Another baseball/physics question

Factual Questions
1

Baseball players and announcers are unanimous in believing that it is easier to hit a HR off a fast ball than a slow pitch. This makes no sense to me for reasons I will explain below. I think that they are using the intuition they get from throwing a ball against a wall, where it is certainly the case that the harder you throw it, the harder it rebounds.

Here is how I see it. There are two issues: energy and momentum. Now a fast ball certainly has more energy, so if you ignore momentum, it looks plausible. A solid wall is rigid and can absorb all the momentum you can throw at it (up to a point, but a thrown ball is nowhere near that point). A batter is not a rigid wall and must absorb and reverse all the momentum of the pitched ball. And the harder it is thrown, the more momentum he must supply. Does anyone understand the physics of this well enough to clarify the situation?

Of course most HRs are hit on fast balls, but that is because fast balls have more readily predicted paths and if your reflexes are good enough, you will pick it up. If your reflexes are not fast enough, you will not be in the major leagues.

2

The batter does not absorb and reverse the energy; the bat does.A springy wooden bat can turn the energy around.

3

Yes. The professionals have no idea what they are talking about and the statistics are conspiring to trick the public.

OK, really though, if you’re having trouble with this. Have a friend hold a basball bat for you. Take a baseball and hold the ball against the bat and let go…it will go nowhere. Now through the ball at the bat lightly making sure your friend holds th abat firmly but without swinging it. The ball will now bounce off a little. Now being careful not to hurt your friend (and he a true friend for helping you do such an obvious experiment), really wing that sucker at the bat and it will bounce back even further, just like throwing at the wall.

Now do you believe the pros?

4

I don’t know the physics, but I’ve played baseball forever. And when you hit a fastball, it just feels right, and the ball seems to go further. When you hit a change/breaking ball, it’s almost a sluggish sort of feeling, unless they leave one up in the zone that you just cream…

5

I couldn’t come close to writing the equations, but the batter swings the bat giving it momentum, and if the ball hits the bat at, or near, the center of mass and the timing is right, the ball’s momentum is easily reversed. The batter hardly feels any reaction at all.

6

By the way, I think a properly hit baseball is hit with the bat traveling down and through the ball, like a golf shot only not so sharply down, which gives it backspin. This backspin gives the ball aerodynamic lift and makes it carry further. It might be easier to give backspin and thus a lot of carry to a fastball than to a slow one.

7

This thread was accidentally started twice, with both copies having responses, so I merged the two threads.

bibliophage
moderator GQ

8

The harder the ball is thrown, the more the ball compresses when striking the bat, storing energy, which is then released as it jumps off the bat.

Of course, the bat’s speed has a lot to add to that energy, too.

9

Weelll, sort of. In almost any collision momentum is conserved.

So we think that that bat/batter/ball system before and after the collision is the same. OK? A batter comes around with his bat. The momenta of the bat, and the batter with his attendant to the Earth are the same, regardless.

And, for the sake of fun argument, let’s say that the ball/bat interface angle is the same.

Then the total impulse needed to send the fastball over the fence is greater than the one needed to send the haging curve outta the yard.

And re Billy’s remarks: rather than have a trusted friend hold the bat mount it sticking out in the air between some bricks. At least that’s what I used. The harder you throw the farther it goes; but it’s going farther towards the catcher, in this case.

I say that partly tongue in cheek. I just don’t think it’s fair to treat the bat/batter as an infinitely rigid object…

Also, in almost any collision energy is not conserved. So a more forcefully thrown ball actually loses more energy coming off of the bat than a smacked hanging curve ball.

Look for this post to get jumped on, but I think I’m right…

10

I am not convinced. I don’t think the momentum of a baseball is trivial. I have seen infielders bowled over by a very hard line drive. Yes a ball that hits a rigid bat will go further the faster it is thrown, but that is not the issue. The bat is not held rigid and the reversal of the momentum vector must come from the swinging bat.

If the reasoning given was correct, it ought to be very difficult, if not impossible, to hit a fungo out of the park. But I regularly see aged coaches hitting fungoes to all parts of the outfield, expending relatively little effort and I have little doubt they could hit it out if they wished.

11

A ball weighs 5 ounces. A bat weighs upwards of 30 ounces, and is actively powered by the batter’s muscles. The momentum of the ball is easily reversed because it is so small in comparison. I doubt the batter even feels much of a blip when they strike the ball.

Interesting thought… what is the speed of the ball as it comes off the bat, in comparison to the speed of the bat itself? I think it it would be easy to say that the ball speed is higher than the bat speed.

By your initial thoughts, the ball would go farthest if it were moving away from the batter when struck. But, if it were moving away at the bat speed, the batter could never get the ball moving at above the bat speed, and the ball would not go far at all. I’ll use that as my starting point. I would say that the ball would go farther if its velocity moved more slowly away from the batter, to being stationary, to moving towards the batter. At some point, of course, this total speed would peak, my intuition says that peak is at a very high speed towards the batter.

Fielders do not get knocked over by the ball, they fall over trying to get out of the way. Batters who get hit by the ball don’t move an inch, catchers are squatting down on their toes and never fall over when they catch the ball or get hit. There’s not enough momentum there.

12

Short answer: the momentum of a baseball, even one on its way to the 410 foot mark, is not that big in comparison to a batter and bat. Say, a 150 pound batter, plus a 2 pound bat, and a 5 ounce ball. That’s a ratio of 486:1, which means if the ball comes in and leaves at 100 mph in opposite directions, the batter gets pushed back at less than 1/2 mph. Now, he’s not just standing there, he’s swinging the bat, his arms, and his torso.
Let’s say that roughly half of his total mass is moving forward. So what happens is that he’s not pushed backwards at all, his swing is just slowed down by 1 mph. [yes, this is very simplified, but gets the idea across].

Plus, the batter is generally braced against the earth (just like the brick wall in your example). So the real limiting factor is whether his feet slip, or there’s enough torque to tip him over.

So the limiting factor in propelling a ball for distance isn’t absorbing the momentum, it’s in getting enough energy to the ball. If the ball has more energy coming in, there’s more available for it going out. Fastballs are easier to hit far than off-speed pitches.

13

I don’t know about ball speed compared to bat speed. However, I do know the results of tests with golf balls. True, they are more resiliant that baseballs but this might give some idea.

Tour pro golfers swing a driver in the 120-130 mph range and the initial velocity of the ball is in the 170-190 mph range with a rotation rate in the vicinity of 2500-3000 rpm.

14

Long answer: A simple model of the ball-bat interaction, like most posters talk about above, doesn’t take into account some of the really important secondary effects. When these are included, you can see that the influence of ball speed is noticeably smaller than that of bat speed, but it’s still there…and probably just enough of an influence to turn a deep centerfield out into an in-the-bleachers homer.

To walk through: The very simplest model of a ball-bat collision would be the collision of two particles. In this case conservation of momentum gives:

V[sub]bat_final[/sub] - V[sub]ball_final[/sub] = e(-V[sub]ball_orig[/sub] - V[sub]bat_orig[/sub])

and

m[sub]bat[/sub]V[sub]bat_final[/sub] + m[sub]ball[/sub]V[sub]ball_final[/sub] = m[sub]bat[/sub]V[sub]bat_orig[/sub] - m[sub]ball[/sub]V[sub]ball_orig[/sub]

where e is the coefficient of restitution (from 0 to 1), and the negative sign in front of V[sub]ball_orig[/sub] accounts for its opposite direction. Combining these two gives:

V[sub]ball_final[/sub] = (e+1)/(1+R)V[sub]bat_orig[/sub] + (e-R)/(1+R)V[sub]ball_orig[/sub]

where R = m[sub]ball[/sub]/m[sub]bat[/sub] (weight ratio). If you examine the coefficients on the two terms, you’ll see that the final ball speed is dependent on the original ball speed, but it’s more sensitive to bat speed [(e+1)>(e-R)].

Unfortunately, the above is not a very good approximation to real life, but I mention it to compare with the expression found in this paper from the American Journal of Physics (pdf). The author of that paper has a great site all about the physics of baseball that covers a number of other issues as well.

Anyway, if you peruse the paper (it’s pretty math-intensive, but the explanatory and discussionary text ain’t so bad), you’ll find that he derives an equation for final ball speed remarkably similar to the simplistic one above, where some of the additional secondary effects are tucked into the parameters in the equation:

V[sub]ball_final[/sub] = (e[sub]eff[/sub]+1)/(1+R[sub]0[/sub])V[sub]bat_orig[/sub] + (e[sub]eff[/sub]-R[sub]0[/sub])/(1+R[sub]0[/sub])V[sub]ball_orig[/sub]

where V[sub]bat_orig[/sub] is the bat speed at the impact point, e[sub]eff[/sub] is an effective coefficient of restitution which takes into account the combined effects of energy dissipation in the ball and vibrational losses in the bat. Also, R[sub]0[/sub] is a ratio of ball mass to “effective” bat mass, which takes into account the collision location distance from the bat center of mass. Note that all of these quantities vary along the length of the bat. In particular, e[sub]eff[/sub] varies from about 0.2 to 0.5 (Fig 12), which makes for a pretty inelastic condition, and R[sub]0[/sub] tends to be somewhat larger than the actual mass ratio (~0.18).

When all this is put together, then, you get something like (taking sort of average numbers):

V[sub]ball_final[/sub] = (0.4+1)/(1+0.2)V[sub]bat_orig[/sub] + (0.4-0.2)/(1+0.2)V[sub]ball_orig[/sub]

and it’s easy to see that, while the original ball speed affects the final speed, the bat speed is something like six or seven times more important. In fact, the actual final ball speed winds up heavily dependent on collision location (Fig 11), with a fairly narrow sweet spot. Within that sweet spot, e[sub]eff[/sub] peaks (at ~0.5), R[sub]0[/sub] is minimal, and final ball velocity is most sensitive to original ball velocity… probably adding enough velocity to turn a deep centerfield out into an in-the-bleachers homer.

15

Good post, zut and I was glad to see the cite brought in aerodynamics. The physics of the bat-ball impact determines the initial conditions of the ball’s trajectory. From then on it is the gradual dissipation of the initial energy through aerodynamics. Getting a lot of lift on the ball so as to get as much carry as possible is a crucial factor. If the the ball is topped even slightly, you get a grounder no matter how fast the ball is going because there is a low initial launch angle and no lift.

16

Thank you zut. Now I am convinced that the conventional wisdom is correct. I am still a bit surprised. Next I will go read that paper. The math per se doesn’t scare me.

17

Well, the math itself isn’t really all that scary, and like I said the text around it is pretty clear. Too, what he’s doing seems conceptually straightforward; however, exactly how he arrives at certain quantities was unlcear to me (for instance, how e[sub]eff[/sub] is calculated, exactly). Still, how often can you find an on-line peer-reviewed journal article that is directly applicable to a GQ?

18

How does vector addition reconcile the fact that the stronger the ball’s vector, the less it counts?

19

It better not, or we’re in for a rough 30,000 years.

20

The idea that the batter supplies all of the momentum/energy to the batted ball ignores basic physics. The combined speed of the bat and ball is what matters. In an elastic collision between two objects, they will rebound away from each other. Both momentum and energy are conserved. If the collision is not perfectly elastic, some of the energy will be turned into heat or some other form of energy (like vibration). The proportion of energy lost is almost completely independent of the total energy - for example, if 50% of the energy is lost to heat in a high-speed collision, 50% will also be lost in a low-speed collision.

In other words, a faster pitch will travel farther when hit than a slower pitch, all other things being equal. In real life, of course, things are never equal. But if a batter hit a 90 MPH fastball and a 75 MPH curveball with exactly the same bat speed, on exactly the same spot on the bat, with the same bat angle, etc., the fastball would go farther, because the same proportion of the pitched ball’s original energy will be retained when the ball flies off the bat. When the ball starts with more energy, it ends with more energy.