What Is the Coefficient of Drag for a Tennis Ball?

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SUMMARY

The coefficient of drag (Cd) for a tennis ball, with a diameter of 6.5 cm and mass of 57 g, is influenced by various factors including surface conditions and Reynolds number (Re). The relationship between Cd and Re is complex, as Cd can vary significantly depending on the object's surface texture and flow conditions. Standard Cd values for smooth objects like spheres may not apply directly to tennis balls due to their rough surface. For accurate calculations, it is essential to consider the specific Reynolds number range and the appropriate reference area for the object.

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  • Understanding of drag coefficients and their significance in fluid dynamics.
  • Familiarity with Reynolds number (Re) and its calculation.
  • Knowledge of the relationship between surface texture and drag in projectiles.
  • Basic principles of fluid mechanics, particularly regarding drag forces.
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  • Research the specific drag coefficient values for various surface textures of tennis balls.
  • Learn about the calculation of Reynolds number for different flow conditions.
  • Explore the relationship between drag coefficients and flow regimes in fluid dynamics.
  • Investigate empirical methods for measuring drag coefficients in experimental setups.
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Students, researchers, and engineers in the fields of fluid dynamics, sports science, and aerodynamics who are interested in the behavior of projectiles like tennis balls in various fluid environments.

Bonulo
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What is the coefficient of drag, Cd, of a tennis ball (roughly d = 6,5 cm, m = 57 g)?

Generally; where do I find such coefficients? I'd like to find coefficients for projectiles in water too - since these are supposedly not the same, because of the water viscosity. I've found the "Drag Coefficient" text on Scienceworld Wolfram, which apparently equals Cd to Re^(-1/2), where Re is the Reynolds number. But the equation from which the text gets Cd has L in it, the "size scale" of the body - which is squared. But in my equation there is no such scale, only the silhouette area A of the body, which isn't squared. Why this difference?

Also - the "length scale" l is in the Cd equation. What is it?

I'm quite puzzled. If the Cd can be found directly from a Reynolds number, which I'm not sure it can, could my problem then be solved?
 
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Reading Wolfram, it specifically quotes beside the use of the size scale of the body L to refer to (Tritton 1988, p. 93). The formula for drag is quite approximatish. From the text I'm familiar with as also found here http://www.lerc.nasa.gov/WWW/K-12/airplane/dragco.html

the L is instead an area A. The choice of this reference area A can vary.

In Wolfram, the Cd being directly a function of the Reynolds number is from successive approximation.
 
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Cd for something like a tennis ball is going to vary greatly due to surface conditions. There are standard Cd values for smooth objects like a sphere, but in your case, they would be introducing more error into your calcuations.

From what I have seen, when there is a nice easy relationship like the one you stated for Re and Cd, that usually means there's a pretty healthy restriction on it's usage. For example, I have a source right here that says the Cd for a sphere is 24/Re. However, that is for Re<1. Probably not very helpful. The Cd is not going to be a nice straight line over a decent range of Re. I do have a chart for the Cd vs. Re of a smooth cylinder and sphere for Re going from 10^-1 to 10^7. I just don't have a way to post it. If you can provide an Re range, I can at least read off some values for you if you think that would help.
 

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