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Calculations of Momentum P=mv, and Kinetic energy KE=(mv^2)/2 of a racket on shuttle

Discussion in 'Badminton Rackets / Equipment' started by visor, May 27, 2013.

  1. visor

    visor Regular Member

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    ​All right fellas, as promised, I've copied over to start a new thread the relevant posts from the "Panhandle grip" thread that was going nowhere (http://www.badmintoncentral.com/for...h-to-adopt-with-Superlite-Rackets)-for-Future)

    At least something good turned out from that "unusual" thread. :D

    [​IMG] [MENTION=86196]craigandy[/MENTION] [MENTION=31680]amleto[/MENTION] [MENTION=54]raymond[/MENTION] [MENTION=89106]vinod81[/MENTION] [MENTION=28441]96382[/MENTION] [MENTION=67305]Line & Length[/MENTION] [MENTION=84118]randomuser[/MENTION] [MENTION=76286]blableblibloblu[/MENTION] [MENTION=64484]j4ckie[/MENTION] [MENTION=47032]MSeeley[/MENTION]

    Please continue discussion... :)

     
  2. visor

    visor Regular Member

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    ......................................


     
  3. DuckFeet

    DuckFeet Regular Member

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    Why has no-one mentioned torque produced from forearm rotation? :)
     
  4. amleto

    amleto Regular Member

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    I have. I suggested that the calculations should include angular momentum and energy components from a very early stage.
     
  5. chrisql

    chrisql New Member

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    agree,I suggested that the calculations should include angular momentum and energy components from a very early stage.[​IMG]
     
  6. vajrasattva

    vajrasattva Regular Member

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    to complicate matters, if you hit the shuttle with a flat board, with a full contact, the transfer is more efficient.

    but when you have strings, the contact area is much smaller, and thus all the momentum of the racquet head is not fully transferred.

    further more, there's no complete transfer of momentum, because the racquet head is not stopped and decelerated to a stand still upon shuttle contact. instead most of the time it still continues moving in the direction it was initially swinging in, except that only some momentum was transferred onto the shuttle.
     
  7. 96382

    96382 Regular Member

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    providing some measurements and numbers:

    '...angular accelerations contribute much more to the speed of the racket tip than the linear
    accelerations.'

    Deflection while smashing:

    'Over a stroke, the maximum strain [...] corresponds to a tip deflection of about 5 cm.'

    Calculated peak acceleration of racket tip: ~800m/s^2

    Estimation for time-window for best momentum transfer (highest racket tip velocity):
    '[...] the window of opportunity where the velocity is close to maximum is only 0.02 s.'

    Of course, this time-window depends on stiffness/weight ditribution/aerodynamics of each racket...


    reference:
    http://link.springer.com/chapter/10.1007/978-2-287-99056-4_69
     
  8. craigandy

    craigandy Regular Member

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    For those interested in the drag/deceleration of a shuttlecock, there is a n interesting piece which can be found here
    http://www.isb.ac.th/HS/JoS/vol4iss2/Papers/1Badminton.pdf

    I did some super vulgar maths with the information found here and in reply to @MSeeley I just don't get why we are even thinking about the initial speed after contact, rather than considering the average speed over the first 3 feet, which would allow us to "compare" to professionals.

    I don't think it is going to make the world of difference, using some super vulgar maths I got about a 8mph reduction over 3feet using our 50mph(racket)/77mph(shuttle speed from racket @Line & Length ).

    I am thinking with everything that is available at the moment, that the best smash power combination for rackets (of the future) are going to be lighter in total but with very high balance points(really head heavy)?? I think though the light materials we construct may not be strong enough yet to handle this.
     
    #8 craigandy, May 28, 2013
    Last edited: May 28, 2013
  9. MSeeley

    MSeeley Regular Member

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    Fair enough. All I want is for people not to start comparing numbers to other numbers that were calculated differently. Now then, over to the rest of you... I haven't done enough study in these areas to contribute unfortunately (I'm a pure mathematician, not mechanic/physicist).
     
  10. visor

    visor Regular Member

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    Re angular component and incomplete transfer of P and KE:

    Since we're only concerned about the instant point of strike/impact when the racket collides with the shuttle, for the purposes of calculating instantaneous "v", I'm thinking we can simplify things a bit if we,

    1. treat it as a linear collision, and
    2. consider conservation of P and KE at that instantaneous point of impact.

    Agreed? :)
    @amleto @j4ckie @randomuser @Line & Length
     
  11. amleto

    amleto Regular Member

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    I think the main reason we don't need to consider the average speed after contact is that we have the (in)famous 421km/h speed which gives very good approx. to instantaneous shuttle speed after contact. Therefore we need not have to consider air resistance on a shuttle to compare numbers.

    For a progression of more accurate models I would consider something like the following stages:

    Most basic:
    Model the racket swing just by the pronation action ie racket is spinning around its own grip.

    Model the shuttle as a ball (no feathers)

    i) consider elastic collisions <-- this will give upper bound results
    ii) consider non-elastic collisions <-- shuttle cork really loses a lot of energy on impact. Consider how hard you have to hit shuttle off a wall to play that training game. Also think of how high a shuttle can fall onto a stationary racket/floor - the shuttle will hardly bounce.

    I would expect there to be a big difference between elastic and non-elastic cases.

    Model Improvement
    add small linear component to racket velocity. This should be relatively simple adjustment.

    Model Improvement
    Model racket swing as a double-pendulum: one for the relatively slow arm, one for the racket.

    Missing physics from the above models (perhaps significant): energy transferred into shuttle rotation
    Missing physics from the above models (probably insignificant): energy loss through sound, shaft flex*.

    Some things you might need to choose or calculate or find empirical evidence of:
    Racket head speed (tip) (cited as ~34m/s elsewhere) - use this to determine angular velocities.
    Mmoment of inertia of a racket.
    Coefficient of restitution for i) string bed, ii) shuttle cork

    Not exhaustive by any means, but that is the kind of progression I think is sensible.

    *probably insignificant if you know (empirically) the correct racket head (angular) speed to choose as an initial condition.
     
  12. amleto

    amleto Regular Member

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    1. Treating it as a linear collision certainly simplifies, yes. The predictions should be compared to other models and empirical results to see if it is a good approximation or not.

    2. Not sure what you're getting at. Conservation of X equations are useful because you can infer some state properties at t = t0 + dt if you know the state at t= t0. So if you're only talking about a single time... conservation of X is not very helpful :p
     
  13. DuckFeet

    DuckFeet Regular Member

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    My mistake. I didn't know angular momentum was torque.

    Form A level maths (calculus?) don't you reduce delta until you are effectively dealing with a straight line? I suppose the contact time is NOT infinitesimally small. I'll leave posting to those more qualified to do so, but reading with interest.
     
  14. amleto

    amleto Regular Member

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    yes, that's the idea if you want to do that. As you point out, though, it may not actually match too well to a badminton swing (or it may).
     
  15. DuckFeet

    DuckFeet Regular Member

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    Question: If I stop the racquet at point of impact, all the momentum goes into the shuttle (if we ignore stringbed/racquet flex)? So stunning rather than playing through the shuttle theoretically gives more shuttle speed? And probably a destoyed elbow, but we'll ignore that. Also this thread has a ton more 'discussion' in it. http://www.badmintoncentral.com/forums/showthread.php/42510-The-Smash
     
  16. craigandy

    craigandy Regular Member

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    Ok, I have been doing the experiment with software. I have some results but I am going to do some more tests as I have been using existing videos and want to try it on my own swing and a few others under the condition that I have set up just to confirm the results.

    If anybody has a side on view of their smash, post it here and I will analyse it and give you a ratio.(The video must include a calibration measurement, but your height will do, everything in shot like racket/shuttle and some space to see the shuttle shoot off.)
    Unfortunately it will not give out readings for really fast smashes due to the frame rate and the fact that you can't really plot it.
     
  17. amleto

    amleto Regular Member

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    Incorrect. How do you think the racket stops? YOU 'take momentum out' of the racket. Or more precisely you apply an opposing to force to what you have already generated.
     
  18. SolsticeOfLight

    SolsticeOfLight Regular Member

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    Also, because the impact is not instantaneous, you (can) continue to apply an accelerating force throughout the impact, and also afterwards. This would have to be taken into account when calculating the momentum after.

    What we really need is a lot of detailed results and a synaptic chip to spew out results :p Not that we'd know the formula, but we'd be able to guess our own smash speeds with it.
     
  19. visor

    visor Regular Member

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    Imho whatever happens to the racket after collision is irrelevant to the shuttle's behaviour and speed because P and KE have already been transferred. You could throw the racket all the way across the court and it still wouldn't matter. ;)
     
  20. visor

    visor Regular Member

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    The laws of conservation can be used for that instant moment at collision.

    In the classical examples of collisions between 2 elastic balls, once the balls have collided and speeds are known, it doesn't matter what you do to the balls after the collision!

    What you do to one ball after the collision will have absolutely zero effect on the other ball.
     
    #20 visor, May 29, 2013
    Last edited: May 29, 2013

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