravity. It eventually reaches its terminal vertical velocity. Now let's add a constant side wind: can this induce a net vertical force and hence increase or decrease the vertical speed of the sphere?

At low Reynolds number I believe the answer is no due to the linearity of the Stokes equation. But what about at finite Reynolds number? Even if the object falling is perfectly symmetrical, the flow around is not, possibly dramatically (e.g. recirculation zone, vortex shedding), and so it would not really surprise me if adding side wind to that situation would have an effect on the vertical direction. But intuitively I can't tell if it would indeed have an effect, and if so if it would speed up or slow down the vertical descent.


1 Answer 1


You can work in the reference frame in which the (constant) crosswind is zero.

In this frame of reference, the speed limit is the same as if there was no crosswind (I am talking only about the speed limit, not how it is reached).

Then, just go back to the original reference frame.

EDIT : Imagine an object falling at a constant speed with no crosswind. Change the frame of reference: you see an object falling at a constant speed with a crosswind. If the solution is unique, it is the right one.

In your example, you talk about a sphere. So orientation in space doesn't matter. You can apply Newton's law to the center of gravity. For any solid, the drag and lift forces depend on the orientation of the solid and it is not obvious that a steady state exists or that the solution is unique. So it's more complicated. Just watch a leaf fall!

  • $\begingroup$ Why would the limit speed be the same? $\endgroup$
    – squille
    Commented Aug 18, 2021 at 9:30
  • $\begingroup$ Because the Newton's law is the same. There is no crosswind in this frame. And the terminal velocity does not depend on the initial conditions. $\endgroup$ Commented Aug 18, 2021 at 10:14
  • $\begingroup$ Choosing another frame of reference doesn't miraculously make the forces exerted by the fluid on the object disappear. An obvious example would be a non-symmetric shape, e.g. an airfoil: it would fall due to gravity, but enough side wind can even make it go up. $\endgroup$
    – squille
    Commented Aug 18, 2021 at 11:46
  • $\begingroup$ I tried to clarify my thinking in the answer. Hope I am not mistaken and that can help you? Keep in mind that I am only talking about the steady state, if there is one. $\endgroup$ Commented Aug 18, 2021 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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