The Physics of the Bunt

Bill James’ explanation was straight and to the point during his NPR interview on September 26, 2011.  “Bunting is usually a waste of time…The home team applauds because they get an out, and the other team applauds because they get a base. So what does that tell you?”  The pages of FanGraphs and The Hardball Times are packed with data-filled articles on the effectiveness (or lack thereof) of the sacrifice bunt.

Let’s take a break from all that and see what physics has to say about the challenges of dropping one down.  Pop quiz!  Which changes the energy of the baseball more, a bunt or a homer?

A homer only changes the speed of the ball from about 90mph coming from the pitcher to about 110mph toward the fence.  This 22% increase in speed amounts to about a 50% increase in the kinetic energy of the ball.  Bunting on the other hand reduces the speed of the 90mph pitch to close to zero resulting in a nearly 100% reduction in the energy of the ball.  The other common method for reducing the energy of the ball is to catch it with the glove.  The similarity is captured in the common description of a bunt as  “catching the ball with the bat.”


Catching a ball with a hard cylinder provides a serious challenge compared to using a 12” soft leather glove.  Figure 1 describes the situation when the ball comes in horizontally.  The ball follows the blue path.  It collides with the bat a distance b below the center of the bat making an angle , q, with the radial line coming directly outward from the center of the bat.  Physics requires the ball to reflect off the bat at almost exactly the same angle.  So, the net deflection of the ball is through an angle of 2q.

To simplify matters, let’s assume that a successful bunt must have a total deflection between 0˚ and 90˚ degrees (q between 0˚ to 45˚).  This is an approximation because bunts hit with slight upward trajectories might work out fine.  Of course, ones hit nearly straight downward will likely be picked up by the catcher and turned into a force play.

Now, imagine the ball coming in straight toward the center of the bat (b=0) and heading directly back from where it came from.  Compare that to a ball hit straight downward so that 2q = 90˚ or q = 45˚.  If we could find this value of b, we would know the allowed range on the bat for a successful bunt.


Remember back to the good old days when you knew what SOHCAHTOA meant?  If you can, then you’ll recall,


b = r sinq.


Since the radius is half the diameter,

b = D/2 sinq.

Using q = 45˚ and the maximum allowed diameter of a bat of 23/4 inches (see Rule 1.10) gives the maximum value of,

b = 0.97in.

The ball must strike the bat within one inch of the center.  For cryin’ out loud, I can’t hit a stationary golf ball consistently to within one inch!  Now we see why this seemingly simple task is actually quite difficult.

We had previously assumed the pitch was coming in horizontally.  Let’s examine the effect of the pitch heading slightly downward as it does in a real game.  Figure 3 shows the ball dropping at an angle a as it collides with the bat.

Finding the fraction of the bat that a good bunt must hit is a nasty bit of geometry which I won’t do here…you’re welcome.  The result is figure 4, a graph of the fraction of the bat diameter within which the ball must strike for a good bunt as it varies with the drop angle.  I guess “vary” is too strong a word.  The line is quite flat, so the drop angle hardly matters.

The graph has implications for the correct pitching strategy versus the bunt.  There is little point to dropping in a 12/6 curveball since the drop angle won’t reduce the portion of the bat that could yield a good bunt.  Actually, the higher drop angle would make it just slightly larger.  In addition, the slower pitch will make it easier for the batter to get the right part of the bat on the ball.  Instead, the pitcher should attempt to make the pitch ride inside so the batter must use a thinner part of the bat thus reducing the margin for error.

For example, the fraction of bat diameter for a zero drop angle from the graph is 0.354.  If the batter can hit the ball on the fat part of the bat we already calculated the 0.97in range.  However, if the pitcher can force the batter to use a part of the bat with only a two inch diameter, then the range drops to  0.71in.

In a previous article, “The Physics of a Foul Down the Line,” I explained how spin is imparted to the ball when it collides with the bat.  Since good bunts hit the bottom half of the bat, they will all leave with top spin.  The ball will pick up additional top spin when it hits the ground.  We’ve all watched the third baseman wait, wait, and wait some more for the ball to roll foul.  The top spin explains why bunts seem to roll away from the plate for an unusually long time before coming to rest.

In the same article, you hopefully learned that the side spin on a ball will tend to make the ball curve toward the foul lines.  Many a grounds crew has been accused of tilting the baseline either into fair or foul territory to encourage bunts to drift one way or the other depending on the proclivities of the home team.   Nonetheless, the side spin on a bunt will always tend to make the ball drift foul.

I was glad to able to call a temporary truce in the Bunt Strategy Wars if only for a short while.  We did make good use of the time, building an understanding of the physics involved.  Now, back to the bunkers!

David Kagan is a physics professor at CSU Chico, and the self-proclaimed "Einstein of the National Pastime." Visit his website, Major League Physics, and follow him on Twitter @DrBaseballPhD.
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9 years ago

Your assertion that a bunt “changes” the energy of the ball more than a home run doesn’t seem quite right. You use the change in speed of the ball (a scalar quantity) to calculate the change in kinetic energy. Aren’t you neglecting the change in direction? After coming off the bat, the ball isn’t moving in the same direction anymore. In that sense, isn’t a home run a much more drastic change in energy than a bunt?

9 years ago
Reply to  Brendan


I thought about that too, but since he said ‘energy’, it makes sense. Energy is a scalar, and is therefore independent of direction. If the ball starts with 10J of energy, and ends up with 12J, then it’s kinetic energy increased by 2J, no matter where the ball was hit.

Now, if he had said ‘momentum’…that’d be a problem.

Dr. Baseball
9 years ago
Reply to  Matt

Thanks for the right answer to my hidden quiz on energy! Isn’t it interesting that the energy change for a homer is fairly small? Perhaps that’s why some folks say to hitters, “Let the pitcher do all the work.” OK – perhaps not….

Alan Nathan
9 years ago

There is a tremendous amount of kinetic energy lost in the ball-bat collision, largely because a baseball is not a superball. When you drop it onto a hard surface, it bounces up to only a relatively small fraction of its initial height. Similarly when the ball bounces off a stationary bat during a bunt. But the bat is also capable of transferring energy to the bat, provided it has energy to transfer; i.e., provided the bat is moving. So, for a home run, the ball actually gains kinetic energy. Lots of energy is lost in the collision but that is more than compensated by the energy transferred by the swinging bat.

9 years ago
Reply to  Alan Nathan

Thank you for the clarification.