Results 52 to 68 of 235
03-07-2013, 09:56 AM #52
03-08-2013, 12:57 AM #53
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?
03-08-2013, 01:08 AM #54
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.
03-08-2013, 11:33 AM #55
(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.
03-08-2013, 11:56 AM #56
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?
Last edited by visor; 03-08-2013 at 11:59 AM.
03-09-2013, 08:22 AM #57
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.
03-09-2013, 07:13 PM #58
03-09-2013, 07:44 PM #59
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).
Last edited by j4ckie; 03-09-2013 at 07:57 PM.
03-09-2013, 08:41 PM #60
03-09-2013, 08:49 PM #61
Dang you're good... and fast.
Shoulda called on you earlier.
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?
03-09-2013, 08:51 PM #62
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
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.
03-09-2013, 08:53 PM #63
03-09-2013, 08:56 PM #64
Wow, this place is getting pretty esoteric! Visor, you have your wish -- I'm out of here, and leave it to you boffins!
03-09-2013, 08:58 PM #65
03-09-2013, 09:02 PM #66
03-09-2013, 09:10 PM #67
03-09-2013, 09:17 PM #68