Download Mathematical Model: Finding the Sweet Spot of a Baseball Bat

Thaksha Kulahan, Bob Tian, Kai Yang
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Mathematical Model:
Finding the Sweet Spot of a Baseball Bat
Thaksha Kulahan
Bob Tian
Kai Yang
University of Toronto
Our approach to the problem is the conservation of energy. The sweet spot should be the
location where maximum energy is transferred to the ball so that the ball attains best
performance. Although hitting at the end gives you the greatest torque that gives you greatest
force and greatest acceleration but high torque also means high rotational force on the handle
which causes the handle to shake. Since hitters are human, their arms cannot be perfectly rigid
when the ball hits at the end of the bat, so there will be some energy loss due to the movement
of the bat because the ball does not hit on the center of percussion of the bat. Also the end of
the bat is an anti-node of the bat which means hitting on the end of bat causes the greatest
vibration, which is another place you lose energy. Given a math model, our goal is to find the
center of percussion where the movement of the bat is minimized and a point near the node of
fundamental mode of the bat where the vibration of the bat is minimized. And hence, the sweet
spot should be in between the center of percussion and the point to minimize vibration so that
the total energy loss it minimized. We will find the sweet spot first in terms of different models
and then we will discuss what corking would do to our model and the impact of different
material of the bat.
I) The location of the sweet spot
1) A simple model
Thaksha Kulahan, Bob Tian, Kai Yang
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We would start with an easier model to illustrate our idea. After a close inspection on how
batter hits the ball, we observe that the pivot of the bat is the wrist of the batter. Since the part
of the bat below the pivot is negligible, we would just consider the part of the baseball bat above
the pivot. Let us take our bat as a uniform cylinder with length L and radius R and density ρ,
centered at the origin which is the pivot.
When the ball strikes the bat, it will cause two motions in the bat depending on the
point it is struck. The rotational velocity rω, and the translational velocity u. The two velocities
are against each other, and so the difference of the two will be the final velocity of v the bat due
to the ball. We can account for the energies lost due to this velocity v. Here, r is the distance of
the centre of gravity to the pivot.
v = u – rω
Differentiate it with respect to time t:
①
d is the distance from the pivot the point of interaction. F is the force between the baseball and
the bat. Torque dF is moment of inertia
times angular acceleration
Hence we have:
②
According to Newton’s second law, where m is the mass of the bat.
③
Sub ②
③ into ①, we have
Integrate on both sides, we have
is the impulse experienced by the baseball.
.
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When the difference between the rotational and transitional velocity is 0, this portion of kinetic
energy is not dissipated. And d* that make v 0 is the distance from the pivot to the center of
percussion. To derive the formula for center of percussion, we let the left hand side to be 0.
Hence,
Calculation of moment of inertia:
where
is the distance from the pivot to a point in the cylinder.
We evaluate the triple integral in cylindrical coordinate,
For a uniform rod the center of gravity would be just at L/2.
The energy lost to shaking the bat Ec by hitting the ball at position x would be:
The node of fundamental frequency of a uniform cylinder would be
second harmonic would be
the node of the
, since the first two harmonics capture the majority of the energy
of the vibration we simply ignore energy of harmonics of higher orders. The mid-point of the
node of fundamental mode and second harmonic is
, but the point to minimize vibration
should be closer to the node of fundamental mode. Hence the point where the vibration of the
bat caused by collision is minimized would be in the interval
really close, we might take the mean of the interval
. Since the boundaries are
as the point to minimize vibration, but
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where exactly the point to minimize vibration of the bat is left to be examined.
Ideally we would be able to figure out mathematically how much energy will be lost if we do not
hit the ball at
but at point x. Suppose we can derive this function f(x), then
We differentiate
to find the critical point where energy lost is minimized. This critical
point is the sweet spot according our mathematical model.
Since f(x) is intractable for now this work cannot be done. But what we are certain according to
our model, the sweet spot lies in the interval of
.
2) A more complex model
To simulate a baseball bat with a cylinder seems to be too inaccurate because a baseball bat has
a much bigger end than the handle. We tried various models, but the only one that is still simple
enough to present analytically is a combination of two cylinders. (actually this model is already
very complicated) The radius of the bottom cylinder is r1 and the radius of the upper cylinder is r2.
The length of the bottom cylinder is a1 and the length of upper cylinder is (a2-a1). The entire
object has uniform density of ρ, centered at the origin which is the pivot. Using the same logic,
we want to figure out the center of percussion and the mode of fundamental frequency, and try
to minimize the energy dissipation, but the calculation is a lot more tedious.
The formula used to calculate center of percussion does not change, to keep consistent we still
use d* to denote center of percussion, r is the distance of the centre of gravity to the pivot, m as
the mass of the object.
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Calculation of moment of inertia:
where
is the distance from the pivot to a point in the cylinder, and V1 is the bottom(RHS)
cylinder while V2 is the upper cylinder.
We evaluate the triple integral in cylindrical coordinate,
For this object, the center of gravity is much harder to calculate.
where
is the distance from the pivot to a point in the cylinder
Where v1 is the bottom cylinder and v2 is the upper cylinder.
With the help of computer, we are still able to calculate the integral,
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m is easy to calculate:
Although tedious we are still able to get the center of percussion which better simulates a
baseball bat:
As long as we hit the baseball at center of percussion, we could minimize the energy loss due to
the translation and rotation of the bat. But unfortunately, we were not able to derive formulas to
find out the nodes of the fundamental mode of this object. But what we are certain is that the
sweet spot should still be somewhere even closer to the center of percussion because now the
mass of the bat is more concentrated on the top and the top is much shorter than the thin
handle.
3) A much more precise numerical model
Thaksha Kulahan, Bob Tian, Kai Yang
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We construct a much more precise numerical model from a 30-inch Little League wood bat. The
barrel is a cylinder with radius of 1 inch and the handle is of radius 0.5 inch. The bat consists of
three parts, the barrel which is 8 inches long, the difference of two cones which is 8 inches long
and a handle which is 14 inches long.
First let us consider the point where vibration is minimized.
Thanks to Dr.Russell1, We have mode shapes for this 30-inch Little League wood bat.
Fundamental bending mode (215 Hz)
Second bending mode (670 Hz)
From his experiment, it is pretty obvious that the node of fundamental bending is at
approximately 7 and that of second bending mode is at 5. We are certain that the point to
minimize vibration is in the interval of [6,7]. Take the mean of this interval,
vibration-minimizing point should be approximately 6.5.
Now we want to calculate the center of percussion of this particular bat. We have seen that
the location of center of percussion does not depend on the material of the bat. Without
losing generality, we assume the density of the bat to be 1 unit.
1 Cited from http://paws.kettering.edu/~drussell/bats-new/bend-sweet.html
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The formula for center of percussion still applies:
Calculation of moment of inertia:
where
is the distance from the pivot to a point in the cylinder, and V1 is the barrel while V2 is
the conical frustum and V3 is the handle part.
To solve the second triple integral we fix z first do the double integral of the circle and then
integrate with respect to z,
Using MATLAB, it is very easy to evaluate these numerical integrations.
I1 = 719.73 I2 = 5287.61 I3 = 17135.84
I = 23143.18
center of gravity is given by the formula:
where
is the distance from the pivot to a point in the cylinder
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It is nearly impossible to solve these integrals analytically but solving them numerically is fairly
simple.
r1 = 77.2012
r2 = 276.5779
r3 = 653.6755
r = r1 + r2 + r3 = 1007.45
That tells us that center of percussion should be around 22.972 inches from the bottom which is
the pivot, which is actually pretty close to the node of the fundamental bending mode which is
approximately 23.5 inches away from the pivot.
The conclusion is that the sweet spot of this particular 30-inch Little League wood bat is about 23
inches away from the pivot which is where you hold the bat. This actually coincides with our
prediction in model1 which is a simple cylinder that the sweet spot lies in
.
II) The effect of corking
Our study shows that corking does have positive effect on sweet spot effect. The positive effect
lies in two reasons. 1) corking might raise the center of percussion so that at the point where
energy loss of the bat is minimized the bat gives a higher torque to enhance acceleration of the
baseball. 2)corking can reduce the mass of the bat to maximize the energy on a bat given the
swing speed—mass equation given by. Terry Bahill23. For the following content, we will calculate
the amount of effect using model1.
1) Corking raises the center of percussion
For simplicity, first let us assume we cork the bat with some substance that has very small density
so that we can completely ignore it.(i.e. hollowing the bat in a certain way)
a) hollow the entire bat with radius
2
A. Terry Bahill and Miguel Morna Freitas, "Two Methods for Recommending Bat Weights," Annals of Biomedical Engineering, 23(4),
436-444 (1995)
3
Robert G. Watts and A. Terry Bahill, Keep Your Eye on the Ball: Curve Balls, Knuckleballs, and Fallacies of Baseball, revised ed. (W. H.
Freeman and Co., 2000)
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Since it is hard to evaluate r, we picked some objects with certain symmetry so that it is easier to
calculate r, for this object here, center of the mass will still be of height
symmetry is.
Let us compare this new center of percussion to the non-corked one,
This proves that corked bat has a higher center of percussion.
b) Increase the radius of hollowness, hollow the entire bat with radius
where the center of
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For this object here, center of the mass will still be of height
where the center of symmetry is.
Let us compare this new center of percussion to a),
This shows that a radially more corked bat might have a higher center of percussion than a
radially less corked bat.
c) Decrease the depth of hollowness, hollow the entire bat with radius
.
where v1 is the entire cylinder while v2 is the hollow part
For this object here, center of the mass is given by the equation:
, but with depth
Thaksha Kulahan, Bob Tian, Kai Yang
center of mass is x =
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.
Let us compare this new center of percussion to b),
This shows that a vertically less corked bat might have a higher center of percussion than a
vertically less corked bat.
d) Cork cylinder b) with substance of density
0. ( 0 <
)
where v1 is the outer part of the cylinder while v2 is the inner cylinder
For this object here, center of the mass is given by
because of the symmetry of the object.
Mass of the object is easy to calculate. It is the sum of the mass of the two components:
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We can use inequality to show the relationship of center of percussion of cylinder b), cylinder d)
and a non-corked cylinder.
the center of percussion of the corked and stuffed cylinder d*stuffed
Hence, we can draw the conclusion that the effect of corking is between non-hollowing and
hollowing completely. Theoretically, rather than corking people should choose to hollow the bat
rather than cork the bat to achieve maximum performance, but in this model we do not consider
how likely the bat is going to break because of hollowing. In reality, People choose to cork a bat
because it is harder to break and also it is less likely to get caught to be cheating.
However, we have seen that corking only raises center of percussion by a very small amount,
of a base ball bat is usually very small and even negligible. This is probably why the alleged
“advantage of corking” has been questioned by so many scientists. One possible drawback is that
they use machines rather than people to swing the bat so that they ignore that corked bats are
easier for people to swing because a corked bat has less mass, and the moment of inertia of the
bat decreases. So the swing speed of a bat could be significantly increased by corking and so is
the energy on the bat. And that is what we are going to discuss next.
2) Corking can increase the total energy on a bat
According to Terry Bahill23 , Given a person, the swing speed of the bat has a negative linear
relation with the mass of the bat. Formula is given as v = -am + b where a and b are coefficients,
m  (0,b/a). (for a power hitter, v= -0.42m + 75 where speed is in mph and weight is in ounces
according to his study.)
Total energy on the bat
is:

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Total energy has a quadratic relation with mass, we differentiate it to find to critical point,
Hence,


The total amount of energy on the bat is maximized when m =
So the conclusion is if the bat does not have the perfect weight for a particular player, corking the
mass to
could help the player to exert maximum power or energy on the bat. But the value of
a and b varies by person. Different age group with different level of baseball skills might have
considerably different values for a and b.
3) A numerical example
Suppose a typical power hitter uses our numerical model3 to bat a baseball. We want to calculate
how much more energy will be transferred to the baseball if we cork the bat with radius 0.5 inch
and length of 8 inches and hit the base ball on the new center of percussion of the bat.
Suppose we fill the bat with cork that has density
of 0.20g/cm3, 4white ash wood has density
of 0.64g/cm3,5. If we convert them into imperial unit,
oz/inch3 ,
oz/inch3.
First we need to determine where the new center of percussion is,
Calculation of moment of inertia:
Where v1 is the entire object and v2 is the corked cylinder.
We did the calculation before, the moment of inertia of the whole bat is I = 23143.18
4
Of´elia Anjos ,Helena Pereira, M. Em´ılia,” Effect of quality, porosity and density on the compression properties” Rosa European
Journal of Wood and Wood Products, 0018-3768 page295-301,2008
5 Cited from http://www.csudh.edu/oliver/chemdata/woods.htm
Thaksha Kulahan, Bob Tian, Kai Yang
we need to calculate
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,
I=
center of gravity is given by the formula:
where v1 is the entire bat and v2 is the corked portion.
We did the calculation of
before,
then we can calculate the new center of percussion,
Unexpectedly, this kind of corking does not improve the performance of the bat. The new center
of percussion decreases and unfortunately does not coincide with the results of analytical
model1. One possible explanation is, although model1 does give us some positive effect of
corking, the improvement is proportional to
. In this numerical example, an approximate
average radius would be
might work, the improvement would be proportional to
inch. Even if a certain way of corking
inch which is safely
negligible. The conclusion is that the improvement of corking is highly questionable although
evidence shows that in some special cases corking does have some small positive effect on
performance. Major league baseball prohibits corking to maintain fairness of a game, but still, the
value of this kind of cheating is highly questionable.
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III) Sweet spot effect between an aluminum bat and a wood bat
In terms of our math models, uniform solid bats of the same shape have the same sweet
spot and the same sweet spot effect because our calculations cancel out the density completely.
But solid aluminum bats are much heavier,
For example, a 30-ounce wood bat would be 126.56 ounces if made by aluminum. According
to swing speed theory, it would decrease the energy on the bat significantly. Hence, the reason
why aluminum bat outperforms wood ones is because of its hollow nature.
We have shown previously, hollowing the bat could raise the center of percussion and thus
enhance the sweet spot effect. And a lighter aluminum bat might have higher swing speed and
hence greater total energy on the bat. In theory, hollow wood bats should work as well as metal
bats but in practice they are much easier to break.
Another possible advantage of a hollow bat is its elastic property. When a ball hits a solid wood
bat, it compresses to nearly half its original diameter, losing up to 75% of its initial energy to
internal friction forces during this compression.6 However, the barrel of an aluminum bat acts
more like a spring. This means the ball is not compressed as much and hence there is less energy
loss due to internal friction. Moreover, the hollow barrel returns energy temporarily stored in it
to the ball.7 In sum, all these effects contribute to minimizing energy loss due to the impact.
6 [5] R.M. Greenwald R.M., L.H. Penna , and J.J. Crisco,"Differences in Batted Ball Speed with Wood and Aluminum Baseball Bats: A
Batting Cage Study," J. Appl. Biomech., 17, 241-252 (2001)
7 J.J. Crisco and R.W. Greenwald, "Metal baseball bats can outperform wood bats with a similar `sweet spot'," Proceedings of the 24th
Annual Meeting, American Society of Biomechanics. Chicago, IL. July 19-22 (2000).