Polarised Rackets

Discussion in 'Racquets' started by Veninga, Sep 21, 2011.

  1. Veninga

    Veninga Rookie

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    Guys,

    Lately there has been much talking about (de)polarised rackets. Depol makes modern spinny way of playing more easy. Pol makes the more traditional more flat way of playing more easy.

    If i understood it right, a depol racket has on both ends of the racket (cap and top) more weight than a pol one. In other words in a polarised racket the weight is more even distributed along the vertical axis.

    The balance rating is of course something else. This number is saying more about the weight distribution in vertical axis. More at one of the other side.

    My questions:
    - what are typical polarised rackets?
    - which numbers on TWU are saying something about (DE)POL

    On the second question. Does plowthrough and or hittingweight in the top of the stringbed compared with the lower part of the stringbed something?

    Anyone?

    Ciao from Holland
     
    #1
  2. spaceman_spiff

    spaceman_spiff Hall of Fame

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    You've got it switched around. A polarized frame will have more weight at the top and down in the handle, whereas a depolarized frame with have more weight concentrated around the neck and lower part of the hoop.

    A good example of a polarized stock frame is the Youtek Extreme Pro. It's relatively light, has a headlight balance, and yet has a rather high swingweight. A good example of a depolarized frame with similar weight and balance is the Youtek Speed MP 18x20. Although it has the same unstrung weight and similar balance as the Extreme Pro, it has a much lower swingweight.

    You can use the hitting weight comparison tool to see the difference between the two. The Extreme Pro has much more weight around the top and sides of the hoop, wherease the Speed has more weight around the bottom of the hoop.
     
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  3. Veninga

    Veninga Rookie

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    oh, yes i swithed them around.

    but thanks for the answer.

    following your answers, a polarised racket has a higher swingweight and a higher hitting weight at the top (when all other specs are more or less the same).

    The SPEED is more depolarised, the EXTREME more polarised and the RADICAL has a pretty high SW compared with the SPEED, but a normal HITTING weight. So its between.

    Clear. Therefore a dunlop 500t is much more polarised than a 300.

    And is SPEED ELITE more polarised than the SPEED 300.
     
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  4. Veninga

    Veninga Rookie

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    little correction. the radical has almost the same hittingweight as the extreme. based on these stats both are polarised. extreme has a more open stringbed, so probably even more spinny.

    on the twu tools, there is also one called racket speed. this one shows the speed when you hit the ball at the top or in de the center. Ill guess these number and the comparison between top/center, would even be a better stat to look at. because it also take the real static weight into account and makes it up to ball speed when. a more polarised racket should have a higher number than a depolarised one.

    the radical scores really low on this particular number. so does all the wilsons and the speed 300. extreme scores higher, babolats also (with exception of the storm). feels like how it should be.

    do you agree?
     
    #4
  5. Bartelby

    Bartelby G.O.A.T.

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    The prestige pro versus the midplus are good examples of the difference.
     
    #5
  6. olliess

    olliess Semi-Pro

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    There have been a few threads where people have computed various numbers to estimate "polarization" -- they might be worth searching for if you're interested in this topic.

    One way to is to divide the moment of inertia of the racquet (swing weight around the balance point) by the mass of the racquet. This number is bigger for more polarized racquets, since their weight is moved out toward the ends.

    By this metric, the racquets that are polarized vs. non polarized tend to be pretty much the ones you think, although there are a few surprises. Also note that I got all the specs from TW play tests where available, and the published specs on TW otherwise.

    YouTek Prestige MP and Mid are not very polarized (.844 and .848 ). Prestige Pro (.855) is only slightly more polarized than those two.

    BLX 6.1 Tour (0.841) is surprisingly even slightly less than Prestige MP, although it's commonly known that the first thing to be done to the 6.1 Tour is to cram lead under the head guard.

    Head IG Speed MP (.863) is nore polar but still not much.

    BLX 6.1 95 18x20 (.874) a bit more still.

    Babolat Pure Drive Roddick GT Plus is a step up in polarization (.912), although some of this is also an effect of the extra length.

    Aero Pro Drive GT (.934) is way up there.

    But the biggest surprise for me anyway is that the most polarized racquets I've seen are not the APDGT, but the YouTek Radical MP (.939) and Radical Pro (.949).


    For comparison, the Pro Staff 85 (.808 ) is the least polarized that I could find specs for.
     
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  7. Veninga

    Veninga Rookie

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    You mean the pro is more polarised than the mp? Because the racket speed in the top compared to the center is higher?
     
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  8. Veninga

    Veninga Rookie

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    Thanks a lot, makes sense. And yes, some surprises there. For example the radical MP. Can you tell me wich stats u mean and where you have find them (in wich tools)?

    And what is the score for the Dunlop Bio 300?
     
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  9. olliess

    olliess Semi-Pro

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    Oops, sorry, I had a typo in the formula when I got those numbers.

    What you need are the racquet mass (m), balance point (x), and swing weight (SW). Then you can get the moment of inertia (I) using
    SW = I + mr^2 = I + m(x - 10 cm)^2. The 10 cm is just because the balance point is measured from the butt end, but the SW is measured 10 cm from the end.

    Then I just divide out the mass and call the ratio I / m the polarization ratio. Maybe it should just be called polarization number or something, since it isn't really a "ratio." (If you use mass in g and swing weight in kg cm^2, you get even funnier units like I did, but you can just multiply everything by 1000 if you prefer). This is one way to try and measure polarization, but it is by no means the only way.

    The idea of polarized racquets is basically to give you big swing weights (and hence more power/spin) with for a given mass / head heaviness, and you can see that reflected in the numbers here.

    Here is a corrected table (accuracy still not guaranteed!):

    Dunlop Bio 300: 0.4208
    Head IG Speed MP: 0.4342
    Pro Staff 6.0 85: 0.4536
    YouTek Prestige MP: 0.4596
    YouTek Prestige Mid: 0.4640
    Bio 300 Tour: 0.4660
    YouTek Prestige Pro: 0.4712
    YouTek Radical Pro: 0.4726
    YouTek Radical MP: 0.4768
    PDR GT Plus: 0.4932
    BLX 6.1 Tour: 0.4997
    BLX 6.1 95 (18x20): 0.5010
    APD GT: 0.5054
    LM Radical MP: 0.5087
     
    Last edited: Sep 21, 2011
    #9
  10. Veninga

    Veninga Rookie

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    Very nice work. Very useful. Hopefully TWU is reading this too and can give the number in their specs.

    I tried to recalculate with your formula...

    Radical MP.
    - swingweight 324
    - weight 312
    - balance 33,2.

    Can you show us the first steps? The I/M is not difficult.. Thanks
     
    #10
  11. Veninga

    Veninga Rookie

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    first steps to calcuate the I.

    Like sw x mass = outcome a
    outcome a / balance = outcome b
    etc.
     
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  12. olliess

    olliess Semi-Pro

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    From the expression above (post #9):

    SW = I + mr^2 = I + m(x - 10 cm)^2,
    so
    I = SW - m(x - 10 cm)^2.

    For your Radical MP,
    I = 324 kg cm^2 - .312 (33.2 cm - 10 cm)^2
    = 156 kg cm^2.

    I / m = 156 kg cm^2 / 312 g * 1000 g/1 kg = 500. cm^2.
    or same as 0.500 in my funny units of 1000 cm^2 (I didn't multiply by the final 1000).


    BTW, I can't take credit for this, as several others have already shown calculations like this. ;)

    Anyway, this is all fun to play with, but you don't want to get too embroiled in the specs. They're just made up measurements that help explain why some racquet A plays differently from some other racquet B, but they don't help you play!
     
    Last edited: Sep 21, 2011
    #12
  13. Veninga

    Veninga Rookie

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    Thanks a lot!! Makes sense now.
     
    #13
  14. guitarplayer

    guitarplayer Guest

    This thread is polarizing.
     
    #14
  15. Veninga

    Veninga Rookie

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    I must admit some outcomes are really surprising. Speed 300 should be less polarised (because advertised as more spinny) than the radical. Same for Extreme. Hmm, doesnt sound like logic?
     
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  16. Veninga

    Veninga Rookie

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    Can you please depolarize:)?
     
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  17. olliess

    olliess Semi-Pro

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    The physics seem to indicate that a racquet with a bigger swing weight will tend to produce more power and spin, given a similar pre-impact velocity.

    Polarized racquets have more swing weight for given a mass and balance, so they should therefore produce better power and spin.

    I didn't do the calculation for the IG Speed 30, only the 18x20. But the 300 (.454) does come out to be slightly more polarized than the 18x20 (.434).

    On the other hand, string pattern is also a big factor. Open string patterns seem to produce more spin and closed string patterns less. The IG Speed 300 has a 16x19 pattern, which should make most of the difference in spin compared to the IG Speed 18x20.
     
    #17
  18. travlerajm

    travlerajm Hall of Fame

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    I think a better measure of the polarization is Ic/(MR^2), where Ic is the recoil weight (moment of inertia about the balance point), M is the mass, and R is the distance from butt to balance point.

    Ic = SW - M(R - 10)^2, where SW is the swingweight abou the 10cm axis.
     
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  19. Veninga

    Veninga Rookie

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    I = SW - m(x - 10 cm)^2.

    its the same?
     
    #19
  20. travlerajm

    travlerajm Hall of Fame

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    Yes. But divide I by MR^2 instead of M, otherwise you are effectively just measuring the balance rather the the degree of polarization.
     
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  21. olliess

    olliess Semi-Pro

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    That method does yield a dimensionless ratio of polarization. On the other hand, dividing by MR^2 highly weights the result toward headlight racquets (with small R), so you get things like PS85 with about the same polarization as APDGT (which doesn't seem correct).

    Dividing by M gives you a measurement which seems to make more sense given our perception of the various racquets. It also gives a result in the same units as R^2, so you might interpret it as a "radius (squared) of polarization."
     
    #21
  22. Veninga

    Veninga Rookie

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    tried this and worked out. something more like we would expect.

    i tried to divide the result with the number of holes in the stringbed (18*20 or 16*19) and then multiply with 10000 to get 'readible' numbers.

    the results.

    head radical 130
    wilson wilson ktour 119
    prince exo red 127
    babolat pure storm gt team 133
    babolat pure storm gt 138
    babolat pure drive gt 151
    dunlop bio 500t 145
    dunlop bio 300 136
    head speed 300 149
    head extreme mp 137
    head speed elite 123
    prince rebel team 95 104
    techni 295 vo2max 146
    techni 290vo2max 148
    wilson six one team 122
    wilson blx pro open 121

    nadal and technifibre scores pretty high. also the bio 500t. Head 300 more spinny than the radical and so. interesting view, isnt it?
     
    #22
  23. travlerajm

    travlerajm Hall of Fame

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    R must be in the formula or else it breaks down at the extremes. But you could argue that the formula should be I/MR, in which case the formula essentially gives you not only a polarization ratio, but also (the period of the racquet pendulum); that is, a measure of how fast the racquet will come around (high polarization ratio would mean slower swingspeed).
     
    #23
  24. olliess

    olliess Semi-Pro

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    It works for the general range of racquets that I've looked at, but I understand that it may not work for all possibilities. I think it's at least sort of reasonable, since the moment of inertia has an R^2 dependency. But if you have an example of a racquet where it clearly "doesn't" work then I'd be interested in examining how it breaks down.

    It doesn't seem like an answer in units of length is quite right either, since the pendulum period is dependent on the square root of length.

    Also, the choice of the 10 cm point as the SW axis is pretty arbitrary, but I don't exactly swing the racquet around the butt end either.

    I think a Rod Cross article on the topic of swing weight and polarization is in order!
     
    #24
  25. Torres

    Torres Banned

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    This thread has gone very, very weird in parts.....

    Anyway getting back to the original question, a PB10 would be an example of a heavily polarized racquet.

    A Dunlop 200 Lite is a example of a heavily depolorized racquet.

    Despite the egghead figures posted above, I wouldn't say that the BLX 6.1 95 18x20 feels particularly polarized. Slightly polarized would be my view.

    I'd also suggest that the weight of the racquet has an influence on how 'polarized' a racquet feels. A heavily polarized 260g stick isn't really going to feel 'polarized' in the way that a PB10 does because there's simply not enough weight at either end to give the stick that 'baton twirling' feel.
     
    Last edited: Sep 21, 2011
    #25
  26. olliess

    olliess Semi-Pro

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    Any point of comparison for the 'eggheads' among us? For example, do you feel it is close to or much less polarized than the 6.1 Tour?

    That's may be true, but it's a bit like saying a racquet might not feel that 'head heavy' if isn't all that heavy. There's still a well-defined number 'points head light/head heavy' which describes the balance of the racquet and not the absolute mass of the head.
     
    #26
  27. Veninga

    Veninga Rookie

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    guys, i tried to combine these figures with the string pattern. Actually i divided the polarisation number by the sring pattern (16*19 etc). Very interesting and logic results:

    head radical 130
    wilson wilson ktour 119
    prince exo red 127
    babolat pure storm gt team 133
    babolat pure storm gt 138
    babolat pure drive gt 151
    dunlop bio 500t 145
    dunlop bio 300 136
    head speed 300 149
    head extreme mp 137
    head speed elite 123
    prince rebel team 95 104
    techni 295 vo2max 146
    techni 290vo2max 148
    wilson six one team 122
    wilson blx pro open 121
     
    #27
  28. Veninga

    Veninga Rookie

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    a great number to compare spinnyness of rackets based on polarisation and string pattern!
     
    #28
  29. travlerajm

    travlerajm Hall of Fame

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    Ok guys. We're all wrong.

    Ic/M doesn't work. And neither does Ic/MR or Ic/MR^2. It can easily be proved that all three of these formulas don't work. For example, if you add weight at 5" or 22" from the butt, it makes a racquet more polarized, but the Ic/M formula suggests it does the opposite.

    So here's the deal: We need to use a formula that reasonably approximates a baseline recoil weight based on the weight and balance. For this, I suggest using the formula from Cross R, "Customising a tennis racket by adding weights," Sports Engineering, 4:1-14, 2001. It assumes that a racket is divided into two beams of equal length but unequal mass.

    The formula for baseline recoil weight = Ic' = (1/12)*(1+7M)RL - MR^2

    Now the polarization formula = the recoil weight divided by the baseline recoil weight = Ic/Ic'

    This formula won't break down, no matter where you add the weight.
     
    Last edited: Sep 23, 2011
    #29
  30. Veninga

    Veninga Rookie

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    The formula for baseline recoil weight is standing above.

    But what is the formula for recoil weight?
     
    #30
  31. olliess

    olliess Semi-Pro

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    Hmm.. But to retain the same balance point, e.g., an even balance, you need to add equal weight at both 5 and 22.

    Imagine that the racquet's total mass is divided into two "virtual" weights, each positioned a distance r from the balance point so that the moment of inertia Ic (the recoil weight) is exactly correct.

    Ic/M tells you that distance r (well, r^2 anyway).

    If you add more pairs of weights to the racquet (so that the balance point remains the same), then these weights have to be further than r from the balance point to make the racquet more polarized. If so, then Ic/M will be bigger than it was before.

    So, at least in this example, Ic/M still doesn't seem to break down.

    I'll think about the 'base recoil weight' some more -- that sounds like it could be useful as well.
     
    #31
  32. pvaudio

    pvaudio Legend

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    This is getting too complicated for its own good.
     
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  33. Veninga

    Veninga Rookie

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    i am getting lost as well a bit..

    can someone make a more or less usefull formula:)?

    on twu they something like a spin window and spinning weight. doesnt say anything about the polarisation of course, but more of it works out in reality.

    is that one useful?
     
    #33
  34. olliess

    olliess Semi-Pro

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    It's really simple. You can think of a tennis racquet like a pair of weights which spin around its balance point, like a weighted baton or a pair of bolas or something. The wider spacing you have been the weights, the more "polar" the racquet seems. That's the number I've been computing. There are other possible ways to think of "polar," some of which travlerajm has brought up, so that's what we're talking about.

    Is this much more complicated than the arcana of subtle differences between differently constructed strings?
     
    #34
  35. Veninga

    Veninga Rookie

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    @olliess as you put is, its not that difficult.

    but how do we get this into usefull numbers? like swingweight or static weight. we all want to figure out which rackets suits which players. At least i do:)

    i am looking for a spinny one, but not too heavy. So i was looking at polarised rackets.

    Does the spinweight of TWU does say somethin useful?
     
    #35
  36. julian

    julian Hall of Fame

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    What is Ic?

    What is Ic?
     
    #36
  37. olliess

    olliess Semi-Pro

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    Ic is the moment of inertia of the racquet. I had called it I earlier, but travlerjajm uses Ic to indicate that this is the moment around the center of mass (the "balance point") of the racquet. A formula for getting I from the swingweight (SW) is shown in post #12 of this thread.
     
    #37
  38. akamc

    akamc New User

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    For me, the most descriptive and useful ratio of polarization would be:

    Racquet swingweight (as defined by the industry) divided by the swingweight of a uniform rod of equal mass around the same point (10cm from the buttcap).

    So, a uniform rod would have a polarization ratio of 1, a "polarized" racquet a ratio > 1, and a "depolarized" racquet a ratio < 1 . Simple enough ?
     
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  39. travlerajm

    travlerajm Hall of Fame

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    Darn it. Just realized I screwed up the algebra somewhere, it's still not quite right. I'll fix this for you guys when I have a little time to scribble the math carefully this weekend.
     
    #39
  40. olliess

    olliess Semi-Pro

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    Sounds good but it has a minor problem: a head heavy racquet will have a higher swingweight than a head light racquet, and therefore a higher 'polarization ratio,' even if they have essentially the same mass distribution, just shifted toward the tip.
     
    #40
  41. akamc

    akamc New User

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    I see what you are saying, but that is exactly the point: a racquet is not symmetrical, we hold it at and swing it from one end (estimated to be 10 cm from the end by industry convention) and not at the geometric center. The two examples you give do not have the same mass distribution relative to that swing point, instead they are more like mirror images, so actually the head heavy one is really more polarized from that point of view.
     
    #41
  42. Veninga

    Veninga Rookie

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    Yes this is simple. Is it logic as well? And do you know how to get the numbers of TWU?
     
    #42
  43. julian

    julian Hall of Fame

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    Multiple problems with this thread

    Hi,
    there are THREE following cases:
    1.lead at both 3 pm and 9 pm
    2.lead at 12 pm
    3.silicon in a handle

    Which one of these THREE your formula should address ?
    I understand that an answer is "YES" for #2 and #3.
    Am I correct?
     
    #43
  44. travlerajm

    travlerajm Hall of Fame

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    Deleted post
     
    Last edited: Sep 24, 2011
    #44
  45. julian

    julian Hall of Fame

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    Please specify whether L is a length of a racket

    Please specify whether L is a length of a racket
    Please explain why Ic' is in a denominator,not in a numerator
     
    #45
  46. olliess

    olliess Semi-Pro

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    That would be fine too, but you can already get that information by just comparing the swingweight to the mass. If you need a single number, you can divide swingweight (in kg cm^-2) by the mass (in grams); multiply by 1.1 if you want a uniform rod of the same mass and length to come out as exactly one. (However, I haven't found any racquets that come out to be less than one).

    Examples:
    Pro Staff 6.0 85:
    swingweight/mass * 1.1 = 329/357 * 1.1 = 1.01.

    AeroPro Drive GT:
    swingweight/mass * 1.1 = 331/320 * 1.1 = 1.14.
     
    #46
  47. travlerajm

    travlerajm Hall of Fame

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    Checked my math and fixed the formula:

    Polarization Index = Ic/Ic' = [Recoil Weight]/[Baseline Recoil Weight] = [SW - M(R - 10)^2]/[(M/6)(6RL - L^2) - MR^2]
    SW = swingweight about 10cm axis in kg-cm^2
    M = mass in kg
    R = balance in cm
    L = length in cm
    Ic' = Baseline Recoil Weight weight estimated from M,R, & L using 2-equal-length beam approximation.

    Again, it is easy to show that the Ic/M formula cannot be correct because it breaks down at extremes (although it can be useful as a first approximation). For example, if you take a frame with specs of 11oz., 13" balance, and 325 Sw, and then add 10oz. at 1" from butt, it becomes extremely polarized, and polarization index almost triples from 1.3 to 3.5. But the Ic/M formula decreases from 512 to 500. I have actually tried adding several ounces to the butt of such a frame, and the result is a frame that feels and plays extremely polarized -- it is almost impossible to hit a flat shot because the dwell time is so long. The impact is extremely soft, and extreme spins come naturally. I loved serving with the racquet because of the extreme weird kicks possible, but the extremely long dwell time means volleys are not very accurate too.
     
    Last edited: Sep 24, 2011
    #47
  48. olliess

    olliess Semi-Pro

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    travlerajm,

    I'm not convinced this example demonstrates a "breakdown" of the formula. Ic/M gives a quantity which has units of length^2 and has a clear physical interpretation: it is the radius^2 of the equivalent system of paired weights which has the same center of mass and moment of inertia as the racquet.

    Let's call the square root of this number the polar radius r' for the moment. The racquet in your example has an r'^2 = I/m = 512 cm^2, or r' = 26.3 cm. This racquet behaves (as far as the measurements we have available) as a pair of half-masses placed at +/- 26.3 cm from the center of mass at 13" = 33.0 cm (the racquet is 4 pts. headlight), or at 59.3 cm and 6.7 cm from the butt.

    By adding a single 10 oz. lead weight (almost as massive as the original racquet) at 1" = 2.54 cm from the butt, you will drastically shift the center of mass. The new balance point will be at 7.28" = 18.5 cm (the racquet will be about 45 points headlight!). As a result, the weight near the butt is only 16.0 cm from the new balance point (compare this to the original configuration, where all the mass was 26.3 cm from the balance point).

    The new I/m = r'^2 is indeed 500 cm^2, which is slightly smaller than the original number of 512 cm^2. Weighted up, the racquet is now the equivalent of two weights which are almost twice as large, but slightly closer together, than in the original configuration. I would therefore argue that the racquet is actually less polarized after the addition of the weight.

    Addendum: If, on the other hand, you add 10 oz. of weight by placing 5 oz. at 1" from the butt and the other 5 oz. as a counterbalance at 25" (so that the balance point is unchanged), then I agree that this racquet will be super-polarized, and indeed for this configuration, r'^2 = I/m = 711 cm^2, which is much larger than the original number of 512 cm^2.
     
    Last edited: Sep 24, 2011
    #48
  49. akamc

    akamc New User

    Joined:
    Mar 20, 2005
    Messages:
    91
    The logic is this: what should we define as neutrally polarized? a uniform rod seems like a good absolute reference (i.e. a theoretical racquet where mass is evenly distributed along its length).

    Dividing an actual racquet swingweight as measured on RDC by the reference swingweight of a uniform rod of equal mass around the same conventionally accepted axis of rotation (10 cm from the grip end) gives you a truly dimensionless ratio which gives you a useful indication of the relative overall distribution of mass about that axis of rotation.

    This is pretty easy to calculate:

    Polarization Index = SW / SW'

    where SW = swingweight of racquet such as provided by TW, in kg-cm2
    SW' = swingweight of uniform rod of equal mass about its center, adjusted using the parallel axis theorem
    = M*L^2/12 + M*(BP-10)^2

    where M = mass of racquet, in kg
    L = Length of racquet, in cm
    BP = Balance point from the buttcap, in cm

    Plug in a few data points from your favorite racquets to try it. It turns out that all racquets are somewhat polarized by this definition, it's just a question of degree. Mine range from 106% to 116% with an average of 110% (which is fairly typical). So all racquets are slightly more like dumbells than uniform beams, but with pro racquets being more extremely so.
     
    #49
  50. olliess

    olliess Semi-Pro

    Joined:
    Nov 25, 2009
    Messages:
    572
    This is interesting; I see now that you are centering the uniform rod around the balance point.

    But does that mean that a 8 pt. (1") head-heavy 27" racquet is being compared against a uniform rod which extends from 1" to 28" from the buttcap position, and a 1" head-light racquet is being compared against a rod which is positioned at -1" to 26"? That shift of the tip should matter a lot due to the r^2 dependence, right?

    Do you find that head light racquets tend to come out rather polarized and head heavy racquets less polarized by your measure? Could you post numbers for a couple popular racquets that could be compared to their "perceived" polarization?
     
    #50

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