Lightest swingweight on ATP tour?

[El_Yotamo]

For sure the deformation of the incoming ball is be considered as wel but has nothing to do with the tennis frame. I d say the less deformation of the ball the more energy of the swing is used for the speed of the ball, while we can speak of "elastic" collision as opposed to inelestic collision. Remember how hot a squasball becomes during play. This is all wasted energy. So loser and or softer strings transfer more energy(momentum) and less change of racquet deformation, do you agree?
The only concept is "dwelltime" i m unsure. Does give loser strings and or softer racquets more dwell time during impact and therefore transfer more "momentum"?

The deformation of the ball does have to do with the frame but to a much smaller part than the strings. The point is the overall deformation of every object in play here anyway. Generally more overall deformation means more dwell time, and the dwell time itself has little to no effect on momentum transfer.

 

[newyorkstadium]

Nishioka. Tested by M&C last year.

Around 313.5 when you add his dampener. Strange, I would not describe his strokes as low powered, as you would imagine. His shots are very modern and spinny, though.

 

[cyanide43]

Holger Rune, up and coming player ranked around 500 also uses 314sw strung.

All a lower swingweight means is you have to compensate for the loss in power with RHS. It does not inherently mean it is more or less suitable to the pro game.

 

[polksio]

Leather grip would only add 3-5 SW pts at max. Nowhere near 15

Leather grip add absolutely 0 SW (zero nyet nada rien)

 

[dr325i]

Leather grip add absolutely 0 SW (zero nyet nada rien)

Do not get so excited if you do not understand what SW is...
Leather grip starts at the center point of rotation (0 cm) however, goes up to 10-15cm on the racket.
that is 5g (approx) = 0.005kg * 10^2cm^2 = 0.5 kg*cm^2.
Leather grip can be heavier, so, it could bring the SW up by 1-2 kg*cm^2. It is very small, however, not ZERO.

 

[polksio]

Do not get so excited if you do not understand what SW is...
Leather grip starts at the center point of rotation (0 cm) however, goes up to 10-15cm on the racket.
that is 5g (approx) = 0.005kg * 10^2cm^2 = 0.5 kg*cm^2.
Leather grip can be heavier, so, it could bring the SW up by 1-2 kg*cm^2. It is very small, however, not ZERO.

We said ADDS, so COMPARED to a synthetic. That's a 7g differential at most. So 3g at around 17cm. You confuse being smart with nitpicking.

 

[dr325i]

We said ADDS, so COMPARED to a synthetic. That's a 7g differential at most. So 3g at around 17cm. You confuse being smart with nitpicking.

you said absolutely zero...which is not true, and nothing "smart" about it.
3g @ 17cm is 1kg*cm^2 (0.9 to be precise)

Yes, you are correct, and I believe I mentioned that, what Leather adds is irrelevant.

 

[polksio]

you said absolutely zero...which is not true, and nothing "smart" about it.
3g @ 17cm is 1kg*cm^2 (0.9 to be precise)

Yes, you are correct, and I believe I mentioned that, what Leather adds is irrelevant.

Wrong.

 

[El_Yotamo]

Leather grip add absolutely 0 SW (zero nyet nada rien)

Did you skip my post or did you ignore it because it was too much to comprehend?

The calculation is simple. Assuming an approximate 10 g difference between leather and synthetic and a handle length of about 20 cm, you get:
0.005*(5^2+5^2)=0.25 kg*cm^2
Very little, not absolute zero though.

 

[El_Yotamo]

Do not get so excited if you do not understand what SW is...
Leather grip starts at the center point of rotation (0 cm) however, goes up to 10-15cm on the racket.
that is 5g (approx) = 0.005kg * 10^2cm^2 = 0.5 kg*cm^2.
Leather grip can be heavier, so, it could bring the SW up by 1-2 kg*cm^2. It is very small, however, not ZERO.

You also seem to have skipped my post... Your calculation error is multiplying by 10^2, remember the axis is at 10 cm so we need to count the distance from it, half of which is below that mark and half above it by my assumption.

 

[polksio]

Did you skip my post or did you ignore it because it was too much to comprehend?

The calculation is simple. Assuming an approximate 10 g difference between leather and synthetic and a handle length of about 20 cm, you get:
0.005*(5^2+5^2)=0.25 kg*cm^2
Very little, not absolute zero though.

It's actually half of that. 0.13 to be **** about it. Or 0 (nyet zero nada rien) if youre a normal human being.
Good day.

 

[El_Yotamo]

It's actually half of that. 0.13 to be **** about it. Or 0 (nyet zero nada rien) if youre a normal human being.
Good day.

Please do explain how it would be half of that. The mass added is 10 g. Half of it (5 g) is below the axis, half of it (5 g) is above. Both are distributed uniformly so are each centered at a distance of 5 cm from the axis. So the calculation is clearly 0.005*5^2+0.005*5^2 which yields 0.25. I'm sorry you can't do basic math.

Last I checked 0 is not equal to 0.25, and I'll explain why this is important because apparently everything has to be spelled out for you. Suppose you add 50 g of mass into the handle, now if you consider that by putting a leather grip you've added exactly 0 SW then adding any mass to the handle will have the same effect. However, suppose this mass is more or less centered at 2 cm from the butt, you've added 0.05*8^2=3.2 kg*cm^2. Might not seem like much (and it isn't) but for many people that's noticeable. And maybe that's a lot of mass to be added to the butt of a racquet, but the point still stands. Saying something is absolute zero is scientifically wrong and misleading even if what you're talking about is quite small. Had you said effectively zero I wouldn't have even replied because you would've been right but unfortunately, as usual people here who have no prior learning are misleading others in a way that I personally have little patience for.

 

[jmacdaununder2]

Nothing like a good 'reductio ad absurdum' to hopefully get your point across.

 

[El_Yotamo]

Nothing like a good 'reductio ad absurdum' to hopefully get your point across.

I've just gotten sick and tired of the "experts" on this forum who think they can write whatever misleading or straight up wrong stuff they want while others of us who do know what they're talking about try to actually help and inform.

 

[Slowtwitcher]

Just a reminder that swingweight is usually given with 3 significant figures, so 0.25 is indeed zero

 

[El_Yotamo]

Just a reminder that swingweight is usually given with 3 significant figures, so 0.25 is indeed zero

You are somewhat correct, I hope you understand it still shows complete misunderstanding to write absolute 0.

Btw, what you wrote isn't completely true. If I had a SW of 312.3 and I add 0.25 then in that case I end up with 312.55 so my 0.25 "equals" 1 by significant figures. So it's either 0 or 1. And in any not it very certainly ain't absolute 0.

 

[Slowtwitcher]

I hope you understand you're being incredibly condescending on a very, very simple question. Do you happen to have an uncle with a high IQ?

 

[El_Yotamo]

You may consider my response condescending, I apologize if you feel offended. I'm here to help as much as I can, but as I've probably written at least 4 times on this thread alone I'm tired of being corrected/disagreed with/whatever you wanna call it by people who clearly don't have the necessary knowledge and who are more likely to mislead others with their erroneous/inaccurate information than help anyone. I don't claim to be smarter than anyone else, and conversely I didn't claim you were stupid, I simply claim to know the scientific/mathematical truth on certain subjects which I've studied and experienced from all angles - theoretically in my engineering degrees, experimentally and in calculations in my degrees as well as in the tennis world and in my lab, and very certainly experientially on the court as a college level tennis player.

Instead of calling people who have knowledge and are trying to share it condescending because they corrected your honest mistake, you could be more appreciative yourself. You said that according to significant figures 0.25 equals 0, I showed you that it could also be 1 by your same argument and explained why calling something absolute 0 and effectively 0 are not the same. I hope you understand I've been trying to help and that you can accept the info I've readily shared rather than trying to justify your error (which all humans make!) by calling me condescending.

If the question was as simple as you're making it out to be explain the amount of people who answered incorrectly before (and even after!) I came in to answer it and set the record straight.

 

[Shaolin]

Anyone else with SW under approx 335?

So far we have

Nishioka
Fritz
Verdasco
Kyrgios
Rublev
Rune

Seems like there are probably more, esp next gen.

 

[Slowtwitcher]

You may consider my response condescending, I apologize if you feel offended. I'm here to help as much as I can, but as I've probably written at least 4 times on this thread alone I'm tired of being corrected/disagreed with/whatever you wanna call it by people who clearly don't have the necessary knowledge and who are more likely to mislead others with their erroneous/inaccurate information than help anyone. I don't claim to be smarter than anyone else, and conversely I didn't claim you were stupid, I simply claim to know the scientific/mathematical truth on certain subjects which I've studied and experienced from all angles - theoretically in my engineering degrees, experimentally and in calculations in my degrees as well as in the tennis world and in my lab, and very certainly experientially on the court as a college level tennis player.

Instead of calling people who have knowledge and are trying to share it condescending because they corrected your honest mistake, you could be more appreciative yourself. You said that according to significant figures 0.25 equals 0, I showed you that it could also be 1 by your same argument and explained why calling something absolute 0 and effectively 0 are not the same. I hope you understand I've been trying to help and that you can accept the info I've readily shared rather than trying to justify your error (which all humans make!) by calling me condescending.

If the question was as simple as you're making it out to be explain the amount of people who answered incorrectly before (and even after!) I came in to answer it and set the record straight.

And you go and double down on your condescension... how many engineering degrees do you want, I got three... is that enough for you?

This is not the time to go all Neil deGrasse Tyson on us. In fact, I'd suspect NdGT would come in and tell you that for all practical purposes, changing your grip weight doesn't change swingweight.

 

[onehandbh]

In a humid climate, my hands and wrist sweat. The moisture gets absorbed by my overgrip and leather grip under it.

How much does this change the SW?

Will the increase in swingweight of my racquet be countered by the loss of hydration in my arm (and less weight?) ?

 

[El_Yotamo]

And you go and double down on your condescension... how many engineering degrees do you want, I got three... is that enough for you?

This is not the time to go all Neil deGrasse Tyson on us. In fact, I'd suspect NdGT would come in and tell you that for all practical purposes, changing your grip weight doesn't change swingweight.

This is the last time I'll respond since I respect the OP and the fact that our discussion is not in line with the thread's subject.

I suppose we're equal on engineering degrees, or at least we will be in about a year's time. Again, not claiming to be above anyone, I never wrote anything incorrect on this thread. Unfortunately you did, I'd expect an engineer like yourself to own up to the mistake rather than dodging it but whaddya know.

I said myself that for all practical purposes the addition of a leather grip does not increase SW. However I defended the fact that generally adding mass to the handle can increase SW and the fact that writing absolute 0 is an erroneous statement, which it still is. As I've written several times at this point I agree that the addition of the leather grip effectively doesn't increase SW.

Call me condescending all you want, I'm doing my best at writing detailed posts in order to inform and help people understand key concepts. Gladly I see many people have liked my posts so I know I'm doing something right even if there's something you specifically don't like about me/it.

 

[haqq777]

In a humid climate, my hands and wrist sweat. The moisture gets absorbed by my overgrip and leather grip under it.

How much does this change the SW?

Will the increase in swingweight of my racquet be countered by the loss of hydration in my arm (and less weight?) ?

 

[Slowtwitcher]

In a humid climate, my hands and wrist sweat. The moisture gets absorbed by my overgrip and leather grip under it.

How much does this change the SW?

0 kg m^2

 

[Slowtwitcher]

I suppose we're equal on engineering degrees, or at least we will be in about a year's time.

Effing grad students

 

[cyanide43]

I hate to bring back the concept of SW1 and SW2 from @travlerajm but that is essentially what we have going on here.

SW1 would be the range of light swingweights from 300s-320s.
Contains:

Holger Rune - 315
Liam Broady - 315
Yoshihito Nishioka - 315
Adrian Mannarino - 325
Nick Kyrgios - 325
Marton Fucsovics - 325
JJ Wolf - 325
Cameron Norrie - 325

Main characteristics from this group of swingweights being that you have to swing faster on the average shot but are awarded extra touch and feel in return for a lighter swinging frame. These swingweights work best for doubles players or those who need the ability to quickly transition from defense to offense.

SW2 would be the modern pro swingweight range, or medium-heavy range from 330s-350s. Many of today's top pros are in this range.
Contains:

John Millman - 335
Miomir Kecmanovic - 335
Frances Tiafoe - 335
Denis Kudla - 335
Karen Khachanov - 340
Roger Federer - 345
Dominic Thiem - 345
Stefanos Tsitipas - 345
Jack Sock - 350
Matteo Berrettini - 355
Juan Martin Del-Potro - 355

With this modern swingweight range, you lose a little bit of acceleration, but gain predictability and don't have to swing as hard on the average ball. It is still possible to transition to offense fairly quickly with these frames. These swingweights combine the best aspects of the maneuverability of the lower swingweights along with the stability of the higher swingweights, and are a reasonable tradeoff for most pros.

Finally, at the top of the modern scale we have the SW3 range, or swingweights from 360s-390s. These swingweights pack a big punch and help hit through the court rather than with angles. A game based on depth (i.e. a clay courter, huge grinder) or on big serves may work best in this range.
Contains:

Zverev - 360
Wawrinka - 360
Djokovic - 360
Nadal - 370
Andujar - 380

The reason I have broken the levels in this way is because a racquet with 30 swingweight points higher actually handles similarly to a racquet with 30 swingweight points below. That is, a racquet with a swingweight of 315 will actually swing somewhat similar to a racquet in the 340s, barring the fact that the racquet in the 340s will be heavier and more difficult to handle.

I've tested this premise using a Gosen swingweight machine and a lot of time on court.

Finally, it's important to remember that a big part of what the best players end up using has to do with the racquet they used growing up, as the swingweights they are comfortable with will react in a predictable manner under critical situations. Don't base your choices based on what the pros use! Do what's most comfortable for you.

 

[Villain]

0 kg m^2

What if he’s wearing wristbands?

 

[El_Yotamo]

What if he’s wearing wristbands?

No no please no more

 

[jmnk]

I've seen too many people not understanding what they're talking about when it comes to SW on this thread to ignore. Here's what you guys should know (at least if you're gonna post about it or apply it to your frame):

The SW is one of the mass moments of inertia of a tennis racquet, which in mechanics is defined very loosely as the resistance to rotational motion. You know how something heavier is more difficult to pick up? Something with a higher moment of inertia is more difficult to rotate about the axis at which the moment of inertia is measured (more on that in a sec). Now, because this is rotational motion and not linear motion, moments of inertia are not measured in grams, kilos, pounds, etc. It's easy to get confused because many of the SW measurements we see have values similar to mass measurements, and the word weight is in the SW name, but in no way is that the actual case. This is important because once you understand units of measurement it's a lot easier to understand the underlying physical concept. The units in which SW is measured are mass*length^2, and in the tennis world we use kg*cm^2. This means that both mass and distance have a role here, because to rotate anything about a given axis you have to apply a force some distance away from the axis (called a moment or torque). Ok so why is all of this important? Let's get back to the particular details invlolving SW and the answers to you guys' posed/potential questions. Anyone who's ever measured the SW of a racquet can by now come to understand that the SW is the moment of inertia about an axis which is located approximately 10 cm from the butt of the racquet. The axis itself runs perpendicular to the handle and parallel to the racquet face. Now remember the units by which we measure SW? All we have to do now is understand the physical implications. For those who really want to know what the moment of inertia is and have some basic knowledge in mathematics and mechanics I recommend looking up "inertia tensor" on google. For our case here, we want to know how adding mass at certain places on a racquet will affect the SW. So our units are kg*cm^2, which means that for any mass added, we multiply the amount of mass added in kg by the distance squared of that mass from our axis in cm. Let's for example consider adding 5 g of mass 50 cm away from the butt. The SW increase will be 0.005*40^2=8 kg*cm^2. It's that simple people. How about our leather grip example? That one's a tiny bit more complicated but it's also pretty easy. When our mass added is distributed and part of it is on the axis we need to split into two distributed masses (one below and one above the axis). So let's say our leather grip adds 10 g distributed evenly from the butt to 20 cm from the butt. We split that into two distributed masses: one from 0 to 10 and another from 10 to 20. Now because the mass is uniformly distributed we treat it as two point masses at 5 cm and 15 cm each with 5 g. Now let's do the math: 0.005*(5^2+5^2)=0.25 kg*cm^2. We did it!

I hate to end this informative post on a douchy note, but I genuinely beg of anyone reading this, please please please don't write information online unless you've done the necessary research/learning and actually know what you're talking about. It pains me that I have to write this because most people don't do it but there are too many people who have no clue what they're talking about that mislead and misinform others. Not necessarily on purpose ofc, but the damage is still done.

@El_Yotamo - I wasn't going to post since it looks like you are trying to help. But it irks me a bit that you while you preach from a high horse you still make a fairly common mistake. And then you also went into that 'my number of engineering degrees outshines your number of engineering degrees' conversation with @Slowtwitcher ...... While your description is in general correct the highlighted part is absolutely not.

What makes you think that for the purpose of analyzing swingweight it is OK to 'split the mass into two distributed masses (one below and one above the axis)? And why is it OK then treat each of those two masses as two point masses at a certain point? With that logic - what would be wrong with treating a grip as a single mass distributed evenly - and therefore treat it as a point mass at 10cm, which obviously would result in 0 added swingweight (since the point mass is exactly at the axis), which is obviously incorrect. Or why not split the mass of the grip into four 'distributed masses' (one 0-5cm, second 5-10cm, third 10-15cm, and the fourth 15-20cm), then to treat each of those as a point mass at 2.5cm, 7.5cm, 12,5cm, and 17.5cm respectively, and calculate swingweight that way. While we are at it - why not split it into 3, or 40 'distributed masses'? Or 400? Or 4000?

If you want to be all analytical and such the proper model is to treat the grip as a thin walled cylinder of a negligible thickness, having a mass of (using your example numbers) m=10grams, and length of l=20cm and radius r=1.5cm (more less for a 4 3/8 grip). And then the MOI around the axis perpendicular to its own axis of symmetry at 10cm from the buttend (which happens to be an axis through the center of mass) is m*r*r/2 + m*l*l/12 = 0.01*1.5*1.5 / 2 + 0.01*20*20/12 =~ 0.3345kg*cm^2

 

[jmnk]

deleted - double post

 

[Aniruddha Phukan]

I've seen too many people not understanding what they're talking about when it comes to SW on this thread to ignore. Here's what you guys should know (at least if you're gonna post about it or apply it to your frame):

The SW is one of the mass moments of inertia of a tennis racquet, which in mechanics is defined very loosely as the resistance to rotational motion. You know how something heavier is more difficult to pick up? Something with a higher moment of inertia is more difficult to rotate about the axis at which the moment of inertia is measured (more on that in a sec). Now, because this is rotational motion and not linear motion, moments of inertia are not measured in grams, kilos, pounds, etc. It's easy to get confused because many of the SW measurements we see have values similar to mass measurements, and the word weight is in the SW name, but in no way is that the actual case. This is important because once you understand units of measurement it's a lot easier to understand the underlying physical concept. The units in which SW is measured are mass*length^2, and in the tennis world we use kg*cm^2. This means that both mass and distance have a role here, because to rotate anything about a given axis you have to apply a force some distance away from the axis (called a moment or torque). Ok so why is all of this important? Let's get back to the particular details invlolving SW and the answers to you guys' posed/potential questions. Anyone who's ever measured the SW of a racquet can by now come to understand that the SW is the moment of inertia about an axis which is located approximately 10 cm from the butt of the racquet. The axis itself runs perpendicular to the handle and parallel to the racquet face. Now remember the units by which we measure SW? All we have to do now is understand the physical implications. For those who really want to know what the moment of inertia is and have some basic knowledge in mathematics and mechanics I recommend looking up "inertia tensor" on google. For our case here, we want to know how adding mass at certain places on a racquet will affect the SW. So our units are kg*cm^2, which means that for any mass added, we multiply the amount of mass added in kg by the distance squared of that mass from our axis in cm. Let's for example consider adding 5 g of mass 50 cm away from the butt. The SW increase will be 0.005*40^2=8 kg*cm^2. It's that simple people. How about our leather grip example? That one's a tiny bit more complicated but it's also pretty easy. When our mass added is distributed and part of it is on the axis we need to split into two distributed masses (one below and one above the axis). So let's say our leather grip adds 10 g distributed evenly from the butt to 20 cm from the butt. We split that into two distributed masses: one from 0 to 10 and another from 10 to 20. Now because the mass is uniformly distributed we treat it as two point masses at 5 cm and 15 cm each with 5 g. Now let's do the math: 0.005*(5^2+5^2)=0.25 kg*cm^2. We did it!

I hate to end this informative post on a douchy note, but I genuinely beg of anyone reading this, please please please don't write information online unless you've done the necessary research/learning and actually know what you're talking about. It pains me that I have to write this because most people don't do it but there are too many people who have no clue what they're talking about that mislead and misinform others. Not necessarily on purpose ofc, but the damage is still done.

Thank you for the detailed post.

One question - why is the axis of rotation considered 10 cm from the butt? Shouldn't it be at the butt itself (or where one holds the racquet)?
That's where the angular movement of the racquet is centered at, right?

 

[MikhailT]

the axis of rotation is more the center of your unit/body anyway. Unless you only rotate your wrist, which you shouldn't be doing...

 

[El_Yotamo]

@El_Yotamo - I wasn't going to post since it looks like you are trying to help. But it irks me a bit that you while you preach from a high horse you still make a fairly common mistake. And then you also went into that 'my number of engineering degrees outshines your number of engineering degrees' conversation with @Slowtwitcher ...... While your description is in general correct the highlighted part is absolutely not.

What makes you think that for the purpose of analyzing swingweight it is OK to 'split the mass into two distributed masses (one below and one above the axis)? And why is it OK then treat each of those two masses as two point masses at a certain point? With that logic - what would be wrong with treating a grip as a single mass distributed evenly - and therefore treat it as a point mass at 10cm, which obviously would result in 0 added swingweight (since the point mass is exactly at the axis), which is obviously incorrect. Or why not split the mass of the grip into four 'distributed masses' (one 0-5cm, second 5-10cm, third 10-15cm, and the fourth 15-20cm), then to treat each of those as a point mass at 2.5cm, 7.5cm, 12,5cm, and 17.5cm respectively, and calculate swingweight that way. While we are at it - why not split it into 3, or 40 'distributed masses'? Or 400? Or 4000?

If you want to be all analytical and such the proper model is to treat the grip as a thin walled cylinder of a negligible thickness, having a mass of (using your example numbers) m=10grams, and length of l=20cm and radius r=1.5cm (more less for a 4 3/8 grip). And then the MOI around the axis perpendicular to its own axis of symmetry at 10cm from the buttend (which happens to be an axis through the center of mass) is m*r*r/2 + m*l*l/12 = 0.01*1.5*1.5 / 2 + 0.01*20*20/12 =~ 0.3345kg*cm^2

Again I apologize if I offended anyone, I used to be a lot less assertive on these forums and would get trampled by people who have no clue what they're talking about. I never claimed to be smarter than anyone else or said my degrees outshined anyone else's I only ever claimed to know what SW is and how to calculate its change given the addition of mass at a certain location on the racquet. My ethos behind my knowledge is the fact that as @Slowtwitcher correctly noted I'm an "effing grad student". I summary I simply claimed to know what a moment of inertia is and explained it to those here who are less knowledgeable in an attempt to help.

The way I learned to calculate the change of a moment of inertia by a point mass is by the calculation ΔI=Δm*r^2, where ΔI is the change in MOI, Δm is the change in mass, and r is the distance from the axis in the perpendicular direction. Now when we put on a uniformly distributed mass we need to either calculate an integral or use our known uniform density function to convert that distributed mass into point mass(es). If this distributed mass doesn't pass through the axis then we can simply take the center of the distributed mass and consider it the point mass as required. You can check that the integral gives the same result. In the case that the distributed mass does pass through the axis in question, the integral needs to be split into two domains because our range of integration passes through the zero point, each of these domains can be assigned a point mass at its center and you can again verify this result if you'd like. So this is why I split into two masses, and not 4 and not 400 and not 4000 and not 4000000. You can split it into as many point masses as you'd like of course as long as none of them correspond to a distributed mass passing through the axis, but I like to do as little calculation as possible and therefore chose just two masses. That's why I think it's ok, because it's the correct method and if you'd like a rigorous proof I'd be delighted to send you one.

I dunno where you got your formula from, but it looks like you tried calculating the MOI of the handle itself rather than the change in MOI due to the addition of mass on the handle. This might work but again I don't know where you got your formula so if you'd like to send me a derivation that'd be great. In any case I agree that I ignored/missed the minor non-vertical contribution of SW which certainly exists because if we said that the SW axis is parallel to the x-axis and that the vertical axis is the z-axis then we are adding mass distanced in the y-direction from the axis.

If I were to calculate the non-vertical contribution I would again use the uniformity of distribution to split the distributed mass into point masses. We'll get that the non-vertical contribution can be again split into two point masses. I just went and measured a handle and while it's approximate I can more or less say from this that the mass distributed uniformly would be centered at a y-distance of about 3 cm from the center so we have two point masses 3 cm from the center each with mass 5 g. So we get a non-vertical contribution of 2*0.005*3^2=0.09. Summing the vertical and non-vertical contributions we get the total change in SW of 0.34 kg*cm^2 which is as close as possible to the exact change in SW as I can get, and seems very similar to what you got so I suppose I can accept your calculation though I would certainly like to make sure I understand exactly what you did there. Of course this is still quite negligible compared to the SW of a tennis racquet as previously discussed, and it's also not absolute zero as similarly discussed.

So, @jmnk and @Slowtwitcher , are we good?

 

[El_Yotamo]

Thank you for the detailed post.

One question - why is the axis of rotation considered 10 cm from the butt? Shouldn't it be at the butt itself (or where one holds the racquet)?
That's where the angular movement of the racquet is centered at, right?

Essentially you're rotating the racquet about where you hold the racquet, so the axis is how far your hand goes up the handle. For most people this is about 10 cm so that's a good enough basis to go with.

 

[dropvolley123]

I've seen too many people not understanding what they're talking about when it comes to SW on this thread to ignore. Here's what you guys should know (at least if you're gonna post about it or apply it to your frame):

The SW is one of the mass moments of inertia of a tennis racquet, which in mechanics is defined very loosely as the resistance to rotational motion. You know how something heavier is more difficult to pick up? Something with a higher moment of inertia is more difficult to rotate about the axis at which the moment of inertia is measured (more on that in a sec). Now, because this is rotational motion and not linear motion, moments of inertia are not measured in grams, kilos, pounds, etc. It's easy to get confused because many of the SW measurements we see have values similar to mass measurements, and the word weight is in the SW name, but in no way is that the actual case. This is important because once you understand units of measurement it's a lot easier to understand the underlying physical concept. The units in which SW is measured are mass*length^2, and in the tennis world we use kg*cm^2. This means that both mass and distance have a role here, because to rotate anything about a given axis you have to apply a force some distance away from the axis (called a moment or torque). Ok so why is all of this important? Let's get back to the particular details invlolving SW and the answers to you guys' posed/potential questions. Anyone who's ever measured the SW of a racquet can by now come to understand that the SW is the moment of inertia about an axis which is located approximately 10 cm from the butt of the racquet. The axis itself runs perpendicular to the handle and parallel to the racquet face. Now remember the units by which we measure SW? All we have to do now is understand the physical implications. For those who really want to know what the moment of inertia is and have some basic knowledge in mathematics and mechanics I recommend looking up "inertia tensor" on google. For our case here, we want to know how adding mass at certain places on a racquet will affect the SW. So our units are kg*cm^2, which means that for any mass added, we multiply the amount of mass added in kg by the distance squared of that mass from our axis in cm. Let's for example consider adding 5 g of mass 50 cm away from the butt. The SW increase will be 0.005*40^2=8 kg*cm^2. It's that simple people. How about our leather grip example? That one's a tiny bit more complicated but it's also pretty easy. When our mass added is distributed and part of it is on the axis we need to split into two distributed masses (one below and one above the axis). So let's say our leather grip adds 10 g distributed evenly from the butt to 20 cm from the butt. We split that into two distributed masses: one from 0 to 10 and another from 10 to 20. Now because the mass is uniformly distributed we treat it as two point masses at 5 cm and 15 cm each with 5 g. Now let's do the math: 0.005*(5^2+5^2)=0.25 kg*cm^2. We did it!

I hate to end this informative post on a douchy note, but I genuinely beg of anyone reading this, please please please don't write information online unless you've done the necessary research/learning and actually know what you're talking about. It pains me that I have to write this because most people don't do it but there are too many people who have no clue what they're talking about that mislead and misinform others. Not necessarily on purpose ofc, but the damage is still done.

I second the opinion!!!

 

[Aniruddha Phukan]

Essentially you're rotating the racquet about where you hold the racquet, so the axis is how far your hand goes up the handle. For most people this is about 10 cm so that's a good enough basis to go with.

Got it. 10 cm is a fair generalization for most.

So if I want to make my racquet feel more 'whippy', the butt would a good place to add some weight, right? Assumption being I'm counter-weighting on the opposite side of the axis.

 

[El_Yotamo]

Got it. 10 cm is a fair generalization for most.

So if I want to make my racquet feel more 'whippy', the butt would a good place to add some weight, right? Assumption being I'm counter-weighting on the opposite side of the axis.

If by whippy you mean easier to swing then it's best not to add any mass at all. If you're looking to counterweight the butt is a good place because mass added there doesn't increase the SW by much relative to other places in the racquet and this will also make the racquet more headlight which can help handling and feel.

 

[Stratotanker]

I hate to bring back the concept of SW1 and SW2 from @travlerajm but that is essentially what we have going on here.

SW1 would be the range of light swingweights from 300s-320s.
Contains:

Holger Rune - 315
Liam Broady - 315
Yoshihito Nishioka - 315
Adrian Mannarino - 325
Nick Kyrgios - 325
Marton Fucsovics - 325
JJ Wolf - 325
Cameron Norrie - 325

Main characteristics from this group of swingweights being that you have to swing faster on the average shot but are awarded extra touch and feel in return for a lighter swinging frame. These swingweights work best for doubles players or those who need the ability to quickly transition from defense to offense.

SW2 would be the modern pro swingweight range, or medium-heavy range from 330s-350s. Many of today's top pros are in this range.
Contains:

John Millman - 335
Miomir Kecmanovic - 335
Frances Tiafoe - 335
Denis Kudla - 335
Karen Khachanov - 340
Roger Federer - 345
Dominic Thiem - 345
Stefanos Tsitipas - 345
Jack Sock - 350
Matteo Berrettini - 355
Juan Martin Del-Potro - 355

With this modern swingweight range, you lose a little bit of acceleration, but gain predictability and don't have to swing as hard on the average ball. It is still possible to transition to offense fairly quickly with these frames. These swingweights combine the best aspects of the maneuverability of the lower swingweights along with the stability of the higher swingweights, and are a reasonable tradeoff for most pros.

Finally, at the top of the modern scale we have the SW3 range, or swingweights from 360s-390s. These swingweights pack a big punch and help hit through the court rather than with angles. A game based on depth (i.e. a clay courter, huge grinder) or on big serves may work best in this range.
Contains:

Zverev - 360
Wawrinka - 360
Djokovic - 360
Nadal - 370
Andujar - 380

The reason I have broken the levels in this way is because a racquet with 30 swingweight points higher actually handles similarly to a racquet with 30 swingweight points below. That is, a racquet with a swingweight of 315 will actually swing somewhat similar to a racquet in the 340s, barring the fact that the racquet in the 340s will be heavier and more difficult to handle.

I've tested this premise using a Gosen swingweight machine and a lot of time on court.

Finally, it's important to remember that a big part of what the best players end up using has to do with the racquet they used growing up, as the swingweights they are comfortable with will react in a predictable manner under critical situations. Don't base your choices based on what the pros use! Do what's most comfortable for you.

I’ve seen several posts from Unstrung Customs on Instagram that outline this concept (that swing weights have a sort of similar, cyclical feel every 30 points).

Is there a (mathematical) explanation to this phenomenon? Or is it just a feel thing? Quite curious.

 

[jmnk]

[...]

The way I learned to calculate the change of a moment of inertia by a point mass is by the calculation ΔI=Δm*r^2, where ΔI is the change in MOI, Δm is the change in mass, and r is the distance from the axis in the perpendicular direction.

of course. That is the very definition of what MOI is. It works for a single theoretical point mass.

Now when we put on a uniformly distributed mass we need to either calculate an integral or use our known uniform density function to convert that distributed mass into point mass(es). If this distributed mass doesn't pass through the axis then we can simply take the center of the distributed mass and consider it the point mass as required. You can check that the integral gives the same result. In the case that the distributed mass does pass through the axis in question, the integral needs to be split into two domains because our range of integration passes through the zero point, each of these domains can be assigned a point mass at its center and you can again verify this result if you'd like. So this is why I split into two masses, and not 4 and not 400 and not 4000 and not 4000000. You can split it into as many point masses as you'd like of course as long as none of them correspond to a distributed mass passing through the axis, but I like to do as little calculation as possible and therefore chose just two masses. That's why I think it's ok, because it's the correct method and if you'd like a rigorous proof I'd be delighted to send you one.

no, it is not OK. Please do send the rigorous proof showing otherwise.

I dunno where you got your formula from,

the formula for MOI of a very thin cylinder: m*r*r/2 + m*l*l/12 can be found in any physics book that deals with mechanics of rotation (for example: https://dynref.engr.illinois.edu/rem.html see 'Hollow cylinder or hoop: moments of inertia' section. It also has the derivation of the formula if you are interested, and yes, integrals are used). It is a bit surprising that you claim to know what sw is, and you claim to have multiple engineering degrees and yet you are unfamiliar with that formula.

but it looks like you tried calculating the MOI of the handle itself rather than the change in MOI due to the addition of mass on the handle.

not at all. Please re-read my post.

This might work but again I don't know where you got your formula so if you'd like to send me a derivation that'd be great. In any case I agree that I ignored/missed the minor non-vertical contribution of SW which certainly exists because if we said that the SW axis is parallel to the x-axis and that the vertical axis is the z-axis then we are adding mass distanced in the y-direction from the axis.

could you please explain what you mean by 'the minor non-vertical contribution of SW'? And define what the x,y, and z axis are?

If I were to calculate the non-vertical contribution I would again use the uniformity of distribution to split the distributed mass into point masses.

you can't do that. Your results will not be correct.

We'll get that the non-vertical contribution can be again split into two point masses. I just went and measured a handle and while it's approximate I can more or less say from this that the mass distributed uniformly would be centered at a y-distance of about 3 cm from the center so we have two point masses 3 cm from the center each with mass 5 g. So we get a non-vertical contribution of 2*0.005*3^2=0.09. Summing the vertical and non-vertical contributions we get the total change in SW of 0.34 kg*cm^2 which is as close as possible to the exact change in SW as I can get, and seems very similar to what you got so I suppose I can accept your calculation though I would certainly like to make sure I understand exactly what you did there. Of course this is still quite negligible compared to the SW of a tennis racquet as previously discussed, and it's also not absolute zero as similarly discussed.

of course the additional SW due to a leather grip is negligible. But if you are attempting to calculate it, as you did, then please use sound physical models and formulas. My problem with your post is not the value itself. It is the incorrect physics yo have used to get that number.

So, @jmnk and @Slowtwitcher , are we good?

I'm good.

 

[Deleted member 768841]

of course. That is the very definition of what MOI is. It works for a single theoretical point mass.


no, it is not OK. Please do send the rigorous proof showing otherwise.


the formula for MOI of a very thin cylinder: m*r*r/2 + m*l*l/12 can be found in any physics book that deals with mechanics of rotation (for example: https://dynref.engr.illinois.edu/rem.html see 'Hollow cylinder or hoop: moments of inertia' section. It also has the derivation of the formula if you are interested, and yes, integrals are used). It is a bit surprising that you claim to know what sw is, and you claim to have multiple engineering degrees and yet you are unfamiliar with that formula.


not at all. Please re-read my post.


could you please explain what you mean by 'the minor non-vertical contribution of SW'? And define what the x,y, and z axis are?


you can't do that. Your results will not be correct.


of course the additional SW due to a leather grip is negligible. But if you are attempting to calculate it, as you did, then please use sound physical models and formulas. My problem with your post is not the value itself. It is the incorrect physics yo have used to get that number.


I'm good.

Wow you guys really are mathematicians of tennis.

 

[Slowtwitcher]

@El_Yotamo you might want to get your money back on one of those engineering degrees, I did those MOI formula deductions in a 2nd year undergrad course

But I'm good. The contribution of grip mass to swingweight is zero, I'm glad we all can agree.

 

[jmnk]

@El_Yotamo you might want to get your money back on one of those engineering degrees, I did those MOI formula deductions in a 2nd year undergrad course

But I'm good. The contribution of grip mass to swingweight is zero, I'm glad we all can agree.

@Slowtwitcher - ok, since I'm equal opportunity bs calling. You actually liked @El_Yotamo posts in question (along with others), and you never called in question his physics/math explanation. You just argued that in practical sense the contribution of a grip to overall sw is zero. So you may want to try to get a refund on your engineering degrees too

 

[Slowtwitcher]

This is how it's done. Grad students do all the work, all the mistakes, you come in at the end and draw the relevant conclusions

 

[El_Yotamo]

of course. That is the very definition of what MOI is. It works for a single theoretical point mass.


no, it is not OK. Please do send the rigorous proof showing otherwise.


the formula for MOI of a very thin cylinder: m*r*r/2 + m*l*l/12 can be found in any physics book that deals with mechanics of rotation (for example: https://dynref.engr.illinois.edu/rem.html see 'Hollow cylinder or hoop: moments of inertia' section. It also has the derivation of the formula if you are interested, and yes, integrals are used). It is a bit surprising that you claim to know what sw is, and you claim to have multiple engineering degrees and yet you are unfamiliar with that formula.


not at all. Please re-read my post.


could you please explain what you mean by 'the minor non-vertical contribution of SW'? And define what the x,y, and z axis are?


you can't do that. Your results will not be correct.


of course the additional SW due to a leather grip is negligible. But if you are attempting to calculate it, as you did, then please use sound physical models and formulas. My problem with your post is not the value itself. It is the incorrect physics yo have used to get that number.


I'm good.

You know what, I admit my mistake, you're right. I went through writing the proof and realized a hole, which showed that for mass added over a relatively small range the change in MOI can be given approximately by my calculation above. I always do this when calculating the changes in SW upon adding mass because I add mass over relatively small ranges and I'm actually quite disappointed with myself for forgetting that it is indeed just an approximation. So thanks, I sort of had it coming tbh.

About the formula you used, I see the derivation and it looks sound. I mean the grip isn't circular in it's cross-section, if anything its cross-section is somewhere between elliptical and octagonal but I'm probably being a little overly critical. In any case we had rigid body dynamics in year two and it was a much more theoretical course so we didn't look at every special case and I certainly didn't go around memorizing formulas.

You may doubt my completion of engineering degrees, thankfully I don't need to prove it to you. If it's worth anything earlier today I actually built a model of a racquet handle on Solidworks to confirm the values attained through our different calculations, and I got something similar (~0.3416 kg*cm^2). Additionally (and not related to MOI) I published my first paper quite recently, if you'd like a copy please send me a message in private.

In any case it is important to make one thing clear to the general tennis community: when customizing a racquet by adding mass (usually lead), it is approximately correct to convert your distributed mass to a point mass, I don't think most tennis players can calculate integrals so this method is easier and quite precise when your mass is added over a range of less than about 10 cm. My main point originally was to point that out, since it is not too difficult to calculate in this method and this can help players who want to be able to customize more precisely.

@Slowtwitcher , you might have something to learn from this "effing grad student" after all, acknowledging one's mistakes and learning from them isn't something everyone can do apparently. That's one of the things my engineering degrees have certainly helped me be able to do, so I definitely won't be asking for a refund.

 

[Slowtwitcher]

@Slowtwitcher , you might have something to learn from this "effing grad student" after all, acknowledging one's mistakes and learning from them isn't something everyone can do apparently. That's one of the things my engineering degrees have certainly helped me be able to do, so I definitely won't be asking for a refund.

Sir, this is a Wendy's

 

[cyanide43]

https://imgur.com/a/zB3bRgB

 

[El_Yotamo]

https://imgur.com/a/zB3bRgB

Was waiting for someone with an RDC! Is the accuracy on that thing 0.1 or 0.5?

 

[Slowtwitcher]

Was waiting for someone with an RDC! Is the accuracy on that thing 0.1 or 0.5?

Uh oh...

 

[Aniruddha Phukan]

If by whippy you mean easier to swing then it's best not to add any mass at all. If you're looking to counterweight the butt is a good place because mass added there doesn't increase the SW by much relative to other places in the racquet and this will also make the racquet more headlight which can help handling and feel.

It is to make it more headlight. I don't have an issue with the weight, but with how clunky and head heavy (relatively) it felt.

I tried it with 1.5 grams in the butt, a world of difference - love it like this. I'll try going for a bit more.

 

[dr325i]

https://imgur.com/a/zB3bRgB

hmmm....post #55