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Author Topic: ping-pong spinny ball dynamics after bounce  (Read 22515 times)
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cam
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on: July 31, 2008, 12:03:14 am »

Hi,

I have read the translation of a chinese national coach speak on a ping-pong forum.

On that translation, I found the information that a ball with topspin loses some rotation when it bounces on the table, but gets increasingly more spin during the flight just after the bounce, until a certain maximum is reached.

I don't really understand this phenomena. I just give it some credit because of the source.

Anyway, if it is true, I would like to see a simulation of it, and make some experiments with the parameters.

The relevance of this simulation is that it can help in the education of advanced table tennis players.

This is my first post here. So, I have a question that is little bit off topic here.

Do you release the source code of your applets for the public?

Thank you very much,

Carlos Mitidieri
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Fu-Kwun Hwang
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Reply #1 on: July 31, 2008, 08:08:20 am »

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You can adjust the initial velocity vx and the ratio of angular velocity to velocity (by R*w/vx slider).
Click play to start the simulation.




Please check out spinny ball after bounce if you want to know the physics invoved (derivation).

The above simulation was created with EJS. The EJS source(xml format)  for all the simulation at this web site are all available. Just click the DOWNLOAD Ejs source link.
you can EITHER:
Install EJS and open the downloaded file to find out the equation used in the simulation. 
OR
EJS will be loaded into your browser with the EJS source if you click load ejs as signed applet.


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Reply #2 on: July 31, 2008, 10:46:05 pm »

Hello Mr. Hwang,

thank you for the nice simulation.

I would like to ask for some extensions.  Smiley

First, I would like to add the bounce with the raquet to the simulation. I mean, to add some graphs where we could observe the value of the forces/speeds developed between the ball and a blocking raquet, all along the trajectory after the bounce. Most interesting would be the relative velocity between the ball and the paddle surfaces. 

So I would like to give some input. Please see below a realistic representation of what happens when the ball impacts the rubber on the raquet. I am cuirous thinking on how this could be abstracted.




And here, a squetch of what I am thinking about:



The red arrows represent the angle with which the ball leaves the raquet. The green arrows are the tangents to the trajectory.

Second, we could have graphs below the simulation showing the (i) initial module of the red vectors (as if a raquet was there with a fixed angle), (ii) of the relative speed between the ball and the raquet surfaces at the contact point (agian, as if a raquet was there with a fixed angle), (iii) variation of the angular speed with time/position.
 
We could also have the choice of fixing the position of the raquet and then run the simulation of the incoming/returning ball. 

I hope I am not being too much eager.

Best regards,

Carlos
« Last Edit: July 31, 2008, 10:53:22 pm by cam » Logged
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Reply #3 on: August 01, 2008, 08:28:58 am »

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Here is the second version:

Click check box to show the angular velocity with respect to the ball (cyan) or the velocity at the bottom of the ball relative to the table (Vx-R*omega: friction force will be in the opposite direction).
You can have backspin or topsion by draging the slider to change the ratio R*w/vx (negative value for backspin).


I am going to create the third version which will impact of the ball with a wall (with different contact angle to simulate the raquet).


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Reply #4 on: August 01, 2008, 11:59:52 pm »

Dear Mr. Hwang,

I have been looking for realistic data from real-world table tennis. These are some facts, that I was wondering if they could be integrated to the model:

- ball diameter: 40 mm
- ball weight: 2,7 g
- coefficient of friction between ball and table: 0,4 to 0,6
- typical ball speed when it reaches the racket: 8 to 12 m/s
- typical ball rotation when it reaches the racket (data here is not too reliable, so I increased the range):  3000 to 9000 rpm 

The newly introduced R*w and Vx-R*w  vectors are very instructive. It would be nice to have them also in the contact point between the ball and the wall.

Best regards.
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Reply #5 on: August 02, 2008, 12:12:48 am »

Here are some interesting links:

http://www.jayandwanda.com/tt/ballspeed_calc1.html
http://www.jayandwanda.com/tt/ttb_calc.html
http://www2.physics.umd.edu/~mfuhrer/course/spr02/AJP/AJP00482.pdf
http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel5/10242/32662/01530584.pdf?arnumber=1530584
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Reply #6 on: August 02, 2008, 03:08:27 pm »

More data:

- ball wall thickness: 0,4 mm
- The ball must rebound on the table to a height of 230-260mm when dropped from a height of 300mm.

This latest results on a vertical coefficient of restitution (COR) of  0,87 to 0,93 (we could well fix this in 0,9)

COR = speed up / speed down  = ((2.g.h2)^1/2) / ((2.g.h1)^1/2)
 
« Last Edit: August 02, 2008, 03:42:51 pm by cam » Logged
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Reply #7 on: August 02, 2008, 09:11:10 pm »

The above simulation is not suitable for pingpong ball according to the data your have provided.
Because I have assume the static friction force is strong enough, so that there is no sliding when the ball touch the table. The above simulation might be more suitable for a hard and elastic superball.

I need to derive new relation for the ping-pong ball according data you have provided.
Air resistance need to be introduced.
And I might not be able to ignore the effect due to the air pressure difference from the rotation of the ball.
This is a complex problem. Because the ball is a 3D object, and force due to the difference in pressure is different at different position (need complicate integration).
But I will ignore this effect to create a simple model first. The size of the ball is small compared to the space it is moving. So I will need to change the view of the simulation, too. I will post it here when it is done.
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Reply #8 on: August 02, 2008, 10:53:20 pm »

Ok, then I should give you a bit more of information:

- for a topspin, initial speed of the ball (when it is hit) can be as high as 30 m/s
- longitudinal dimension of the table is 274 cm (two halves of 137 cm)
- the players will be typically 4 to 8 m apart from each other

Please ask if you need some missing information.
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Reply #9 on: August 02, 2008, 11:27:02 pm »

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And I might not be able to ignore the effect due to the air pressure difference from the rotation of the ball.

You are right. Magnus effect plays a major role on ping-pong physics. But as you said, it could be left out for simplicity on first version.

Thank you for your interest, Mr. Hwang. Smiley
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Reply #10 on: August 04, 2008, 05:20:47 pm »

Ok, then I should give you a bit more of information:

- for a topspin, initial speed of the ball (when it is hit) can be as high as 30 m/s
- longitudinal dimension of the table is 274 cm (two halves of 137 cm)
- the players will be typically 4 to 8 m apart from each other

Please ask if you need some missing information.
The radius for the ball is only 2cm, and the distance between player is 4-8 meter apart.
The ball will become a tiny dot if everything are drawed to their size with the same proportion.

Do you have any suggestion how to show everything on the computer screen at the same time?
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Reply #11 on: August 04, 2008, 07:14:48 pm »

Could we could have a "translating window", moving with the same horizontal speed as the ball?

Or if not possible, a "stationary window" showing only a selectable part of the trajectory?

Best regards.
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Reply #12 on: August 06, 2008, 01:21:57 pm »

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I have created another version(5) according to your model. I have been playing with this new model for several days.
But I feel something did not look correct so that I did not post it earlier. May be you can provide more input to modify the model.

New model used in the simulation:
 vy' = -COR *vy; when the ball touch table.
 then, the change of momentum in vertical direction is dPy=m*(vy'-vy)= -m*(COR+1)*vy;
 The Normal force N= dPy/dt=-m*(COR+1)*vy/dt;
 The friction force fr= u*N = -u*m*(COR+1)*vy/dt;
 Change of momentum in horizontal direction is dPx= fr*dt
 So vx'= vx+dPx/m =vx -u*(COR+1)*vy;
 
 Torque = R*fr = I *dw/dt
 so dw=R*fr*dt/I = R*dPx/ I

Air resistance is also included in the model. F= 0.5*rho*v2*A/2= -0.143 (vx*vx+vy*vy);





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