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The bounce of a tennis ball (clay v. hard court) |
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| Feb18-10, 06:39 AM | #1 |
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The bounce of a tennis ball (clay v. hard court)
I'm trying to understand why tennis balls have a higher bounce on clay courts than on hard courts.
I understand that the amount of friction is greater on clay so as to slow the ball's vertical speed down and create a steeper angle of reflection. But I don't see how that explains why the coefficient of restitution is higher on clay (0.85) than on hard court (0.80). From what I understand, that means that a ball dropped on to a clay court will retain more of it's speed after the bounce (85 percent) than a ball dropped on to a hard court (80 percent). I understand why grass courts have the lowest coefficient of restitution (0.75). They are soft and so I suppose they 'withhold' more of the energy of the dropping ball. But aren't clay courts softer than hard courts? A ball dropping on a clay court sometimes even make a little 'dent' in the court surface, whereas hard courts (like Rebound Ace) should work precisely to re-bound the ball due to it's cement/rubber surface. No? Anyway, I hope someone is able to clarify this for me. |
| Feb18-10, 08:25 AM | #2 |
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The higher the friction of the surface of the court, the more the spin is increased when it impacts with the court, and the increase in angular momentum comes at the expense of linear momentum so the ball bounces at a more veritcal angle than it would on a court with less friction.
In the case of hard courts, there's some type of elastic material under or mixed in with the surface that affects the bounce. I recall that the Austalian open courts are different than the US courts, but don't remember which one has the more elastic surface and which one is "faster". |
| Feb18-10, 11:12 AM | #3 |
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Right, I'm with you that far. The increased friction of clay courts help to 'redirect' the forward momentum of the ball upward causing a higher bounce (sorry for the lay terms, I'm no physicist).
But what if the ball has no forward momentum? Let me explain what I fail to understand: I'm reading Rod Cross' and Crawford Lindsey's "Technical Tennis" and it says that "the rules of tennis says a tennis ball must bounce vertically at about three-fourths of its vertical drop speed". Further: "The ratio of the ball's vertical speed after the bounce to that before is known as the 'coefficient of restitution (COR)'". Ok, so the COR varies, depending on the surface. It's 0.75 for grass, 0.8 for hard court, 0.85 for clay. But these numbers are valid, are they not, even if I simply drop a ball from a meter's height, straight, in a 90 degree angle on to the court? And if I do so, there is no friction involved. The ball simply bounces straight up and down. So how, without friction, can it be that a clay court has a higher COR? Is the surface 'harder' than hard court? Does clay have more of a trampoline effect? How is that possible when hard courts are made of rubber while clay courts are made of crushed brick? Hope you can help me out here, I'm desperate to understand this :-) |
| Feb18-10, 12:13 PM | #4 |
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The bounce of a tennis ball (clay v. hard court)
Hi hrhaahr! Welcome to PF!
![]() From http://en.wikipedia.org/wiki/Tennis_court#Clay_courts …Clay courts Main article: Clay court Clay courts are made of crushed shale, stone or brick. The red clay is slower[clarification needed] than the green, Har-Tru, "North-American" clay. The French Open uses clay courts, unlike the other three Grand Slam tournaments of each year. Clay courts slow down the ball and produce a high bounce[citation needed] when compared to grass courts or hard courts. This is because clay courts have more grab and when the ball lands there is more friction pushing against the ball's horizontal path, therefore slowing it and creating a higher bounce. For this reason, the clay court takes away some advantage of big serves, which makes it hard for serve-based players to dominate on the surface. |
| Feb18-10, 02:43 PM | #5 |
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ITF rules state that a vertical drop from 100 inches onto a concrete base, for a "type 2" ball, should result in the ball bouncing 53 to 58 inches, a much lower COR than you're quoting. See "standard" ball at this web site: http://www.itftennis.com/technical/e...alls/index.asp http://www.itftennis.com/technical/e...alls/index.asp |
| Feb19-10, 03:02 AM | #6 |
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So I guess the question I'm still left with is this: If I drop a standard tennis ball directly on to a clay court and I do the same thing on a hard court with an identical ball from the same height, which ball would bounce higher? And, most importantly, why? |
| Feb19-10, 01:57 PM | #7 |
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| Oct7-10, 03:21 PM | #8 |
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If you dropped a tennis ball onto an ideal hard surface -- one with a CoR of 1.00 -- the rebound would be higher, and the result would give you the CoR of the ball alone. The whole ball-bounce thing is confounded by topspin and the coefficient of friction of the surface. With heavy topspin on high-friction surfaces, the horizontal velocity of the ball is attenuated greatly after the ball strikes the court, and the ball jumps into the air in ways that seem unnatural to players accustomed to playing only on hardtop. |
| Oct7-10, 03:46 PM | #9 |
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Another thing worth adding -- which actually might go further to answering the original question -- is that a tennis ball that strikes a "slow" surface obliquely can have a CoR as high as 0.9, which means slow courts can be "fast" in the vertical direction!
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