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SystemicAnomaly
09-13-2009, 02:05 AM
These are excerpts from another (25 page) thread in this forum

... The radar gun should be picking up the speed of the ball at its maximum velocity -- this would happen a short time after it comes off racket, since it takes some finite amount of time for the ball to accelerate up to that max velocity. Someone in another thread claimed that this was a few feet (nearly 1 meter) after contact -- maybe, maybe not...

Sorry, but that is a physical impossibility (unless you stick a burning rocket to the ball). It can *not* accelerate after it has left the contact with the racket strings. Acceleration needs some kind of force operating on the ball. The only force in effect on a ball after it has left the strings is the drag from the wind causing it to slow down...


Nope, you are wrong about this. What you suggest is an impossibility.

Prior to contact, the ball speed is nearly zero (very slow, anyway). The force of the impact accelerates the ball, therefore the ball is initially accelerating as it comes off the racket. Acceleration, by definition, means that the ball is increasing its velocity. The ball cannot achieve it maximum speed instantaneously - it takes a brief, finite amount of time to accelerate up to that speed.

For a ball to go from zero (essentially) up to a faster speed, let's say 120 mph, cannot possibly happen in zero time -- it takes a finite amount of time as I stated above.

SystemicAnomaly
09-13-2009, 02:07 AM
What??? the ball accelerates after impact????? Physically impossible. The ball accelerates "during" direct contact with the string bed. This happens in fractions of a second. After the ball leaves the string bed, the only two forces acting with the ball are gravity and air friction, so it immediately starts decelerating after contact.

Exactly. The ball isn't going to speed up after it leaves the string bed unless it has some sort of on board jet pack. It reaches maximum velocity the moment the fuzz leaves the strings and the drag slows it down thereafter.

That "jetpack" may very well be the stored energy in the ball. Energy is delivered to the ball by its impact with the moving stringbed of the racket. As the ball compresses, I believe that it stores some of this energy. Some of the energy may also be lost in the form of heat (of compression). The rubber, felt, and compressed air inside the ball all experience a slight increase in temperature. Not sure, but perhaps, the heat may even contribute to the jetpack effect.

Some energy is also stored in the stringbed during impact. I have heard the the frame does not really contribute any energy to the ball since it does not recoil until the ball has already left the stringbed. Is this true?

I'm saying that the ball accelerates due to the impact. As the ball starts to touch the strings, its velocity is nearly zero. At that point, the impact area of the stringbed is going at some speed -- the racket may or may not be accelerating at this point. As the ball makes contact with the strings, the ball compresses and the stringbed deforms. This stringbed/ball interaction will affect the racket head speed a little bit -- conservation of momentum, I believe.

As I understand it, the ball is on the strings for just a few milliseconds, perhaps as much as 10 ms. While the ball is on the stringbed, the ball and the racket are going at the same speed, more or less -- different parts of racket and various parts of the ball are actually going at somewhat different speeds. Since the ball deforms (compresses) and subsequently regains it original shape (almost), the front of the ball and the back of the ball are not really going at the same speed.

As long as the ball is on the stringbed, the average speed of the ball is pretty much going at the same "local" speed of the racket in that area. As the stringbed recovers from the impact, it release some stored energy and starts to push the ball away from it.

At some point the ball speed exceeds the speed of the ball/racket combination. Exactly when doe this happen? Altho' it may only be milliseconds, or even microseconds, I believe that it is possible for the ball to accelerate somewhat as it is leaving the strings or, for a time, albeit a very brief time, after it has left the stringbed.


I haven't taken physics in a while, but this sounds absolutely insane. The ball will not accelerate once it has left contact with the racquet (heck, even when some of the hairs are still touching the strings).

Am I crazy? Misinformed?

^ NOPE, absolutely right....
thats highschool physics, if not elementary physics........

and the most ridiculous thing is that Sys wants to keep on at it....

SystemicAnomaly
09-13-2009, 02:08 AM
http://wings.avkids.com/Tennis/Project/serve-01.html

http://wings.avkids.com/Tennis/Images/Set_2060.jpg

SystemicAnomaly
09-13-2009, 02:11 AM
Take a very close look at the spacing of the ball snapshots. Last year, in another thread in this forum, I made the claim that the ball continues to accelerate for a (very) short time after leaving the stringbed before it begins to decelerate (due to air drag). Several posters were quick to ridicule this "absurd" idea. The graphic above would appear to validate my claim.

Look closely at the spacing of the first 5 ball snapshots. The spacing between the 1st 2 balls show that the ball is just starting to move forward & has not yet achieve its max velocity. The distance between ball #2 and ball #3 is noticeably greater. What is quite interesting is that the distance between ball #3 and ball #4 is even greater than the previous interval. The intervals are that one are smaller which would indicate that the ball is starting to slow down slightly.

Ball #1 is undoubtedly on the stringbed. Ball #2 may (or may not) still be on the stringbed -- my guess is that it is. Ball #3 cannot possibly be on the stringbed any longer (judging from the trajectory of the racquet head). This would strongly suggest that the ball is still accelerating for a short time after leaving the stringbed.

Please review this and let me know what you think about this evidence.

WBF
09-13-2009, 05:30 AM
Thanks for sending this Systemic, interesting stuff.

I do question how long the ball is in contact with the stringbed though. I think it is at least plausible that it remains in contact throughout, but I feel a bit sheepish about my incredulous response ;-)

dozu
09-13-2009, 05:52 AM
If the max speed is indeed a foot or so forward of the contact point, then I'd say it's probably due to the ball compression.

at the point when the ball leaves the racket, the back half of the ball is caved in due to compression..... this is the 'stored energy'.... because once the 'caved in' part recovers to it's normal position, it is actually a backward move in reference to the center of gravity of the ball, and therefore propelling the center of gravity FORWARD.

this is like a kid sitting on the shopping cart by himself, rocking back and forth, he can actually making the entire shopping cart move without anybody else pushing.

pabletion
09-13-2009, 08:52 AM
Take a very close look at the spacing of the ball snapshots. Last year, in another thread in this forum, I made the claim that the ball continues to accelerate for a (very) short time after leaving the stringbed before it begins to decelerate (due to air drag). Several posters were quick to ridicule this "absurd" idea. The graphic above would appear to validate my claim.

Look closely at the spacing of the first 5 ball snapshots. The spacing between the 1st 2 balls show that the ball is just starting to move forward & has not yet achieve its max velocity. The distance between ball #2 and ball #3 is noticeably greater. What is quite interesting is that the distance between ball #3 and ball #4 is even greater than the previous interval. The intervals are that one are smaller which would indicate that the ball is starting to slow down slightly.

Ball #1 is undoubtedly on the stringbed. Ball #2 may (or may not) still be on the stringbed -- my guess is that it is. Ball #3 cannot possibly be on the stringbed any longer (judging from the trajectory of the racquet head). This would strongly suggest that the ball is still accelerating for a short time after leaving the stringbed.

Please review this and let me know what you think about this evidence.

Well, I wasnt trying to ridicule ya, if it sounded like that I apologize, i just cant conceive it, no matter how much I picture it.

The graph is still not very clear to me, cause, in the first place, it shows the racquet too far back, its even angleded back, while the photo shows the racqet past that vertical line so, my guess is, where those two snapshots of the ball are, when they seem to accelerate, is during the contact of the ball with the racquet, and on that 2nd ball, its where the ball finally leaves the stringbed. So with that, I still say that deceleration immediately starts.

Keep in mind that the energy transfer from the racquets kinetic energy, the string, and the ball compression and restitution is very violent and happens very very rapidly, in micro seconds, so that results in an abbrupt burst. The balls initial speed is very high, but immediately after it leaves the stringbed (the strings are tensed and deformed back, and then slingshot forward, and the ball deforms and springs back to original form, during contact) it is only governed by air friction and gravity.

Im not a physics expert, but thats what Ive learned and what I can picture. Maybe theres more to this than what I can see, but still what you have provided seems very confusing and not too accurate. What are you experties on the subjetc?

pabletion
09-13-2009, 09:00 AM
http://img149.imageshack.us/img149/5251/set2060.jpg

This is what I mean: theres a very short time period when the racquet is in contact with the ball, while strings are deforming and coming back, and ball also, and then the ball explodes away from the racquet face, my guess is, this graph is missing that portion where the racquet is more forward, during that short period of time where the ball appears to accelerate.
The red line indicates what im talking about.

Steady Eddy
09-13-2009, 09:31 AM
So I'm recalling the time I went to a mall that was having a fast serve contest. One guy liked to stand far away from the radar gun, he felt that this gave his ball more time to pick up speed. Are you guys saying that he was right?

gastro54
09-13-2009, 09:51 AM
There is absolutely 0 chance of the ball accelerating after leaving the stringbed. ALL of the acceleration occurs while the ball is in contact with the racquet. To say otherwise violates the most fundamental laws of physics.

Even if the ball was still compressed after leaving the racquet, any expansion of the ball could not possibly contribute to ball acceleration... Why? Newton's 3rd law of motion: every action has an equal and opposite reaction. Midair ball expansion is an internal process resulting in a net force of 0 on the ball, so this cannot accelerate it further.

meowmix
09-13-2009, 10:43 AM
That "jetpack" may very well be the stored energy in the ball. Energy is delivered to the ball by its impact with the moving stringbed of the racket. As the ball compresses, I believe that it stores some of this energy. Some of the energy may also be lost in the form of heat (of compression). The rubber, felt, and compressed air inside the ball all experience a slight increase in temperature. Not sure, but perhaps, the heat may even contribute to the jetpack effect.

Some energy is also stored in the stringbed during impact. I have heard the the frame does not really contribute any energy to the ball since it does not recoil until the ball has already left the stringbed. Is this true?

I'm saying that the ball accelerates due to the impact. As the ball starts to touch the strings, its velocity is nearly zero. At that point, the impact area of the stringbed is going at some speed -- the racket may or may not be accelerating at this point. As the ball makes contact with the strings, the ball compresses and the stringbed deforms. This stringbed/ball interaction will affect the racket head speed a little bit -- conservation of momentum, I believe.

As I understand it, the ball is on the strings for just a few milliseconds, perhaps as much as 10 ms. While the ball is on the stringbed, the ball and the racket are going at the same speed, more or less -- different parts of racket and various parts of the ball are actually going at somewhat different speeds. Since the ball deforms (compresses) and subsequently regains it original shape (almost), the front of the ball and the back of the ball are not really going at the same speed.

As long as the ball is on the stringbed, the average speed of the ball is pretty much going at the same "local" speed of the racket in that area. As the stringbed recovers from the impact, it release some stored energy and starts to push the ball away from it.

At some point the ball speed exceeds the speed of the ball/racket combination. Exactly when doe this happen? Altho' it may only be milliseconds, or even microseconds, I believe that it is possible for the ball to accelerate somewhat as it is leaving the strings or, for a time, albeit a very brief time, after it has left the stringbed.

I'm going to have to agree. You're all forgetting one very simple principle: changes in acceleration take time. You can compare hitting a shot to firing a bullet. With the bullet, the acceleration is positive only as long as the bullet is in the barrel. As soon as the bullet leaves the barrel, acceleration drops to 0 (and then becomes negative due to friction). However, we all know that velocity graphs are not straight lines, and that for acceleration to drop down to 0, there must be a small period time between when the bullet leaves the barrel and when the bullet's acceleration drops to 0. Same thing must occur with tennis shots. According to (very) elementary physics, there must be positive acceleration for a very small period of time after the ball leaves the stringbed. Granted, this period of time can probably be measured in nanoseconds, but it's still a period of time.

And about ball compression... that does nothing. As somebody earlier said, every action has an equal and opposite reaction, and ma must equal ma.

TenniseaWilliams
09-13-2009, 11:28 AM
So I'm recalling the time I went to a mall that was having a fast serve contest. One guy liked to stand far away from the radar gun, he felt that this gave his ball more time to pick up speed. Are you guys saying that he was right?

No. If he understood how radar guns worked, he could get a higher (more accurate) score by decreasing the transmission angle between the gun and the ball travel. Distance would add air drag.

To the OP, I am sorry to inform you that you are in violation of Newton's first law of motion if you expect the ball to increase speed without an accelerating force present. Speed and acceleration are tightly coupled in (and to) time. Acceleration is nothing more than the rate of change in speed, and speed is nothing more than the rate of change in position.
Acceleration has likely dropped to zero with the strings still in the last phases of contact. http://j.photos.cx/rulebreaker-684.gif

To meowmix, changes in acceleration don't take time to occur, acceleration describes the rate of change of speed over a time period. If you did something to place a constant acceleration on an object, (an effect,not a cause) it would build up velocity over time. At the end of the ball/racquet contact, there isn't any extra air/gas pressure dissipating like there is at the end of a gun muzzle.

YULitle
09-13-2009, 11:45 AM
TW & SE, do you know what they call the rate in change of acceleration? :D Just a tid bit of trivia: A jerk.





But yes, there is no acceleration after impact, except in the y direction, but that isn't enough to compensate the loss due to wind resistance. Balls don't just get up and go on their own.

Furthermore, IF IT COULD, which it can't, it wouldn't be "violating physics", it would be either challenging our understanding of physics or our understanding of the situation (more likely the latter.)

mtommer
09-13-2009, 12:01 PM
Even if a ball is uncompressing after leaving the strings of the racquet it still can't result in additional energy to further accelerate the ball as there is nothing for that compression to "push" against thereby redirecting the energy into a forward vector (very simply - stopping the compression movement of the ball back towards server so the energy has to go somewhere). The ball would merely continues to elongate while continuing to slow down due to air friction, friction if the resistance of the ball to compress and elongate, and gravity.

Even a topspin groundstroke does not accelerate the ball as it bounces. What happens is that the ball is suddenly "stopped" and potential energy rises beyond the forces currently "holding" the ball resulting in a the potential energy being converted suddenly back into kinetic energy. Although the ball "accelerates" it doesn't go faster than the moment before the ball impacts the ground. It just goes faster than the "stopping" moment of impact. The result is that the ball has continued to decrease speed overall as it left the racquet. It's just not a smooth linear deceleration.

El Diablo
09-13-2009, 12:02 PM
Lets simplify with an analogy. A figure skater doing a spin with his arms extended pulls his arms in closer to his body and his spin speed accelerates. No external force is acting on the skater, but his acceleration is consistent with the conservation of angular momentum. Angular momentum is not exactly the issue with the tennis ball, but as it decompresses it becomes more aerodynamic.

jonnythan
09-13-2009, 12:05 PM
The ball cannot - and does not - continue accelerating after it has left the strings of the racquet.

Sorry SA. You're just not understanding the basic, elementary physics that are involved here.

You use lots of words but it's clear that even first-semester physics is above your head. I suggest you listen to the people who understand the forces involved better than you do.

mtommer
09-13-2009, 12:07 PM
You can compare hitting a shot to firing a bullet. With the bullet, the acceleration is positive only as long as the bullet is in the barrel...

...due to the explosive forces acting on the rear of the bullet as they too have to exit the barrel. In other words, they do not stop acting on the bullet until the bullet leaves the barrel and the forces can leave in other directions of less resistance than the back of the bullet. Once those forces stop imparting their energy onto the bullet it begins to slow down, just like once a ball leaves it's source of acceleration it starts to slow down.

YULitle
09-13-2009, 12:29 PM
Lets simplify with an analogy. A figure skater doing a spin with his arms extended pulls his arms in closer to his body and his spin speed accelerates. No external force is acting on the skater, but his acceleration is consistent with the conservation of angular momentum. Angular momentum is not exactly the issue with the tennis ball, but as it decompresses it becomes more aerodynamic.

That is not an accurate analogy. Anything in motion the suddenly becomes more aerodynamic, tennis ball or otherwise, doesn't start to accelerate. It may reduce it's rate of deccelartion, but that's not the same thing as increasing velocity.

jonnythan
09-13-2009, 12:34 PM
Increasing the aerodynamic properties of a moving object is simply a different way of saying decreasing its drag.

Drag slows down a moving object. You can reduce drag, but you can't reverse it without adding an external force.

SystemicAnomaly
09-13-2009, 12:41 PM
...

Furthermore, IF IT COULD, which it can't, it wouldn't be "violating physics", it would be either challenging our understanding of physics or our understanding of the situation (more likely the latter.)

Of course it does not actually violate physics. That was just a convenient title used to grab your attention. If violates our rudimentary understanding of physics, not the physics itself (but if I used that as a title for my post, it would be way too long).


The ball cannot - and does not - continue accelerating after it has left the strings of the racquet.

Sorry SA. You're just not understanding the basic, elementary physics that are involved here.

You use lots of words but it's clear that even first-semester physics is above your head. I suggest you listen to the people who understand the forces involved better than you do.

This goes beyond a basic understanding of elementary physics. Some of what I wrote is speculation to be sure. Note that I've actually taken several years of engineering physics in school, including classes in Statics and Dynamics.

What do you think about the point that meowmix makes in post #11? This is the very thing that I was thinking when I made my original claim.

TenniseaWilliams
09-13-2009, 12:45 PM
TW & SE, do you know what they call the rate in change of acceleration? :D Just a tid bit of trivia: A jerk.



:D

I remember the good old days, entropy today just isn't what it used to be.

jonnythan
09-13-2009, 12:46 PM
This goes beyond a basic understanding of elementary physics. Some of what I wrote is speculation to be sure. Note that I've actually taken several years of engineering physics in school, including classes in Statics and Dynamics.

What do you think about the point that meowmix makes in post #11? This is the very thing that I was thinking when I made my original claim.

It really doesn't, and I'm not sure how you can say that it does when you clearly don't have a decent understanding of elementary physics.

Meowmix's point doesn't apply. The acceleration in a bullet is different - the gases are still pushing the bullet for a short period of time after it leaves the barrel, so acceleration drops abruptly but smoothly.

The same is true of the tennis strings - the ball is accelerating while it is in contact with the strings. However, the acceleration is at a maximum at a certain point and then begins to drop while the ball is still touching the strings. The acceleration will drop to 0 while the ball is still in contact with the strings.

It's not much different from dropping a tennis ball on concrete. The ball doesn't continue to accelerate once it has left the ground, even for a few nanoseconds.

If you seriously took a bunch of university-level physics classes and still can't understand this (and say things like "conservation of momentum, I believe") then you need to either get your money back or pull your books back out because it's been too long.

SystemicAnomaly
09-13-2009, 12:56 PM
Thanks for sending this Systemic, interesting stuff.

I do question how long the ball is in contact with the stringbed though. I think it is at least plausible that it remains in contact throughout, but I feel a bit sheepish about my incredulous response ;-)

In their books on tennis science, physicists Rod Cross & Howard Brody indicate that the ball is in contact with the stringbed for 4 to 5 ms.


...

The graph is still not very clear to me, cause, in the first place, it shows the racquet too far back, its even angled back...

The racquet was apparently "photoshopped" into the picture to provide a reference. I believe that the data points are all valid tho'. Note that the ball toss data points were eliminated from most of the photos (If you look at the study, you will see that they did leave in the ball toss data points in 1 or 2 of the photos).

SystemicAnomaly
09-13-2009, 01:05 PM
It really doesn't, and I'm not sure how you can say that it does when you clearly don't have a decent understanding of elementary physics...

You don't need to be condescending or insult people to make your points. I understand a lot more than you give me credit for. (You know, I could turn this around and say that this level of physics is over your rudimentary understanding of the subject -- but I won't).

The bullet analogy is spot on.

The ball cannot instantaneously achieve its max velocity. While the ball is directly on the stringbed, the ball speed and the local racquet head speed must be going at the same speed, more or less., for several ms. That is why they are in contact. However, the exit speed of the ball is about 40-50% greater than the local racquet speed. How do you account for that? Does the ball instantaneously jump from the local racquet speed to its higher speed in zero time?

.

El Diablo
09-13-2009, 01:09 PM
The stringbed is moving faster than the racquet head. The speed of the stringbed (AT THE POINT where it is deformed by the ball) is the racquet head speed plus the speed of the stringbed's rebound from deformation.

stormholloway
09-13-2009, 01:12 PM
This was a topic I remember discussing in the past, and it was also one that should have been left alone. There's no point in resurrecting an argument where you were dead wrong.

After the ball leaves the strings it cannot accelerate. Acceleration takes place while the ball is still on the string. If the ball is accelerating while it's off the strings, then what force is being applied to it to make it increase in velocity?

It's a silly argument, if not stupid.

El Diablo
09-13-2009, 01:13 PM
Analogy for relativity fans....I am on a train moving 40 mph. I am running from the back of the train toward the front at 2 mph. My speed relative to the ground is 42 mph. (OK relativity fans, I know I'm being Newtonian here -- it is not EXACTLY 42 mph.) Similarly the rebounding point of the stringbed that contacted the racquet is moving faster than the racquet.

SystemicAnomaly
09-13-2009, 01:16 PM
This was a topic I remember discussing in the past, and it was also one that should have been left alone. There's no point in resurrecting an argument where you were dead wrong.

After the ball leaves the strings it cannot accelerate. Acceleration takes place while the ball is still on the string. If the ball is accelerating while it's off the strings, then what force is being applied to it to make it increase in velocity?

It's a silly argument, if not stupid.

More insults. If you are not going to contribute, then don't bother to reply.

I already made some speculations as to that force.

SystemicAnomaly
09-13-2009, 01:25 PM
The stringbed is moving faster than the racquet head. The speed of the stringbed (AT THE POINT where it is deformed by the ball) is the racquet head speed plus the speed of the stringbed's rebound from deformation.

This is a good point. Actually, I would say that the deformed stringbed is moving slower than the surrounding area as the ball first makes contact. It is storing some energy as it deforms. As it recovers, it is probably moving faster than the surrounding area.

How about the ball? It loses some energy in the form of heat as it deforms. Does it also store some of the energy by compressing?

How about a ball-court interaction (~ 4ms) ? When a ball bounces doesn't its velocity drop to 0. When it comes off the ground its speed increases. Yet the ground does not really move any faster that it was before.
.

stormholloway
09-13-2009, 01:41 PM
More insults. If you are not going to contribute, then don't bother to reply.

I already made some speculations as to that force.

I explained myself in my post. Get over it. If you don't like my calling the argument silly and stupid then ignore it. It's not an insult. I didn't say YOU were stupid, did I?

Again: acceleration occurs WHILE force is being applied. Unless you're throwing sudden gusts of wind into the equation you are flat wrong.

SystemicAnomaly
09-13-2009, 01:51 PM
For those of you who insist that the ball does not accelerate once it has left the stringbed, how do you account for the photographic evidence? Is it a lie? An illusion? Looks very real to me.

SystemicAnomaly
09-13-2009, 02:04 PM
...

Meowmix's point doesn't apply. The acceleration in a bullet is different - the gases are still pushing the bullet for a short period of time after it leaves the barrel, so acceleration drops abruptly but smoothly.

The same is true of the tennis strings - the ball is accelerating while it is in contact with the strings. However, the acceleration is at a maximum at a certain point and then begins to drop while the ball is still touching the strings. The acceleration will drop to 0 while the ball is still in contact with the strings.

It's not much different from dropping a tennis ball on concrete. The ball doesn't continue to accelerate once it has left the ground, even for a few nanoseconds...

The ball experience radical compression when it contact the stringbed or the ground. The compressed gases inside the ball experience further compression as a result. As the ball is leaving the stringbed, those gases are expanding. Doesn't that provide some impetus to further accelerate the ball.

How about spring compression. When a compressed spring is released, the stored energy causes the spring to accelerate. Isn't the same thing happening here?

When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?

stormholloway
09-13-2009, 02:07 PM
For those of you who insist that the ball does not accelerate once it has left the stringbed, how do you account for the photographic evidence? Is it a lie? An illusion? Looks very real to me.

Haven't seen it.

SystemicAnomaly
09-13-2009, 02:15 PM
Haven't seen it.

Post #3 and #4 of this thread.

YULitle
09-13-2009, 02:17 PM
The ball experience radical compression when it contact the stringbed or the ground. The compressed gases inside the ball experience further compression as a result. As the ball is leaving the stringbed, those gases are expanding. Doesn't that provide some impetus to further accelerate the ball.

How about spring compression. When a compressed spring is released, the stored energy causes the spring to accelerate. Isn't the same thing happening here?

When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?

No, because the spring has something to push off of. When the ball is in mid-air, it has nothing to push off of, and therefor does not accelerate when the ball expands, just as a compressed spring would if it were released in mid-air.

It accelerates after the bounce because it has something to exchange energy with (the ground.) However, as soon as it's airborne, it starts to decelerate again.

YULitle
09-13-2009, 02:19 PM
Post #3 and #4 of this thread.

That was corrected for you on post #8.

mtommer
09-13-2009, 02:42 PM
For those of you who insist that the ball does not accelerate once it has left the stringbed, how do you account for the photographic evidence? Is it a lie? An illusion? Looks very real to me.

The hand and racquet isn't steady at a single pivot point. They are both moving forward as the ball is hit. The blue and yellow dots still overlap showing the ball is still on the string bed thereby causing the acceleration you see.

TenniseaWilliams
09-13-2009, 03:07 PM
Another source of error in the diagram could be caused by the timing of the samples. By examining the position of the racquet and the ball at time intervals, you have effectively digitized the path of both.

The Nyquist–Shannon sampling theorem demonstrates that you would need to sample at twice the maximum rate of target change to ensure that a reasonably accurate copy could be recreated from the information gathered.

The fewer the samples in comparison to how fast the object being studied changes, the more possible shapes can fit the pattern information generated.

jonnythan
09-13-2009, 04:00 PM
(You know, I could turn this around and say that this level of physics is over your rudimentary understanding of the subject -- but I won't

You could, but you'd be woefully incorrect :)

The ball cannot instantaneously achieve its max velocity. While the ball is directly on the stringbed, the ball speed and the local racquet head speed must be going at the same speed, more or less., for several ms. That is why they are in contact. However, the exit speed of the ball is about 40-50% greater than the local racquet speed. How do you account for that? Does the ball instantaneously jump from the local racquet speed to its higher speed in zero time?

.

No one is saying it instantaneously achieves its max velocity. The ball spends a period of time on the stringbed, which is of course not a rigid structure.

The ball comes off the racquet face faster than the racquet head because the strings move faster than the racquet itself. They flex and rebound. That's the entire reason to have flexible strings, and the entire reason that strings at a lower tension give more power - they flex more.

None of this is instantaneous. The ball makes contact with and deforms the strings just like a spring. They rebound, pushing the ball. It's not a long period of time, but it's absolutely more than zero.

The simple fact is that the ball does not accelerate once it is no longer experiencing a force imparted by the strings. Acceleration requires force. Without anything imparting a force on the ball, there is zero acceleration. The acceleration goes from 0 to some maximum value back down to zero, all while the ball is in contact with the strings.

ubermeyer
09-13-2009, 04:07 PM
it accelerates during impact, unless there is a tailwind

TonyB
09-13-2009, 05:53 PM
Disclaimer: I haven't read through all the posts in this thread, so I might be accidentally repeating what someone else already said. But so be it.


I think S.A. has a problem understanding what the terms "acceleration" and "velocity" mean.

The ball has ZERO acceleration the instant that it leaves the strings. Period.

To accelerate an object requires an external force. Once the ball leaves the strings, there is no external force acting on the ball. Therefore no acceleration.

The ball cannot increase in velocity against drag without at least some amount of acceleration. But since there is ZERO acceleration after the ball leaves the strings, the ball cannot increase in velocity, either.

The point of maximum velocity of the ball is the same point as that where the acceleration becomes zero: immediately upon leaving the strings.

There's no voodoo physics involved here. It's actually pretty simple.

Ucantplay2much
09-13-2009, 06:19 PM
As the ball is leaving the stringbed, those gases are expanding. Doesn't that provide some impetus to further accelerate the ball.

No, because those forces are essentially acting on the ball evenly across the entire inner surface area of the ball. Since it is pushing in all directions, the forces cancel each other out.

How about spring compression. When a compressed spring is released, the stored energy causes the spring to accelerate. Isn't the same thing happening here?

Maximum ACCELERATION occurs right after the spring reaches its stopping point in compression. Coming off a racquet, the rate of acceleration decreases as the strings extend and the ball decompresses. Acceleration reaches zero when the force being applied by the strings and the ball decompressing becomes lower than the drag that the air is exerting against the ball.

When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero?

Only if you drop it completely vertically. When the ball contacts the ground, it never stops moving forward. Its spin speed is altered when the force exerted by the surface of the court exceeds the pressure of the air resistance on the ball.


Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?

No to both questions :) Peak exit speed is reached as soon as the force being exerted by the ball against the surface is lower than the force of the air resistance.

FYI: The ability of the ball to re-accelerate itself after impact is called the "coefficient of restitution." For a tennis ball it is roughly 40% of its initial momentum. The USTA and other tennis-governing bodies have specific rules as to how much the ball will bounce when dropped from a certain height. A sampling of "x" number of balls must meet minimum and maximum standards for energy return.

autumn_leaf
09-13-2009, 06:23 PM
Disclaimer: I haven't read through all the posts in this thread, so I might be accidentally repeating what someone else already said. But so be it.


I think S.A. has a problem understanding what the terms "acceleration" and "velocity" mean.

The ball has ZERO acceleration the instant that it leaves the strings. Period.

To accelerate an object requires an external force. Once the ball leaves the strings, there is no external force acting on the ball. Therefore no acceleration.

The ball cannot increase in velocity against drag without at least some amount of acceleration. But since there is ZERO acceleration after the ball leaves the strings, the ball cannot increase in velocity, either.

The point of maximum velocity of the ball is the same point as that where the acceleration becomes zero: immediately upon leaving the strings.

There's no voodoo physics involved here. It's actually pretty simple.

well this is totally over my head. no wonder i got a D in physics x_X

mtommer
09-13-2009, 08:14 PM
well this is totally over my head. no wonder i got a D in physics x_X

Acceleration is an increase of speed over some distance. A car traveling at 500 mph at a steady speed is not accelerating. If it speeds up to 501 mph it has accelerating 1 mph. It's velocity is still high but acceleration is not.

When a tennis ball leaves the strings of a racquet is has reached the highest speed it can possibly be at. From that point on it slows down which is the opposite of acceleration. In a completely frictionless vaccum the ball would never increase it's speed after leaving the racquet because there is no force acting on it to increase the speed (acceleration). A ball cannot produce it's own power (thus accelerate) under any circumstances. It can only sit there until acted upon by an outside force.

YULitle
09-13-2009, 08:17 PM
Acceleration is an increase of speed over some distance. A car traveling at 500 mph at a steady speed is not accelerating. If it speeds up to 501 mph it has accelerating 1 mph. It's velocity is still high but acceleration is not.

When a tennis ball leaves the strings of a racquet is has reached the highest speed it can possibly be at. From that point on it slows down which is the opposite of acceleration. In a completely frictionless vaccum the ball would never increase it's speed after leaving the racquet because there is no force acting on it to increase the speed (acceleration). A ball cannot produce it's own power (thus accelerate) under any circumstances. It can only sit there until acted upon by an outside force.


Actually, that's average acceleration. :D Acceleration is simply change in velocity.

SystemicAnomaly
09-14-2009, 01:49 AM
Disclaimer: I haven't read through all the posts in this thread, so I might be accidentally repeating what someone else already said. But so be it.

I think S.A. has a problem understanding what the terms "acceleration" and "velocity" mean...

I am well aware of what the terms "acceleration" and "velocity" mean. If you had bothered to actually read my posts, you should be able to see that I refer to acceleration as a change in velocity (either in magnitude or direction).

I refer to the observed acceleration as a "violation of physics". In reality, it is an "apparent" violation, not a real one. It seems to violate what we all learned in physics classes. Without falling back on what we think we know about collisions and velocity/acceleration, how do you account for the photographic evidence presented in post #3 (and post #4)?

SystemicAnomaly
09-14-2009, 01:54 AM
No, because those forces are essentially acting on the ball evenly across the entire inner surface area of the ball. Since it is pushing in all directions, the forces cancel each other out...

Yes, that occurred to me right after I suggested that as a possible explanation in that earlier post but I decided to let is stand to see what type of responses I would get from the suggestion. I'll have to look at your other comments at a time other than 3:00 in the am when my brain is a bit fresher.

SystemicAnomaly
09-14-2009, 02:57 AM
The hand and racquet isn't steady at a single pivot point. They are both moving forward as the ball is hit. The blue and yellow dots still overlap showing the ball is still on the string bed thereby causing the acceleration you see.

No, they don't actually overlap. Count the dots. That apparent intersection of blue & yellow dots is not real. The ball is off the strings after ball #2.


That was corrected for you on post #8.

And I responded to that in post #23. It still doesn't explain why the distance between ball #3 and #4 is greater than between #2 and #3.

OTOH, I'll accept what you say about the spring (this very thing also occurred to me as well).

TonyB
09-14-2009, 04:05 AM
how do you account for the photographic evidence presented in post #3 (and post #4)?


I don't have to. First off, that's not "photographic evidence", that's a ROUGH, computer-rendered graphic that doesn't show any direct evidence of acceleration or velocity. You might better say that it's "for reference only."

I think you're making way too much out of that rough computer graphic. Look at the fundamentals of acceleration and velocity and you'll be better off. There's no magic here. There's no mystery. There's just a bad computer-rendered image that you seem to want to cling to for hope in defending your erroneous position.

Let's move on.

jonnythan
09-14-2009, 04:55 AM
how do you account for the photographic evidence presented in post #3 (and post #4)?

There is no photographic evidence presented in post #3.

That is a drawing.

SystemicAnomaly
09-14-2009, 05:23 AM
There is no photographic evidence presented in post #3.

That is a drawing.

It is a undoubtedly a computer-generated image from data taken from high speed cameras. Only the racquet image was drawn in.

jonnythan
09-14-2009, 05:31 AM
You can't take any information whatsoever from that computer-generated image. It contains no information that is relevant to the discussion here.

Imagine for a minute that it were 100% completely accurate - that each "ball" image were an accurate representation of the location of the ball 100 ms after the previous image.

The image falls apart for this discussion. The image of the ball at the location of racquet strike could have easily been hanging in the air for 85 ms before actually getting struck by the racquet, leaving the next ball image to represent only 10 ms of actual motion (assuming a fairly arbitrary ball-to-racquet contact time of 5 ms).

You're really, really hung up on that computer drawing. Don't be. That's not "photographic evidence." I'm sorry, there is absolutely no doubt that you are completely mistaken. I really suggest that you listen to the people who clearly understand the physics better than you do.

SystemicAnomaly
09-14-2009, 05:34 AM
...

There's no voodoo physics involved here. It's actually pretty simple.

It's a lot more complex that you'd have us believe. I've just posed the question to a couple of physics PhD's and am awaiting details. I've gotten some preliminary feedback from one who indicates that it is not simple at all -- certainly not as simple as our college physics classes would have us believe.

jonnythan
09-14-2009, 05:38 AM
It's a lot more complex that you'd have us believe. I've just posed the question to a couple of physics PhD's and am awaiting details. I've gotten some preliminary feedback from one who indicates that it is not simple at all -- certainly not as simple as our college physics classes would have us believe.

There are a lot of things going on in the situation. You have strings deforming, friction between strings, head flexing, friction between the strings and the head, tennis ball compression and spin, yadda yadda yadda. There are a lot of interactions happening very very quickly.

But there is one thing that is not happening. The ball is not accelerating after it leaves the string bed.

Actually, that's false. The ball does accelerate after it leaves the string bed - but it's accelerating in the form of slowing down due to drag.

SystemicAnomaly
09-14-2009, 05:42 AM
...

Imagine for a minute that it were 100% completely accurate - that each "ball" image were an accurate representation of the location of the ball 100 ms after the previous image.

The image falls apart for this discussion. The image of the ball at the location of racquet strike could have easily been hanging in the air for 85 ms before actually getting struck by the racquet, leaving the next ball image to represent only 10 ms of actual motion (assuming a fairly arbitrary ball-to-racquet contact time of 5 ms).

You're really, really hung up on that computer drawing. Don't be. That's not "photographic evidence." I'm sorry, there is absolutely no doubt that you are completely mistaken. I really suggest that you listen to the .

As i mentioned in one of my posts, the ball toss images were eliminated from most of the examples. If you look at the link of the study, there are some additional images for different servers on subsequent pages of that presentation. At least one of them shows the ball toss images that were eliminated from the other examples.

I understand the physics that those "people who clearly understand the physics better than you do". I learned the very same things. I am saying that there is more to the story than the simplistic view of collisions that we all learned in our college physics classes.

jonnythan
09-14-2009, 05:45 AM
As i mentioned in one of my posts, the ball toss images were eliminated from most of the examples. If you look at the link of the study, there are some additional images for different servers on subsequent pages of that presentation. At least one of them shows the ball toss images that were eliminated from the other examples.

Whatever they "eliminated," if the ball-at-racquet image is a representation of the exact moment that the racquet begins to make contact with the ball, the "100 ms" that occurs before the next ball image necessarily includes all the time that the ball actually spent accelerating while in contact with the strings. So therefore the average ball velocity was significantly lower in that 100 ms span than the subsequent one.

Again, the "photographic evidence" you keep talking about is not proof of anything at all. Sorry.

I understand the physics that those "people who clearly understand the physics better than you do". I learned the very same things. I am saying that there is more to the story than the simplistic view of collisions that we all learned in our college physics classes.

You have no idea what physics classes I took or what external physics study I have done, yet you seem pretty sure that you know everything I know...

sureshs
09-14-2009, 09:47 AM
Disclaimer: I haven't read through all the posts in this thread, so I might be accidentally repeating what someone else already said. But so be it.


I think S.A. has a problem understanding what the terms "acceleration" and "velocity" mean.

The ball has ZERO acceleration the instant that it leaves the strings. Period.

To accelerate an object requires an external force. Once the ball leaves the strings, there is no external force acting on the ball. Therefore no acceleration.

The ball cannot increase in velocity against drag without at least some amount of acceleration. But since there is ZERO acceleration after the ball leaves the strings, the ball cannot increase in velocity, either.

The point of maximum velocity of the ball is the same point as that where the acceleration becomes zero: immediately upon leaving the strings.

There's no voodoo physics involved here. It's actually pretty simple.

Agreed. We are talking about contact forces here. When there is no more contact, there is no more force. When there is no force, there is no acceleration due to it. The ball can only decelerate due to gravity and air drag. The force can still be "remembered" as deformation of the ball and its subsequent reshaping, but, overall, the ball cannot increase its speed once it has left the strings.

raiden031
09-14-2009, 09:55 AM
F = M * A. A = F / M. If F = 0, then A must be zero.

mtommer
09-14-2009, 06:59 PM
Actually, that's average acceleration. :D Acceleration is simply change in velocity.

Hah! Nice catch YU, nice catch! :oops:

gastro54
09-14-2009, 10:04 PM
The ball experience radical compression when it contact the stringbed or the ground. The compressed gases inside the ball experience further compression as a result. As the ball is leaving the stringbed, those gases are expanding. Doesn't that provide some impetus to further accelerate the ball.
No. See Newton's 3rd law of motion. How is a ball expanding in midair going to have anything to "push" against to cause it to accelerate?

How about spring compression. When a compressed spring is released, the stored energy causes the spring to accelerate. Isn't the same thing happening here?

No. A compressed string shoots up when in contact with the ground because the ground is exerting an external force on the string. Suspend a coiled spring midair, release it evenly from both sides and it wont go anywhere except outwards to its uncoiled length.

When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?
No. The velocity is not zero when in contact with the ground. It is still moving in a horizontal direction. Only the vertical component of the velocity is zero at lowest point in the bounce.

jonnythan
09-15-2009, 05:27 AM
"When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?"

That's a very telling question. It shows that you're thinking in completely rigid structures.

A falling tennis ball that impacts the ground stays in contact with the ground for a brief period of time. During this period of time, the tennis ball compresses, rebounds, and then leaves the ground. The entire time the ball is in contact with the ground, the ground is exerting an upward force on it and causing it to accelerate upwards.

The acceleration on the ball goes smoothly from 0 just before impact to some maximum value then back down to 0 just as it leaves the ground. The velocity on the ball goes smoothly from some maximum value just before impact to 0 to some maximum value (of just slightly lower magnitude than the previous maximum, but in the opposite direction) as it leaves contact with the ground. The center of mass of the ball continues to move downward, then upward, while the bottom of the ball is contacting the ground.

None of this is instantaneous. We're not talking about long periods of time, but they are periods of time. These are first-semester mechanics concepts.

raiden031
09-15-2009, 05:31 AM
"When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?"

That's a very telling question. It shows that you're thinking in completely rigid structures.

A falling tennis ball that impacts the ground stays in contact with the ground for a brief period of time. During this period of time, the tennis ball compresses, rebounds, and then leaves the ground. The entire time the ball is in contact with the ground, the ground is exerting an upward force on it and causing it to accelerate upwards.

The acceleration on the ball goes smoothly from 0 just before impact to some maximum value then back down to 0 just as it leaves the ground. The velocity on the ball goes smoothly from some maximum value just before impact to 0 to some maximum value (of just slightly lower magnitude than the previous maximum, but in the opposite direction) as it leaves contact with the ground. The center of mass of the ball continues to move downward, then upward, while the bottom of the ball is contacting the ground.

None of this is instantaneous. We're not talking about long periods of time, but they are periods of time. These are first-semester mechanics concepts.

First-semester yet you think that a ball dropping and bouncing has ZERO acceleration? Gravity is always acting on the ball.

Yes there is force acting on the ball by the ground that is not instantaneous. But as soon as the ball leaves the ground the force stops and the only remaining force is gravity.

jonnythan
09-15-2009, 06:27 AM
I was talking about only the upward acceleration on the ball bouncing due to the ground. You're right, I was completely ignoring gravity. Sorry I wasn't more clear. I'm home sick from work and I was still in bed when I wrote that.

sureshs
09-15-2009, 06:31 AM
F = M * A. A = F / M. If F = 0, then A must be zero.

Only if M is not 0

raiden031
09-15-2009, 06:47 AM
Only if M is not 0

If M = 0, then there is no ball. :)

Djokovicfan4life
09-15-2009, 07:49 AM
My head hurts.

AAAA
09-15-2009, 09:29 AM
The ball cannot instantaneously reach the speed measured by the radar gun. There must be an acceleration phase after the ball is hit when serving.

jonnythan
09-15-2009, 09:32 AM
The ball cannot instantaneously reach the speed measured by the radar gun. There must be an acceleration phase after the ball is hit when serving.

The ball is accelerating while it is being hit. It does not accelerate before nor after.

YULitle
09-15-2009, 09:32 AM
The ball cannot instantaneously reach the speed measured by the radar gun. There must be an acceleration phase after the ball is hit when serving.

The acceleration phase is the time it spends on the racquet face.

raiden031
09-15-2009, 09:34 AM
The ball cannot instantaneously reach the speed measured by the radar gun. There must be an acceleration phase after the ball is hit when serving.

There is acceleration due to the acceleration caused by the racquet pushing the ball as well as acceleration caused by the ball compressing and then de-compressing against the racquet as it pushes off. So yeah this acceleration takes a fraction of a second. However as soon as the ball and racquet are separated, acceleration in horizontal direction becomes 0 (except for acceleration in the negative direction caused by wind resistance).

jonnythan
09-15-2009, 09:36 AM
http://www.youtube.com/watch?v=-99QU0i-XTg&feature=PlayList&p=47B21F50560833F8&playnext=1&playnext_from=PL&index=12

That's a high speed video of a tennis ball hitting a racquet.

YULitle
09-15-2009, 09:39 AM
http://www.youtube.com/watch?v=-99QU0i-XTg&feature=PlayList&p=47B21F50560833F8&playnext=1&playnext_from=PL&index=12

That's a high speed video of a tennis ball hitting a racquet.

And this is a puppy...

http://fabulousblueporcupine.files.wordpress.com/2008/05/puppy1.jpg

jonnythan
09-15-2009, 09:40 AM
:lol:

Just a helpful video to hit home the fact that the ball spends time on the strings and that's where the acceleration occurs.

mtommer
09-15-2009, 09:45 AM
There is acceleration due to the acceleration caused by the racquet pushing the ball as well as acceleration caused by the ball compressing and then de-compressing against the racquet as it pushes off.

That's like saying a car accelerates after you take your foot of the gas because the tires are compressing and decompressing. The simple fact is that a tennis ball is not made of material that has the ability to absorb, store and then release energy like a battery. If a ball could accelerate on it's own it could do so with the help of a racquet. All you would need to do is bounce the ball a few times and it would accelerate. In fact, you could stop the ball just after the ball bounces, hold it in your hand, the ball would fly off for just a bit.

raiden031
09-15-2009, 09:50 AM
That's like saying a car accelerates after you take your foot of the gas because the tires are compressing and decompressing. The simple fact is that a tennis ball is not made of material that has the ability to absorb, store and then release energy like a battery. If a ball could accelerate on it's own it could do so with the help of a racquet. All you would need to do is bounce the ball a few times and it would accelerate. In fact, you could stop the ball just after the ball bounces, hold it in your hand, the ball would fly off for just a bit.

Don't be foolish. If you drop a billiards ball on the floor, what happens? It will bounce like 1/2". If you drop a tennis ball, it will bounce like 2 feet.

Its not that the ball can accelerate on its own, but the momentum of the ball hitting the surface causes it to compress which increases potential energy that is then released, causing force to suddenly act in the opposite direction.

AAAA
09-15-2009, 09:53 AM
There is acceleration due to the acceleration caused by the racquet pushing the ball as well as acceleration caused by the ball compressing and then de-compressing against the racquet as it pushes off. So yeah this acceleration takes a fraction of a second. However as soon as the ball and racquet are separated, acceleration in horizontal direction becomes 0 (except for acceleration in the negative direction caused by wind resistance).

It's not my argument, I only have basic physics knowledge but I'm still interested. At what point does the acceleration become deceleration? Immediately after racquet and ball impact like a few inches or atleast a few metres after the ball has been struck? Intuitively I'm thinking the length of the acceleration period increases for harder hit serves.

raiden031
09-15-2009, 09:57 AM
It's not my argument, I only have basic physics knowledge but I'm still interested. At what point does the acceleration become deceleration? Immediately after racquet and ball impact like a few inches or atleast a few metres after the ball has been struck?

I am not a physicist either but took some physics classes back in the day. A free moving object has no deceleration in the horizontal direction (if it was in a vacuum). According to one of Newton's laws, Objects in motion stay in motion. The only reason the ball slows down is wind resistance. So the instant the ball leaves the racquet the wind resistance is the only force acting on it horizontally.

gastro54
09-15-2009, 10:06 AM
First-semester yet you think that a ball dropping and bouncing has ZERO acceleration? Gravity is always acting on the ball.
A ball dropping and bouncing DOES have 0 acceleration but only at the instant when the downward force of gravity on the ball equals the upward force of the ground on the ball.

It's not my argument, I only have basic physics knowledge but I'm still interested. At what point does the acceleration become deceleration? Immediately after racquet and ball impact like a few inches or atleast a few metres after the ball has been struck? Intuitively I'm thinking the length of the acceleration period increases for harder hit serves.
The acceleration becomes negative ("deceleration") due to air resistance the instant the racquet stops exerting a force on the ball (when the ball is no longer on the strings).

AAAA
09-15-2009, 10:12 AM
Thanks guys for the answers.

sureshs
09-15-2009, 10:47 AM
It's not my argument, I only have basic physics knowledge but I'm still interested. At what point does the acceleration become deceleration? Immediately after racquet and ball impact like a few inches or atleast a few metres after the ball has been struck? Intuitively I'm thinking the length of the acceleration period increases for harder hit serves.

Immediately. Gravity is acting all the time and there is air all around. Neither of them "know" when and where the ball was struck.

mtommer
09-15-2009, 12:35 PM
Don't be foolish. If you drop a billiards ball on the floor, what happens? It will bounce like 1/2". If you drop a tennis ball, it will bounce like 2 feet.

Its not that the ball can accelerate on its own, but the momentum of the ball hitting the surface causes it to compress which increases potential energy that is then released, causing force to suddenly act in the opposite direction.

No one's being silly. If you drop a ball does it bounce back up to the same height or beyond the original height? No, it's lesser. It's lesser because the compression doesn't add any force to what gravity provided while it was dropped. If you lose the energy source you lose anything to provide more acceleration. THE COMPRESSION DOES NOT ADD FORCE to provide additional acceleration. The strength of the materials to resist change is greater than gravity's "ability" to keep pulling the ball down so the potential energy gets redirected so to speak.........redirected NOT added to and then redirected.

This is simple physics. The potential energy increases while the kinetic energy decreases. Then the potential energy decreases and the kinetic energy increases MINUS THE ENERGY DUE TO FRICTION ETC.! This leaves you with "less" energy, not more. Without additional energy there can be no acceleration. If this were not so then the compression and decompression of a car's tire would accelerate the tire thereby accelerating the car.

diredesire
09-15-2009, 02:33 PM
this thread is awesome.

As far as this being very complicated, it's as simple as the others say it is. F = ma, and once you remove the force upon the ball, a = 0.

If we're really interested in why the "photographic" ("computer generated") "evidence" in post 3 is "proof":
http://wings.avkids.com/Tennis/Images/Set_2060.jpg

You make one terrible, terrible assumption. That the racquet ball is "accelerating" away from the frame because the racquet is still in this photo.

As is stated in the article that is being referenced, the tip of the frame is what is 'drawing" the blue line. Line the tip of the racquet up with the next ball silhouette. You'll notice that it is darn close to the next blue mark. Assuming the grip is a pivot point, the ball is STILL ON THE STRINGS. (or, admittedly, just leaving it, but that doesn't PROVE any sort of acceleration.) You assume that the drawing is 2D, and it's not.

Sorry, but IMHO, this IS a simple open and shut case.

raiden031
09-15-2009, 02:50 PM
No one's is being silly. If you drop a ball does it bounce back up to the same height or beyond the original height? No, it's lesser. It's lesser because the compression doesn't add any force to what gravity provided while it was dropped. If you lose the energy source you lose anything to provide more acceleration. THE COMPRESSION DOES NOT ADD FORCE to provide additional acceleration. The strength of the materials to resist change is greater than gravity's "ability" to keep pulling the ball down so the potential energy gets redirected so to speak.........redirected NOT added to and then redirected.

This is simple physics. The potential energy increases while the kinetic energy decreases. Then the potential energy decreases and the kinetic energy increases MINUS THE ENERGY DUE TO FRICTION ETC.! This leaves you with "less" energy, not more. Without additional energy there can be no acceleration. If this were not so then the compression and decompression of a car's tire would accelerate the tire thereby accelerating the car.

Yes you are talking about an inelastic collision, where energy is lost during the collision. That is why each bounce is lower than the previous until the ball stops bouncing. If gravity is the ONLY force exerted on the ball, then why does it bounce up? It can only bounce up if the net force (ie. M*A) is in the upward direction. So the change in energy from potential to kinetic and vice versa every time the ball hits the surface causes acceleration to change as well. Although what happens to the energy within this system is not really that relevant to this argument.

mtommer
09-15-2009, 06:54 PM
Yes you are talking about an inelastic collision, where energy is lost during the collision. That is why each bounce is lower than the previous until the ball stops bouncing. If gravity is the ONLY force exerted on the ball, then why does it bounce up? It can only bounce up if the net force (ie. M*A) is in the upward direction. So the change in energy from potential to kinetic and vice versa every time the ball hits the surface causes acceleration to change as well. Although what happens to the energy within this system is not really that relevant to this argument.

It causes acceleration to change. It doesn't cause an increase in acceleration beyond the top speed produced after force has been applied INITIALLY. This is what is being discussed. Does the ball go FASTER after it leaves the strings of the racquet. The answer is that it can't.

Bud
09-15-2009, 07:48 PM
The ball cannot instantaneously reach the speed measured by the radar gun. There must be an acceleration phase after the ball is hit when serving.

The ball is accelerating while it is being hit. It does not accelerate before nor after.

The speed gun measures velocity, not acceleration. As the ball leaves the racquet/strings (which provide the force), acceleration quickly decreases to zero and velocity very quickly maxes out... then decreases gradually.

If we could measure the fastest serve speed at it's maximum velocity... it would be much higher than 150 MPH.


Question for all:

Approximately how far is the ball from the contact point (racquet strings) when serve speed (velocity) is measured/calculated?

It would be interesting to see a chart of acceleration and velocity versus time... while also measuring the amount of time the ball stays on the stringbed of the racquet (t0 would be the moment the strings first contact the ball). It would have to be measured in very small units of time.

I'm assuming acceleration would max out while the ball is still in contact with the strings while velocity would max out very shortly after the ball leaves the strings.

raiden031
09-16-2009, 04:28 AM
It causes acceleration to change. It doesn't cause an increase in acceleration beyond the top speed produced after force has been applied INITIALLY. This is what is being discussed. Does the ball go FASTER after it leaves the strings of the racquet. The answer is that it can't.

I never said the ball goes faster after it leaves the string bed. I agree with everyone that says there is zero force/acceleration on the ball after it leaves the stringbed. My only point was that the compression of the ball and stringbed adds additional force to the ball and its not just the mere collision of a moving racquet and ball. As an example, try feeding someone a billiards ball who is holding a frying pan and the resultant shot will be very weak.

And back to what we are arguing, if you drop a tennis ball from shoulder height, the acceleration of the ball right before contact is -9.8m/s^2. During contact there must be an upward force (acceleration) greater than 9.8m/s^2 in order to launch the ball upward on the bounce. So even though energy is lost in the system, that doesn't matter because energy is a function of force and distance.

sureshs
09-16-2009, 04:39 AM
The speed gun measures velocity, not acceleration. As the ball leaves the racquet/strings (which provide the force), acceleration quickly decreases to zero and velocity very quickly maxes out... then decreases gradually.

If we could measure the fastest serve speed at it's maximum velocity... it would be much higher than 150 MPH.


Question for all:

Approximately how far is the ball from the contact point (racquet strings) when serve speed (velocity) is measured/calculated?

It would be interesting to see a chart of acceleration and velocity versus time... while also measuring the amount of time the ball stays on the stringbed of the racquet (t0 would be the moment the strings first contact the ball). It would have to be measured in very small units of time.

I'm assuming acceleration would max out while the ball is still in contact with the strings while velocity would max out very shortly after the ball leaves the strings.

The velocity is maxing while contact is being made (during the dwell time). However, the velocity cannot increase from the moment it leaves the strings. It doesn't matter if the interval is a millisecond or an hour - there cannot be an increase in velocity once it is gone from the strings.

sureshs
09-16-2009, 04:43 AM
I never said the ball goes faster after it leaves the string bed. I agree with everyone that says there is zero force/acceleration on the ball after it leaves the stringbed. My only point was that the compression of the ball and stringbed adds additional force to the ball and its not just the mere collision of a moving racquet and ball. As an example, try feeding someone a billiards ball who is holding a frying pan and the resultant shot will be very weak.

And back to what we are arguing, if you drop a tennis ball from shoulder height, the acceleration of the ball right before contact is -9.8m/s^2. During contact there must be an upward force (acceleration) greater than 9.8m/s^2 in order to launch the ball upward on the bounce. So even though energy is lost in the system, that doesn't matter because energy is a function of force and distance.

The upward force on the ball by the ground is exactly equal to the downward force of the ball on the ground. Newton's third law, and from this can be derived the conservation of linear momentum. Whether it is an inelastic or elastic collision, this is always true.

raiden031
09-16-2009, 06:04 AM
The upward force on the ball by the ground is exactly equal to the downward force of the ball on the ground. Newton's third law, and from this can be derived the conservation of linear momentum. Whether it is an inelastic or elastic collision, this is always true.

Regardless of how you look at it, there are 2 forces acting on the ball during contact:

1. gravity which is equal in both directions (ball attracting earth, earth attracting ball)

2. deformation of the ball where energy is converted from kinetic to potential and then back to kinetic energy. This force is greater than #1 (in both directions), causing the ball to bounce back up.

mtommer
09-16-2009, 08:03 AM
I never said the ball goes faster after it leaves the string bed.

But the OP does. That's the only issue here that, I do believe, is under contention. It appears we are in agreement. :D

sureshs
09-16-2009, 08:56 AM
But the OP does. That's the only issue here that, I do believe, is under contention. It appears we are in agreement. :D

It is very counterintuitive. That is why people get confused. The casual way to think is to assume that anything that is thrown increases in speed till it gradually slows down.

J011yroger
09-19-2009, 06:37 AM
SA, got your e-mail, and haven't had a chance to reply to the thread.

Read the whole thing, and would like to know for certain that the ball was still not on the stringbed when the 2nd data point was recorded.

That is a very bold assumption.

Did they also mention the time between data points?

That would be very helpful.

J

AAAA
09-19-2009, 06:47 AM
The speed gun measures velocity, not acceleration. As the ball leaves the racquet/strings (which provide the force), acceleration quickly decreases to zero and velocity very quickly maxes out... then decreases gradually.



That's why I wrote

'........instantaneously reach the speed measured by the radar gun'

and not velocity or acceleration. I just wanted to know when the acceleration phase ended. Interesting discussion fwiw.

FlamEnemY
09-19-2009, 07:04 AM
It's not my argument, I only have basic physics knowledge but I'm still interested. At what point does the acceleration become deceleration? Immediately after racquet and ball impact like a few inches or atleast a few metres after the ball has been struck? Intuitively I'm thinking the length of the acceleration period increases for harder hit serves.

The acceleration drops immediately after the ball leaves the strings. Actually it immediately becomes negative because of the air friction. There can only be positive acceleration when there's an external force 'pushing' the ball, and the only external force in this case is the racquet.

jonnythan
09-19-2009, 08:18 AM
SA, got your e-mail, and haven't had a chance to reply to the thread.

Read the whole thing, and would like to know for certain that the ball was still not on the stringbed when the 2nd data point was recorded.

That is a very bold assumption.

Did they also mention the time between data points?

That would be very helpful.

J

They didn't mention any of that data because the graphic was in no way trying to represent the speed or acceleration of the ball. It was illustrating racuqet head path.

AAAA
09-19-2009, 10:17 AM
The acceleration drops immediately after the ball leaves the strings. Actually it immediately becomes negative because of the air friction. There can only be positive acceleration when there's an external force 'pushing' the ball, and the only external force in this case is the racquet.

Other people have said as much. Interesting thinking that the ball goes from the relative stand-still of the ball toss to over 120mph in the space of a feet or even less as the ball is being hit.

FlamEnemY
09-19-2009, 11:02 AM
^^ True that.
http://www.racquettech.com/store/learningcenter/lc_ballspeeds.html
This may be helpful :)

topher.juan
09-21-2009, 04:03 PM
My brother and I talked about this for a while, I started of the opinion that it's impossible for it to accelerate after leaving the strings as energy would need to be stored and converted into speed, he argued otherwise saying it takes time to reach maximum speed, regardless of being off the strings. The ball obviously cannot accelerate after contact, that is to say: you hit it, it goes slow then suddenly speeds up on it's own. Any object put into motion takes time to get to full speed, it's not instant, so what's being argued here is how long does this acceleration from take... A 130mph serve, is it 130mph when leaving the strings or is it in the process of accelerating to 130mph after leaving the strings (by leaving the strings, i mean the most minute fraction of having left the strings).. I think that is what's in question? My brain tells me it takes a very brief moment after leaving the strings to reach maximum speed, I don't think the acceleration period occurs entirely "on the strings", that is clearly where force is applied, but acceleration is put into the ball and briefly done after it leaves the strings... but I have no scientific backup to back this up, should it be true, I'm curious how long distance/time is it to maximum speed. :P

jonnythan
09-21-2009, 04:22 PM
Topher, you're missing a critical fact: the ball spends a period of time on the stings. The stringbed and the ball both flex a fair bit.

You're right that you have no scientific reasons to back up what you're claiming. Suffice to say, you are wrong and you should review this thread.

jimbo333
09-21-2009, 04:26 PM
What a Superb thread!!!

So does anyone actually know the absolute maximum speed of a tennis ball as it just leaves the strings from a serve?

Is it as much as 200mph for the fastest ever serve?

Are new radar guns or other methods needed to measure this?

jonnythan
09-21-2009, 04:32 PM
What a Superb thread!!!

So does anyone actually know the absolute maximum speed of a tennis ball as it just leaves the strings from a serve?

Is it as much as 200mph for the fastest ever serve?

Are new radar guns or other methods needed to measure this?

I'd guess that the radar guns already do a good job of measuring the maximum speed. Radar can handle objects moving much faster than 150 mph, and I don't doubt that their software is set up to simply display the maximum speed recorded during the serve.

The maximum speed recorded will, of course, be the speed of the tennis balls as it leaves the stringbed. So it's very doubtful that "actual" serve speeds are significantly different from what we already measure, let alone 33-50% higher.

meltphace 6
09-21-2009, 04:45 PM
a = F / m

Sorry, OP, there's no F after the ball left the string bed.

meltphace 6
09-21-2009, 04:49 PM
Of course air friction causes F to become negative after the ball left the string bed.

meltphace 6
09-21-2009, 05:11 PM
OK, case was closed already. Sorry for my redundant posts.

diredesire
09-26-2009, 10:02 AM
That's why I wrote

'........instantaneously reach the speed measured by the radar gun'

and not velocity or acceleration. I just wanted to know when the acceleration phase ended. Interesting discussion fwiw.

It really, really ends after it leaves the racquet. It's not even close to instantaneous, and I'm still unsure why this point is being argued. But yes, this is an interesting discussion. The ball spends a few ms on the strings (ON THE STRINGS), thus on the racquet, the acceleration is being given to the ball at this time, and the "proof" graphic above is seriously misleading.

SA, got your e-mail, and haven't had a chance to reply to the thread.

Read the whole thing, and would like to know for certain that the ball was still not on the stringbed when the 2nd data point was recorded.

That is a very bold assumption.

Did they also mention the time between data points?

That would be very helpful.

J
IMHO, there is no other explanation, and it is silly to assume that the racquet is stationary anyways. Even if the ball ISN'T on the strings, it would have just recently left the strings, which doesn't disprove anything. The point of my previous post is to point out that the racquet tip is moving along the path of the blue dotted line, which rules out the argument that the ball is accelerating from the racquet. The racquet is moving.

Other people have said as much. Interesting thinking that the ball goes from the relative stand-still of the ball toss to over 120mph in the space of a feet or even less as the ball is being hit.
It is interesting, but not entirely hard (at all) to believe. I'd imagine a sling shot could do a similar amount of acceleration in a similar amount of distance.

My brother and I talked about this for a while, I started of the opinion that it's impossible for it to accelerate after leaving the strings as energy would need to be stored and converted into speed, he argued otherwise saying it takes time to reach maximum speed, regardless of being off the strings. The ball obviously cannot accelerate after contact, that is to say: you hit it, it goes slow then suddenly speeds up on it's own. Any object put into motion takes time to get to full speed, it's not instant, so what's being argued here is how long does this acceleration from take... A 130mph serve, is it 130mph when leaving the strings or is it in the process of accelerating to 130mph after leaving the strings (by leaving the strings, i mean the most minute fraction of having left the strings).. I think that is what's in question? My brain tells me it takes a very brief moment after leaving the strings to reach maximum speed, I don't think the acceleration period occurs entirely "on the strings", that is clearly where force is applied, but acceleration is put into the ball and briefly done after it leaves the strings... but I have no scientific backup to back this up, should it be true, I'm curious how long distance/time is it to maximum speed. :P

It isn't true, it is at maximum velocity when it leaves the strings. If it ISN'T at maximum velocity, there is a force being applied that is outside of the racquet/ball system.

Take a look at these photos:
http://2008.usopen.org/images/pics/large/b_0907_043_Murray.jpg
http://www.hdtennis.com/images/matt/ball_and_string_deformation.jpg
http://essentialtennis.com/gearReviewBlog/wp-content/uploads/2009/08/ball-impacting-strings1-275x300.jpg

The ball gets compressed, and rebounded off in an incredibly fast amount of time, a few milliseconds. We like to think of our string beds as a rigid plane, but it's anything but. If we had some really good, really close up footage of a ball being struck, I think we'd all believe this to be true. Yes, it happens fast, but it really is just simple physics. For those of who you are saying that there is a minute amount of acceleration after the ball leaves the string bed, give a reasonable explanation of how this would possibly occur. Unless I'm missing an extremely obvious missing force, it just isn't happening.

(The ball doesn't "isntantly" hit the strings and rebound, there IS time for acceleration to occur... while the ball is on the strings)

jonnythan
09-26-2009, 07:28 PM
Man, if you think a few ms is a short amount of time, think about the amount of time a neutron spends accelerating from 0 to a million meters per second in a hydrogen bomb, or the amount of time a new Intel processor would take to add 2 + 2.

diredesire
09-28-2009, 12:00 PM
Man, if you think a few ms is a short amount of time, think about the amount of time a neutron spends accelerating from 0 to a million meters per second in a hydrogen bomb, or the amount of time a new Intel processor would take to add 2 + 2.

I hope you're not referring specifically to me, i know "relative" when I see it. Neither of those interactions has a human body directly involved with production, so I fail to see the relationship to the thread...

jonnythan
09-28-2009, 12:16 PM
I hope you're not referring specifically to me, i know "relative" when I see it. Neither of those interactions has a human body directly involved with production, so I fail to see the relationship to the thread...

My comment was inspired by a phrase in your post, but I was just speaking in general. It's clear you have a grasp on what's going on. It's equally clear that a number of posters seem to think that the racquet-ball interaction happens instantaneously and occupies basically zero time. That's troubling.

spacediver
09-28-2009, 12:49 PM
I think there is at least one situation where the ball could accelerate after leaving the strings.

That would be the case where the decompressing tennis ball pushes against air molecules that are at a higher pressure behind the ball than they are in front.

However, in reality the pressure is always highest in front of the ball, and there is a low pressure zone behind it, so if anything, the decompressing ball may actually end up slowing it down, since the decompressive force pushes against the air molecules in front of the ball.

You could conceivably contrive a situation where it would continue to accelerate after leaving the strings, if you had a device pumping air molecules against the back of the ball at such a rate that the pressure behind the ball was greater than the pressure in front of the ball. That way, the decompression of the ball could push against these air molecules and speed it up. However, note that in this case, the ball would continue to accelerate after leaving the strings regardless of decompression, since the air pump would need to be pumping air molecules at a velocity faster than the velocity of the racquet in order to achieve this higher air pressure behind the ball. In this case, the ball would have accelerated to its maximum speed regardless of the racquet or decompression. In fact, it is questionable whether the racquet would even be able to touch the ball if the ball is being moved faster than the racquet head.

In a zero gravity vacuum, the ball will not change velocity after leaving the string bed, unless it somehow expells some of its mass mid flight.

sureshs
09-28-2009, 03:32 PM
Another situation is if a meteorite happened to hit the ball after it left the strings.

Or a huge body shot past the earth, exerting lots of gravity on the ball.

Or the ball had some chemical inside which cause a reaction and it produced an explosion and accelerated the ball

TonyB
09-28-2009, 03:36 PM
Why is this thread still alive?

jimbo333
09-28-2009, 03:46 PM
In a zero gravity vacuum, the ball will not change velocity after leaving the string bed, unless it somehow expells some of its mass mid flight.

Well I think I was playing in a zero gravity vacuum last week:)

(certainly seemed like it at least, his serves were ridiculous)

raiden031
09-28-2009, 06:40 PM
In a zero gravity vacuum, the ball will not change velocity after leaving the string bed, unless it somehow expells some of its mass mid flight.

Why not? Gravity has nothing to do with the horizontal velocity of the ball, unless you're talkng about the effect that gravity has on the surrounding air. The ball would still slow down from the friction with the air, even though it may never hit the ground (although that depends on the angle the racquet hits the ball during contact).

YULitle
09-28-2009, 07:03 PM
Why not? Gravity has nothing to do with the horizontal velocity of the ball, unless you're talkng about the effect that gravity has on the surrounding air. The ball would still slow down from the friction with the air, even though it may never hit the ground (although that depends on the angle the racquet hits the ball during contact).

vacuum = no air

raiden031
09-28-2009, 07:26 PM
vacuum = no air

You're right, nevermind.