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

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

​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/foru...ts)-for-Future)

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

@craigandy @amleto @raymond @vinod81 @96382 @Line & Length @randomuser @blableblibloblu @j4ckie @MSeeley

Originally Posted by craigandy
Sorry not big on my physics
What is this referring to in real life? What I assumed you are trying to say is that m1 and v1 are referring to the racket and m2 and v2 would refer to the shuttle? But when I used the formula I got ridiculous figure for shuttle speed(v2?)

For instance if I were to swing a 88g(0.088kg) racket at say a lowly 50mph(22.35m/s). Using a shuttle weight in the average of the rules about 5g(0.005kg)

so
m1 = 0.088
v1 = 22.35
m2 = 0.005

0.088*22.35 = 0.005*v2
1.96 = 0.005*v2
v2 = 393.36m/s which = about 880mph that's pretty fast.

Obviously I am getting the wrong end of the stick can you correct me please with regards to your equation meanings.
Originally Posted by craigandy
If i square rooted m2 the equation made sense or am I just going further off track?
Originally Posted by Cheung
M1 should be much smaller as it's the head of the racquet. There's also loss of energy due to sound generation, suboptimal impact, damping etc (not sure what exactly what 'etc' also includes).
Originally Posted by craigandy
Even give the lighter weight of head only, the figures still crazy. What is this formula? and how is it related/useful for badminton calculations?
Originally Posted by visor
If you use conservation
of kinetic energy equation, then 37g head, 50mph swing speed, 5g shuttle, then shuttle speed is 136mph. Assuming no other losses from friction, sound generation, etc as Cheung mentioned.
Originally Posted by amleto
It's conservation of momentum, so you have to take the system as a whole

ie the system is the racket + the shuttle

If you hit the shuttle from a dangling string (ie the shuttle is not moving before you hit it), then you get this equation

p_racket_before = p_racket_after + p_shuttle_after

(p = momentum = mv)

So you can't say anything about the shuttle speed unless you know how much the racket slowed down after impact.
Originally Posted by amleto
only if all of the KE from the racket transfers to the shuttle, which it doesn't - not by a long way.
Originally Posted by craigandy
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?
Originally Posted by raymond
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.
Originally Posted by craigandy
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/foru...-good-smash-is on post #8
Originally Posted by vinod81
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)
Originally Posted by craigandy
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.
Originally Posted by visor
Imo, we should use KE... because even Yonex uses KE, instead of P.
Originally Posted by amleto
because it is being mis-applied.
Originally Posted by craigandy
That's reeeaaly helpful. I am under no illusions that I must be missing something. What? is the question.
Originally Posted by vinod81
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.
Originally Posted by amleto
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.
Originally Posted by visor
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.
Originally Posted by amleto
use the equations here for v1 & v2.

http://en.wikipedia.org/wiki/Elastic_collision

I'm sure you can fiddle the shuttle and racket masses to make things seem plausible.
Originally Posted by amleto
both are equally valid. But you do need to know what you're doing before you star plugging numbers in to equations.
Originally Posted by craigandy
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.
Originally Posted by amleto
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.
Originally Posted by craigandy
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.
Originally Posted by amleto
I don't know how else to say, all of the calculations done here are not correct because they assume all of the energy/momentum from the racket goes into the shuttle. Some have seen that the answers are ridiculous when using the full racket mass so they think the solution is to pick a number at random (the mass of the racket head), and try again and see what comes out.

It doesn't matter what numbers are picked, the problem that everyone seems intent on solving is the wrong one!

I did point those interested in the correct direction by linking to the elastic collisions wiki page. No one cared to plug their numbers into the equations I suggested, though.
Originally Posted by Cheung
LOL, And Yonex could be wrong..
Originally Posted by visor
Nah... with their r&d, I doubt that we know any better than them...
Originally Posted by 96382
Someone with some free PhD time could do a finite elements simulation for that situation ;-)
Paper? (Aerodynamics - p.53 http://tinyurl.com/b28x6em)
Originally Posted by 96382
Originally Posted by craigandy
Say it by showing me the proper calculations with numbers and an explanation along with it or wait untill someone else can. Thanks
Originally Posted by amleto
Originally Posted by craigandy
I can't see any numbers relating to badminton or any explanations about the numbers used with regards to badminton on that page. It's just a wiki page with elastic collision equations. How do I know what numbers to enter into those equations? you said I got the wrong numbers so whats the point. No help. Thanks anyway appreciate it.

2. ......................................

Originally Posted by craigandy
Say it by showing me the proper calculations with numbers and an explanation along with it or wait untill someone else can. Thanks
Originally Posted by amleto
Originally Posted by amleto
You read the wiki section - it explains what goes where...

Attachment 142207
Actually you can say m2 is the shuttle, and u2 = 0 to simplify things.

It's still not that accurate, though because the racket motion is not really linear - angular momentum should be considered.
Originally Posted by Line & Length
Given the constraints of the principle of conservation of linear momentum (PCLM) and assuming that the racket is very heavy relative to the shuttle and that the shuttle isn't moving appreciably horizontally, then a purely elastic collision would result in a shuttle speed of twice the racket head speed.

Given that any collision is not going to be fully elastic & that the racket head isn't infinity massive relative to the shuttle, I think 70mph off a head-speed of 50mph is plausible.

As the racket head gets lighter, then the head speed will increase. However, due to the PCLM, the 'speed multiplication' ratio will worsen. I would suggest that there's an optimum head weight which will differ for each player. Too light and there will be insufficient momentum to transfer much impulse. Too heavy & the head speed will suffer.
Originally Posted by amleto
Given that a racket swing isn't linear by a long shot, conservation of linear momentum alone doesn't do you much good.
Originally Posted by visor
this panhandle grip discussion is not going anywhere if you don't recognize that proper badminton stroke technique require forearm pronation/supination (hence the basic grip)

... you can't be using wrist flexion/extension (as you would with panhandle grip) all the time because that would cause injury to your wrist over time

don't be like those beginner aunts and uncles wearing wrist and elbow tensors that i see at the courts using this improper stroke... you're smarter than that... you're on this BC forum, being helped out by so many coaches and advanced players... please just think perhaps all these players who are trying to help you may know something you don't

i admire your enthusiasm shown in your posts, but really you're trying to reinvent the wheel without having any knowledge of the basics

now can we please get back to the KE and P calculations?
Originally Posted by amleto
there is a similar example to what would be suitable for us here at about 5 minutes
Originally Posted by randomuser
actual physicist here

finally someone who got it correct.
the absolute limit to the shuttle's speed is twice the racket head speed at the moment of contact (neglecting the shuttle's initial speed, which is an order of magnitude lower than the racket head speed for overhead shots)

in practice the ratio is not 2 but probably around 1.6 I guess, due to various things that I can describe in detail if anyone wants.

anyway, using "superlite" rackets does not increase the maximum possible shuttle speed.
why?
1. the most important factor to shuttle speed is the racket head speed at contact (which of course depends on power and technique). other factors like string tension just change the ratio of ~1.6 slightly.
2. below a certain point, using lighter rackets does not really increase your maximum swing speed because your hand simply cannot rotate faster.
3. normal weight rackets are already below this point.

as far as shuttle speed is concerned, the potential benefit to lighter rackets is for overhead shots with less swing than a full smash (i.e. half smash, some clears). for these, the constraint is how fast you can accelerate and light rackets have an obvious advantage for that.

this isn't taking into account how control and consistency changes for lighter rackets, but those are more subjective.
Originally Posted by visor
^^ Very interesting... learn something new everyday!
http://curricula2.mit.edu/pivot/book...l?acode=0x0200

So the max speed of the shuttle is 2x racket head speed, and in reality less due to loss of energy and momentum in the collision.

So if someone who has not yet reached his physical max racket head speed and acceleration with a certain racket, by switching to a lighter racket in both total wt and swing wt, he can swing faster to get a 1.6x faster shuttle speed.

Time for me to try out the Arc FB to see if I'm near my max speed limit yet...
Originally Posted by amleto
How embarrassing for you that you missed the post underneath what you quote, which proves that that particular maths makes assumptions which are not true for badminton.

Ho hum.
Originally Posted by craigandy
Yes finally a physicist.
So you say you can get about 80Mph 0f 50Mph swing. That's seems too high, more details would be great.

Can you also inform me what mass would be used when trying to figure out formula's regarding badminton. What Mass is taken into consideration with a say horizontal(overhead) swing forward like a drive.(hope you understand that)
Originally Posted by craigandy
Also why do you think Yonex use KE instead of momentum thanks!
Originally Posted by amleto
Gosh, things are out of hand, aren't they?
1) the racket swing in badminton is not linear - there is little point in trying to use linear mechanics to describe it. Ever heard of the computer phrase "garbage in, garbage out"? Well that's is what is happening here by trying to describe a non-linear problem with (only) linear mechanics.

You want to know what masses are involved? Sure - racket with strings: 80 ~ 100g, shuttle: 5g.
Sorry, no-one is interested in the rotational dynamics so there seems no need to mention a racket's moment of inertia (the rotational equivalent of mass).

Why do Yonex use KE instead of P? That's just an awful question. You can't use one without the other! Seriously, they are both inherent physical properties of a system. One is not superior to the other.
Originally Posted by craigandy
Yip out of hand, and sorry for the awful questions. Just trying to get a grip on it.
Plugged the figures into the other equation you gave btw and got 98mph which is even closer
Originally Posted by amleto
Probably meaningless - as I've been trying to explain, the equation is not applicable to a badminton swing anyway.*

*edit: the equation on its own is not a reasonable description
Originally Posted by craigandy
Only because we were using 50mph racket speed as the example.
If you were to use a faster swing then 100mph shuttle speed is fine.

For example if you where to swing at 150mph by my calculations the shuttle speed could be about 450mph jk
Originally Posted by MSeeley
What I mean to say is that if 50mph swing speed gave 100mph shuttle speed (I know it doesn't), how do we still know that is too fast? Without considering what that would turn into over the next few feet, how can we judge if it is correct? It may be for example, that a 50mph swing speed creates a 70mph average speed over 3 feet, which is suitably pathetic compared to a professional, but the "calculation" was 100mph which is "too high"?

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.
Originally Posted by visor
Re measuring smash speeds, commonly it's done with radar guns just like in tennis. This is easily accessible and done cost effectively, especially in tournaments.

But the problem in badminton is that the shuttle decelerates rapidly immediately after collision
(due to air drag on the skirt) , so this method's measurements will yield numbers that will be inevitably less than using high speed video under controlled conditions to analyze frame by frame to calculate instantaneous speed off the racket in the first few inches after collision. Eg. Yonex video of TBH 421km/h smash. And video of FHF smash compared to a beginner/intermediate player.
Originally Posted by visor
And I would add that for the purposes of theoretical calculations of smash speed, the high speed video measurement is obviously more accurate and true to reality.
Originally Posted by Line & Length
I'm actually a Mechanical Engineer, not a physicist.

Whilst the entirety of an overhead swing is far from linear, contact only occurs over a small period (few ms?). During that short time, the racket head may well be accelerating in different directions (i.e. towards the center of rotation), but its velocity will be nigh-linear. I would therefore argue that the PCLM is a relevant approximation at the point of contact.

The other point I'd make is with regard to the effective mass of the racket head. Whilst rackets are typically 80-90g, not all of this is located within the head. Indeed, the further from the contact that the mass is located, then less it would influence the impulse transferred to the shuttle.

The other point is that the collision isn't elastic. Using http://en.wikipedia.org/wiki/Inelastic_collision, replacing 'a' with 'r' (for racket), replacing 'b' with 's' (for shuttle) and setting u_s to 0, we get the following equation:

v_s/u_r = (C + 1)*m_r/(m_r+m_s)

Therefore, assuming a coefficient of restitution (C) of 0.8, shuttle mass of 4.9g and racket head mass of 30g, I get a ratio of 1.54. Therefore, a swing speed of 50mph would result in a shuttle speed of approx. 77mph. If this sounds 'high', remember that this is the instantaneous speed after contact. Even after a few feet, this speed will have come down due to the air resistance of the shuttle.

As racket head mass rises, the speed ratio will rise, approaching (but not exceeding) (C+1). However, as the weight increases, it is a fair assumption that the player can't generate the same u_r. Therefore, I would maintain that there is some optimum weight between getting a good u_r and a good v_s/u_r.
Originally Posted by craigandy
Thank you! very well explained, really clear much appreciated!

I had a look and I also see they assume a cor of .85 for all tennis rackets(regardless of tension or strings for purpose)http://en.wikipedia.org/wiki/Coefficient_of_restitution so 0.8 sounds good to me.

Right now all I need is a really good camera calibrated with some lines on a wall, some software so i can slow down the pic, and a bunch of different weighted rackets. An hour to swing them all as fast as i can, then a bit of paper and this equation! Then i will have the perfect weighted racket for me.... for a smash at least
Originally Posted by craigandy
Thank you! very well explained, really clear much appreciated!

I had a look and I also see they assume a cor of .85 for all tennis rackets(regardless of tension or strings for purpose)http://en.wikipedia.org/wiki/Coefficient_of_restitution so 0.8 sounds good to me.

Right now all I need is a really good camera calibrated with some lines on a wall, some software so i can slow down the pic, and a bunch of different weighted rackets. An hour to swing them all as fast as i can, then a bit of paper and this equation! Then i will have the perfect weighted racket for me.... for a smash at least
Originally Posted by blableblibloblu
haha I love the general idea. I'm saying that it just can't be solved by taking one formula and putting up 2 speeds and 2 masses and that it also requires a new thread

I ran out of time while typing my long post and I admit I didn't properly say what I thought.
Originally Posted by visor
yes, agreed... i'll try to get a mod to move the appropriate posts over to a new thread...
Originally Posted by visor
Can we please not feed this troll or whatever? It's not going anywhere. If we're going to talk about MV calculations, I'll move the appropriate threads.

3. Why has no-one mentioned torque produced from forearm rotation?

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

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agree,I suggested that the calculations should include angular momentum and energy components from a very early stage.

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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.

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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:

8. For those interested in the drag/deceleration of a shuttlecock, there is a n interesting piece which can be found here

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.
Last edited by craigandy; 05-28-2013 at 10:56 AM.

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Originally Posted by craigandy
I don't think it is going to make the world of difference
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. 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

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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.

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Originally Posted by visor
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
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

13. Originally Posted by amleto
I have. I suggested that the calculations should include angular momentum and energy components from a very early stage.
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.

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Originally Posted by DuckFeet
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.
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. 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/foru...2510-The-Smash

16. 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.

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Originally Posted by DuckFeet
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/foru...2510-The-Smash
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.

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