The Qualities of a Racquet, Their Causes, and What You (Probably) Need

[Amone]

Because I don't have a set format, I'll start where all good arguments should start: My premises. I make a few very basic assumptions that I won't explain, such as that racquets exist, etc etc. However, here are the important premises:

1. That all the qualities of a racquet can be determined by certain information about the racquet, wether that information is published or not.
2. That there are a set number of important 'intangibles' about a racquet, which can also be simplified to their parts. These include:
   • Maneuverability
   • Power Level
   • Stability
   • Control
   • Comfort
3. That for a given swingspeed and racquet, there is a specific power
   • That power (momentum, and angular in the case of a racquet) is derived of two things: Head Speed (or more accurately, contact point speed) and Swingweight.
4. That some specs (most notably length) can be simplified into more basic parts.

This list is subject to change at any time, and will be noted at the end of this post when it does so.

Maneuverability is one of the many issues discussed widely on this forum.

• There is no part of maneuverability involved in beam width, as some have decided to believe.
• Maneuverability is comprised of two parts:
  • Linear, which is effected by static mass
  • Rotational, which is effected by swingweight along two axes (plural axis): Longitudinal (published swingweight) and Latitudinal (twistweight, which is not published)
• The higher the weight (static or rotational) the lower the maneuverability. Generally, rotational is by far more important than static.
  • Longitudinal swingweight is what it sounds like: swinging the racquet. It swings, however, around 3 axes: the shoulder, the elbow, and the core.
  • Latitudinal swingweight is also what it sounds like: twisting the racquet. It rarely matters at the baseline, where the racquet face is set long before the stroke begins, but matters at the net considerably, where one has to change the angle of the racquet face to adjust to changing conditions.

Power Level is an important trait to many, and one of the central issues involved in the 'SW2' concept, which is garnering (regardless of your position on it) quite a lot of attention.

• As I clarified earlier, though simplified, initial power level depends on angular momentum at impact... in other words, angular velocity, and swingweight.Final power, however, is based on both initial power and stability.
• For the sake of making use of this, I'll explain the central point of SW2 with one of these bullet points.
  • For a given power level, two factors matter. Initial racquet velocity, and recoil velocity, or 'stability.'
  • Those two factors split up into four parts themselves: Swingweight around the swinging axis (wherever you swing the racquet), Incoming racquet head speed, Balance Point, and Swingweight around the Balance Point (AKA Recoilweight). I can mathematically prove this, but at this point I'll spare you.
    • Increased I(10) /swingweight/ results in increased recoil.
    • Higher Balance Point results in decreased recoil.
    • Higher racquet head speed results in increased recoil.
    • Increased I(r) /recoilweight/ results in decreased recoil.
  • Lead in the hoop raises the balance point, increases I(10), decreases racquet head speed, and has a varying effect on I(r), depending on where the balance point is.
  • The final ball velocity, or 'power,' depends on the incoming velocity, and the recoil velocity, as I already clarified. SW2 decreases incoming velocity while decreasing recoil. Therefor, the same sum can be gotten in at least two different ways.
    • Basically, this is a subtraction problem. In fact, not basically... it is a subtraction problem. V(i)-V(r), to put it simply. Say you want (in simple math) to get the answer "3:" You might start with 8, and subtract five, or with 7 and subtract 4, etc...
  • The idea that you get more spin this way, I admit to being unable to completely explain at this time.
• An obvious factor in power level is flex. A stiffer frame bends less on impact, ergo loses less energy.
• Finally, string pattern is a small factor, due to the trampolining effect of the strings.

Ah, Stability. This factor is so important, it effects most of the others, too (Power, Control, and Comfort... In short, there is only one that isn't directly related). However, that means that most of it has already be explained. Therefor, we'll just recap here, unless it hasn't mattered:

• Twistweight is one third of stability, it's the difficulty in twisting the head. It varies based on how much lead is in the 3 and 9 (basically). Higher is more stable.
  • It is worth noting that head width, given the same distribution of mass, effects twistweight directly. Wider heads are more stable on off-centre hits.
• Recoilweight is the second third, it is how much the racquet recoils when hit (higher means less recoil). It can be derived by published measurements: Swingweight, Mass, and Balance. Generally, if both other measures are held static: More Swingweight means more Recoilweight, Higher Mass (remember, this is assuming the same swingweight, which is generally not a safe assumption) means lower Recoilweight, and Higher Balance Point means lower Recoilweight.
• The final third, and probably the least important, is static mass itself. Not only does it play a part in Recoilweight derivation, it's important for blocked shots, such as volleys and (some) service returns.

Control generally comes from three things:

• Swing Speed: Lower is better. If you tend to swing too fast, then your chances of mis-hitting are increased, as are your chances of guiding the ball improperly even for a properly struck shot. In short: You get less margin for error in your swing, past a certain point of swing speed.
• Spin: Accidental spin notwithstanding, spins provide depth control primarily, and directional control secondarily, mainly on sidespin shots.
• Trampoline Effect: This is a function of string length (wide or long headed racquets) and tension. This tends to effect both depth and directional control, because a deeper trampoline tends to result in more ball speed, and also tends to make the reflection of the ball more pronounced.

Finally, we come to Comfort. So indescribable, yet so important. However, it's important to try to figure out all this... stuff, so we must try to figure out what changes it.

• The materials a racquet is made out of will have a tendency to effect how it feels, regardless of flex or weight. Even though there are likely some wood racquets as stiff as the 200g Max, they certainly don't feel the same... just for one example.
• Stability is a major factor in comfort. How comfortable, after all, is it to have a racquet torque or recoil out of your hand? This carries with it most everything that stability entails.
• Finally, flex is a major factor in comfort. How many have commented that Babolat frames are uncomfortable, because they're too stiff, or that wooden frames are so comfortable because they're flexy (and stable, but that's another issue)?

I apologize, I definitely sputtered after Power Level... I will probably need to fix this all later, or make additions. Unlike re-wording or additions to my premises, however, those points will be added in seperate posts.

Good night,
Nathan Holden, AKA Amone

 

[anirut]

Thanks for the explanation, Amone.

... but I guess I must get my college physics book ...

 

[Noveson]

Yeah I definitely don't understand most of that. I'm only a sophmore Why does all of that matter though, why do we need to derive power, comfort, ect....To me it is just like percieved swingweight, how it feels is what matters.

 

[mileslong]

you lost me at angular...

 

[travlerajm]

I would take an entire book to comment on all of this (and I may in fact write one). But for today I'll just comment on one part at a time. I'll start by chiming in on my take on control, which is a little different:

Control comprises 3 essential elements:
1) Directional accuracy.
2) Control of the power level.
3) Control of the spin level.

In order to execute a groundstroke consistently and have the ball land at the proper depth, you must have command of all 3 of these elements.

Element 1: directional control.

Factors that increase directional control are stiffness, twistweight, recoilweight, and hitting weight. I like to define hitting weight as the first moment about an axis 4cm from the butt = M*(R - 4cm), because it is this property that has the most influence on stability. A racquet with higher hitting weight will feel most stable on volleys. Adding weight to the butt will destabilize your racquet and decrease directional control, even though it will increase swingweight adn recoil weight.

Also, essential to directional control is a constant racquetface angle through the contact zone. I.e., you must keep the plane the same. And, you must make contact near the center of the stringbed for maximum directional control.

A factor that people don't think about is that the spin-friendliness, or "bite" of a racquet affects directional control. Bite and directional control are inversely related. The more bite you have, the more your shot will be at the mercy of your opponent's shot's rpm. Stringing tighter will reduce bite, thus increasing directional control even for shots hitting the center of the stringbed.

Twistweight is highly important for directional control. Some people incorrectly assume that twistweight is only important on offcenter hits, and that if you hit the sweet spot, then low twistweight doesn't matter. This is untrue. When your opponent's shot has lots of topspin rpm, a ball impacting the sweetspot will embed into the stringbed and apply a torque to the racquet that twists your racquet more open. So if you use a 9-ounce racquet with low twistweight, it will be very difficult to volley a heavy spinning passing shot because the spin will make you pop balls up. But with higher twistweight, your racquet will twist less, even on impacts in the center of the sweet spot.

Recoil weight comes into play mostly on volleys. A higher recoil weight will cause the racquet to twist less about it's center of mass. Recoil weight mostly affects control of power level, but it alsi affects directional control in the same way that twistweight does. If you play against a player with a big kick serve or slice serve, the sidespin of the ball will cause the ball's impact to twist your racquet back, so you will tend to miss your returns high and left (assuming the big server is righthanded) unless your racquet is stable enough.

Stiffness is important for directional control because a ball will rebound at different directions on off center hits if the frame is too flexible.

I'll save the other 2 elements of control for another post.

 

[oldguysrule]

Amone,
Nice post...loooong, but nice. You put a lot of work into it and I commend you for your understanding and insights. Please don't take this as a slight, but, for me, a discussion of racquets at this level holds no interest for me. Actually, it might be of interest if I had time to delve into it. The funny thing is, folks around here think I know a lot about racquets. Little do they know...

I like to pick up a racquet and hit a tennis ball with it. If the tennis ball does what I want it to do, then I figure the racquet must be pretty good. I have opinions regarding weight, balance, headsize, SW, etc., but that's about it. The only thing that really matters to me is what happens to the ball when I hit it, how does the racquet feel, and will I be able to play with it with no pain.

One of these days, maybe one of your SW2 racquets will make it into my neighborhood and I will have a chance to hit with it. Until then, I will just have to be content with what I have. Good luck with your exploration...I see a high school research paper on the physics of racquets in your future.

Oh, don't forget to have fun playing tennis. It is a great game.

 

[Amone]

Amone,
Nice post...loooong, but nice. You put a lot of work into it and I commend you for your understanding and insights. Please don't take this as a slight, but, for me, a discussion of racquets at this level holds no interest for me. Actually, it might be of interest if I had time to delve into it. The funny thing is, folks around here think I know a lot about racquets. Little do they know...

I like to pick up a racquet and hit a tennis ball with it. If the tennis ball does what I want it to do, then I figure the racquet must be pretty good. I have opinions regarding weight, balance, headsize, SW, etc., but that's about it. The only thing that really matters to me is what happens to the ball when I hit it, how does the racquet feel, and will I be able to play with it with no pain.

One of these days, maybe one of your SW2 racquets will make it into my neighborhood and I will have a chance to hit with it. Until then, I will just have to be content with what I have. Good luck with your exploration...I see a high school research paper on the physics of racquets in your future.

Oh, don't forget to have fun playing tennis. It is a great game.

No offense taken. This is just what I have fun with, and I know it's above most people's heads and/or interests. *shrug* The problem is, I got sidetracked in the middle (bet you can spot it, it's right after the SW2 part) and when I came back I didn't have my heart in it.

This sort of discussion... eh. I get digressive, because I have focus problems anyways, and when I see something else to say, I just say it. My intention, though I-- excessively-- digressed, was to make an assertion, and then prove it... what I ended up doing was just sort of having fun explaining why things were the way they were, though I did one thing I had found important to do originally, which was to clear up some points where people seemed to have a whole lot of misconceptions, and to explain (I tried to use simple terms at times) SW2, since there is a lot of disagreement about how it's actually possible.

 

[haerdalis]

Higher Mass (remember, this is assuming the same swingweight, which is generally not a safe assumption) means lower Recoilweight
Nathan Holden, AKA Amone

The other way around?
Edit. Sorry i misread.
Ok but I dont get why higher sw means more recoil?

 

[Amone]

The other way around?
Edit. Sorry i misread.
Ok but I dont get why higher sw means more recoil?

Dammit, Haer! I start to take a TW break, and you ask me a question that I really want to answer!

This is going to get technical, because I can't explain it any less so.

Well, basically, you've got to have conservation of momentum. In short: p(i) = p(f). Initial momentum is made up of two parts, racquet momentum, and ball momentum, as does final momentum. So, instead, we have p(i,r) + p(i,b) = p(f,r) + p(f,b). For simplicity's sake, we'll say that we return the ball with same velocity as when it came to us. The ball's motion is linear, so p(b)=m(b)*v(b). If m(b) is the same, and v(b) is the same in the opposite direction, then we know that p(i,b)= -p(f,b)=p(b). Ergo, p(i,r)+p(b)=p(f,r)-p(b). Simplified, 2p(b)+p(i,r)=p(f,r)
/This is not what I expected to happen... I think I might've done it wrong before. But I'll keep going, and we'll see what happens./
However, the racquet's motion, and therefor momentum, is not linear. It's rotational. In that case, L=rp or L=Iw. L is the angular momentum, w is the angular speed. Plugging in that equation, we get:

2p(b)+L(i,r)/r(10)=L(i,r)/r(cm)

In this case, r(10) is the distance of the impact from the swinging axis, and r(cm) is the distance of the impact from the balance point. We don't know L, so we have to derive that a little more, to:

2*p(b)+I(10)*w(i,r)/r(10)=I(cm)*w(f,r)/r(cm)

We need to solve for w(f,r) so we multiply by r(cm) and divide by I(cm). Finally, we're finished, because w(f,r) is the amount of recoil you get: you want it as low as possible.

2*p(b)*r(cm)/I(cm) + I(10)*w(i,r)*r(cm)/r(10)*I(cm) = w(f,r)

If we look, we've got 7 variables. w(f,r) is the seventh, though that is what's being solved for, so we'll ignore it.


1 - Momentum of the ball increases recoil.
2 - Distance of impact from balance point increases recoil.
3 - Recoilweight decreases recoil.
4 - Swingweight increases recoil.
5 - Initial Racquet Head Speed increases recoil.
6 - Distance of Impact from the Swinging Axis decreases recoil.

 

[retrowagen]

Nathan,

Good stuff thus far. A couple issues jumped out which I'd like to discuss.

In your synopsis of manoeuverability, you mention that the rotational component is divided between two axes, longitudinal and rotational, to which I agree; however, I'd like to argue that there is a third axis (like on an airplane) which would be a sort of yaw motion around the center of gravity. Imagine it as the racket sitting flat on the ground (e.g., plan view) and it rotating like a helicopter propeller around its cg. This yaw axis is felt during play as how well the racket can snap to a "heads up" volley position.

Further in regards manoeuverability, I do believe that the width of a racket beam in the hoop area does bear some consideration. If weight/mass being completely equal between a "thin" beam and a "fat" beam, and the overall racket head width being equal (thus from a purely mathematical point of view, the rotational masses being equal), the fat beamed racket will shank more balls across its girth, and likely hit my leg on some swings, so from a purely practical point of view the fat racket is, in my mind, less manoeuverable.

Ball control is subservient to the three factors you mentioned, but you forgot a huge factor: contact area on the ball. The more points of contact (area) the racket exerts on the ball, the better it can transfer its kinetic energy to the ball, manipulating the dynamics of the ball. That is why it is generally felt that denser string patterns offer better ball control than do more open ones. However, the geometric regularity of the string "grids" also factors into this. I believe the wisdom of the ages has shown that perfectly square-shaped grids offer the best, most regular interface to the ball.


The three attributes I look for in a racket (as a moderately accomplished player) are:
1.) Ball Control
2.) Manoeuverability with a certain minimum total overall racquet weight (more than 11 ounces for me, by the way)
3.) Feel
Having played for 25 years, I have concluded that if a racket offers me anything beyond these three basics, it's probably going to interfere with my established game.

 

[haerdalis]

Dammit, Haer! I start to take a TW break, and you ask me a question that I really want to answer!

This is going to get technical, because I can't explain it any less so.

Well, basically, you've got to have conservation of momentum. In short: p(i) = p(f). Initial momentum is made up of two parts, racquet momentum, and ball momentum, as does final momentum. So, instead, we have p(i,r) + p(i,b) = p(f,r) + p(f,b). For simplicity's sake, we'll say that we return the ball with same velocity as when it came to us. The ball's motion is linear, so p(b)=m(b)*v(b). If m(b) is the same, and v(b) is the same in the opposite direction, then we know that p(i,b)= -p(f,b)=p(b). Ergo, p(i,r)+p(b)=p(f,r)-p(b). Simplified, 2p(b)+p(i,r)=p(f,r)
/This is not what I expected to happen... I think I might've done it wrong before. But I'll keep going, and we'll see what happens./
However, the racquet's motion, and therefor momentum, is not linear. It's rotational. In that case, L=rp or L=Iw. L is the angular momentum, w is the angular speed. Plugging in that equation, we get:

2p(b)+L(i,r)/r(10)=L(i,r)/r(cm)

In this case, r(10) is the distance of the impact from the swinging axis, and r(cm) is the distance of the impact from the balance point. We don't know L, so we have to derive that a little more, to:

2*p(b)+I(10)*w(i,r)/r(10)=I(cm)*w(f,r)/r(cm)

We need to solve for w(f,r) so we multiply by r(cm) and divide by I(cm). Finally, we're finished, because w(f,r) is the amount of recoil you get: you want it as low as possible.

2*p(b)*r(cm)/I(cm) + I(10)*w(i,r)*r(cm)/r(10)*I(cm) = w(f,r)

If we look, we've got 7 variables. w(f,r) is the seventh, though that is what's being solved for, so we'll ignore it.


1 - Momentum of the ball increases recoil.
2 - Distance of impact from balance point increases recoil.
3 - Recoilweight decreases recoil.
4 - Swingweight increases recoil.
5 - Initial Racquet Head Speed increases recoil.
6 - Distance of Impact from the Swinging Axis decreases recoil.

I dont like this part: Well, basically, you've got to have conservation of momentum. In short: p(i) = p(f). Initial momentum is made up of two parts, racquet momentum, and ball momentum, as does final momentum. So, instead, we have p(i,r) + p(i,b) = p(f,r) + p(f,b). For simplicity's sake, we'll say that we return the ball with same velocity as when it came to us. The ball's motion is linear, so p(b)=m(b)*v(b). If m(b) is the same, and v(b) is the same in the opposite direction, then we know that p(i,b)= -p(f,b)=p(b). Ergo, p(i,r)+p(b)=p(f,r)-p(b). Simplified, 2p(b)+p(i,r)=p(f,r)

Momentum of the racquet decreases when the ball is struck.
So p(i,r) - p(i,b) = p(f,r) + p(f,b). Assuming the numeric value of the balls momentum is unchanged we then get p(i,r) = p(f,r) + 2p(,b). Which makes a lot more sense to me.

 

[dave333]

agh, little numbers scare me!!!! Too much time w/ em in my algebra 2 honors class.

 

[Amone]

Nathan,

Good stuff thus far. A couple issues jumped out which I'd like to discuss.

In your synopsis of manoeuverability, you mention that the rotational component is divided between two axes, longitudinal and rotational, to which I agree; however, I'd like to argue that there is a third axis (like on an airplane) which would be a sort of yaw motion around the center of gravity. Imagine it as the racket sitting flat on the ground (e.g., plan view) and it rotating like a helicopter propeller around its cg. This yaw axis is felt during play as how well the racket can snap to a "heads up" volley position.

Further in regards manoeuverability, I do believe that the width of a racket beam in the hoop area does bear some consideration. If weight/mass being completely equal between a "thin" beam and a "fat" beam, and the overall racket head width being equal (thus from a purely mathematical point of view, the rotational masses being equal), the fat beamed racket will shank more balls across its girth, and likely hit my leg on some swings, so from a purely practical point of view the fat racket is, in my mind, less manoeuverable.

Ball control is subservient to the three factors you mentioned, but you forgot a huge factor: contact area on the ball. The more points of contact (area) the racket exerts on the ball, the better it can transfer its kinetic energy to the ball, manipulating the dynamics of the ball. That is why it is generally felt that denser string patterns offer better ball control than do more open ones. However, the geometric regularity of the string "grids" also factors into this. I believe the wisdom of the ages has shown that perfectly square-shaped grids offer the best, most regular interface to the ball.


The three attributes I look for in a racket (as a moderately accomplished player) are:
1.) Ball Control
2.) Manoeuverability with a certain minimum total overall racquet weight (more than 11 ounces for me, by the way)
3.) Feel
Having played for 25 years, I have concluded that if a racket offers me anything beyond these three basics, it's probably going to interfere with my established game.

You're right about the control issue, relating to strings and all... I almost never think about them, and know less (than almost nothing), but I had to mention something, so I threw up what I did know even though it was incomplete.

You're right that there is a third axis entirely, but it's generally almost identical to swingweight (less than 5% difference, I'm told) and if it's that little difference, I figure it's more-or-less negligible. However, beam width has very little to do with manoeuverability... (did I spell that right? Apparently I don't know how to. ^_^; because the closest measure you can get to that is Area Moment of Inertia, which is to say, how much surface area there is. A thicker beam by even 1 cm, isn't that much in the scale of most things.

 

[retrowagen]

You're right that there is a third axis entirely, but it's generally almost identical to swingweight (less than 5% difference, I'm told) and if it's that little difference, I figure it's more-or-less negligible.

It ties in nicely to where the racket masses are concentrated. On a headlight racket, the "yaw" is easily initiated; on a head-heavy racket, it is not. I perceive this as being a huge factor in manoeuverability, and somewhat similar to overall longitudinal rotational (total swingweight) consideration.

However, beam width has very little to do with manoeuverability... (did I spell that right? Apparently I don't know how to. ^_^; because the closest measure you can get to that is Area Moment of Inertia, which is to say, how much surface area there is. A thicker beam by even 1 cm, isn't that much in the scale of most things.

I'll agree that on paper, there's no difference. But when my topspin or slice strokes are applied to a ball and its deflection on the strings and manipulation by the super widebody racket cause it to collide with the wide beam and be clubbed over the fence (mishit!), then it's functionally not a manoeuverable racket, since it can't get out of its own way, so to speak. We all want a racket that wields like a cutlass, not a box kite!

 

[haerdalis]

So where are we on this? I just cannot buy that higher sw means more recoil. It definitely feels like more recoilweight to me.

 

[anirut]

Please continue the discussions ... thanks, guys.

 

[Amone]

So where are we on this? I just cannot buy that higher sw means more recoil. It definitely feels like more recoilweight to me.

It doesn't, necessarily. I'm assuming that SW2 has a lot to do with increasing Swingweight, until you stop getting that extra recoilweight-power out of it. Basically, I chose to make swingweight a stand-alone from recoilweight, so what we're looking at here is the effect of swingweight on power. How different modifications of a racquet will effect its power, wasn't what I was looking for. Just what the numbers, which everyone has so much debate over, actually mean.

But conceptually, it should make plenty of sense. You need conservation of momentum, and a bigger object has more. For a given ball speed (both inc and out, which was another assumption I made in my calculations, for ease of use) more sw means more rw, all other things remaining equal.

EDIT: I'll write it all out for you during school, Haer, and show you later.

 

[Kevo]

[*]That for a given swingspeed and racquet, there is a specific power

• That power (momentum, and angular in the case of a racquet) is derived of two things: Head Speed (or more accurately, contact point speed) and Swingweight.

I'd have to disagree with this. For one thing, swing weight does not equate to that portion of the frames weight that interacts with the ball. For another thing, different strokes are going to have various amounts of weight behind them. I think the term typically used on these boards is hitting weight. Also, you make no mention of frame stiffness, and this certainly factors into a frames power. Especially for lighter frames. I like your initiative in putting this all together and trying to collect this info in one place. Power is a really tough subject though, along with many of the other things described here. For me though the most important thing is feel. I haven't met a frame yet that was underpowered.

 

[Kevo]

Well, basically, you've got to have conservation of momentum.

Don't forget that the ball loses a significant portion of it's energy to compression. That will explain some loss of momentum that would lessen the racquets recoil.

 

[haerdalis]

Energy loss yes. Momentum loss no.

 

[sureshs]

Dude, L = r x p is a vector cross product. You can't get p by "dividing" L by r, unless r and p are perpendicular.

Your combining of linear and angular formulas also appears highly suspicious to me. I know you love the analytical stuff, but it has to be correct.

 

[Amone]

Dude, L = r x p is a vector cross product. You can't get p by "dividing" L by r, unless r and p are perpendicular.

Your combining of linear and angular formulas also appears highly suspicious to me. I know you love the analytical stuff, but it has to be correct.

You're right, that it does need to be perpendicular. However, if I were to show you the diagrammed formulas I worked with (I lost focus a few times so I needed drawings to keep my mind in line), you'd note that I did use perpendicular angles.

You're not necessarily so close, however, on the comment that the two aren't interchangeable in their way. You can't calculate a rotational momentum with linear equations, but you can use the final (instantanious) linear momentum, derived from rotational ones. After all, that is exactly what happens.