Mathematical racquet profiling

will1087

Rookie
Great website, very informative.

However, i might recommend posting a location on your website (if you have, i apologize for being blind) and how shipping/handling is handled for people who dont live close enough to visit.
 

anirut

Legend
Greg, nice one. I'm sure travler will jump in with you.

... And I'll keep my weather eye open ...
 

ohplease

Professional
Greg Raven said:
I've attempted to use published racquet specs and a bit of math to develop visual profiles for tennis racquets, with mixed results. I cordially invite mathematicians and physicists to visit my results page and let me know your thoughts.

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.
 

katastrof

Rookie
Greg Raven said:
I've attempted to use published racquet specs and a bit of math to develop visual profiles for tennis racquets, with mixed results. I cordially invite mathematicians and physicists to visit my results page and let me know your thoughts.

http://www.hdtennis.com/grs/racquet_profiles.html
What you are after is very interesting, I think.

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

SW2.JPG


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?).
 

travlerajm

Talk Tennis Guru
I routinely use numerical solutions to this problem in order to match weight distributions for different starting frames. But I haven't attempted an analytical approach. I'll take a look at it when I have a chance.
 

Greg Raven

Semi-Pro
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.
 

Greg Raven

Semi-Pro
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.
 

Greg Raven

Semi-Pro
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.
 

ohplease

Professional
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

Semi-Pro
Wow, that'll be fun. You'll have to calculate COP, and recoil weight for each point of interest, as part of calculating hitting weight. I've got too much to do right now to bang it out, but let me know if you get bogged down -- I'll take a swing at it.
 

oldguysrule

Semi-Pro
It seems there are racquets that have roughly the same static weight and same balance point but different swing weights. Is this the type of difference that you are trying to analyze? I would be curious to see what the curves would look like from these two racquets. An example would be the n6.1 95 and the 6.0 95. Weight = 12.2/12.3, balance = 10/10, SW = 330/317. (These are from memory so might be off a bit).
 

ohplease

Professional
Greg Raven said:
Wow, that'll be fun. You'll have to calculate COP, and recoil weight for each point of interest, as part of calculating hitting weight. I've got too much to do right now to bang it out, but let me know if you get bogged down -- I'll take a swing at it.

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.
 

Greg Raven

Semi-Pro
Unfortunately, I don't think averages are going to get the job done. What we are looking for is a concentration of mass in an individual frame. I plotted hitting weight against length for the original three racquets in my test, and came up with about the same results. You can see the chart here:

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.
 

ohplease

Professional
Greg Raven said:
Unfortunately, I don't think averages are going to get the job done. What we are looking for is a concentration of mass in an individual frame. I plotted hitting weight against length for the original three racquets in my test, and came up with about the same results. You can see the chart here:

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.
 

travlerajm

Talk Tennis Guru
I numerically calculated m(x) for every 1-in. segment of the POG OS.

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:

POG OS calculated mass distribution:
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

Hall of Fame
Though I'm not much of a thinker myself, I was doing some reading the other day, and came across the term 'polarization index,' which basically meant 'percent of polarization.' It was defined as RecoilWeight over StaticWeight. Though you would not be able to tell the exact distribution with that, you could get a general idea, am I right? And that also gives you a clue as to what polarization does; as it increases for a constant weight it means less recoil (and therefor more swingweight as well)
 

travlerajm

Talk Tennis Guru
Amone said:
Though I'm not much of a thinker myself, I was doing some reading the other day, and came across the term 'polarization index,' which basically meant 'percent of polarization.' It was defined as RecoilWeight over StaticWeight. Though you would not be able to tell the exact distribution with that, you could get a general idea, am I right? And that also gives you a clue as to what polarization does; as it increases for a constant weight it means less recoil (and therefor more swingweight as well)

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

gregraven

Semi-Pro
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.
 

travlerajm

Talk Tennis Guru
I'm just using excel without using the macro language too.

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.

x ..... m(x)
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
 

Greg Raven

Semi-Pro
If these calculations are correct, it doesn't say much for the perimeter weighting system, huh?

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

travlerajm

Talk Tennis Guru
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.
 

travlerajm

Talk Tennis Guru
Greg Raven said:
If these calculations are correct, it doesn't say much for the perimeter weighting system, huh?

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.
 

travlerajm

Talk Tennis Guru
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. (delta = 0.001 = sum of squares of differences)

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
 

akamc

New User
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 ?
 

katastrof

Rookie
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. (delta = 0.001 = sum of squares of differences)

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.
 

travlerajm

Talk Tennis Guru
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.
 

ohplease

Professional
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. (delta = 0.001 = sum of squares of differences)

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
 

Amone

Hall of Fame
I believe I have found a way to map out the weight of each specific segment.

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.
 

man-walking

Semi-Pro
LOL, maybe just me but this method of calculating the weight distribution (basically based only by balance and swingweight) can't never detect such point to point precision.
It seems to me just a bilinear way to guesstimate the distribution.
Greg Raven said:
If these calculations are correct, it doesn't say much for the perimeter weighting system, huh?
Greg, don't you realize that you have to dissect every inch of the frame to really do this?
 

Greg Raven

Semi-Pro
Perhaps you missed this, from the link I provided:

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.
 

travlerajm

Talk Tennis Guru
I've played around with this a little. I've found that the solutions I posted are not unique - i.e., the solution has a bit of "play" in it. I'm currently investigating what happens when I use a higher resolution (more segments), as I suspect that this will eliminate some of the "play" by adding more constraints.

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.
 

ohplease

Professional
Well, I'll say this much - taking the hitting weight at measured increments along a racket's length pretty much reduces to multidimensional analysis. In other words, it's all just vectors of numbers.

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.
 

travlerajm

Talk Tennis Guru
I just discovered that an applet is not necessary to automate the solution. I automated the solution using Microsoft Excel Solver.

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...
 

Greg Raven

Semi-Pro
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.
 

Amone

Hall of Fame
Well, lay-up's not that complex. So you'd be telling me that the difference is in resin, materials, and flex. This is believable, but that's also a quantifiable item, so instead, what might happen when we classified them beforehand by flex and/or material?

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?
 

ohplease

Professional
Amone said:
Well, lay-up's not that complex. So you'd be telling me that the difference is in resin, materials, and flex. This is believable, but that's also a quantifiable item, so instead, what might happen when we classified them beforehand by flex and/or material?

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.
 

Amone

Hall of Fame
Well, I mean, consider that the flexiest racquet in that basic weight class is 58, and the highest is about 72. That's a difference of 14. To have a difference of 2-4 between those three, is a big difference. About 15-30%, and if that were the difference between points I won one day and another, I'd say I played much better.
 

travlerajm

Talk Tennis Guru
One thought: perhaps measuring the balance at a 45-degree angle would provide additional constraints? I might take a look at how the math works out.
 

Bolt

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

Amone

Hall of Fame
travlerajm said:
One thought: perhaps measuring the balance at a 45-degree angle would provide additional constraints? I might take a look at how the math works out.
What do you mean, Traveler?
 

travlerajm

Talk Tennis Guru
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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