Bats and Balls

Cecil's Columns/Staff Reports
1

I think the staffer dropped the ballon this one, by not actually answering the question. He comments the relationship between force, mass and acceleration without, and the tradeoff between them in terms of how force gets applied to the ball, and then blithely states that heavier bats are better.

Taking this to the extreme, if you imagine a thousand pound bat then hitting a ptiched ball would be basically the same as throwing against a wall, which would clearly not give it as much momentum as people get when they actually hit the ball.
I think the following is a better answer.

If the batter applies a fixed force (F) over a fixed distance (d), then regardless of the mass of the bat (M), the bat at the time of collision will have energy E=F*d
Similarly the energy e=(mv^2)/2 of the ball will be fixed depending on its mass (m) and starting velocity(v) . To maximize the ending velocity of the ball, the optimal situation would be that after the collision the bat is stationary so that all energy has transferred to the ball.

So that in the end the ball has energy = (Fd+(mv^2)/2), so its ending velocity will be v’=Sqrt(2Fd/m+v^2), and its change in momentum will be m(v’+v). In order for momentum to be conserved, this must be equal to the original momentum of the bat, which is equal to Sqrt(2EM).

So M=m^2*(v’+v)^2/(2*E) is the optimal bat weight.

So if we set reasonable values for F=80 kg*m/s^2, d=0.5m , v=45 m/s and m = .16kg

We get that the optimal bat weight is around 2.9kg (damn heavy bat). But with a stronger hitter or a slower balls lighter bats will be better.

2

I think there is a very important timing issue.
When the ball meets bat, it deforms, absorbing energy. The bat slows a bit. Energy is dissipated through the arms of the batter via shock waves.
Ideally, the bat continues with the least amount of deceleration, or if timing is right, maybe increasing acceleration. At some point the ball begins regaining it’s shape, bouncing off the bat due to that reshaping.
The best transfer of energy should be bat speed keeping ahead of the ball reforming it’s shape. ( if that speed is even possible )
The bat pushing even after the bounce.

If the bat hits with too much force, the ball might be permanently deformed, lessening the rebound off the bat. Absorbing too much of the bat energy, with no return.

At least that is how I visualize it, with my math handicap.

3
4

Is there a paragraph missing from this column? Because seriously, at present it’s just plain bad writing.

It just goes from inconclusive background to The Answer with no explanation of how the answer is derived.

5

The staffer’s response is naive in the extreme. The response does not come close to answering the question as posed. I cannot answer it precisely myself, but I do have a better appreciation of the issues involved.

F=ma has very little to do with it. The real issue is how much impulse is transferred to the ball during the time it is in contact with the bat. That determines its change in momentum. To get this, you have to integrate the force between bat and ball over the time while they are in contact. Yes, that force does depend at any instant on the rates at which the ball is accelerating and the bat is decelerating; but getting at the actual values of this force is the hard part. It is definitely not constant since both the ball and the bat are deforming. This is actually a very difficult problem to model in exact detail. It is especially difficult since the question leaves open the issue of how the mass of the bat is distributed. It would be more reasonable to assume a particular bat shape and uniform density. Then there are additional issues about its flexibility, compressibility, etc. Such issues are not 100% predictable for all balls either.

Consider extreme cases of bat density: Too heavy, and it will be moving so slowly when the ball comes that the situation will be like the ball bouncing off of a stationary object. Too light, and the bat will be moving so fast that the impact with the ball will destroy the bat.

6

The stuff on the deformation of the ball, and the force application rate, is mostly irrelevent: the situation is described, as above, by total energy and total momentum, and can be calculated from initial and final states, ignoring the rate of change during the collision.

However, the calculation needs to be corrected for energy loss, which I believe is significant for Baseball. I’m thinking that there is something like 75% energy loss. If you drop a baseball on a hard floor, how high does it bounce? That should give the other limit for the range (a baseball bat is not hard, so the actual number will be between the uncorrected and first-order corrected values)

7

That’s not exactly correct, depending in what is meant by “constant torque”. I think the staffers answered the question taking that top mean a constant angular acceleration. If you can accelerate the bat at a constant rate independent of bat mass, then indeed a more massive bat will give more energy and momentum.

However, or you take “constant torque” to mean the “angular force analog”, which is what torque is, then ability to accelerate the bat is a crucial element.

I’m not seeing the problem. They explain the general principles at play, then assume bats are pretty optimized.

That’s my complaint. They answered the wrong question. They talk about accelerating the bat, not how the bat and ball interact.

8

I have no idea how you derive this from the column in question.

The first three paras explain F=ma, and the fourth states that increasing and decreasing mass has an inverse effect on acceleration. Which means presumably there is a trade off, but how to optomise in the face of that trade-off isn’t dealt with. At all.

Then *BAM *para five just vomits out an answer.

9

Column on the front page again.

Princhester, you state that how to optimize isn’t dealt with at all. You are correct in that they don’t address the right question.

As I stated, SDStaff Ken makes this statement:

This, I take it, states his underlying interpretation of the question,

If you assume “fixed torque T” to mean “same angular acceleration”, then his answer that more mass equals a farther hit is the direct result, no other optimization factor is required.

But I think “with a fixed torque T” refers to “angular force analog” as I called it.

Torque (T) = Mass (m) x Angular Acceleration (w)

Where the interplay comes into it is momentum transfer and energy transfer, which are governed by the two basic equations

momentum(M) = mass(m) x velocity(v)

and

Kinetic energy (KE) = 1/2 x mass(m) x [velocity(v)][sup]2[/sup]

Of course for the momentum transfer you have to consider the losses from inelasticity of the system (i.e. ball, bat wobble, batter arms, etc).

However, if you look at the kinetic energy equation, you see that velocity[sup]2[/sup] term.

Thus, if the angular force is constant, then an increase in mass will decrease the velocity of the bat on impact with the ball.

Because in the kinetic energy equation mass affects the energy linearly, but velocity affects by the square, I believe that a lighter bat will allow a faster swing, thus more velocity on impact, thus a greater increase in KE than increasing the mass, which reduces the velocity on impact.

I’m sure a numerical answer would likely include the calculus that DrHow mentions, but analytically I believe this captures the essence of the effects as pure factors.

And as SDStaff Ken states, in the real world, batter’s abilities are limited by their ability to see the ball coming and judge when to swing as well as other factors. Thus any particular batter may not do better with reducing his bat weight. A heavier bat may help some and hurt others.

10

I want to take exception to one statement in the column:

McGwire’s biceps are better explained with one word: steroids.

11

NM wrong thread

12

Nope. Energy in this situation is a product of force and distance, not mass

The energy of a given mass increases non-linearly with velocity, as you say. However, where the velocity results from a given force over a particular distance, the decrease in velocity with an increase in mass is correspondingly non-linear, and consequently the resulting kinetic energy of the mass stays the same.

If that seems counter intuitive, bear in mind that as mass increases, acceleration decreases, but over the same distance the mass is accelerating for a longer time because it takes longer to cover the distance. This doesn’t cancel out the decrease in velocity due to increased mass, but it does mean it doesn’t decrease linearly.

The resultant velocity of a mass subject to a given force varies linearly with mass and time not distance.

The equations:

F=force. We are assuming per your example that this is fixed
m=mass. The bat.
a=acceleration
v=velocity
d=distance. I’m assuming the swing distance from shoulder to impact stays constant

F=ma or a=F/m
v=at or
d=1/2at[sup]2[/sup] or t=√(2d/a)
E=1/2mv[sup]2[/sup]

Combining the first two equations obviously gives you:

v=tF/m

Combining that with the third equation gets you:

v=(F/m)√(2d/a)
or
v= √(2dF[sup]2[/sup]/am[sup]2[/sup])

Now we substitute out acceleration from that equation by combining it with the first equation again and get:

v=√(2dF[sup]2[/sup]m/m[sup]2[/sup]F)
or
v=√(2dF/m)

Then combining that with the fourth equation you get:

E=1/2m(√(2dF/m))[sup]2[/sup]
or
E=Fd

All else being equal, a change in mass does not alter the resultant energy of a mass accelerated by a given force over a given distance.

13

I’m sorry but the answer in the staff report seems perfectly correct to me, because the answer is this line:

“… a light bat or a heavy bat is best depends on the skills and strength of the batter.”

Well, okay, there’s a grammatical error. But the idea is correct; the ideal weight of the bat is dependent upon the batter. If the bat is too light, it will not impart as much force as it could; there is a point at which you can’t swing it any faster no matter how light it is. But if it is too heavy the batter will not be able to swing it fast enough to maximize the force.

I don’t know what else could have been said without getting into complications that aren’t relevant to the question that was asked.

14

Your answer is an illustration of why it’s such a crap report. You’ve put it better - in about half the words - than the report. Most of the report is irrelevant waffle, given the answer in the line you give.

It really reads as if the writer started writing technically about the mechanics of the situation, then realised none of it was leading anywhere, so just finished off with “eh, it’s too hard, the answer is that it depends.” The first four paras could have been cut without affecting the ultimate answer given.

15

Force has nothing to do with this question, it’s all about momentum. Specifically the conservation of momentum. Momentum is the mass times velocity of objects involved in a collision. The mass of the bat times it’s velocity + the mass of the ball times it’s velocity before the collision will equal the mass of the bat times it’s velocity + the mass of the ball times it’s velocity after the collision. So all else being equal, a heavier bat will always, always impart a higher velocity to the ball.

Of course a heavier bat is harder to swing at a higher velocity than a lighter one, but that’s really not the question.

16

Actually, the question says “equal torque”, so the difficulty of swinging the heavier bat is definitely a factor.

17

As JWK says it is precisely an element of the question and consequently what you say is not wrong but isn’t an answer.

18

Princhester, thank you for the calculations. So if the force is the same, then the change in mass and change in velocity offset each other for the energy balance, due to the time element of the motion.

In regards to your original point, SDStaff Ken gave two parts to his “answer”. The first part was if we assume that baseball is played by machines, which he discounted in the previous paragraph, but for that answer he assumed that “fixed torque” meant the velocity would be the same regardless of the mass of the bat. In that case, a heavier bat is always better. If the machine can swing a truck at the same speed it swings a toothpick, then more mass is better.

His second part is to say that the answer in indeterminate in a real world case. Here, he basically discounts the fixed torque requirement.

In other words, he fails to answer the intent of the question. It is a crap answer, but the answer he “vomits out” comes from his assumption “the game were played strictly by laboratory robots”. He interprets “fixed torque” to mean the speed of the bat is the same, not that the force applied to the bat is the same.

No, it only appears like a grammatical error when you extract part of the quote and show it independently. The actual quote is:

“… whether a light bat or a heavy bat is best depends on the skills and strength of the batter.”

Whether makes all the difference.