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Drag Conceptual Question

  1. Jul 21, 2011 #1
    If a sphere is moving through air.
    Will it have more drag if it is spinning while moving or just moving ?
     
  2. jcsd
  3. Jul 21, 2011 #2
    The side that is spinning into the direction of the ball's motion will have more friction than the side that's spinning away from from this direction (since it's moving faster with respect to the air). Since drag rises with speed squared, the side spinning into the direction of motion will have gained more drag than the other side will have lost, and therefore the total drag will be greater.
     
  4. Jul 21, 2011 #3
    I can't remember the answer but I've seen this done for a circle using complex potential for a viscous fluid, by adding a cylinder plus a circulation to a uniform flow. I don't know if there's a drag but I'm pretty sure there's a lift.

    For a sphere I have no idea but would guess the answer is in the same direction as the 2D case. Would be interested in any other replies.

    If you're integrating drag (speed^2) over a surface, you've just assumed that the size of the two surfaces are equal. What if the higher drag surface is smaller?
     
    Last edited: Jul 21, 2011
  5. Jul 21, 2011 #4
    It's a sphere...how would one surface be smaller than the other?
     
  6. Jul 21, 2011 #5

    boneh3ad

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    One side could be "smaller" due to the fact that spin will change the location of both the stagnation point and the two separation points.
     
    Last edited: Jul 21, 2011
  7. Jul 21, 2011 #6
    I admit my answer might be overly simplistic. Air and the effects of drag probably behave differently at different speeds and turbulence. Still, my guess is that for the most part the spinning ball would be slower.
     
  8. Jul 21, 2011 #7

    boneh3ad

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    Assuming the flow is symmetric (it isn't), you can do some back of the envelope calculations that would imply that on a cylinder, very roughly:

    [tex]D \propto u_{\theta}^{2} + \omega^{2} R^{2}[/tex]

    It would be similar to that of a sphere. In other words, the drag would likely be slightly higher.
     
  9. Jul 21, 2011 #8

    Drakkith

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    Does a spinning sphere cause other aerodynamic effects that would affect the amount of drag?
     
  10. Jul 21, 2011 #9

    boneh3ad

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    Yes. It would change the location of the forward stagnation point as well as the separation lines as opposed to a sphere that isn't spinning. It would also change the character of the von Kármán vortex shedding.
     
  11. Jul 21, 2011 #10

    Drakkith

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    So the answer to the OP's question is probably much more complicated than simply drag then?
     
  12. Jul 21, 2011 #11

    boneh3ad

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    Of course, but unless one of us wants to sit down and mesh a sphere and then let a computer do a 1-day, 3-D CFD problem, I would imagine the simplified answer would probably help the OP at least somewhat.

    Doing the potential flow with a sphere and rotation would be a reasonable approximation if you want to ignore viscous effects such as viscous drag and separation, but that is where most of the drag is likely coming from.
     
  13. Jul 21, 2011 #12

    Drakkith

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    Sorry, I didn't mean to imply that we could ignore the drag, but merely was asking whether it would be "reduced" or whatever by the spinning.
     
  14. Jul 21, 2011 #13

    boneh3ad

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    I have been at work for far too long today so I probably just worded that kind of strangely.

    The point is you could definitely get a simplified answer but I can't say exactly how accurate it would be because separation is a very important factor here.
     
  15. Jul 22, 2011 #14
    Something I don't understand is this, how can you have a stagnation point on the surface. We are working with viscous fluids so surely the non-slip condition must mean the streamline at the boundary is a circle (or the stream plane is the surface of a sphere, in the 3D case), otherwise the fluid is clearly slipping against the rotating sphere.

    It highlights something I never knew I didn't understand! We always ended up with stagnation points when dealing with complex potentials, but always used non-slip conditions when, instead, solving the Navier-Stokes equation.
     
  16. Jul 22, 2011 #15

    boneh3ad

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    Just because the velocity against the surface is the same as that surface doesn't mean that stagnation points have no meaning. If you really want to think about it, it would be an attachment line in this case rather than a point, and it woul be the line above which the flow moves up and below which the flow moves down just above the surface.
     
  17. Jul 22, 2011 #16

    Andy Resnick

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  18. Jul 22, 2011 #17
    I have a feeling it may be the same. The rotation breaks the vertical symmetry because the stagnation points move, but I don't believe it breaks the upstream/downstream symmetry (apart from the direction of the flow, which shouldn't affect the following argument). To find drag you're integrating pressure times the normal vector around the surface of the sphere, and taking the horizontal component. Pressure is not affected by direction of flow, only speed (therefore it is upstream/downstream symmetric), so an integration of the horizontal component of the normal times pressure will be zero due to cancellation of the upstream/downstream hemispheres.
     
  19. Jul 22, 2011 #18

    boneh3ad

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    That has to do with a force perpendicular to the direction of motion (lift), which is not drag. Drag is parallel to motion.

    However, the drag actually is different as you said originally.

    In an inviscid sense, this is correct. However, in a viscous flow, you would have separation which would mean even the original case isn't upstream-downstream symmetric. Adding rotation would change the upper and lower separation points, meaning that in all likelihood, the drag would change.

    Again, that will only be the case in an inviscid flow. The separation breaks this symmetry. Using your logic, you could also argue that a cylinder moving steadily through air has no drag on it, which is untrue. This is commonly called D'Alembert's paradox ([URL [Broken]).[/URL] Taking viscosity into account greatly complicates the matter, but makes it correct.
     
    Last edited by a moderator: May 5, 2017
  20. Jul 22, 2011 #19
    Ok, thanks for the clarification. I went searching for something on this,

    Found a nice bit of relevant material: http://pof.aip.org/resource/1/phfle6/v11/i11/p3312_s1 [Broken]

    Edit- I'm amazed such a simple question has such a complicated answer- in 2d!
     
    Last edited by a moderator: May 5, 2017
  21. Jul 22, 2011 #20

    boneh3ad

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    I haven't read that paper yet but if you quote is correct, it tells me that the increased drag predicted by inviscid theory is offset plus some by the apparently smaller separated region behind the cylinder/sphere.
     
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