Panhandle Grip to Play (New Approach to adopt with Superlite Rackets) for Future

Discussion in 'Techniques / Training' started by Superzoom, May 12, 2013.

  1. amleto

    amleto Regular Member

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    only if all of the KE from the racket transfers to the shuttle, which it doesn't - not by a long way.
     
  2. craigandy

    craigandy Regular Member

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    Yes that is exactly correct 136mph. When you take away all of Cheung's suggestions this formula correlates perfectly with real life and badminton, but what is this other formula and it's meanings from superzoom?
    With only racket head his formula still states 360mph shuttle speed?
     
    #42 craigandy, May 16, 2013
    Last edited: May 16, 2013
  3. raymond

    raymond Regular Member

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    Complications

    With the momentum equation - I vague recall that it is for elastic collisions, as someone here (Cheung?) pointed out other form of energies involved (string tension that's related to string bed deformation comes to mind).

    The "m" in both this and kinetic energy is also tricky. The racket is not a "free body". It's connected to the player's hand, which is connected to the player's body that has its own mass M, which is also moving. Of course, we can talk about a lighter or heavier racket in the hand of the same player, in an attempt to neutralize this element. Still, it's not clear what percentage of the connected body part would play into the equation, to the point to make head light/heavy less relevant.

    Also, someone here suggested that with k = 1/2 mv**2, a lighter racket would favor more energy transfer. That's not clear to me either. A lighter racket would reduce "m", and potentially increase "v". While "v" would increase in magnitude of a square, it's not clear if a loss of mass would translate to "more than compensating" gain in "v". How much faster can one swing with a lighter racket?? Some experiments might be needed.

    If a lighter racket comes with a more flexible shaft, maybe another mechanism is also involved in pre-storing energy on a "deformed" shaft. For that, another analysis may be needed.
     
  4. craigandy

    craigandy Regular Member

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    Yonex claimed you can swing flash boost 20mph faster than conventional racket (I guessed at 85g?) My calculation obviously ignore and assume a lot of things but can be found herehttp://www.badmintoncentral.com/for...-Weight-of-racket-determine-how-good-smash-is on post #8
     
  5. vinod81

    vinod81 Regular Member

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    I think the equaltion has to be: M1 * V1 + M2 * V2 = M1 * V1' + M2 * V2'

    V1' and V2' are the velocities after the collsion.
    Please note: since the shuttle change directions after the collssion, V2 and V2' will have opposite signs(+/-)
    It is improtant to have V1 - V1' large (shuttle square to the raquet, sweet sport etc)
     
  6. Stratlover

    Stratlover Regular Member

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    While some people may teach shifting towards panhandle for smashing, I would not recommend it unless you are taking it lower and farther in front than you should be, resulting in more of a powerful drive than a smash. Almost every professional player uses a grip more towards basic while smashing, again, in order to get the downwards angle. They prioritize high contact point over the shuttle being very in front of their body. For shifting towards panhandle when clearing, that's not necessary especially for attacking clears, but may help for high defensive ones.

    You can see Lin Dan use a slightly more panhandle grip when he clears as you can see his wrist flex at the end of the stroke. However, for smashes he goes to basic to bring the shuttle down, unless the shuttle is a little low, in which case he maintains his clearing grip and hits a flatter smash. I've seen pros hit drops either way, depending on their preference.
     
  7. craigandy

    craigandy Regular Member

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    I don't think it would make much of a difference given the shuttle can be falling vertical before a smash therefore carrying no speed to take into consideration really. Still this formula does not seem to work for me.
     
  8. visor

    visor Regular Member

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    Imo, we should use KE... because even Yonex uses KE, instead of P.
     
  9. amleto

    amleto Regular Member

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    because it is being mis-applied.
     
  10. craigandy

    craigandy Regular Member

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    That's reeeaaly helpful:rolleyes:. I am under no illusions that I must be missing something. What? is the question.
     
  11. vinod81

    vinod81 Regular Member

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    Agree that V2 can be assumed to be zero.
    What about V1'. The raquet continue to tavel in the same direction alsmost at the same speed even after the collission. So V1' will only be slightly smaller than V1.
     
  12. amleto

    amleto Regular Member

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    I've already pointed out the logic error that people have been caught out by, trying to convert ALL the momentum/energy of the racket into the shuttle.

    Everyone is also missing the fact that throughout the swing, the player is accelerating the racket so the system of racket + shuttle that people are considering is subject to external forces that invalidate all the maths + conclusions.
     
  13. visor

    visor Regular Member

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    But it doesn't matter what happens throughout the swing, it's exactly at that point of shuttle and racket contact that we're concerned about with either P or KE exchanged and hence their respective conservation equations... and I still prefer KE, as does Yonex apparently. :p
     
    #53 visor, May 16, 2013
    Last edited: May 16, 2013
  14. amleto

    amleto Regular Member

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  15. amleto

    amleto Regular Member

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    both are equally valid. But you do need to know what you're doing before you star plugging numbers in to equations.
     
  16. craigandy

    craigandy Regular Member

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    Still not picking up what your putting down.
    I don't think anybody was trying to convert all momentum, just using what's available to get ball park figures. Friction,repulsion, absorpsion, etc can be subtracted after using either formula and affect it the same? I have not read anybody disagreeing with this or not getting it. What still remains is that the ke formula seems to give a good ball park figure but the momentum one doesn't.
     
    #56 craigandy, May 16, 2013
    Last edited: May 16, 2013
  17. raymond

    raymond Regular Member

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    I'm getting a feeling that we're sidetracked... :rolleyes:

    Pronation should help generate more power. However, as some have pointed out, panhandled grip would restrict the amount of rotation one can have.

    For the same reason, but with a completely different effect, panhandled grip would allow full supination to be applied without sending the shuttle outside the court. So even with overhead shots, panhandle works much better with backhand shots.
     
  18. Gollum

    Gollum Regular Member

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    This is not something I've observed. If you have an example, I'd be interested to see it. :)


    That's not the idea. The idea is to take your ideal contact point -- which is only slightly farther in front than a clear -- and adjust your grip just slightly to avoid slicing the shuttle. It is very subtle -- so subtle that it's extremely difficult to tell the difference when watching players, even if you have slow motion video.

    In order to hit with maximum power, you need to have a (roughly) neutral wrist at impact. That's why the contact point for a smash should be brought slightly in front of the contact point for a clear.

    If a smash uses the same contact point as a clear, you have to bend your wrist downwards (i.e. flex) in order to get the racket pointing down. Doing so loses power, and this is one reason that a behind-the-body stick smash is less powerful than a power smash.

    Of course, this lowers your contact point back to what it would have been in the first place, so you don't gain any height advantage in the end. And the difference in height between the contact point of a smash and a clear is minimal anyway: about an inch.

    Whenever the contact point changes, the grip should change (all other things being equal). This applies to every shot. On the forehand side, farther in front = more panhandle, farther behind = more thumb grip. On the backhand side, it's the opposite.

    Large changes in the contact point require large changes in the grip. Small changes in the contact point require small changes in the grip.


    Certainly the grip is much, much closer to basic than panhandle. I'm not sure how you can tell exactly what grip the professional players are using -- we are talking about a very small difference.


    Of course -- but the contact point is still not the same as a clear.


    That's definitely the wrong way around.

    A punch clear is an example of a shot where the wrist can be used deceptively. You can start with a smash (or fast drop) contact point, and then bend back (extend) the wrist to angle the racket more upwards.

    This gives you the option of winding up for a smash, deliberately slowing down your swing so that your opponent thinks you're playing a drop shot, and then finishing with a punch clear.

    This means you're hitting with an extended wrist and a partial panhandle grip. This results in a significant loss of power, but that's okay because you're playing a shot that needs little power.
     
    #58 Gollum, May 17, 2013
    Last edited: May 17, 2013
  19. amleto

    amleto Regular Member

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    I had a quick scan and couldn't see the KE calculations that give a good result (and that I haven't already pointed out as being wrong). Can you point me towards it? Thanks

    I don't think anybody was trying to convert all momentum,...
    That's what you did in post 35 and why the answer you get is so crazy.
     
    #59 amleto, May 17, 2013
    Last edited: May 17, 2013
  20. craigandy

    craigandy Regular Member

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    I said I used the equation for momentum but in no way did i think the figure would be even vaguely correct. I just know when you used the same figures in both equations momentum came out 360mph and as visor calculated 136mph using the ke equation. Also I posted a link to other ke calculations done. Now I would guess if I did a 50mph swing in real life the shuttle would probably go about 70mph?
    So i can see 360mph is a long way off 70mph but 136mph is not(I know you lose a lot of speed given other factors).
    I think you are getting the wrong end of the stick though. I can only assume I must have the maths way wrong, I am not disagreeing. I was just looking for some more informed answers since it was raised as an argument in this thread.
     

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