I saw a thread about the weight of impact. It should be force of impact.

[brokenball]

The weight/mass of a tt ball doesn't change. However, the force of impact does as a function of mainly relative impact speed but there are some other factors.
So what is the force of impact if the ball hits the paddle with a relative speed of 10 m/s?
Some assumptions are necessary, state them.
The answer will surprise you.

Many people make a big deal about the flex or vibration of a paddle/blade. How much can a TT ball impact bend a blade. I have a high speed video of a TT ball hitting my toxic 5 and it wobbles visibly a lot but my Firewall+ doesn't seem to notice the impact or the vibrations are so small that it effectively makes no difference.

 

[Zwill]

Well if I assume that the deacceleration time is 0.01s, then it would be 2.7N.
So basically the lower the deacceleration time the ball has with the racket the larger the impact force will be.
In this case the ball and racket contact time or their elasticity will influence the impact force greatly given the velocity and the ball mass remains the same.

Actually I don't agree that the ball mass doesn't change it changes with use and its elasticity as well, so the same ball will feel lighter after several hours of use.

 

[brokenball]

Well if I assume that the deacceleration time is 0.01s, then it would be 2.7N.
So basically the lower the deacceleration time the ball has with the racket the larger the impact force will be.
In this case the ball and racket contact time or their elasticity will influence the impact force greatly given the velocity and the ball mass remains the same.

Actually I don't agree that the ball mass doesn't change it changes with use and its elasticity as well, so the same ball will feel lighter after several hours of use.

Not close for three obvious reasons.
1 you are assuming the ball is decelerating linearly. I doesn't. At first contact there is no force when the ball first touches the rubber and it increases as the ball penetrate the rubber.
2. During that time, 0.01s, how far the ball penetrate into the rubber?
3. The dwell or contact time would be very long 10 milliseconds for just the decel time is a long time.

 

[Zwill]

Not close for three obvious reasons.
1 you are assuming the ball is decelerating linearly. I doesn't. At first contact there is no force when the ball first touches the rubber and it increases as the ball penetrate the rubber.
2. During that time, 0.01s, how far the ball penetrate into the rubber?
3. The dwell or contact time would be very long 10 milliseconds for just the decel time is a long time.

1, sure, the ball's contact area will change as it deforms, but you are assuming there is a rubber. Why does there need to be a rubber? Are you saying I cannot look at this problem without a rubber? Why?
2, If there is no rubber it doesn't penetrate it, can we agree?
3, Well you said assumptions are necessary and should be stated, I stated 10ms, I can state 1ms and then we're talking about 27N, still assuming there is no rubber and contact is linear.

 

[brokenball]

1, sure, the ball's contact area will change as it deforms, but you are assuming there is a rubber. Why does there need to be a rubber? Are you saying I cannot look at this problem without a rubber? Why?
2, If there is no rubber it doesn't penetrate it, can we agree?
3, Well you said assumptions are necessary and should be stated, I stated 10ms, I can state 1ms and then we're talking about 27N, still assuming there is no rubber and contact is linear.

1. OK, but you didn't state your assumption and you are also assuming that pip don't compress.
2. Does it bend the paddle like my Toxic 5 or assuming that nothing is deforming?
3. Yes, you can state the contact time is 1 ms. Is that decelerating and accelerating time? If it is just decelerating time then how far does the ball move in the 1 ms it is slowing down? I thought you said there is no rubber so the blade must be pushed back how far? Your point 2 and 3 are at odds.

27N is getting closer. That is like have a 2.7 kg weight on the blade. Wow! Who da thunk it? I was using 10m/s^2 as 1 g even though it is roughly about 9.8g. I am doing the math in my head.

Keep going, but think about it.

I was thinking the rubber should be a standard 2mm thick but you can assume what you want as long it as it consistent with the points above.

 

[Zwill]

1. OK, but you didn't state your assumption and you are also assuming that pip don't compress.
2. Does it bend the paddle like my Toxic 5 or assuming that nothing is deforming?
3. Yes, you can state the contact time is 1 ms. Is that decelerating and accelerating time? If it is just decelerating time then how far does the ball move in the 1 ms it is slowing down? I thought you said there is no rubber so the blade must be pushed back how far? Your point 2 and 3 are at odds.

27N is getting closer. That is like have a 2.7 kg weight on the blade. Wow! Who da thunk it? I was using 10m/s^2 as 1 g even though it is roughly about 9.8g. I am doing the math in my head.

Keep going, but think about it.

I was thinking the rubber should be a standard 2mm thick but you can assume what you want as long it as it consistent with the points above.

1, What pip? The ball and blade must compress to some degree. If there were rubbers of course it would compress.
2, Is this a trick question? If nothing is deforming then of course nothing is bending. But then the deacceleration is instantaneous. But blades do flex and deform, but the degree also depends on where the ball hits the racket.
3, Well just for the sake of simplicity only the deacceleration time. During that 1ms the ball should move 5mm. Now that will be ball deformation, blade deformation and blade flex. If there is rubber then of course rubber compression as well. It's quite a lot if you think about it. Obviously, all of these happen at the same time and said 5mm is shared between all participants.

27N is a lot, and 10m/s is not even that high of a velocity. I'm pretty sure in some cases I hit 25m/s during gameplay, that would be 67,5N force. Just by thinking about it like this I really don't want to put 6.75kg weight on my blade, especially not on the very tip.

If you introduce rubbers it gets more complicated, but 2.0mm is only the sponge, a typical 2.0mm thick sponged sheet is about 3.4-3.5mm thick, and only if we assume you hit dead flat. If you hit with a 45 degree angle I think that thickness changes to squareroot24,5mm (~4.95mm) if we take 3.5mm.

I can kinda guess the contact time is not 1ms, especially not with rubbers since that 67.5N feels very unnatural. Not sure my blade would handle that force without breaking and surely that would twist the racket out of my hand.

 

[UpSideDownCarl]

If this is your subject, instead of asking people who aren't going to know to guess, why don't you just explain the subject matter?

 

[dingyibvs]

I've seen dwell time from studies as anywhere between 1 to 4 ms, which would include both acceleration and deceleration time of course. That would translate to quite a bit of force on the racket!

 

[igorponger]

Momentary force on the ball impact would be 80 Newton units on punching shots. In side-to-side comparison, maximum thumb pressure is about 65 N.
This is to say you should not make use of the trashy balls sold at discount price.

Be happy.

 

[lodro]

If this is your subject, instead of asking people who aren't going to know to guess, why don't you just explain the subject matter?

......and where would be the fun in that ??🤣
🤣🤣🤣

 

[igorponger]

ABS BALLS TO SAVE MY MONEY.

The highest resistance to shock is the best praised virtue with the abs balls of *** Star grade,
A thicker outer shell of .35 mm on the ABS balls is to count for a small percentage of ball crackage in play.

IT IS amazing how much pressure on the ball yet can't destroy the ball under heavy impact. A botlle volumed of 5 litres water is the exact equivalent to the uploaded force in laboratory tests for the ball geometrical hardness. Amazingly high Young's modulus we have now got with the newly developed composition of ABS material. Far better to celluloid material.

 

[lodro]

ABS BALLS TO SAVE MY MONEY.

The highest resistance to shock is the best praised virtue with the abs balls of *** Star grade,
A thicker outer shell of .35 mm on the ABS balls is to count for a small percentage of ball crackage in play.

IT IS amazing how much pressure on the ball yet can't destroy the ball under heavy impact. A botlle volumed of 5 litres water is the exact equivalent to the uploaded force in laboratory tests for the ball geometrical hardness. Amazingly high Jung modulus of the ABS plastic, indeed. Far better to celluloid material.

Yes, the Balls are very strong but often they are not very round

 

[Tony's Table Tennis]

I've seen dwell time from studies

popcorn out when I saw the word "dwell time" on a BB thread.

 

[lodro]

popcorn out when I saw the word "dwell time" on a BB thread.

i could not believe my eyes either

 

[Tony's Table Tennis]

i could not believe my eyes either

yeah
since dwell time is only a myth from few centuries ago

 

[Der_Echte]

Grip pressure and change of grip pressure at impact is game changer.

 

[Zwill]

yeah
since dwell time is only a myth from few centuries ago

I think we shouldn't dwell on it too much.

 

[ttarc]

Assuming a coefficient of restitution of 90% (blade with rubbers) and a 4 ms contact time.
For the velocity of the ball after impact we get:
cor = 90%, m_racket = 0.18 kg, m_ball = 0.0027 kg, v_racket_0 = 0 m/s, v_ball_0 = 10 m/s, m_total = 0.1827 kg

v_ball_1 = (m_ball * v_ball_0 + m_racket * v_racket_0 + m_racket * cor * (v_racket_0 - v_ball_0)) / m_total = -8.719 m/s

v_racket_1 = (m_ball * v_ball_0 + m_racket * v_racket_0 + m_ball * cor * (v_ball_0 - v_racket_0)) / m_total = 0.2807 m/s

For the average force we get: F_average t = m (v_ball_0 - v_ball_1) so
F_average = (0.0027 kg * (10 m/s - (-8.719 m/s))) / 0.004 s = 12.63 N
or for the racket F_average = (0.18 kg * (0 m/s - 0.2807 m/s)) / 0.004 s = -12.63 N

 

[Tony's Table Tennis]

Assuming a coefficient of restitution of 90% (blade with rubbers) and a 4 ms contact time.
For the velocity of the ball after impact we get:
cor = 90%, m_racket = 0.18 kg, m_ball = 0.0027 kg, v_racket_0 = 0 m/s, v_ball_0 = 10 m/s, m_total = 0.1827 kg

v_ball_1 = (m_ball * v_ball_0 + m_racket * v_racket_0 + m_racket * cor * (v_racket_0 - v_ball_0)) / m_total = -8.719 m/s

v_racket_1 = (m_ball * v_ball_0 + m_racket * v_racket_0 + m_ball * cor * (v_ball_0 - v_racket_0)) / m_total = 0.2807 m/s

For the average force we get: F_average t = m (v_ball_0 - v_ball_1) so
F_average = (0.0027 kg * (10 m/s - (-8.719 m/s))) / 0.004 s = 12.63 N
or for the racket F_average = (0.18 kg * (0 m/s - 0.2807 m/s)) / 0.004 s = -12.63 N

alright.
there is where I say, no engrish

 

[Tony's Table Tennis]

I think we shouldn't dwell on it too much.