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    Default "Faster" swing = more power? opinions please

    Ok, so I've been reading that the pronation and supination is the main movement/action to generate power/speed through impact.

    So, would swinging faster always equate to more power? I have a feeling it does, but when I use a heavy racket (HH or not 2U heavy 3U) and hit, it will seem to produce a stronger shot. Of course physics comes into play with moving a heavier mass to transfer but was wondering if a lighter racket (still 3U but easier to swing) moving "faster" would also produce a powerful shot?

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    It should depend on how much stiff the racket is...

    I mean, flexible rackets tend to "flex" with the swing and this gave extra power to the shuttle on the impact, but if the movement is done "too" quickly the speed is somehow nullified by this "flex effect".

    On the contrary, stiffer rackets tend to be less powerful but a good-made quick swing is able to get the best from them

    To summarize...

    Flexible rackets ----> More speed ----> less power (or equal power)
    Stiff rackets ----> More speed ----> more power

    It should be like this
    hope it helps ^_^

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    Based on conservation of momentum, m1v1 = m2v2 (m is mass, v is velocity),
    for same racket (m1) and same shuttle (m2), the faster u can swing (higher v1) will produce higher speed of shuttle (v2) ideally.

    So i think this is a trade-off case. When u use heavier racket (higher m1), ur swing speed required will be lower than when u use lighter racket to produce same smashing power (speed of shuttle, v2).

    I hope i explain it clear enough. Correct me if i am wrong.

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    If we're considering only racket weight you're definitely right

    However, i believe that stiffness plays an important role too, as i've explain

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    Ya. The 4 key parameters when deciding which racket to buy. Racket weight, shaft stiffness, head heavy/balance/light, and the last one your badminton skill or can say as your playing style.

    Actually the string and tension, also does significant effect to it also. So hard to consider all factors ya.

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    Thanks for the replies, I guess trying to figure out if most things being equal:

    shuttle mass, swing speed, stiffness, string tension would the shear weight of the heavier racket always produce a stronger shot versus a lighter version?

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    yes, but you have to consider that the heavier racket requires more strenght to generate the same swing speed than a lighter one.

    So, with your conditions (and with the same weight distribution too), the heavier racket will generate more power but the player will get tired sooner xD

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    Quote Originally Posted by PapA_xlonG View Post
    Based on conservation of momentum, m1v1 = m2v2 (m is mass, v is velocity),
    for same racket (m1) and same shuttle (m2), the faster u can swing (higher v1) will produce higher speed of shuttle (v2) ideally.

    So i think this is a trade-off case. When u use heavier racket (higher m1), ur swing speed required will be lower than when u use lighter racket to produce same smashing power (speed of shuttle, v2).

    I hope i explain it clear enough. Correct me if i am wrong.
    Just to clarify since the first object, the racket, does not stop and therefore transfer all it's momemtum the equation is actually...

    m1v1 = m2v2 + m1v3

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    in general, yes. the faster you can swing a racket (which proper form etc.), theoretically, you should be able to generate more power. the faster you swing, the more torque one can generate = more power.

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    Quote Originally Posted by druss View Post
    Just to clarify since the first object, the racket, does not stop and therefore transfer all it's momemtum the equation is actually...

    m1v1 = m2v2 + m1v3
    not that simple
    energy transfer from string to shuttle isn't like 2 colliding billard balls

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    Yes, more power can only come from either faster racket speed or heavier racket. Of course, a stronger player can get both. But given the same player, it depends on two factors. If using a heavier racket, whether he has the strength to wield it effectively without getting overly tired; and if using a faster racket, whether he has the right technique and timing to hit it right. So, as you see, there is no free lunch. It really depends on the strength, stroke, tecnique, and timing of your swing.

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    Quote Originally Posted by cooler View Post
    not that simple
    energy transfer from string to shuttle isn't like 2 colliding billard balls
    I know it's not... but it's a simple enough approximation and closer than the original posted equation.

    Conservation of Energy would be better but also much more complicated.
    Last edited by druss; 02-24-2010 at 05:21 PM.

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    Quote Originally Posted by druss View Post
    I know it's not... but it's a simple enough approximation and closer than the original posted equation.

    Conservation of Energy would be better but also much more complicated.
    that momentum balance equation is based on conservation of energy
    Last edited by cooler; 02-24-2010 at 05:48 PM.

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    Quote Originally Posted by Boaster View Post
    Ok, so I've been reading that the pronation and supination is the main movement/action to generate power/speed through impact.

    So, would swinging faster always equate to more power? I have a feeling it does, but when I use a heavy racket (HH or not 2U heavy 3U) and hit, it will seem to produce a stronger shot. Of course physics comes into play with moving a heavier mass to transfer but was wondering if a lighter racket (still 3U but easier to swing) moving "faster" would also produce a powerful shot?
    since pros use basically similar rackets as what we can buy in the store, if u want to hit as hard as the pros, speed is the main variable u can change for more power.

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    Quote Originally Posted by cooler View Post
    that momentum balance equation is based on conservation of energy
    Actually, the momemtum equation is based on force not energy.... the conservation of energy is a scalar equation since it does not involve direction while momentum does which makes momentum a vector, as is force.

    Conservation of momentum is actually based on Newton's third law and is derived by taking the law the next step.

    F = ma where a = v/t

    So F = m x v/t if you consider that the time is the same and that it's a closed system then you end up with

    m1v1 = m2v2 which is the current conservation of momentum equation.

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    actually it is the force of the acceleration of the racket head generated at the moment of impact and not velocity. I would say 65% wrist and 35% forearm strength , some more technical players would be around 75% wrist and 25% forearm and they tend to smash harder and more accutate as well.

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    Quote Originally Posted by druss View Post
    Actually, the momemtum equation is based on force not energy.... the conservation of energy is a scalar equation since it does not involve direction while momentum does which makes momentum a vector, as is force.

    Conservation of momentum is actually based on Newton's third law and is derived by taking the law the next step.

    F = ma where a = v/t

    So F = m x v/t if you consider that the time is the same and that it's a closed system then you end up with

    m1v1 = m2v2 which is the current conservation of momentum equation.
    yes, i should had said conservation of momentum as mv is momentum. However, when i said conservation of energy it is not wrong. In fact your posted conservation of momentum equation m1v1=m2v2 is derived from conservation on energy principle for ideal or near ideal condition but in real world, perfect ideal condition is very rare so it was safe for me to say conservation of energy

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