How to measure the distribution of weight across a solid object

[milowie]

Hello,

I have a tennis racquet that has a certain weight distribution that makes it play very well. Due to quality control issues from the manufacturer my second racquet of same make and model has a very different weight profile, making matching difficult.

I'd like to determine the weight distribution of the target racquet on an inch-by-inch-segment basis. I'll do the same with the second racquet, adding weight where needed to match the target racquet (the second racquet is far lighter than the target, so this should generally work- and I recognize there may be cases where I would need to remove weight from the second racquet; I'm ok with this problem).

How do I measurement the weight distribution per one-inch segment?

Thanks in advance...

 

[Orodruin]

You cannot. At least not without taking it apart and considering it a rigid body. The properties you can measure are the centre of mass and moment of inertia. You are probably more interested in the first, but the second may also play a role. Finding the center of mass should be the simplest. It is the point on the racquet where you can balance it on a finger. Moment of inertia would be more difficult to measure.

 

[Nidum]

If the rackets are made from materials which always have the same properties then you should be able to get some useful information by measuring the rackets systematically and noting where there are any differences between them .

If you have access to CAD you could prepare comparative drawings based on the dimensions .

 

[sophiecentaur]

I guess the question arises about what you actually want to achieve with these measurements. Is it in order to select a near-identical racquet or just to mimic your present 'good' one by modifying another one? Or would you like to measure the properties of a new racket?

If you want to assess and characterise a new racket as easily as possible and non-invasively then it may be easier to do some entirely different measurements and with different parameters. You can find the CM, easily enough by hanging it on a fine string or laying on a long knife edge. You can find its Moment of Inertia by measuring its angular acceleration under a given torque (three axes), which would be more difficult of course but quite doable with a simple rotating jig and falling weights. Those four parameters would probably be enough to get a close match between two racquets without actually playing with them.
It may be unsafe to assume that the various distinct parts of the racquet are of consistent internal construction but it would be possible to add mass to certain parts of the racquet to produce the same CM and MI. The MI of the whole (around a particular axis) would be the same as the sum of the MI of the original plus the MI of the added parts but, of course, there are many possible solutions to the problem of where to distribute the added mass to produce the same net MI. The higher order moments are relevant when actually striking the ball.
Stringing and the strength of the loop (?) that the strings are supported on will also make a difference

 

[sandy stone]

How about this? Balance the racket on a knife edge - now you know the center of gravity. Next tack a thread to the end of the racket and suspend a lightweight cup from the string. Now move the racket one inch in the direction away from the thread and cup, still resting on the knife edge. Add water to the cup until the racket is balanced again. Repeat the process an inch at a time, and the amount of water you add to the cup each time should give you an indication of the mass distribution of the racket. Does that sound right?

 

[milowie]

@Orodruin I've attempted to match the racquets by adding weight to the second racquet so that (1) balance points (CM) match, (2) static weights match, and (3) swingweights (MI) match (hanging from the top string and measuring the time it takes to complete a full swing, using an online tool for the calculation). There is still a significant difference in how the racquets feel.

@Nidum the materials can be assumed to be identical, but I believe the manufacturer places metal weights in the frame at the time of molding to achieve a certain weight setup, and this process is imprecise leading to the primary reason for the differences (though unequal material distribution is probably contributing). How would you "measure the racquets systematically" to determine the differences?

@sophiecentaur I'm trying to add weight to a second racquet so it matches the first. I've matched CM and MI but, as you note, this does not mean the weights are equally placed - and I think that is the root of the difference, so I'm trying to match the actual weight distribution "inch-by-inch"

@sandy stone I've thought about this but wasn't sure how to go about it, or if it would work... once you have the weight difference, how would you calculate how much weight to add for that inch segment to achieve the match? and would you not have to go back through all prior one inch segments and adjust, as their balance is now affected, as well?

One more point - I don't need to know the actual weight per-inch segment, I just need to know they are the same between the two racquets. I tried stabilizing one end of the racquet and moving a scale along the length of the racquet, moving in one-inch increments. This is essentially the same approach as the @sandy stone water in the cup method.

Doing this with both racquets, I can see variations in the weight reported on the scale, which are presumably the areas where the weights differ (see attached screenshot of results). I just don't know how to calculate how much weight to add to achieve an end to end match. Would this approach work?

 

[jbriggs444]

How about this? Balance the racket on a knife edge - now you know the center of gravity. Next tack a thread to the end of the racket and suspend a lightweight cup from the string. Now move the racket one inch in the direction away from the thread and cup, still resting on the knife edge. Add water to the cup until the racket is balanced again. Repeat the process an inch at a time, and the amount of water you add to the cup each time should give you an indication of the mass distribution of the racket. Does that sound right?

Take two hypothetical objects. The first is a thin rod of mass m and length l with the mass evenly distributed all along its length. The second is a massless rod of length l with two small weights of mass m/2 attached at its endpoints.

Will the above method distinguish between the two rods?

Claim: The only thing you can measure in the above manner is the mass of the rod and its center of gravity. (Four independent values).

You could distinguish between the above rods by measuring their moments of inertia. But suppose we slide the two attached masses to a position $\frac{L}{\sqrt{12}}$ from the center of the rod. Now the moment of inertia is $\frac{mL^2}{12}$, just like the rod with a uniform mass distribution.

The basic problem you are up against is that there are lots of different mass distributions that are equivalent with respect to their total mass, center of mass position and moment of inertia tensor. If all you are doing is measuring mass, balance points and comparing angular momentum with spin rate, you do not get enough information to uniquely identify a possible matching mass distribution.

 

[milowie]

Take two hypothetical objects. The first is a thin rod of mass m and length l with the mass evenly distributed all along its length. The second is a massless rod of length l with two small weights of mass m/2 attached at its endpoints.

Will the above method distinguish between the two rods?

Claim: The only thing you can measure in the above manner is the mass of the rod and its center of gravity. (Four independent values).

You could distinguish between the above rods by measuring their moments of inertia. But suppose we slide the two attached masses to a position $\frac{L}{\sqrt{12}}$ from the center of the rod. Now the moment of inertia is $\frac{mL^2}{12}$, just like the rod with a uniform mass distribution.

The basic problem you are up against is that there are lots of different mass distributions that are equivalent with respect to their total mass, center of mass position and moment of inertia tensor. If all you are doing is measuring mass, balance points and comparing angular momentum with spin rate, you do not get enough information to uniquely identify a possible matching mass distribution.

Would CM remain the same, as well?

Although MI is now the same, would you not see the deviations in weight distribution if you were take measurements in one-inch increments (using either the water cup method or the moving scale method I described in my last post)? If so, is there a way to translate that to a formula for changing the mass distribution of the first rod to match the second (assume the mass of that first rod is pliable / movable)?

If that won't work, what would?

thanks.

 

[jbriggs444]

Although MI is now the same, would you not see the deviations in weight distribution if you were take measurements in one-inch increments (using either the water cup method or the moving scale method I described in my last post)?

No. You would not see any deviations.

That's how center of gravity works. For point-of-balance purposes, it does not matter how mass is distributed. The only thing that matters is the total mass and center of gravity.

 

[milowie]

No. You would not see any deviations.

That's how center of gravity works. For point-of-balance purposes, it does not matter how mass is distributed. The only thing that matters is the total mass and center of gravity.

Thanks @jbriggs444 - if I understand your example correctly, would there not be a difference in the following example (I hope the ascii diagrams translate ok)?

~ : weightless one-inch segment
$ : m/2
= : m/27
^ : fulcrum

(tennis racquet is 27 inches)

R1 (27 segments)
===========================

R2 (27 segments)
$~~~~~~~~~~~~~~~~~~~~~~~~~$

Move $ to L/√12 from center of R2 = 27/√12 = 7.8"

~~~~~$~~~~~~~~~~~~~~~$~~~~~

Then place a fulcrum at equal positions ^ for each R1 and R2:

R1
===========================
------^

R2
~~~~~$~~~~~~~~~~~~~~~$~~~~~
------^

Would there not be a difference that can be measured, given there is a counter-balance to the left of the fulcrum in R1 but none in R2?

 

[jbriggs444]

Would there not be a difference that can be measured, given there is a counter-balance to the left of the fulcrum in R1 but none in R2?

Asked and answered. No. No difference whatsoever.

Why not do the math? Pick a fulcrum position. Compute the torque due to gravity for both rods about that position.

 

[milowie]

What if we use the "cup of water" counterbalance method and then compare the weight needed to counter-balance each rod...

Using the method described in this article, it seems a different mass is required to counter-balance R1 (47.6g) vs R2 (45.9g), giving us a method of detecting the difference in mass distribution... would this not work?

How to Calculate Counterbalance Weights
https://sciencing.com/calculate-counterbalance-weights-10022172.html

R1
===========================
------^
Length: 27"
Fulcrum position from left edge: 5"
Mass of R1: 27g
Mass left of fulcrum: 5g
Mass right of fulcrum: 22g
Counter-balance position: 5" from fulcrum

torque left of fulcrum: 15
torque right of fulcrum: 253
mass needed to counter-balance: 47.6g

R2
~~~~~$~~~~~~~~~~~~~~~$~~~~~
------^
Length: 27"
Fulcrum position from left edge: 5"
Mass placed on rod: m1 13.5g, m2 13.5g
Mass placement from center point: 7.8"
m1 distance from fulcrum: 0.7"
m2 distance from fulcrum: 16.3"
Counter-balance position: 5" from fulcrum

torque left of fulcrum: 0
torque right of fulcrum: 229.5
mass needed to counter-balance: 45.9g

 

[Orodruin]

What if we use the "cup of water"

It will not work. The only thing it does for you is to tell you when you get the exact torque that cancels the torque from gravity. This is uniquely given by the position of the CoM and the total mass.

 

[jbriggs444]

Fulcrum position from left edge: 5"
Mass of R1: 27g
Mass left of fulcrum: 5g
Mass right of fulcrum: 22g
Counter-balance position: 5" from fulcrum

torque left of fulcrum: 15

You have 5 grams force an average of 2.5 inches from the fulcrum and you claim a torque of 15 gram-force-inches. The correct product would be 12.5 gram-force-inches.

[The units are cringe-worthy]

 

[milowie]

You have 5 grams force an average of 2.5 inches from the fulcrum and you claim a torque of 15 gram-force-inches. The correct product would be 12.5 gram-force-inches.

[The units are cringe-worthy]

(22g x 11in) - (5g x 2.5in) = 229.5 = 45.9g counter-balance = same as R2; so the counter-balances would match.

I'm sure you know a lot about the subject matter; too bad you aren't able to share it more fully, and in a more positive way...

 

[milowie]

It will not work. The only thing it does for you is to tell you when you get the exact torque that cancels the torque from gravity. This is uniquely given by the position of the CoM and the total mass.

Got it, thanks. Maybe this was already answered indirectly - if static weight, CM, and MI are the same between the two racquets can the distribution of mass still be different?

If so, any other ideas on how to approach this? I saw one suggestion of doing a CT scan...

 

[Orodruin]

if static weight, CM, and MI are the same between the two racquets can the distribution of mass still be different?

Yes, this is perfectly possible.

If so, any other ideas on how to approach this? I saw one suggestion of doing a CT scan...

The CT will not reveal the mass distribution and would be significant overkill. Do your racquets have the same weight? If not, they are going to feel different solely based on this, even if the distribution is the same - the moment of inertia depends directly on the mass as well.

 

[milowie]

Yes, this is perfectly possible.

Very helpful, confirms what I'm experiencing must be the distribution.

Do your racquets have the same weight?

Yes, I've matched their static weight, CM and, as best I can, their MI (manually timing the racquet as a swinging pendulum, using an app that does the same, and having the racquets measured on a special tennis racquet swingweight (MI) machine). Everything I've read about racquet matching says match those three specs and the racquets will be feel / play the same - but, in this case, they don't!

 

[Orodruin]

Treating the racquets as rigid bodies, they should then be completely equivalent. The only things left are then not distribution related but related to issues that are not rigid body properties, such as the racquets’ flexibility etc. You will most likely not be able to fix this by changing the mass distribution, which is a rigid body property.

 

[milowie]

Treating the racquets as rigid bodies, they should then be completely equivalent.

Maybe I misunderstood you? Can you clarify your statement quoted above? If those three specs match, then the racquets should be equivalent? (I took your "yes, perfectly possible" response to mean they can be different even though those three specs match)

I've had the stiffness measured, as well (same tennis machine) - the stiffness ratings match, so that shouldn't be a factor. And it's not about how the racquet feels at impact, it's about how it feels when swung.

 

[jbriggs444]

Maybe I misunderstood you? Can you clarify your statement quoted above? If those three specs match, then the racquets should be equivalent? (I took your "yes, perfectly possible" response to mean they can be different even though those three specs match)

Two rackets with two different mass distributions can have the same mass, same center of mass and same moment of inertia.

 

[Orodruin]

Maybe I misunderstood you? Can you clarify your statement quoted above? If those three specs match, then the racquets should be equivalent? (I took your "yes, perfectly possible" response to mean they can be different even though those three specs match)

They can have different weight distributions, but from the point of view of forces acting on them the exact distribution will not matter as long as they can be considered rigid bodies. Only the center of mass position and the moment of inertia are of relevance for the rigid body treatment. Anything else you experience will have to do with properties that are not related to rigid body mechanics. Your exact distribution together with the flexibility of the racquet will for example affect the allowed eigenfrequencies of the racquet (not saying that it is that, just an example of a non-rigid body property).

I've had the stiffness measured, as well (same tennis machine) - the stiffness ratings match, so that shouldn't be a factor. And it's not about how the racquet feels at impact, it's about how it feels when swung

From a mechanical viewpoint, they should feel exactly the same. There remains psychological and ”feel” issues. Did you make a blind test? Did you use the same type of grip tape on both? (In both cases new to avoid wear bias.)

 

[milowie]

Fascinating.

I have not performed a true blind test, and I regularly change the overgrip on the handle- so, yes, both racquets have been compared with fresh new overgrips.

Though I haven't performed a true blind test, the difference is pretty stark - the target racquet feels "heavier" in terms of lag... when I serve, feeling the increased stress on my shoulder, rotator, and effort to square the racquet to the ball; on forehands, feeling how much the racquet lags in my wrist; on volleys at net, in effort it takes to move the racquet into position.

I spent a sustained period of time playing only racquet #2, as a kind of reset to make sure it wasn't something psychological. Returning to test racquet #1 the difference in feel was immediately obvious.

 

[milowie]

btw - I'm sure the cm measurements are accurate - I've been very precise with them.

perhaps it's an issue with the MI measurement... thought about this, but was skeptical that I would be able to notice minor inaccuracies (was equally diligent in these measurements).

 

[sophiecentaur]

If so, any other ideas on how to approach this? I saw one suggestion of doing a CT scan...

This is a really complicated business - I now realize.
The mention of "rigid body" made me think. The stiffness is of equal importance to the mass distribution. If you were to apply an impulse force at various points along the racquet, the initial displacement would be affected by a combination of the local mass and modulus. Looking at the very first response to the impulse would eliminate the effect of any standing waves and resonance of the whole racquet. Your initial description of the situation was that the second racquet happens to be lighter. What if the second racquet were heavier? Two racquets would have different profiles along their length and you could add mass or stiffness to each section to get a match.

 

[milowie]

I can see there may be some confusion regarding my initial post. I had R1 and R2 matched in weight, balance (CM), and swingweight (MI) (specs below). Although specs matched, they clearly did not feel the same (in the ways noted prior). I recast the problem as a clean slate, with R1 as the target and R2 in stock form, to take a fresh look at how folks in this forum may approach the goal.

I have also adjusted the weights of R2 further, by feel, to try to get a match - and did add too much weight, making it too heavy, as part of this (to test the boundaries). I wasn't able to find the sweetspot to match the two.

You mention "impulse force" when exploring the role of stiffness - and more broadly, the behavior of the rigid body. Perhaps this is already clear and considered, but the feeling is different independent of actual impact... simply swinging the racquet in the air, even without much force, the difference can be felt.

Specs (after matching):

R1
352.67g - static weight
328 kg-cm^2 - swingweight (MI)
31.12cm - balance point from tip of racquet (CM)
66RA - stiffness (measurement of flex on longitudinal axis)

R2
352.95g - static weight
328 kg-cm^2 - swingweight (MI)
31.12cm - balance point from tip of racquet (CM)
66RA - stiffness (measurement of flex on longitudinal axis)

If there is no way to feel the difference in distribution of mass and the racquets must feel "equivalent" with the above specs, then the measurements must be incorrect. Static weight and balance are very easy to measure, and validate. I would rule those out. I would discount the role of stiffness, given this isn't about impact (but you tell me if that's reasonable).

Seems likely then, given the initial assumption above, the swingweight measurement must be inaccurate? I've had varying results, based on how it is measured (even different swingweight measurement machines differ in their results, compared to each other). I figured as long as I use the same machine, it doesn't really matter what the actual number is - as long as it is consistent across measurements. Maybe it's not.

FWIW - here's the prescribed "do it at home" method to measure swingweight:
http://twu.tennis-warehouse.com/learning_center/howto_swingweight.php