#### Greg Raven

##### Semi-Pro

http://www.hdtennis.com/grs/racquet_profiles.html

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- Thread starter Greg Raven
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http://www.hdtennis.com/grs/racquet_profiles.html

Greg Raven said:

http://www.hdtennis.com/grs/racquet_profiles.html

Interesting approach. However, I don't think measured curves of swingweight vs. pivot point will be quite as smooth as those calculated.

How do auto manufacturers measure weight distribution in performance cars, when they're striving for 50/50? There's got to be a way to do this, other than laboriously weighing rackets balanced at various points along its length.

What you are after is very interesting, I think.Greg Raven said:

http://www.hdtennis.com/grs/racquet_profiles.html

Let me try put the main question you are posing into mathematical perspective.

As known, the Swingweight of a racquet measured around axis of rotation "A" centimeters from butt, is given by

Where "x" is distance from butt, and "m(x)" denotes the mass located at x centimeters from butt (the mass density of the racquet, which is what you are after). SW is given as a function of the axis of rotation it is measured around.

What you are trying to answer is the following: If I know the function SW(A), and I do (this is what you plot using the parallel axis theorem & the published swingweight), can I solve for the function "m(x)"? In other words, does the above equation have a UNIQUE solution in m(x)? And if so, what is "m(x)"?

From what I gather, the solution is not straightforward. I just learned that the above equation is what mathematicians call a "Fredholm equation of the first kind". If you search for it, you'll find that the solution is known for very similar functions to the above, but I couldn't (yet) find a solution that is applicable to the above. I'll work on it a bit, since it would be great to be able to say something about the mass distribution of a frame (without cutting it to 1000 pieces, of course).

Also I am hoping more math buffs would jump in to say more (any math Ph.Ds around?).

ohplease said:Interesting approach. However, I don't think measured curves of swingweight vs. pivot point will be quite as smooth as those calculated.

How do auto manufacturers measure weight distribution in performance cars, when they're striving for 50/50? There's got to be a way to do this, other than laboriously weighing rackets balanced at various points along its length.

Actually, Crawford Lindsey and I discussed this project some months ago, and we concluded that the curves would be smooth and not very revealing. That seems to be what we got. In the formula, the distance from the balance point to the axis is adjusted as the axis point is changed, which lead us to expect a smooth curve.

As for weight distribution, other than a couple of Porsches (the 914 with a lower polar moment and the 928 with a high polar moment), there are not many cars with a 50/50 weight distribution. It's difficult and expensive to achieve. Also, while auto manufacturers do pay some attention to weight distribution, the calculation of the roll center is much more important, IIRC.

adely said:Try adding lead to see if your formula works!

That's the next step. I've got a couple of frame-only racquets to which I can attach massive amounts of mass to see what happens.

katastrof said:[...]From what I gather, the solution is not straightforward. I just learned that the above equation is what mathematicians call a "Fredholm equation of the first kind".[...]

From my understanding of the physics involved, the reduced-down version* of the Parallel Axis Theorem should work in a fairly straightforward manner for this analysis. I'm just not certain that this approach will yield the information I was hoping.

*By "reduced down," I mean that the standard formula for the Parallel Axis Theorem has been mathematically simplified for the specific task at hand, as shown on Wilmot McCutchen's Racquet Research website.

Oh Greg you shouldn't have!Greg Raven said:

http://www.hdtennis.com/grs/racquet_profiles.html

Greg Raven said:Actually, Crawford Lindsey and I discussed this project some months ago, and we concluded that the curves would be smooth and not very revealing. That seems to be what we got. In the formula, the distance from the balance point to the axis is adjusted as the axis point is changed, which lead us to expect a smooth curve.

As for weight distribution, other than a couple of Porsches (the 914 with a lower polar moment and the 928 with a high polar moment), there are not many cars with a 50/50 weight distribution. It's difficult and expensive to achieve. Also, while auto manufacturers do pay some attention to weight distribution, the calculation of the roll center is much more important, IIRC.

Actually, it just occured to me that graphing hitting weight vs. length would pretty much exactly tell us the mass distribution for any particular frame. In my recollection, that formula (which I can't recall off the top of my head) accounts for swingweight (at 10cm), balance, and place on the racket face at which one makes contact. That contact point is expressed in my lead tape calculator as center of precussion, but if we instead treat that as a continuous variable along a racket's entire length, we should get the mass curve.

I'll see if I can't punch out a few graphs if I have some time.

Greg Raven said:

Just doing it all quick and dirty like, where swingweight was held constant (which means recoil weight was held constant) and I calculated the hitting weight in 5 cm increments. Then, I normalized the hitting weights as a percentage of total mass - which in turn tells us what percentage of mass is in each 5 cm segment for each racket.

After all that, I take an average percentage for each segment across all rackets. Think of this as the typical weight distribution for rackets being sold today. Finally, I run a correlation coefficient on the series, which gives me the top 10 rackets nearest this typical modern weight distribution:

pure control zylon plus

lm radical mp

catapult v1 mp

dnx 8

o3 white

warrior os

lm instinct

ki15 pse

rq 7

fxp radical os

I'm not sure if swingweight should vary or not, as rackets are pretty much meant to be held in one place. But still - that's a nicely balanced collection of frames. Beginners to advanced should be able to pick up any one of those and have some fun.

Now, what if do the same thing, but this time we want the correlation coefficient relative to the prototypical player's frame - the PS 6.0 85? Well then:

tour 10 mid v

ti-80

pro x1

c10 pro

nxg mp

rdx 500 mp

rdx 500 mid

fxp radical tour

rdx 500 hd

lm prestige mp

No surprises there. Ah, but what if we do the same thing with the nouveau classic Pure Drive?

head classic tour

aero tour

aero tour plus

warrior mp

aeropro drive

hotmelt 200g

i.x3 mp

ki15 pse

mp tour-1 90

tour 9 v 18x20

Now that is a strange list. But the mass distribution curves seem to indicate that these frames tend to have more mass in the middle, around the throat or shoulders.

What about Kaptain Karl and his preferred ProKennex 7g?

tour 95

ki5x

o3 tour

lm prestige mid

ki5

m speed 105

m speed 98

tour 10 mp v

dnx 10 mp

core 1 no. 6 (w/bumper)

Some interesting results. I'm not sure it's at all definitive, however. As an experiment, someone can always suggest a racket they're very familiar with, and I'll come back with what correlates according to this method. Maybe the results will make sense, maybe not.

http://www.hdtennis.com/grs/racquet_profiles.html

When I have some time, I'll take a frame, measure it, add mass to it, re-measure, and then see if the location of the additional mass appears on one of the charts. At this point, I'm less than confident that it will.

It appears that the only ways to know if there is a concentration of mass in a racquet frame are: 1) lay up the frame yourself, 2) add a concentration of mass to the frame, and 3) section the frame, weigh the pieces, and compare against an average mass for such a section of that frame.

Greg Raven said:

http://www.hdtennis.com/grs/racquet_profiles.html

When I have some time, I'll take a frame, measure it, add mass to it, re-measure, and then see if the location of the additional mass appears on one of the charts. At this point, I'm less than confident that it will.

It appears that the only ways to know if there is a concentration of mass in a racquet frame are: 1) lay up the frame yourself, 2) add a concentration of mass to the frame, and 3) section the frame, weigh the pieces, and compare against an average mass for such a section of that frame.

To clarify: the averages part was just to figure out the typical hitting weight as a percentage of total racket weight at a given location - across all segments, across a sample of rackets. For example, at the tip segment, the average is about 25%, give or take, slightly more 5 cm down, 100% at the balance point. That series (25, 28, 30, 35, 50, 100 vs. 23, 30, 33, 40, 60, 100 - for example) is somewhat unique to each racket model. What I'm doing is seeing which curves are most similar to one another.

I think you're asking a slightly different location - i.e. where are mass concentrations, if any?

I'm saying if there's a frame out there that I like, what other frames have mass distributions (or at least, hitting weight distributions) that are somewhat similar? The approach is providing some food for thought, at the very least.

Consider another example - someone mentioned the nSixOne 95 vs. the PS 6.0 95. At least according to this method, there does appear to be a slight movement of mass towards the tip and handle in the nCode. How much does that matter? I don't know - but it'd be interesting if anyone has noted such a difference in a side-by-side playtest.

I did it by first applying the parallel axis theorem to get I(x), just like Greg did. Then I generated another curve for I(x) by integrating the swingweight contributions of the 27 length segments, using constant m(x) = 1/27 of the total mass as a starting point. Next I found the sum of the squares of the differences between I(x) for the 2 curves. By iteratively adjusting the weight of each of the mass segments until the sum of the squares of the differences was minimized, I was able to solve for m(x). The result was interesting.

These results are given as normalized weights, with 1/27 of the total mass = 1:

The mass of the first segment that contained the buttcap = ~2.3

Each segment in the rest of the handle = ~1.3

The mass of the throat segments = ~0.6

The mass of the segments in the bottom of the hoop = ~1.0

The mass of the last segment at the top of the hoop with the bumper guard = ~1.3

The mass of the segments in the middle of the hoop follow a smooth trough-shaped curve, dipping to a minimum = ~0.8

I haven't quite finished the iterations, so these are approximate numbers for now. I'll post the exact solution for all segments tomorrow when I get a little time to finish the process.

I did the iterations manually, but it wouldn't be that difficult to write a macro that automates the iteration loops. And we could use smaller segments then too. That way we could use this method to solve for the exact mass distribution of any frame, as long as we know the weight, balance, and swingweight. I'm not sure whether there is only one unique solution, but the solution is starting to take a shape that makes sense based on the racquet geometry. I'm not a skilled programmer, but I hope someone else is interested in trying this. Ohplease? Greg?

Amone said:

You're on the right track. But increased polarization means more recoil, not less.

travlerajm said:[...] I'm not a skilled programmer, but I hope someone else is interested in trying this. Ohplease? Greg?

I'm not a programmer, either. I do just about everything in Excel. Also, I haven't touched calculus since I "learned" it in high school, about 35 years ago.

If you send me the (simplified) formulae, I might be able to create a spreadsheet.

Here's the numerically solved result for the weight distribution of the PS 85. The first column numbers are x in inches, where x is the center of mass of each element. The second column are the normalized weights of each element, where 1/27 of the total mass = 1. What I find interesting is that the weight density design in the hoop appears to be varied to give constant weight distribution as a function of x, in contrast to the POG OS, where the hoop appears to have constant density all throughout the hoop.

0.5 ..... 1.9525

1.5 ..... 1.270

2.5 ..... 1.271

3.5 ..... 1.258

4.5 ..... 1.260

5.5 ..... 1.249

6.5 ..... 1.210

7.5 ..... 0.910

8.5 ..... 0.900

9.5 ..... 0.827

10.5 ... 0.789

11.5 ... 0.630

12.5 ... 0.569

13.5 ... 0.520

14.5 ... 0.912

15.5 ... 0.945

16.5 ... 0.946

17.5 ... 0.949

18.5 ... 0.949

19.5 ... 0.951

20.5 ... 0.951

21.5 ... 0.952

22.5 ... 0.951

23.5 ... 0.950

24.5 ... 0.950

25.5 ... 0.960

26.5 ... 1.0200

Would it be possible for you to send me a copy of your Excel file?

I also find it interesting that the tip of the hoop doesn't register higher in this profile.

Greg Raven said:I also find it interesting that the tip of the hoop doesn't register higher in this profile.

Actually, I finishing solving for the POG OS, and it turns out that the main difference between the POG OS and the PS 85 is that the POG has more weight at the tip and more weight in the buttcap area.

Greg Raven said:

Would it be possible for you to send me a copy of your Excel file?

Greg,

If you can provide an e-mail address that I can send it to, I'll send you the file. Once you have the spreadsheet, you can change the spec inputs and use it to solve for any frame. I wasn't able to use the contact feature on your website without knowing the e-mail address.

As for the PWS, it may possibly show up if I solve it to further accuracy, but it's safe to say that the weight there is not very significant. The result I posted for the PS was solved to delta = 0.83, where delta is the sum of the squares of the differences. I have now solved the POG to delta = 0.017.

Also, it's possible (and likely) that there is more than one unique solution to the weight distribution when solved this way. I believe that if I divided the racquet into more elements (instead of only 27), it would reduce the number of alternate solutions, and make it easier to see small weight concentrations like the PWS.

Edit: I edited the result above for the PS85, now solved to delta = 0.001. And reposted below compared side-by-side to POG with delta = 0.001.

Greg Raven said:Travlerajm,

You can use either:

raven at corax dot org

or:

greg at racquettech dot com

I look forward to hearing from you.

Hi Greg. I just sent you the files.

0.5 ..... 1.9525 .... 2.3178

1.5 ..... 1.270 ...... 1.341

2.5 ..... 1.271 ...... 1.311

3.5 ..... 1.258 ...... 1.230

4.5 ..... 1.260 ...... 1.230

5.5 ..... 1.249 ...... 1.210

6.5 ..... 1.210 ...... 1.190

7.5 ..... 0.910 ...... 1.131

8.5 ..... 0.900 ...... 0.861

9.5 ..... 0.827 ...... 0.840

10.5 ... 0.789 ...... 0.610

11.5 ... 0.630 ...... 0.480

12.5 ... 0.569 ...... 0.478

13.5 ... 0.520 ...... 0.4685

14.5 ... 0.912 ...... 0.930

15.5 ... 0.945 ...... 0.916

16.5 ... 0.946 ...... 0.911

17.5 ... 0.949 ...... 0.911

18.5 ... 0.949 ...... 0.907

19.5 ... 0.951 ...... 0.897

20.5 ... 0.951 ...... 0.883

21.5 ... 0.952 ...... 0.864

22.5 ... 0.951 ...... 0.851

23.5 ... 0.950 ...... 0.848

24.5 ... 0.950 ...... 0.930

25.5 ... 0.960 ...... 1.0614

26.5 ... 1.0200 .... 1.3934

travlerajm said:Here are side-by-side numerically solved normalized weight distributions for the POG OS and PS85. Note that the POG has more weight at the butt and the tip, while the PS has more weight in the mid-handle, throat, and sides of the hoop.

Interesting thread and good comparison, but are these numbers for strung or unstrung racquets ?

travlerajm said:

x ..... PS85 m(x) ... POG OS m(x)

0.5 ..... 1.9525 .... 2.3178

1.5 ..... 1.270 ...... 1.341

2.5 ..... 1.271 ...... 1.311

3.5 ..... 1.258 ...... 1.230

4.5 ..... 1.260 ...... 1.230

5.5 ..... 1.249 ...... 1.210

6.5 ..... 1.210 ...... 1.190

7.5 ..... 0.910 ...... 1.131

8.5 ..... 0.900 ...... 0.861

9.5 ..... 0.827 ...... 0.840

10.5 ... 0.789 ...... 0.610

11.5 ... 0.630 ...... 0.480

12.5 ... 0.569 ...... 0.478

13.5 ... 0.520 ...... 0.4685

14.5 ... 0.912 ...... 0.930

15.5 ... 0.945 ...... 0.916

16.5 ... 0.946 ...... 0.911

17.5 ... 0.949 ...... 0.911

18.5 ... 0.949 ...... 0.907

19.5 ... 0.951 ...... 0.897

20.5 ... 0.951 ...... 0.883

21.5 ... 0.952 ...... 0.864

22.5 ... 0.951 ...... 0.851

23.5 ... 0.950 ...... 0.848

24.5 ... 0.950 ...... 0.930

25.5 ... 0.960 ...... 1.0614

26.5 ... 1.0200 .... 1.3934

Good job, travler. Well done.

Btw, how come you come up with these numbers MANUALLY? Sounds like a lot of effort to me.

katastrof said:Good job, travler. Well done.

Btw, how come you come up with these numbers MANUALLY? Sounds like a lot of effort to me.

Ultimately, I'd like a to write a program that automates the process. Unfortunately, it's been a long time since I've done any programming. Of course, with the spreadsheet, it's partially automated already.

akamc said:Interesting thread and good comparison, but are these numbers for strung or unstrung racquets ?

The numbers are for strung TW specs.

travlerajm said:

x ..... PS85 m(x) ... POG OS m(x)

0.5 ..... 1.9525 .... 2.3178

1.5 ..... 1.270 ...... 1.341

2.5 ..... 1.271 ...... 1.311

3.5 ..... 1.258 ...... 1.230

4.5 ..... 1.260 ...... 1.230

5.5 ..... 1.249 ...... 1.210

6.5 ..... 1.210 ...... 1.190

7.5 ..... 0.910 ...... 1.131

8.5 ..... 0.900 ...... 0.861

9.5 ..... 0.827 ...... 0.840

10.5 ... 0.789 ...... 0.610

11.5 ... 0.630 ...... 0.480

12.5 ... 0.569 ...... 0.478

13.5 ... 0.520 ...... 0.4685

14.5 ... 0.912 ...... 0.930

15.5 ... 0.945 ...... 0.916

16.5 ... 0.946 ...... 0.911

17.5 ... 0.949 ...... 0.911

18.5 ... 0.949 ...... 0.907

19.5 ... 0.951 ...... 0.897

20.5 ... 0.951 ...... 0.883

21.5 ... 0.952 ...... 0.864

22.5 ... 0.951 ...... 0.851

23.5 ... 0.950 ...... 0.848

24.5 ... 0.950 ...... 0.930

25.5 ... 0.960 ...... 1.0614

26.5 ... 1.0200 .... 1.3934

Using the hitting weight as a function of length approach discussed earlier, I'm seeing that this indeed mostly correct. Those numbers seem to indicate that relative to the POG 110, the PS 6.0 85 has more of its hitting weight/length concentrated from about 12-16 inches up from the butt - around the shoulders and lower head

I started with the original equation, I(a)=ABS(I(10)+M*[2*(10-a)*r-(10-a)^2])

I plugged in the data for my DNX10, to get some real numbers.

I then modified it, by changing M to be M/137 (as I was using .5 cm increments) and solving for M. This resulted in M(seg)=ABS(135*[(I(a)-I(10))/(2*[10-a]*r-[10-a]^2)])

where M(seg) is the weight of each segment.

The numbers totalled to 4059.509, so they were obviously off. However, they seemed to have a pattern to them: a pattern that apex'd about 1 cm high of the balance point. This made me question, why not just make it a ratio?? So I did.

M(g)=100*[M(seg)*M/Sum(M(seg))]

where Sum(M(seg))=4059.509, and M(g) is the true mass of the segment in grams.

That made some sense! They all added up to exactly what they should: 340g.

Then, for some added flair, I made another ratio to get the %age of each .5 cm segment, using the same format as before:

M(%)=M(seg)*100/SUM(M(seg))

M(%) is the percentage of the distrobution in each place.

I realise that I made a number of those variables up, so if you would like I could send anyone who wants it a copy of the Excel book I did this work in.

(I'll save you 135 sets of numbers from posting here)

To wrap up any confusions:

M was in kg, r in cm, a in cm, otherwise the numbers always seem to come out funny. Watching your units helps.

However, this does suffer from the same smootheness as the original equation, so who knows.

It seems to me just a bilinear way to guesstimate the distribution.

Greg, don't you realize that you have to dissect every inch of the frame to really do this?Greg Raven said:If these calculations are correct, it doesn't say much for the perimeter weighting system, huh?

One of the posters on the Talk Tennis bulletin board (fora generously provided by Tennis Warehouse), made a great deal of explaining racquet differences in part by stating that such and such a racquet had more mass at the bottom of the hoop, and that other racquets had concentrations of mass elsewhere. These comments seemed very odd to me, not the least because I doubted that the poster had any idea what he was talking about, and also because it seemed that the only way to know about concentrations of mass in a racquet would be to section the racquet and weigh each section -- a procedure that would require sectioning at least two of each racquet, to avoid missed data points due to the mass lost in the sectioning process.

However, it's intriguing to think that perhaps one could get a fairly accurate approximation of mass distribution without having to destroy two racquets.

I'm also looking into other ways that I can add another constraint to reduce the degrees of freedom without having to cut off a piece of the racquet and weigh it.

As such, you can either compare the trend (which would in this case be a relative concentration in hitting mass along the length - regardless of mass or swingweight) or compare the degree of similarity or distance between the two vectors (which would attempt to "match" two frames' specs). For example, consider the pure drive, again.

These rackets are similar in trend (note differences in mass, swingweight, balance):

classic tour

aero tour

pure control mp

pro tour

diablo xp mp

aero tour plus

m-fil 200 plus

warrior mp

aeropro drive

hotmelt 200g

These rackets are similar in distance (note tight ranges in specs):

pro tour

laver type s

tour 9 v 18x20

laver type sx

pure control zylon plus

rdx 300 mp

o3 white

nPro Surge

dnx 8

nPro

In the end, however, I'm beginning to think that weight distribution is nice to see (I might still build an applet), but I'm still not convinced that it's all that useful. Consider the flexpoint prestige mp, DNX 10 mp, & mSpeed 98. Very similar specs, trends, distances - and yet, I bet people could easily tell the difference in a side-by-side play test.

As I suspected, there are infinite solutions. And unfortunately, using more segments does not constrain the solution. So additional constraints would be required to find a unique solution.

But I wonder, can we find a creative way to take additional measurements without dissecting the racquet? I have an idea that I'm working on...

ohplease said:[...]In the end, however, I'm beginning to think that weight distribution is nice to see (I might still build an applet), but I'm still not convinced that it's all that useful. Consider the flexpoint prestige mp, DNX 10 mp, & mSpeed 98. Very similar specs, trends, distances - and yet, I bet people could easily tell the difference in a side-by-side play test.

This could be from different mass distribution, but more likely it would be because of differences in the layup.

The Voelkl uses Graphite (with some DNX mutation), and Fibreglass, flexing in at 63.

The Head uses LiquidMetal Titanium and Graphite, Flexing at 65.

The Fischer uses Fibreglass, Graphite, and Piezotech fibre (ala Intelligence series) and a flex of 59.

There's a number of big differences there, and it's no mystery that they'd be different feeling. But the idea is that they probably have similar characteristics, though I'm hard pressed to figure out what they are. I'm just a talkative student in this post, y'know?

Amone said:

The Voelkl uses Graphite (with some DNX mutation), and Fibreglass, flexing in at 63.

The Head uses LiquidMetal Titanium and Graphite, Flexing at 65.

The Fischer uses Fibreglass, Graphite, and Piezotech fibre (ala Intelligence series) and a flex of 59.

There's a number of big differences there, and it's no mystery that they'd be different feeling. But the idea is that they probably have similar characteristics, though I'm hard pressed to figure out what they are. I'm just a talkative student in this post, y'know?

Ah, but just as mass distribution can vary along a racket's length - so can flex. personally , I wouldn't call those big differences, but I could see how someone else might.

What do you mean, Traveler?travlerajm said:

Bolt said:travlerajm, could you send your spreadsheet to my email address (scotthigg at yahoo dot com) so I can follow along?

I just sent you the latest version.

Amone said:What do you mean, Traveler?

Additional mass distribution information could be obtained by balancing about different axes, instead of the axis that is perpendicular to the longitudinal axis.

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