What is the Lift Coefficient of a Basketball?

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How can I find an equation to calculate the lift coefficient on a basketball at different launch velocities?

I looked at different papers explaining the physics of basketball throws but all i could find was an equation for the drag coefficient: -mv +1/2 ρ A Cd v^2 = mv
 

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  • #2
HallsofIvy
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Why would a basket ball have any lift?
 
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I thought lift force was created with bill spin. I'm confused now..
 
  • #4
HallsofIvy
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Yes, it can. But you hadn't said anything about "spin" before, just "different launch velocities".
 
  • #5
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Sorry.

Basically I need to develop a trajectory model for 3-point shot in basketball with a given launch, angle and spin rate. I cannot find an equation for the lift coefficient.
 
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CWatters
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  • #8
boneh3ad
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Lift due to spin is the Magnus effect and it's the only possible source of lift on a basketball. Now, I haven't done the math, but my intuition tells me it is likely a very small effect given the low spin rate and low velocity of the ball.
 
  • #9
A.T.
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Do basket balls travel or spin fast enough?
Depends where you throw it:

 
  • #10
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The rotational lift force for a rotating and translating sphere is given by the Rubinow-Keller equation (low Reynolds approximation):
[itex]F_{\omega} = \omega U \cdot \pi r^3 \rho[/itex]
Compare this to the force due to gravity:
[itex]F = \frac{4}{3}g \cdot \pi r^3 \rho[/itex]

Even if the spin rate is low, if the ball has sufficient velocity the force will be of the same order of magnitude as the gravity force. The terminal velocity of a basketball with a drag coefficient of 0.5 is around 20 m/s. At this velocity, the angular velocity that makes the Magnus force equal to the gravity force (and also equal to the drag force in the direction opposite of the gravity vector) is 0.65 rad/s, so only about 0.1 revolution per second.
 
  • #11
boneh3ad
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The rotational lift force for a rotating and translating sphere is given by the Rubinow-Keller equation (low Reynolds approximation):
[itex]F_{\omega} = \omega U \cdot \pi r^3 \rho[/itex]
Compare this to the force due to gravity:
[itex]F = \frac{4}{3}g \cdot \pi r^3 \rho[/itex]

Even if the spin rate is low, if the ball has sufficient velocity the force will be of the same order of magnitude as the gravity force. The terminal velocity of a basketball with a drag coefficient of 0.5 is around 20 m/s. At this velocity, the angular velocity that makes the Magnus force equal to the gravity force (and also equal to the drag force in the direction opposite of the gravity vector) is 0.65 rad/s, so only about 0.1 revolution per second.
By this you are claiming that I can go grab a basketball and throw at 2 m/s and it will stay floating in the air without falling if I put 1 rev/sec on it. Shoot, I could throw it at 4 m/s at the same rotation rate and it will curve up. Go test that in your driveway and tell me what you find.

Rubinow and Keller did their analysis for small particulate matter in the wind. That low Reynolds number approximation was made in that context and is basically the same as the low Reynolds number in Stokes flow. In other words, that equation is not valid for a basketball. The Reynolds number is way too high by several orders of magnitude.
 
  • #12
256bits
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Most likely, the lift for a basketball with back spin, would have to be determined experimentally. A basketball does not have a smooth surface, for which most of the equations are tailored.

Players put a backspin ( a forward spin would tend to drop the ball ) for several reasons - it hangs in the air a little while longer, a slows down a bit so that at the basket it is dropping a bit more vertical into the hoop, and to futuristically anticipate the trajectory of the ball. In fact, with no spin, ( which is probably difficult to do for basketball shoot ) the trajectory of the ball may deviate from what one expects due to the air playing on one side and then the other. As said these are probably a minute part of the game for basketball.

In fact, one is not trying to outwit the loop as one would do for a goalie, pitcher, or opposite team as when playing soccor, baseball, or volleyball. One is trying to put a ball into a little bit larger loop, often from some distance away.
The best part then, and most important part of the game, is scoring points when hitting the loop front end or the backboard. The backspining ball, when hitting the top front of the loop, will naturally deflect upwards and possibly drop into the basket; or deflect down into the basket when hitting the backboard. More points scored with backspin than with forward spin or no spin. As such, the thrower can be a little bit less than perfect than attempting to put the ball through the loop, than he would have to be without the ball touching anything.
 
  • #13
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By this you are claiming that I can go grab a basketball and throw at 2 m/s and it will stay floating in the air without falling if I put 1 rev/sec on it. Shoot, I could throw it at 4 m/s at the same rotation rate and it will curve up. Go test that in your driveway and tell me what you find.
No no, there is also something called drag, please see this explanation: https://en.wikipedia.org/wiki/Drag_equation
I'm just teasing you of course. ClapClap already mentioned the drag coefficient.

Rubinow and Keller did their analysis for small particulate matter in the wind. That low Reynolds number approximation was made in that context and is basically the same as the low Reynolds number in Stokes flow. In other words, that equation is not valid for a basketball. The Reynolds number is way too high by several orders of magnitude.
Absolutely true, the Rubinow-Keller equation is not valid for high Reynolds numbers, but the OP asked for an equation. This is the simplest equation that shows the effect of rotation on a sphere and it can be derived analytically. I based my reply on the estimated level of knowledge of the original poster, which is why I did not give more complicated but more accurate equations. Of course, the Rubinow-Keller equation gets a correction factor for high Reynolds numbers, in the same way as the Stokes drag equation gets a correction term for high Reynolds numbers. But it will basically be the same equation. You can get very accurate correlations up to Re=10,000 or so, but for higher Re, the surface roughness becomes more important. You need a correlation based on detailed measurements of the basketball to predict the transition to turbulent separation and the drag collapse.

I hope this is more clear.
 
  • #14
boneh3ad
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No no, there is also something called drag, please see this explanation: https://en.wikipedia.org/wiki/Drag_equation
I'm just teasing you of course. ClapClap already mentioned the drag coefficient.
Of course the drag will only serve to slow it down. If you really threw it double the speed to stay level like I said, even drag wouldn't slow it down fast enough to prevent a noticeable rise. If the Rubinow-Keller numbers were correct here. The point of my post is to illustrate the value of going through a bit of a sanity check to see if the answer is reasonable. In this case, the answer was not and that should be pretty easy to see just by looking at a scenario that you can easily achieve in your driveway your numbers say would happen but experience says it will not.

Absolutely true, the Rubinow-Keller equation is not valid for high Reynolds numbers, but the OP asked for an equation. This is the simplest equation that shows the effect of rotation on a sphere and it can be derived analytically. I based my reply on the estimated level of knowledge of the original poster, which is why I did not give more complicated but more accurate equations.
I would argue that, when considering that the OP likely is not all that familiar with fluid dynamics, it is important not to quote equations just because they exist if they do not apply. A cursory knowledge of the topic is likely to lead to misuse of the formulae due to not understanding why they don't apply, so just quoting Rubinow-Keller because it is the only closed-form solution doesn't make it useful for the OP. It just makes it more likely that the OP will misuse the equation.

Of course, the Rubinow-Keller equation gets a correction factor for high Reynolds numbers, in the same way as the Stokes drag equation gets a correction term for high Reynolds numbers. But it will basically be the same equation. You can get very accurate correlations up to Re=10,000 or so
At the 20 m/s you quoted, ##Re = O(10^5)## for a basketball. I calculated roughly ##3\times 10^5##, so that is already several times larger than your quoted limit of the corrected Rubinow-Keller equation, which isn't even the one you quoted.

but for higher Re, the surface roughness becomes more important. You need a correlation based on detailed measurements of the basketball to predict the transition to turbulent separation and the drag collapse.
Honestly, I'd bet that just the potential flow approximation, while obviously wrong, would give a better approximation in this case than Keller-Rubinow. It obviously won't take separation into account (or viscosity at all except for the spin rate), but it will probably give something that is at least the right order of magnitude. That may be good enough for the OP.
 

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