Magnus effect, what is the correct formula?

  • #1
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Main Question or Discussion Point

So I've been investigating the magnus effect for I guess 2 weeks, but I've not been able to find a formula that would seem to be the correct formula. I did find some formula, but I have my doubts about the completeness/correctness of all of these formulas. Note that I am trying to simulate what would happen to a golf ball, so the formulas I found that were for calculating cylinders I have not included.

Fm = π2 * ρ * ω * v * r3

My main issue with this formula is that the lift coefficient is not included in this formula, meaning it doesn't matter if the ball is smooth or (as in the case with golf balls) dimpled.

Fm = S*(ω x v)

I found this formula here http://farside.ph.utexas.edu/teaching/329/lectures/node43.html but it was not noted what S was, only a way to calculate it for the example of baseballs.

Fm = ½ * CL * ρ * A * v2 (ω x v)

I found this formula a lot, but the thing that bugs me with this one is that the cross product of ω and v is not a cross product of the vectors, but of the unit vectors, meaning that the last part is just to calculate the direction of the force. This would not be a problem if that was not the only place ω is used in that formula. Meaning that according to that formula ω is only used to calculate the direction, not the actual size of the force. I find it highly unlikely that the speed of which the ball is rotating does not affect the force, only the fact it is or is not rotating.

I could be wrong of course about these formulas. Or I could be interpreting them wrong. If any of you actually know something about the magnus force that could help me, I'd love for you to reply to this topic and maybe help me out, since I'm stuck at the moment.
 

Answers and Replies

  • #2
rcgldr
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ω is not a unit vector. Assume the ball is moving left to right with back spin, ω is a vector pointed towards the observer, v is a vector pointed to the right, and (ω x v) is a vector pointed up with a magnitude of |ω| |v|.
 
  • #3
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but then the units wouldn't match up again. Right now it's:

Kg*m/s2 = kg/m3 * m2 * m2/s2 = kg*m4/m3*s2 = Kg*m/s2

If you would add ω and v as normal vectors, not unit vectors.
 
  • #4
rcgldr
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So that should have been ##(\hat{\omega} \ x \ \hat{v})## ?

Apparently the spin affects coefficient of lift:

http://spiff.rit.edu/richmond/baseball/traj/traj.html

That article was about baseballs. The spin effect is more evident in the case of ping pong balls.
 
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  • #5
boneh3ad
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As with all topics in aerodynamics, things are typically complicated here. For example, lift (and drag) coefficients are typically an empirical (or model) representation of a much more complicated phenomenon. In this case, the rotation rate would be wrapped up into the lift coefficient. On the other hand, as with the first equation, the lift coefficient need not be used. The first equation represents the lift on a ball due to the Magnus effect ignoring viscosity (if my memory serves me) so it is perfectly valid as an approximation. It won't account for separation effects, and therefore for golf ball dimples, though.

If you want to include all the physics then you are just out of luck in finding a simple formula because one simply doesn't exist. You would have to flat out solve the Navier-Stokes equations, and any simplifications you make from there will be removing at the very least a little bit of the physics for the sake of practicality.
 
  • #6
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mm6c5w.jpg
23sviwx.png
mm6c5w.jpg
23sviwx.png
but then the units wouldn't match up again. Right now it's:

Kg*m/s2 = kg/m3 * m2 * m2/s2 = kg*m4/m3*s2 = Kg*m/s2

If you would add ω and v as normal vectors, not unit vectors.
Well folks, I was going to open a new thread to ask this but I found this one.

I ask here because the question has not been properly answered, that is what I think.

The dimensional analysis of the equation is what bothers me.

Fm = ½ * CL * ρ * A * v2 (ω x v)

The term Fm= ½ * Cl * ρ * A * v2 is already dimensionally correct. We have:
Force= (Mass * Lenght)/Time2 = Density (Mass/Length3) * Area (Lenght2)* Speed(Lenght2/Time2) = Mass * Lenght / Time2

And this result raises the question on this term:

## \vec w \times \vec v ##

If it is dimensionless we are saying the lift force is not dependent on w modulus which is absurd.

If ##\vec w## and ##\vec v## have their dimensions we get:

## \vec w \times \vec v ## = IwI * IvI* sin (90) * unit vector.

Since w= (1/time) and v = lenght/time I think w * v = Lenght/Time 2

Force = (Mass * Lenght)/ (Time2) * Lenght/Time 2 = Mass Lenght2 / Time4

Which is not correct.

So what am I doing wrong?

In this article:
http://spiff.rit.edu/richmond/baseball/traj/traj.html#spin
The force is given by:
mm6c5w.jpg


I found the same formula in another article:
https://www.researchgate.net/public...r_a_spinning_tennis_ball_I_The_service_stroke
Guess what, now it is different:

23sviwx.png


Can anybody clarify what is the correct dimensional analysis of this equation?

Thanks.
 

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  • #7
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I found on another book FL does not depend explicitly on w, it does just because CL is a function of w.

I hope this helps others.
 

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