## Wind resistance on a rapidly rotating flywheel

I am simulating a rapidly rotating (horizontal axis, if that makes a difference) flywheel. The software isn't too sophisticated w.r.t. wind resistance, unfortunately, and seems to use the diameter of the rotating body as one of the main parameters, in addition to the velocity of the rotating body. Consequently, I get pretty much the same results for a rotating disk with diameter (D) and velocity (v) as I do for my rotating flywheel with the same overall diameter and velocity.

Intuitively, the rotating thin-walled and narrow cylinder-shaped flywheel, with its lesser surface area, would be less effected by wind resistance than a solid disk of the same thickness and diameter, right? Unless there are odd turbulence effects with the cylinder shape that come into play; anyone know?

Anyway, the equation used in the software is F= -D[a(v) + b(v^2)], where a and b are the only values I can modify in that equation. Also, I can totally turn off wind resistance, but that seems to give pretty unrealistic results, and can vary the wind speed and angle w.r.t. the rotating object, although in my case, I have left these 0 to simplify things (I am not sure how these are factored in, exactly, for fixed position objects, especially round, rotating ones).

Anyway, anyone have any ideas on what I should set a and b to in F= -D[a(v) + b(v^2)] for my flywheel, or have knowledge of how wind resistance affects this sort of rotating shape? Thanks.
 PhysOrg.com physics news on PhysOrg.com >> Kenneth Wilson, Nobel winner for physics, dies>> Two collider research teams find evidence of new particle Zc(3900)>> Scientists make first direct images of topological insulator's edge currents
 Sorry, I've stopped at the second paragraph, I don't understand the difference in the two shapes. One is a cylinder shape, and another is a solid disk? Is one hollow?
 Google found this which has equations for the drag on a flywheel of thickness h. See equation 3.140.. http://books.google.co.uk/books?id=Z...stance&f=false To work it out for a hollow cylinder perhaps you could work out the drag for two solid flywheels one with radius Router and the other Rinner. Subtract one from the other to account for the hole.

## Wind resistance on a rapidly rotating flywheel

MikeyW - think wagon wheel without much in the way of spokes vs. solid disk, or bike wheel vs solid disk. Since the bike/wagon wheel has less surface area, I would assume it is less impacted by wind resistance if other factors like turbulence aren't an issue.

 Quote by CWatters Google found this which has equations for the drag on a flywheel of thickness h. See equation 3.140.. http://books.google.co.uk/books?id=Z...stance&f=false To work it out for a hollow cylinder perhaps you could work out the drag for two solid flywheels one with radius Router and the other Rinner. Subtract one from the other to account for the hole.
Thanks, Cwatters. I will see if the reference helps and can be adapted to my software's limited capabilities in this area. Nice find.

Recognitions:
[QUOTE=CWatters;4003990]
 Quote by tadietz MikeyW - think wagon wheel without much in the way of spokes vs. solid disk, or bike wheel vs solid disk. Since the bike/wagon wheel has less surface area, I would assume it is less impacted by wind resistance if other factors like turbulence aren't an issue.
Before you jump to that conclusion, ask yourself why bikes for indoor velodrome racing have solid wheels rather than spokes.

(Of course the reason solid wheels are a bad idea for outdoor riding is the increased area affected by crosswinds, which makes the bike harder to steer).

I have no idea where the formula in your Google book reference came from, but most reputable empirical formulas for fluid dynamics use nondimensional quantites (Reynolds number, Prandtl number, etc) rather than units-dependent parameters like r and ω. It's quite likely the "magic constant" 0.04 in the formula is also units dependent, so be careful!

[QUOTE=AlephZero;4004126]
 Quote by CWatters Before you jump to that conclusion, ask yourself why bikes for indoor velodrome racing have solid wheels rather than spokes. (Of course the reason solid wheels are a bad idea for outdoor riding is the increased area affected by crosswinds, which makes the bike harder to steer). I have no idea where the formula in your Google book reference came from, but most reputable empirical formulas for fluid dynamics use nondimensional quantites (Reynolds number, Prandtl number, etc) rather than units-dependent parameters like r and ω. It's quite likely the "magic constant" 0.04 in the formula is also units dependent, so be careful!
Well, bike wheels typically have a lot of thin spokes (though many are going to carbon-fiber wheels with thicker and fewer spokes...), so I don't know the whole set of issues there.

In my simulation (for now), I sort-of ignore the need to connect the rim to the axle for various reasons I won't go into here, and so having non-existent material factored into a wind resistance equation - just because of the diameter of the flywheel rim - is not going to give me valid results.

I may attempt to see if I can find some fluid dynamics references, as you suggest, although those models tend to be something that would be hard to adapt to the limited parameters I can modify in my simulation software.

I will say that maybe it might be a valid approach to calculate the differences in wind resistance between what the software would give for a disk with the maximum flywheel diameter and one with the diameter of the inner rim of my flywheel. I MIGHT be able to find parameters I could plug in to the software that give me a similar value using this approach, but I will need to play around with it.

Thanks everyone.

 Similar discussions for: Wind resistance on a rapidly rotating flywheel Thread Forum Replies Introductory Physics Homework 1 General Physics 5 Introductory Physics Homework 8 General Physics 1 Introductory Physics Homework 1