# Thread: Head weight: an easier, simpler appoximation of swing weight

1. Originally Posted by visor
Update from the latest visit, with bare wts and bare bp calculated.

 Arc FB 73.1g 32.6g 301mm Arc 11 86.8g 38.1g 296mm MX70 86.9g 38.7g 301mm MXJJS 4u 84.6g 37.2g 297mm BSLYD 4u 84.3g 36.3g 291mm BSLYD 3u 90.1g 40.0g 300mm TK8000 89.4g 39.3g 297mm NR700FX 83.7g 36,4g 293mm NR700RP 88.4g 38.3g 292mm
Interesting notes:

- dang that 3u bslyd is a beast! more so than tk8000. i had dry swings with all of the rackets and this one takes the cake in swing wt! could've passed for a training racket!

- obviously arc fb felt like a fly swatter on dry swinging...
this ARC FB with only 73.1 g, and BP at 301mm...isn't this the ultimate racket for XD? light weight and a monster for killer smash?

2. Visor, how does knowing head weight benefit players?

For example, since in your data your BS 12 and your MX 80 head weights seems to be equal, the BS 12 must be more powerful then? (Due to it being more aerodynamic thus can be swung faster). Am I correct in assuming this?

3. There's only 0.2g difference, not significant. But even then, we have to talk about specific copies of rackets because there's variations. Not all mx80 or bs12 have those same specs.

That's why I emphasizing the importance of bringing a scale and a ruler and measuring or calculating head wt of the rackets at the store when choosing a specific one that you prefer, instead of leaving it to chance.

4. Hi Visor,

(head wt / total wt) = (bp / 675mm)

Can you describe head weight - what does this value represent, the weight of what? ... or from where to where?
If you string and grip your racket and then cut it with a saw at fulcrum point, remove the piece of handle and then weight the rest - that would be your swing weight. If you come up with a formula for that, you might get Nobel prize for physics.

5. @meteor

Swing wt is a dynamic measurement on an expensive machine costing thousands afaik. There is no cutting of rackets or handles.

But for my informally made up head wt, I wanted an easier (and cheaper!) method of measuring or estimating swing wt statically. If you look at the pics on page 1, you'll see how I do it. And when I analyzed the data, I saw the relationship in that equation.

So yes, if you cut the racket at the bp and weigh the top half, that will be exactly the same as my measurement of head wt. Isn't math and physics wonderful?

6. Originally Posted by visor
@meteor

Swing wt is a dynamic measurement on an expensive machine costing thousands afaik. There is no cutting of rackets or handles.

But for my informally made up head wt, I wanted an easier (and cheaper!) method of measuring or estimating swing wt statically. If you look at the pics on page 1, you'll see how I do it. And when I analyzed the data, I saw the relationship in that equation.

So yes, if you cut the racket at the bp and weigh the top half, that will be exactly the same as my measurement of head wt. Isn't math and physics wonderful?
It is wonderful for those who understand it... not me though )
Just to point out that the upper side of the bp weighs exactly half of the overall weight (as much as the other half below) it doesn't depict the head load at all. Also there is a problem with the weighing methodology you use: the handle/grip contributes too to the presumed "static head weight" or if the handle support is placed at fulcrum point, the handle segment below fulcrum counterweighs the reading.

7. @meteor
Dang... I'm gonna have to make a diagram to explain this concept so that it's easier to grasp.

@j4ckie
IIRC, You're good with physics. Help me out here please.
Am I making sense? And how can I make it easier to understand?

8. @visor
Ya, the equation is correct. Simple mechanics, actually....what you do is basically to mount the racket on 2 bearings. Ideally, you do it at the ends and parallel to the ground. Mechanics (and common sense) suggest that the full weight of the racket has to be born by those two bearings, thus if you measure the weight on one, you know the weight on the other (=full weight - measured weight).

The BP is now easily calculable: BP = ("visor's head wt" / full weight) * racket length (usually 675mm).

Or, if you measured at the bottom instead of the head, BP = ( (full wt - bottom wt) / full wt) * racket length

Please note that all "visor's head weight" does is unify weight and BP to form a somewhat more reliable reference. It does not and can not take into account
a) the stiffness, which will influence how the racket feels on swinging, with softer shafts making the racket feel heavier as they allow the head to drag behind more and create a bigger moment of inertia, and
b) the distribution of the weight, which will influence both how heavy it feels on swinging (with a more extreme distribution feeling heavier by [theoretically] up to 33% with the same wt and BP) and how solid/heavy it feels on contact with the shuttle (with a more extreme distribution allowing a more solid head and thus more solid/heavy impact feel).

Hope this helps.

PS: reading this thread, I have to add c) the "kickpoint", i.e. the flexing behavior of the shaft. A tapered shaft will behave differently depending on how it is tapered (thick bottom will result in a more rigid feel, less bending and lower torsional rigidity of the head, thin bottom will result in more bending and better torsional rigidity).

9. Originally Posted by j4ckie
----snip----

PS: reading this thread, I have to add c) the "kickpoint", i.e. the flexing behavior of the shaft. A tapered shaft will behave differently depending on how it is tapered (thick bottom will result in a more rigid feel, less bending and lower torsional rigidity of the head, thin bottom will result in more bending and better torsional rigidity).

@visor

10. @j4ckie

Dang you're good... and fast.
Shoulda called on you earlier.

Many thanks!

Is there a diagram that we can draw to show the mechanics involved so that it can be easily visualized by others who don't intuitively see it as we do?

11. Originally Posted by visor
@kwun @vajrasattva @demolidor @cobalt

No one is interested? Or is this too complicated or arcane? Or useless?

It's not complicated and it's very useful, I think. Just ratios and proportions.

Just instinctively assume:

(head wt / total wt) = (bp / 675mm)

675mm being racket length.

And you can see the relationship between head wt and bp. Hence if you know any two measures, you can calculate the third.

Pretty nifty huh?

No one interested in this or is this already well known on badminton central?
It's very basic and can be derived from equivalence of torques. Someone suggested it was worthy of a Phd, but I'm hoping that was tongue in cheek.

Here is the derivation:

What you have is a 'rod', with mass m, and length l, on a pivot at one end. The torque, T, on this system is given by
T = Force x distance from pivot
T = mgb --(1) where b is the distance from pivot to the CoG / balance point, g is gravity.

By measuring this torque via scales at the free end of the rod we can deduce that
T = L m' g --(2)
where T is the same torque as above, L is the length of the rod, and m' is some pseudo mass (our head weight)

Hence (compare (1) & (2))
Lm'g = mgb

therefore
m'/m = b/L

Now you may call the rod a badminton racket WLOG (without loss of generality) and we get your equation ie
(head wt / total wt) = (bp / 675mm).

So yes, the equation is sound.

12. Originally Posted by cobalt
Stop going off topic re flex points and whatnot you!
Or else I'll report you to a moderator.
Oh wait... you are a moderator!

13. Wow, this place is getting pretty esoteric! Visor, you have your wish -- I'm out of here, and leave it to you boffins!

14. @amleto

Thanx!
But having read that over twice, I'm afraid I can't visualize
it. Is there a diagram that can show it visually the forces and lengths?

15. Originally Posted by SSSSNT
Visor, how does knowing head weight benefit players?

For example, since in your data your BS 12 and your MX 80 head weights seems to be equal, the BS 12 must be more powerful then? (Due to it being more aerodynamic thus can be swung faster). Am I correct in assuming this?
It doesn't, really. It's just another static measurement like BP that isn't related to rotational movement so it doesn't accurately describe the feel of a swing where you should instead consider the moment of inertia.

16. Originally Posted by visor
@amleto

Thanx!
But having read that over twice, I'm afraid I can't visualize
it. Is there a diagram that can show it visually the forces and lengths?
Hope you can read my scrawl

17. Originally Posted by amleto
It doesn't, really. It's just another static measurement like BP that isn't related to rotational movement so it doesn't accurately describe the feel of a swing where you should instead consider the moment of inertia.
Wouldn't the moment of inertia also be directly proportional to the "head wt"?

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