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dwelltime |
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wturber
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Not even nearly most of that energy. It transfers just a small fraction of it. Otherwise, the racket would nearly stop during the collision. |
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Jay Turberville
www.jayandwanda.com Hardbat: Nittaku Resist w/ Dr. Evil or Friendship 802-40 OX |
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igorponger
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LESS DWELL TIME, MORE SPEED. (and vice versa).
Some technical guidance for those looking to get the ball bouncing FASTER off the racket upon impact: --Take rubbers with the minimum possible DT (springy, high resilience rubbers) A physics' law behind this advice is that the longer DT corresponds to less Force produced on the impact ---> less speed of the ball after the impact. Equation to describe the effect F=mV/ t m --Mass of the ball V -velocity of incoming ball t --dwell time F impact force on the ball PS// Mass of the racket has no perseptible effect on the bounce speed of the ball. Edited by igorponger - 09/20/2013 at 9:21pm |
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tt4me
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zeio, what part is non sense? You aren't clear and neither is AgentHex. You and AgentHex are talking past each other because you are talking about what happens to the ball and AgentHex is talking about what is happening to the system which includes the ball and paddle and everything else that is moving.
I will nit pick. In the motion of what? The ball or the paddle? Obviously the paddle or that is what you meant to say. Most of the stuff you say is basic except.
WRONG!!!!Don't give us any of the condescending sighs until you prove it. We are waiting. It should be easy.
I have got it but I am not sure about you.
In general I think you have the right idea. If the dead time is longer there will be less force but then this lessor force has more time to accelerate the ball because of the longer dwell time. You can't simply say that less force results is less speed. You must take into account the time the force is applied. The integral of force over time is called the impulse. Edited by tt4me - 09/20/2013 at 2:23pm |
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zeio
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Note the phrase "during collision". There is a reason for that. Real collisions are not instantaneous even though they are often treated that way where the force is infinite and duration is zero. The paddle and ball do not separate at an instant. They remain in contact and travel at the same velocity for a certain amount of time during the swing, albeit extremely brief, not unlike what happens during a perfectly inelastic collision, before separating. It is during this phase of collision that most of the kinetic energy of the paddle is transferred to the ball. |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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wturber
Premier Member Joined: 10/28/2008 Location: United States Status: Offline Points: 3899 |
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This has nothing to do with the near instantaneous nature of the collision. If most of the kinetic energy of the paddle was transferred to the ball, the paddle would slow down rather dramatically and the ball would travel a LOT faster than it does (or perhaps shatter or deform dramatically) But those things don't happen. Only a small fraction of the kinetic energy of the paddle is ever transferred. Most of the energy is retained in the racket as evidenced by the fact that it doesn't slow down much. Edited by wturber - 09/20/2013 at 7:06pm |
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Jay Turberville
www.jayandwanda.com Hardbat: Nittaku Resist w/ Dr. Evil or Friendship 802-40 OX |
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AgentHEX
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A 2.7 g ball is less than 10% over the mass of the 2.5, whereas frontal area diff is >10%. Not sure why you're arguing aero if you can't figure out how basic drag works. The COR is a measure of speed, not "inertia". At least last time you got the units right.
Still haven't figured out what happened last time? You either didn't understand what was said or were pedantic in the worse sense: incorrectly. What's quite funny is that not even linking to your hero JRSDallas helped. The worst part is that you're stuck in this situation where corrections such as above won't even help because you have no idea what to even look at. "Work"? Really? How exactly do you figure to measure or present this in any meaningful way rather than dropping technical words you can't grasp?
OMG LOL. They're the same numbers in different representation. Completely clueless.
OMG, LOL. So are you a computer to only accept binary answers? . To provide some clue as to what backtracking is, this is a proper example:
So now "work" is about getting up to that speed, which is the only factor that matter, thus making said references to work irrelevant? Not your earlier theories about how work is changing energy or whatever during the dwell? Which is it?
How so? Be specific so there's more meat to mock. |
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AgentHEX
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I'm not sure what you're trying to correct. A ratio is a unitless number, in this case the aftereffect of a specific collision. What physical manifestation of the blade or whatever is it supposed to be of? |
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AgentHEX
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I think it should be clear by now what zeio means, and there's no need for us to dwell on the issue any more than the ball on the bat. Some actually meaningful considerations of how this relates to practical TT equipment: A faster blade will only increase COR in the normal direction, whereas tangential should remain exclusively matter of the rubber. This would suggest that a player "upgrading" blade will tend to produce more speed and less spin for similar stroke, which would put more pressure on accuracy of blade angle. Better players can probably compensate for this with greater handspeed, ie. more brush, but this would be problematic for less technically skilled. The only way to correct the ratio "imbalance" would be to get rubber with very elastic tangential CORs, which might exacerbate the problem for those without already quite good control. Edited by AgentHEX - 09/20/2013 at 7:53pm |
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tt4me
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Jay said what I would have said but.
I don't understand why zeio used the word most instead of all if he is only talking about the energy transferred during dwell time. The only time energy is transferred to the ball is when the ball is struck or when the ball is tossed during the serve and we are not talking about serves. At what other time would energy be transferred to the ball? |
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JacekGM
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If all kinetic energy was transferred to the ball the racket would have to freeze in the air, I presume.
Semantics, I guess...
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(1) Juic SBA (Fl, 85 g) with Bluefire JP3 (red max) on FH and 0.6 mm DR N Desperado on BH; (2) Yinhe T7 (Fl, 87 g) with Bluefire M3 (red 2.0) on FH and 0.6 mm 755 on BH.
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AgentHEX
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Magic. |
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zeio
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Yes, it does have to do. Check out the section for "Kinetic energy of colliding bodies" of thissite. They have a diagram showing the collision of two identical blocks moving on a smooth floor, A and B. Block A moves right with velocity VA towards block B with velocity VB, where VA > VB. Block A eventually catches up to block B and the two collide, with VA slowing down and VB speeding up. Part of the kinetic energy is converted to elastic potential energy in the process. Once VA matches VB, the two blocks move in conjunction at velocity V for a certain distance. The process then reverts as the elastic potential energy gets converted back to kinetic energy and block B separates from block A. As the above shown, block A starts out with more energy than block B before the collision, but at the end of it block B carries just as much less the energy loss. That's what I meant when I said "...most of that energy..." |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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tt4me
Gold Member Joined: 01/17/2013 Location: RC Poverty Zone Status: Offline Points: 1019 |
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Jay is right and you are wrong on this. Show us the math!
That is a very good site. Do you understand it?
Do the math and show us. I repeat, Jay is right. Since you will not I can and will. It is easy. The paddle would have to slow down by 1/sqrt(2) or to 0.707 of its initial speed to lose half of its energy and we all know that the paddle doesn't slow down that much during the contact time so how do you justify your statement? Not all the energy the paddle loses is converted to ball kinetic energy. |
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AgentHEX
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Generally in school you had those people who did well by grasping the concepts taught, and those who got their grades by memorizing the text and how to turn the crank on test problems.
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zeio
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Suppose the setup is the one in my signature: Paddle mass m1 = 185g, Ball mass m2 = 2.7g Before collision, Paddle initial velocity = 20m/s, Ball initial velocity = -10m/s After collision, Paddle final velocity = 19.8m/s, Solve for Ball final velocity = ? Solve for initial and final kinetic energy = ?, ? Compare the ratio of KEi and KEf = ? Initial momentum = Final momentum m1v1i + m2v2i = m1v1f + m2v2f 185g*20m/s + 2.7g*-10m/s = 185g*19.8m/s + 2.7g*v2f 3.7kg m/s - .027kg m/s = 3.663kg m/s + 2.7g*v2f v2f = 3.7037 Initial kinetic energy 1/2m1v1i^2 + 1/2m2v2i^2 185g/2*20m/s^2 + 2.7g/2*-10m/s^2 37J + .135J = 37.135J Final kinetic energy 1/2m1v1^2 + 1/2m2v2f^2 185g/2*19.8m/s^2 + 2.7g/2*1m/s^2 36.2637J + .01851J = 36.2822J KEf/KEi 36.2822J/37.135J = .977 97.7% of the kinetic energy of the paddle has been transferred to the ball during the collision. See now? Edited by zeio - 09/21/2013 at 1:22am |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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AgentHEX
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LOLWUT I stand corrected. Not all the latter group learned how to turn the crank. |
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zeio
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Now I gotta catch the women's world cup.
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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tt4me
Gold Member Joined: 01/17/2013 Location: RC Poverty Zone Status: Offline Points: 1019 |
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Wow, I would be very disappoint if I hit the ball with a paddle speed of 20m/s and it went back at only 3.7m/s. Doesn't it seem a little strange to you that the final velocity of the ball is slower than that of the paddle? This doesn't make sense.
Why did you use 1m/s for v2f when you calculated 3.7037 m/s for v2f? Other than that, this part is correct given your nonsensical assumptions.
This is the ratio of the total kinetic energy after to the total kinetic energy before. OK.
No! from you own calculations you can see the the paddle still retains 98% of its kinetic energy v1f/v1i=36.2637J/37J=0.98 The paddle has lost little of its kinetic energy. I repeat, Jay is right!!! You should have sanity checked what you wrote before going to watch TT. Edited by tt4me - 09/21/2013 at 6:54am |
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tt4me
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Can't you see that zeio is trying to weasel word out of his embarrassing predicament? It is true that given his assumptions that the ball and paddle system retain 97.7% of its initial kinetic energy but it is not true that that 97.7% of the energy of the paddle has been transferred to the ball.
WHAT??? How can you be corrected by something that is wrong? I am still waiting for someone that can 'turn the crank'. This is turning the crank. The first part of the document derives the speed after impact using two equations to solve two unknowns that are the final velocities of the paddle and ball. The two equations are the conservation of momentum and the definition of the coefficient of restitution. I assumed a coefficient of restitution of 0.5 which is typical. Notice that the paddle slows down a little more than in zeio's example but the ball does go faster than the paddle as it should. |
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mercuur
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I was referring to a bat in a vice (to exclude that energy is increased by an impulse).
Cor=rel speed after/rel speed before right. So what it says behind the equal sign "=" is what it is ? Cor would just be a naming then that can be exchanged for any other name or letter. rel speed after /rel speed before is a ratio of change. Change inplies time so this ratio nor a number for this can be without time as dimension. Change to time ; C/T.. Or should cor be interpreted T*C/T = C as interpretation ; no time no change ? Instead of "rel speed before" and "- after" Both V's can be adressed with time coordinates t. Dwell as period T=1 is the natural time unit for any particular bounce then. This unit varies with the duration but for a particular case it,s always a perfect constant. Clocktime has clockunits so T=1 (dwelltime) and T=n (sec) both correspond to the same period. |
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tt4me
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mercuur, it is hard to tell what you are trying to say. You are misusing terms.
Different people use different symbols for the coefficient of restitution. Putting the paddle in a vise that is fixed simplifies the calculation but it isn't necessary if you do all the calculations relative to the paddle. For instance, in the example above the paddle velocity can be subtracted from the ball velocity to get a paddle relative velocity. In the example above the relative velocity of the ball before impact is -10-20=-30. The relative velocity of the ball after impact is 34.35269-19.35269=15. Notice the relative velocity is the opposite and half and opposite of the incoming velocity as one would expect when the coefficient of restitution is 0.5. Now lets relate this to the dwell time. In this example it should be obvious that since the incoming velocity relative to the blade is twice that of the outgoing velocity, the time to stop the ball relative to the paddle is about 1/2 the time to accelerate the ball again. If the COR was 0.3333 the time to accelerate would be 3 times longer than the stop time so a lower COR does increase dwell time as people have stated above. The extreme case occurs when the COR is 0 so the impact is inelastic, like hitting a ball of putty, so the dwell time is infinite. |
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zeio
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The ball final velocity depends on a large extent of the paddle final velocity. The more the paddle slows down, the more the ball speeds up. Consider the following: Paddle mass m1 = 185g, Ball mass m2 = 2.7g Before collision, Paddle initial velocity = 20m/s, Ball initial velocity = -10m/s After collision, Paddle final velocity = 19.5m/s, Solve for Ball final velocity = ? Initial momentum = Final momentum m1v1i + m2v2i = m1v1f + m2v2f 185g*20m/s + 2.7g*-10m/s = 185g*19.5m/s + 2.7g*v2f 3.7kg m/s - .027kg m/s = 3.6075kg m/s + 2.7g*v2f v2f = 24.2593m/s For perfectly elastic collision, v2f would become 49.1369m/s, where v1f = 19.1369m/s. Happy now? Give me a break. Had you bothered to verify the calculation by hand, you'd have known it was a TYPO! Let's do the math here. For v2f = 1m/s, 1/2m2v2f^2 = 2.7g/2*1m/s^2 = .00135J. Now consider the phase of that collision where the paddle and ball move along at the same velocity vf: Initial momentum = Final momentum m1v1i + m2v2i = m1vf + m2vf 185g*20m/s + 2.7g*-10m/s = (185g + 2.7g)vf (3.7kg m/s - .027kg m/s)/(187.7g) = vf vf = 19.5685m/s Initial kinetic energy = same as before = 37.135. Final kinetic energy (1/2m1 + 1/2m2)vf^2 (185g/2 + 2.7g/2)19.5685m/s^2 35.9374J During that time, most of the paddle's energy is transferred to the ball through "bonding". However, the paddle doesn't stop because its momentum is conserved. |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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mercuur
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It,s pretending to take C=0,5 or any other particular value and then pretend you (or I or others) know by that C=0,5 for value. That,s what I understand. I.m not even starting then to go into equations and calculations. Then from a random chosen value and a random rel V before explain rel v out from that while it needs this value first for knowing C as ratio to begin with. Or pretend to know V before ánd - after by giving two example values and then calculate a value for their ratio as C and use that for the bounce. Picking one or two random numbers outside the bounce and use this as information for the bounce and/or pick an example ratio for the bounce and use this as information for the bounce. V4 and V3 doesn,t allow this type of easy skipping as for schoolwork. V after/ - before seems to allow this and repeat the bounce over and over again. V before is a V after during any bounce and V after is a future V for the period before any bounce. When this is confusing then better so as mistaking randomchoice numbers for information or knowledge (I think your fooling yourself with this). On itself I wouldn,t have a problem with this but then do it from a first fixed dwelltime. Choose any fixed T for dwell and work from there. This choice can also be symbol T ( Tdwell =T then) but problem for that can be that it can take any value later on and then it,s all vague (until it has an outcome as some rabbit from some hat prooving there was a rabbit in the hat). So better pick a fixed value in millisec. Then you may pick two values for rel V before and after.. This example then has a delta Vrel =zero, neg or pos and a period T to combine it with for an overall decelleration (or accelleration) over the period of dwell between before and after. Under these circumstances your welcome to try explain again. A deceleration and an accellertion inside the period for dwell must then correspond to this overall outcome and effect to both sides (accelleration also relative). Edited by mercuur - 09/21/2013 at 4:03pm |
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AgentHEX
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You see, there's being right, and there's wrong, and frankly the gap isn't as large as most suppose. People who're more right or more wrong can discuss and easily come to the correct conclusions. There's no shame in being wrong. But then there's also not even wrong, where someone doesn't know what's going on and therefore can't qualify for being wrong about it. The gap between not even wrong, vs. either right or wrong is not so easily bridged. What's devastating funny is that he actually calculated the KE for both items before AND after. Yet more comedic still is that he went back to "fix" it not once but twice from 97 to 96% and then back to 97%. It's worth noting again my comparison of pnatchwey and zeio linked by the latter himself above, except pnatchwey was more just wrong but really stubborn about it. This here is far worse but fortunately far more comedic than anything from that prior situation that I've read. It's also worth mentioning that not even wrong doesn't preclude someone from being accidentally correct akin to a broken clock as evident from the pnatch dogpile.
Let me explain what I mean above. School exams usually consist of two types of question: rhetoric and applications. Students are meant to understand ideas conceptually to answer the former, and apply this understanding to solve that latter. However it's also possible to "study to the test" by simply memorizing the words to answer the first and learning a few proscribed mechanical processes for the set types of trivial problems found on tests. I've always thought lessor students need both of these skills to get a passing grade, but apparently I was wrong because zeio can't seem to do the latter correctly at all. |
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tt4me
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NO!!! There are two unknowns. The final paddle and ball velocities. The initial ball and paddle velocity plus the coefficient of restitution determine the velocities of the ball and paddle after impact. See my example. I can simply change the coefficient of restitution from 0 to 1 to match the results you got below.
This is better. So did you just pick lower v1f until v2f>v1f?
I will be when you answer my question about what determines the speed of the paddle after impact. I looks like you just picked speeds, using my hint, until you found a result that makes sense.
Your numbers are right but what you are saying about transferring energy is wrong because the paddle retains over 90% of it energy. Therefore the paddle can't be transferring most of its energy. This is the point Jay was making. |
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tt4me
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pnachtwey was questioning zeio's assertion that the dwell time is in the 4 millisecond range and we now know the dwell time is closer to what pnachtwey claimed and it was zeio that was wrong. |
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zeio
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The lower and upper limit of the slowdown is 19.1m/s and 19.5m/s. I chose 19.8 randomly just to see how much velocity would affect the ratio of kinetic energy. Then I knew you would pick on it and went to change it. That's where the KE ratio .967 came from. But I forgot to change the rest of the value before updating the post. And then it was time for the world cup. I said to hell with it and reverted the ratio. Your hint? I chose 19.5m/s because that's the value where both paddle and ball move together with the same kinetic energy during the collision, which is what I've always been saying. What you and Jay refer to is after the collision where the two separate with their own share of that energy. |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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AgentHEX
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I'm curious whether zeio has figured out what's going on and just posting garbage to cover it up, or actually believes that's a coherent reply to tt4me.
Quoted again for posterity. Edited by AgentHEX - 09/22/2013 at 1:51am |
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mercuur
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Intrinsic time it was and possible intrinsic math unit for time I was thinking.
I stumbled on this a few years ago but couldn,t google it until I remembered the word intrinsic to use for finding it back. http://www.vip.ocsnet.net/~ancient/Constants-Version%20Postprint.pdf.. For some Zeio support . A workingline or line of motion is not a euclid space. It has no coordinates or point zero for coordinate. Needs to use a space for the system (can be one dimensional on the workingline but free to move to the workingline). Not tying it solidly to the tabe, floor or a workingline without coordinates. In that case the paddle would move from the neg sector to the positive part or vice versa passing through this point. This would count for an all sudden turn of direction for the paddle while the ball stays in the same sector. That heavily intereferes then with "relV before/rel V after" as much higher or lower (dependant wether the bat starts in pos or neg secor.) Edited by mercuur - 09/22/2013 at 9:14am |
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zeio
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I say it's the other way around. That number is valid because it's during the compression phase of the collision where the ball is under acceleration in the opposite direction by the paddle. Without realizing that, tt4me made the "not even wrong" comment that the ball "went back at only 3.7m/s." Still in ignorance, he even went on to show how could "turn the crank" with the speed-after-impact formula, again without realizing that equation applies to the situation after the collision. As an aside, remember your "not even wrong" comment about dwell time? It is unfortunately the lower the harder you hit(Figure 15). p.s. added link to the "not even wrong" comment Edited by zeio - 09/22/2013 at 1:14pm |
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Viscaria FL - 91g
+ Neo H3 2.15 Blk - 44.5g(55.3g uncut bare) + Hexer HD 2.1 Red - 49.3g(68.5g 〃 〃) = 184.8g |
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