Who Cares about Recoil Weight?

[esm]

I've never heard of 'pickup weight,' so thanks for the heads-up @Grafil Injection. It's curiously relevant. My theory Is that recoil weight, as a Moment of Inertia measure, most affects the beginning your swing, where we use a dynamic slot to pickup momentum. The fact that the racquet head flips around behind the hand means that here's where changes in the center of mass make the racquet feel anywhere from too maneuverable in the slot to too laggy coming out of it.

This idea is corollary to Crawford Lindsey's research into how swingweight affects the relative speed of the forearm and racquet during whip-through into ball contact later in the swing. If I'm right, this means both measures are essential, but to different stages in the swing. It also menas that you can use them together to give you perfectly maneuverable slots and perfectly whippy angles of impact.

Back to pickup weight, as a combination of weight and balance alone (vs. RW, which also includes SW) it's similar to @Irvin's idea that weight x balance = torque may be the best measure of racquet maneuverability.

So I have been fine tuning my RW for a couple of months now.
According to the chart my suggested RW is in the 158 range and I have been using between 158 - 160 for quite a long time and have always felt something isn’t just quit right.
So on an off chance/in an effort to reduce weights (static, SW and TW), I landed on RW157.xx
For some reason everything felt just abit better, with greater results.
QQ - in my case, what’d be (any) difference between a night low RW157.xx and high RW157.xx?

I have sent up a racquet with the similar SW and balance and v minor increase in the static weight, but it landed on mid RW155.xx. Going to give it a whirl and see if I get a chance later today.

My aim is to settle on. Setup that is as low in static weight and SW that I am comfortable with and balance at 32.5x. I felt it’d give me the best performance advantage and I won’t get too fatigued at the need of a league match session. Lol

 

[Brando]

RW does not include SW, it’s the other way around. SW = RW + mr^2

Thanks for the correction, @Irvin. You are technically correct, RW = MOI = Σ mass x (distance of each particle from the balance point)²
...none of which has anything to do with swingweight.

What I could've more accurately said is that we geeks typically use swingweight to functionally calculate recoil weight, as in:
Recoil Weight = swingweight - (mass in kg x (balance point in cm - 10)²)

 

[Irvin]

Thanks for the correction, @Irvin. You are technically correct, RW = MOI = Σ mass x (distance of each particle from the balance point)²
...none of which has anything to do with swingweight.

What I could've more accurately said is that we geeks typically use swingweight to functionally calculate recoil weight, as in:
Recoil Weight = swingweight - (mass in kg x (balance point in cm - 10)²)

All that is correct but swing weight and recoil weight are both the same thing. Through the use the parallel axis theorem one cand convert the RW at the COM to another axis. The axis used to SW is 10 cm which is the basis for:
SW = RW + mr^2 or RW = SW - mr^2

 

[ZIMING]

All that is correct but swing weight and recoil weight are both the same thing. Through the use the parallel axis theorem one cand convert the RW at the COM to another axis. The axis used to SW is 10 cm which is the basis for:
SW = RW + mr^2 or RW = SW - mr^2

no, they are not the same thing. the devil is in Mass x (BP-10)², where there is a square thing which makes the formula non linear

 

[tele]

no, they are not the same thing. the devil is in Mass x (BP-10)², where there is a square thing which makes the formula non linear

I think he meant that both are moments of inertia, just about different axes.

 

[Irvin]

no, they are not the same thing. the devil is in Mass x (BP-10)², where there is a square thing which makes the formula non linear

It does not matter if you use the 10 cm axis for SW or the 0 cm axis for I (MGR/I,) if the RW goes up or down x points the SW and I will go up or down x points (assuming weight and balance stay the same.) If you ever want to increase SW by 10 points but maintain the present balance, increase RW 10 points adding the same amount of torque on each side of the BP. This will get you pretty close.

EDIT: I use this method often when customizing. First I add mass to match balance and weight. If I need to increase SW I divide the mass in half and place it above and below the initial point equidistantly. This does not change the mass or balance but increases RW, SW, and I.

 

[tele]

It does not matter if you use the 10 cm axis for SW or the 0 cm axis for I (MGR/I,) if the RW goes up or down x points the SW and I will go up or down x points (assuming weight and balance stay the same.)

I am not sure I understand this point. If I add the same amount of weight at noon, RW, SW, an I should not all be affected by the same number of kg/cm^2.

 

[Irvin]

I am not sure I understand this point. If I add the same amount of weight at noon, RW, SW, an I should not all be affected by the same number of kg/cm^2.

That’s true because you are increasing the balance point and not just the SW, RW, and I. If the weight are balance stay the same only the inertia increases.

EDIT: SW = RW + mr^2 Therefore is mr^2 remains the same (especially r) any increase in RW directly increases SW.

 

[Irvin]

@tele assume you add 3 g at 34 cm to match a rackets weight and balance you want. Then you want to increase the SW, RW, or I by 2 points.

I = mr^2
2 = .003r^2
2/.003 = r^2
SQRT(666.667) = r
25.82 = r

So you place 1.5 g at 34-25.82=8.18 cm and 1.5 g at 34+25.82=59.82 cm. You will increase inertia by 2 points and not change the weight or balance from the torque you had originally with 3 g at 34 cm.

 

[tele]

@tele assume you add 3 g at 34 cm to match a rackets weight and balance you want. Then you want to increase the SW, RW, or I by 2 points.

I = mr^2
2 = .003r^2
2/.003 = r^2
SQRT(666.667) = r
25.82 = r

So you place 1.5 g at 34-25.82=8.18 cm and 1.5 g at 34+25.82=59.82 cm. You will increase inertia by 2 points and not change the weight or balance from the torque you had originally with 3 g at 34 cm.

I still think I am not following you. In the case you pointed out above, I and SW will be affected differently than RW and each other, even though weight and balance remain the same. Assuming I did not make mistakes in my calculations, SW would go up by (.0015*1.82^2)+(.0015*49.82^2)=3.73, and I would increase by (0.0015*8.18^2)+ (0.0015*59.82^2)=5.47, right?

 

[Irvin]

I and SW will be affected differently than RW and each other, even though weight and balance remain the same.

I (or SW) = RW + mr^2 if mr^2 remains the same increases in I, SW, or RW are all equal always.

EDIT: if the value of mr^2 is 155 Kgcm^2 then SW = RW + 155. If RW changes x points so does SW.

 

[tele]

I (or SW) = RW + mr^2 if mr^2 remains the same increases in I, SW, or RW are all equal always.

But in your example, mass is not remaining the same. You were talking about adding weight.

edit: also, r in the formula above refers to the distance from the com to a different axis of rotation, whereas the r in your example was used to talk about the distance from the com to where weight was being added

edit2: i looked at your example again to make sure I had not missed something. so your point is that if you redistribute weight that already exists(i.e. no change in mass) in a racquet such that the balance point also remains the same, sw and rw will increase by the same number of points? If so, I that makes sense based on the parallel axis theorem equation. However, rw and sw start being affected differently as soon as you start *adding* weight, even if you keep the balance constant.

 

[Irvin]

But in your example, mass is not remaining the same. You were talking about adding weight.

In my example in post 1,159 mass remained constant

also, r in the formula above refers to the distance from the com to a different axis of rotation, whereas the r in your example was used to talk about the distance from the com to where weight was being added

That is true. If r is the point where weight is located and you divided that weight so the torque above that point is the same and the torque below then mas and balance does not change.

i looked at your example again to make sure I had not missed something. so your point is that if you redistribute weight that already exists(i.e. no change in mass) in a racquet such that the balance point also remains the same, sw and rw will increase by the same number of points? If so, I that makes sense based on the parallel axis theorem equation.

True but it has nothing to do with the parallel axis theorem.

Assume you have a racket and you added 3 g at 34 cm. That adds 102 gcm of torque. If you added 1.5 g at 10 cm and 1.5 g at 58 cm you would also add ((1.5*10)+(1.5*58))=102 gcm. As long and you use 3 g divided up so the total torque is 102 gcm mas and balance does not change.

However, rw and sw start being affected differently as soon as you start *adding* weight, even if you keep the balance constant.

If mass and balance remain the same they don’t. 3 g is the same as 1.5+1.5 g.

 

[tele]

True but it has nothing to do with the parallel axis theorem.

It has nothing to do with the equation? If you.keep the mass and balance constant but shift the weight around, both rw and sw increase by the same amount, but mr^2(with r being the distance from the com to the sw axis) stays the same because thebalance does not change so the com is still thesame distance from the sw axis. Is I=rw+mr^2 not the parallel axis theorem equation?

 

[tele]

If mass and balance remain the same they don’t. 3 g is the same as 1.5+1.5 g.

right, but my point was that if you *add* that wright instead of just moving it around, rw and sw will be affected differently. If I had（edit:i meant to say "add") 3g at the com, rw remains constant but sw Increases. If I *add* 1.5g at the two locations in your example, RW and SW go up by different amounts. If I *transfer* equal amounts of weight from the com to those locations, rw and.sw go up by the same amount.

 

[Irvin]

Is I=rw+mr^2 not the parallel axis theorem equation?

Yes but I was taking about splitting the weight so m and r remain constant. RW and SW increase as you separate the mass farther apart. That is PAT

 

[tele]

Yes but I was taking about splitting the weight so m and r remain constant. RW and SW increase as you separate the mass farther apart. That is PAT

I was talking about that too in the post you quoted. Either way, it seems like we do not have a disagreement about the physics.

 

[ZIMING]

plow thru is important to consider in modding. if stability is not enough return on power serve will be awfully terrible.

I had head heavy issue on my PA plus so I removed bumper guard. Return was always not satisfactory with or without bumper guard. Then I added about 4g at both 3/9, and about 5g inside buttcap if I remember right. Problem then solved while other parameters are still good to me.

 

[Sampo]

Hello, @Brando!
Over the past few months I've done several tests regarding RW with various different rackets. And I've always get indications that if I customize any racket to get closer to the RW indicated for my height (1,76 m), those rackets always improve significantly. My best racket (#1) has 323 grams, 321 SW, 32,1 BP and 14,0 TW (RW = 163 and MgRI = 20,5).
My problem now is: when I try (and I’ve tryed a lot) to match the racket #2 (same model as #1), the best especification it can get is 322 grams, 319 SW, 32,2 BP and 14,3 TW (RW = 160 and MgRI = 20,6). But the racket #2 still performs much, much worse (especially heavy weight and unforgiven) than #1.
My question is: would the higher TW of racket #2 makes that much of a difference, to the point of "compensating" a little for its lower SW ? (i.e.: for the same weight and BP, would higher SW and lower TW equals to lower SW and higher TW ?) Could you please help me with this ?

Thank you very much for all your work! It's wonderful!