Going to be a tough one if that is the case.
So how's life Amone?
Edit: or should I say stubborn ass?
Lower Balance Point, Lower Power: Explained! ((Basic Physics Edition))
- Thread starter Amone
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Going to be a tough one if that is the case.
So how's life Amone?
Edit: or should I say stubborn ass?
Amone
Hall of Fame
Lolz @ 'stubborn ass' comment, it's just great. Aside from the very tiring loss today-- could've kept it going, I'm sure, brought up new points... then again, coulda killed myself, didn't-- it's just peachy keen. If I think about it, it's not so great, but I am enjoying myself nonetheless. Yourself?
I find several problems. In the first post, accln was 10 ms^-2, which I assume is an approximation to 9.8. Why would g be the acceleration of a horizontal ball?
Second problem is linear motion formulas are being used while the racquet tip motion is an arc.
Thirdly, in these calculations, moment of inertia and angular accln need to be used, as the racquet and ball are not point masses at this intimate scale of contact.
Something is quite not right. Very HL racquets don't put much mass behind the ball, so I don't think they are good for baseline topspin shots. Quite opposite to your results.
Second problem is linear motion formulas are being used while the racquet tip motion is an arc.
Thirdly, in these calculations, moment of inertia and angular accln need to be used, as the racquet and ball are not point masses at this intimate scale of contact.
Something is quite not right. Very HL racquets don't put much mass behind the ball, so I don't think they are good for baseline topspin shots. Quite opposite to your results.
Amone
Hall of Fame
Well, the acceleration is a vector. Assuming there's no wind resistance, I said, so there is no acceleration in the ball. There was some impulse, now the ball's moving at a constant velocity horizontally, and vertically it's moving at 10 m/s^2, assuming the ball is falling. If it was rising, it would be -10 m/s^2, because it would be decelerating upwards. You can get the final vector like the hypotenuse of a triangle. a^2 + b^2 = c^2, but if b = 0, then a = c, see? c is the actual acceleration, in two dimensions.
In some cases, I used rotational formulas. In some, I used linear ones for good reasons. In others, I was being stupid. I answered this in reverse, and noticed on the issue after this one.
Well, the problem is, Moment of Inertia is something like the torque you put on something from the axis of rotation. They are not the same, just in case I misspoke, but the basic idea is similar. Since I never assumed an axis at which the racquet was being rotated from that point, but instead explained where my axis was, beyond the point of the buttcap (where the racquet is being physically moved by the hand) so I used torque. The angular acceleration, I believe I did use. Perhaps I removed it-- no, it's there, but I converted it to linear acceleration. I'm so ******** sometimes! You got me there, I forgot to use that bit there.
What are you talking about? My results showed that a racquet with a lower balance was less powerful. I think you are agreeing with me and just confused. Maybe I'm misreading.
---
Drat, I said I was gonna stop doing this...
Then what did this mean in the OP:
I hope that this and this alone has shown that the ball applies a greater torque to the racquet with a lower balance point.
If it applies greater torque, why is it less powerful, since racquet must be applying same in the opposite sense?
Chauvalito
Hall of Fame
I have not read all of the posts...but from what I have read, the majority of you have provided degrees of support for your opinions; just wanted to say thank you, because we probably all have read claims in threads for which the poster will give no support/porof/argument, even when asked.
My only thoughts are that I hope we can keep things constructive so that we can all learn something.
have a good night all (its a bit late here).
My only thoughts are that I hope we can keep things constructive so that we can all learn something.
have a good night all (its a bit late here).
travlerajm
Talk Tennis Guru
I didn't vote in the poll, because the correct answer was not available. The correct answer is "sometimes yes, sometimes no."
At low swingweights (340 kg-cm^2 or less), lowering the balance point by adding incremental mass to the butt reduces power, while lowering the balance point by adding incremental mass to the top of the handle increases power.
These effects at low swingweights are due mostly to changes in the apparent coefficient of resititution.
At high swingweights (360 kg-cm^2 or more), lowering the balance point by adding mass anywhere on the handle will increase power.
This effect at high swingweight is mostly due to the increase in racquet velocity (much like adding a mass near the pivot of a metronome increases the oscillation frequency).
At low swingweights (340 kg-cm^2 or less), lowering the balance point by adding incremental mass to the butt reduces power, while lowering the balance point by adding incremental mass to the top of the handle increases power.
These effects at low swingweights are due mostly to changes in the apparent coefficient of resititution.
At high swingweights (360 kg-cm^2 or more), lowering the balance point by adding mass anywhere on the handle will increase power.
This effect at high swingweight is mostly due to the increase in racquet velocity (much like adding a mass near the pivot of a metronome increases the oscillation frequency).
How does added weight increase velocity? The racquet is inanimate it cannot produce more velocity.
The metronome example cannot be right. There is no way that ADDING mass/inertia to something is going to increase it's velocity. Are you talking about when you slide the mass down toward the base of a metronome? This would be moving the mass not adding mass, big difference. Moving mass down a metronome to increase it's frequency is analagous to making a racquet more head light. And by that same token inertia would be reduced, reducing energy given to the ball and in effect lowering power.
Now you could argue all day about which is better, high swing speed low inertia, or low sping speed high inertia? But that would be all subjective because the net energy transfered to the ball would be the same.
travlerajm
Talk Tennis Guru
The laws of physics tell us that adding mass near the pivot point of a physical pendulum does in fact increase its velocity.
At low swingweight (~330), the racquet velocity depends mostly on how fast a player tries to swing the racquet. But this is not true at higher swingweights.
A racquet at high swingweight (~370) behaves on a groundstroke much like a physical pendulum, with a natural swing frequency that is almost independent of the strength of the player (but dependent on the length of the player's arm). Let's do the math now.
The formula for the frequency of a pendulum:
F = 1/(2*pi*sqrt(I/(m*g*L)))
F = frequency
I = moment of inertia about axis of rotation
m = mass
g = acceleration of gravity
L = distance between center of mass and axis of rotation
If a racquet has a swingweight of 370 kg-cm^2, a mass of 360g, a balance R of 33 cm, and the axis of rotation is 20 cm beyond the butt, then by the parallel axis theorem:
I = 370 + 0.360*(2*(33 - 10)*(20 + 10) + (20 + 10)^2) = 1191 kg-cm^2
g = 980 cm/s^2
L = 20 + 33 = 53 cm
F = 1/(2*pi*sqrt(1191/(0.360*980*53))) = 0.6306 Hz
Now add 10g to the handle at 1 cm from the butt.
New specs:
I = 1191 + 0.010*(20 + 1)^2 = 1195 kg-cm^2
m = 360 + 10 = 370g
R = (33*360 + 1*10)/370 = 32.14 cm
L = 20 + 32.14 = 52.14 cm
F = 1/(2*pi*sqrt(1195/(0.370*980*52.14))) = 0.6330 Hz
0.6330/0.6306 = 1.004
So adding 10g to the butt of the racquet with these specs increases the angular velocity by approximately 0.4%. This doesn't sound like much, but this change in swing velocity has an effect on the power level that is on the same order as changing string tension by a few pounds. You will definitely notice it.
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T
topspin_17
Guest
Sorry about this but I'm sorta a beginner at all this torque n motion engineering stuff cos im only 13 (even though I'm one of the top students in my grade), but does this mean that a higher racket-headlightedness mean that the racket provides more control? and does this mean that if a racket is really headlight already, can I make it more powerful by adding weight to the inside hoop of the racket (12 and 3 and 9 o'clock etc.)?
and an extra question: what university degree category does this stuff fall into? eg. engineering or physics
and an extra question: what university degree category does this stuff fall into? eg. engineering or physics
anirut
Legend
This is about physics. Engineering is physics (and other disciplines) applied.
T
topspin_17
Guest
Ok I see now... Thanks once again Anirut...
Amone
Hall of Fame
The laws of physics tell us that adding mass near the pivot point of a physical pendulum does in fact increase its velocity.
At low swingweight (~330), the racquet velocity depends mostly on how fast a player tries to swing the racquet. But this is not true at higher swingweights.
A racquet at high swingweight (~370) behaves on a groundstroke much like a physical pendulum, with a natural swing frequency that is almost independent of the strength of the player (but dependent on the length of the player's arm). Let's do the math now.
The formula for the frequency of a pendulum:
F = 1/(2*pi*sqrt(I/(m*g*L)))
F = frequency
I = moment of inertia about axis of rotation
m = mass
g = acceleration of gravity
L = distance between center of mass and axis of rotation
If a racquet has a swingweight of 370 kg-cm^2, a mass of 360g, a balance R of 33 cm, and the axis of rotation is 20 cm beyond the butt, then by the parallel axis theorem:
I = 370 + 0.360*(2*(33 - 10)*(20 + 10) + (20 + 10)^2) = 1191 kg-cm^2
g = 980 cm/s^2
L = 20 + 33 = 53 cm
F = 1/(2*pi*sqrt(1191/(0.360*980*53))) = 0.6306 Hz
Now add 10g to the handle at 1 cm from the butt.
New specs:
I = 1191 + 0.010*(20 + 1)^2 = 1195 kg-cm^2
m = 360 + 10 = 370g
R = (33*360 + 1*10)/370 = 32.14 cm
L = 20 + 32.14 = 52.14 cm
F = 1/(2*pi*sqrt(1195/(0.370*980*52.14))) = 0.6330 Hz
0.6330/0.6306 = 1.004
So adding 10g to the butt of the racquet with these specs increases the angular velocity by approximately 0.4%. This doesn't sound like much, but this change in swing velocity has an effect on the power level that is on the same order as changing string tension by a few pounds. You will definitely notice it.
Travler... I feel like a tool whenever you open your mouth. However, thanks for the information! I think I'm gonna use it, haha.
By the way, [read: edit] I noticed you used an O3 Tour OS. Could I get some thoughts? I've been wanting to be interested, but nobody else seems to know anything about it at all. I'd like to get an idea of how it's base specs go together, because I can get what I want out of a racquet within reason. My DNX has some things I would like to do outside reason, so I have started eyeing some other racquets, and that was one.
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The laws of physics tell us that adding mass near the pivot point of a physical pendulum does in fact increase its velocity.
At low swingweight (~330), the racquet velocity depends mostly on how fast a player tries to swing the racquet. But this is not true at higher swingweights.
A racquet at high swingweight (~370) behaves on a groundstroke much like a physical pendulum, with a natural swing frequency that is almost independent of the strength of the player (but dependent on the length of the player's arm). Let's do the math now.
The formula for the frequency of a pendulum:
F = 1/(2*pi*sqrt(I/(m*g*L)))
F = frequency
I = moment of inertia about axis of rotation
m = mass
g = acceleration of gravity
L = distance between center of mass and axis of rotation
If a racquet has a swingweight of 370 kg-cm^2, a mass of 360g, a balance R of 33 cm, and the axis of rotation is 20 cm beyond the butt, then by the parallel axis theorem:
I = 370 + 0.360*(2*(33 - 10)*(20 + 10) + (20 + 10)^2) = 1191 kg-cm^2
g = 980 cm/s^2
L = 20 + 33 = 53 cm
F = 1/(2*pi*sqrt(1191/(0.360*980*53))) = 0.6306 Hz
Now add 10g to the handle at 1 cm from the butt.
New specs:
I = 1191 + 0.010*(20 + 1)^2 = 1195 kg-cm^2
m = 360 + 10 = 370g
R = (33*360 + 1*10)/370 = 32.14 cm
L = 20 + 32.14 = 52.14 cm
F = 1/(2*pi*sqrt(1195/(0.370*980*52.14))) = 0.6330 Hz
0.6330/0.6306 = 1.004
So adding 10g to the butt of the racquet with these specs increases the angular velocity by approximately 0.4%. This doesn't sound like much, but this change in swing velocity has an effect on the power level that is on the same order as changing string tension by a few pounds. You will definitely notice it.
What about calculating the length of the arm? would it be lower swingweights for longer arms and higher swinweights for shorter? or vice versa?
jackson vile
G.O.A.T.
IMO you have to consider where you are playing from most also, IE basliners enjoy extended length because they can have a higher SW with less weight.
So the more you come to the net or just plain want stability then you add more weight reguardless of SW in that case.
The point here is that you want to keep that racket as effective for you game as possbile while keeping the weight as low as possbile
joeyscl
Rookie
"This is about physics. Engineering is physics (and other disciplines) applied."
LOL, this is what We always associate Engineering with and what I always thought. But if I were to define Engineering i would mention something about how it is a Method of problem solving ETC... CUZ Just recent i Stumbled upon the word
Bioengineering
Genetic Engineering
...i dont think this has ANYTHING to do with Physics, lmao
Anyway, This Balance point and Mass distribution stuff Definetly has to do with has to do with Rotational Momemtum (and of course, more momemtum= faster ball)
LOL, this is what We always associate Engineering with and what I always thought. But if I were to define Engineering i would mention something about how it is a Method of problem solving ETC... CUZ Just recent i Stumbled upon the word
Bioengineering
Genetic Engineering
...i dont think this has ANYTHING to do with Physics, lmao
Anyway, This Balance point and Mass distribution stuff Definetly has to do with has to do with Rotational Momemtum (and of course, more momemtum= faster ball)
