Bahill
11/14/13
1
A new formulation for computing batted

ball speed
Terry Bahill
http://ww
w.sie.arizona.edu/sysengr/
Spring
20
10

2012
Problem statement
De
termine the speed of the batted

ball using only Newton’s principles.
This paper presents a model for bat

ball
collisions. Unlike previous models, this model, for use by
students of the science of baseball, uses simple Newtonian equations to explain several collision
configurations.
My original intention was to add an equation for the conservation of energy to the
previous literature on bat

ball collisions. We had three eq
uations in three unknowns. A
dding a fourth
equation could
not possibly cause a problem, b
ecause f
or the past 300 years everyone has agreed that the
conservation equations are consistent. But I found out that they are slightly off, because they are not
theoretical
equations, they are empirical
equations. In
particular,
the coefficient of restitution an
d the
kinetic energy lost in a collision a
re empirical, not theoretical: a
nd there are discrepancies. My goal is to
produce equations that satisfy configuration 2c
(described below)
.
And then, b
y setting certain
parameters to zero we should be able to get
all of the previous equations.
Right now, my problem is that
equation (2) is only valid for central direct collisions. I need an equation for the kinetic
energy
lost in an
eccentric collision.
I thought that the book by Brach would be all en
compassing, but
his equations did
not work
for me.
Outline of this Endeavour
Configuration
1 is a head

on collision at the center of mass of the bat
; there is n
o spin on the ball
.
It u
ses
conservation of
linear momentum
and
the
coeffic
ient of restitution
.
Configuration
2a
is a
collision
at the sweet spot of the bat with no spin on the ball
.
This is the s
ame as
Configuration 1
,
but
it moves the
collision to
the
sweet spot and
adds collision impulses.
Configuration
2b
is a
collision
at the sweet spot of the bat
with no
s
pin on the ball
.
This is the same
as
configuration
2a
,
but
it adds
con
servation of energy.
Configuration 2c
is a
collision
at the
sweet spot of the bat with
spin on the ball
.
This is the same
as
configuration
2b
,
but
it adds
spin on the ball
and conservation
of angular momentum.
Configuration 3
is a
collision
at the sweet spot
of the bat,
but above the ho
rizontal axis of the bat
.
This is
the same
as
configuration
2c
,
but
it adds
that the
bat hits the bottom part of the ball
, and an equation
f
or
bat twist
.
Co
nfiguration 4
is a
n oblique collision
at the sweet spot
,
but above
the horizontal axis of the bat.
This is
the same
as
configuration
3
,
but
it adds
that the bat is rotated short of (or beyond) the y

axis
at the time
of
the
collision
.
Configuration 5
is a
collision
at the center of mass
of the bat,
but above
the horizontal axis of the bat
.
This is the same as
configuration
1, but it adds spin on the ball and an offset between the bat and ball
vertical velocities at the time of collision
.
Bahill
11/14/13
2
The purpose of this paper is to creat
e a model for batted

ball collisions
using fundamental principles of
Newtonian mechanics.
First,
we
note that f
orce, velocity,
acceleration
, impulse and momentum
are
vector quantities
.
Newton’s first
model
states that e
very object
moving with a constant velocity will
continue to move with that constant velocity unless acted upon by an external force.
This concept is
called
inertia.
Newton’s second
model
states that a
force acting on an object
produces an ac
celeration in
accordance with
the
equation
F ma
.
This
model
is often stated
as;
applying an impulse
to an object
will change its
momentum.
Newton’s third
model
states that f
or every action (force) there is an equal
and opposite
reaction
: this is the basis of conservation of momentum.
These three
model
s give rise to the
conservation laws that state, “
Linear
m
omentum,
angular momentum and
energy and
cannot be created
or destroyed.”
I call these models (not laws) because they have b
een refined many times in the last 300
years. They only help to explain physical phenomenon. They have been and will be
further
refined as we
learn more about the universe.
Figure 1.
When a batter swings a bat, t
here is a body rotation,
body
, and another rotation about the pivot
point between the wrists,
wrists
.
Bat

ball collisions
There are several
simple configurations
of bat

ball collisions.
If the duration of the collision is short
and
the are
a of collision is small
,
then
it is called an impact.
For the past
century,
people
have
describe
d bat

ball impacts with the following three variables
(1)
planar or nonplanar, (2)
collinear

direct,
parallel

direct
or oblique and (3) central or eccentric
.
If the equations of motion require description in three

dimensional space, then the
impact
is
nonplanar
.
Otherwise,
if the equations of motion can be described
in
a two

dimensional plane
, then the impact
is
planar
.
For a nonplanar
impact
between two rigid
bodies,
there is a common tangent plane that is perpendicular to the radius of curvature of each object at the
point of contact. The normal vector is perpendicular to this plane at this contact point.
For
a planar
impact
there is a
common
tangent lin
e and
the
line perpendicular to it at this
point of contact
is called the
line of impact
.
An impact is
collinear

direct
if the velocities of both centers of mass are collinear with
the line of impact (either in the same direction, the opposite direction, or with zero velocity). An impact
is
parallel

direct
if the velocities of both centers of mass are parallel with the lin
e of impact.
Else if one
(
or both
)
of the bodies move
s
along a line that is not
co
lli
near
with the line of impact, then the impact is
Bahill
11/14/13
3
oblique
.
An impact is
central
if the centers of mass of the bodies are on the line of impact, otherwise the
impact is
eccentric
.
These terms are useful because they predict the complexity of the equations of motion.
Collinear

direct,
central impacts are the simplest, because all motions are along the same
axis and there are no impulse
torques
.
Nonplanar, oblique
,
eccentr
ic impacts have the most complicated equations.
These descriptions
also help a person to determine the types of analyses that will be necessary to study a certain collision
configuration
.
Central impacts allow the equations of motion for the normal and tan
gential directions to
be decoupled.
Footnote: If one of the colliding objects is a plane, then its center of mass velocity is defined as the
direction of the normal.
We will now describe five
simple configurations of bat

ball collisions
.
Bahill
11/14/13
4
Configuration
1 is a head

on collision at the center of mass of the bat; there is no spin on the ball
. This
type of analysis was done by
Bahill and Karnavas
[
1989]. It uses conservation of linear momentum
(equation 4) and the coefficient of restitution (equation 5).
hoc
m
knob
pivot
cm
ss
d
knob

pivot
d
pivot

cm
d
cm

ss
d
pivot

ss
d
ss

end
planar
,
collinear

direct
,
central impact at the center of mass
(
for a right

handed batter
)
x
y
x
z
Figure 2.
Configuration 1 a planar,
collinear

direct, central impact at the cm.
Configuration 1 is a head

on collision at the center of mass of the bat, Figure 2. This bat

ball collision is
a planar, collinear

direct, centr
al impact.
The impact is planar, because the equations are in the x

y plane.
The impact is collinear

direct, because
both the bat and
the ball
are
moving along the x

axis
at
the time of the impact. This means that the initial tangential (y

axis) velocity
components
are zero.
The impact is central, because the line of impact passes through the center of mass of both the
ball and the bat.
This type of collision would produce a line drive back to the pitcher. The equations for this type of
impact will be pres
ented in an early section of this paper.
Bahill
11/14/13
5
Configuration 2a is a collision at the sweet spot of the bat with no spin on the ball
. This type of analysis
was done by Watts and Bahill
[
1990
and 2000
]. This is the same a
s Configuration 1, but it moves the
collisio
n to the sweet spot and adds c
ollision impulses (equation 6).
hoss
Configuration 2b is a collision at the sweet spot of th
e bat with no spin on the ball.
This section contains
original equations. This is the same as configuration 2a, but it adds
conservation of energy (equation 3).
hossce
Configuration 2c is a collision at the sweet spot of the bat with spin on the ball. This section contains
original equations. This is the same as configuration 2b, but it adds spin on the ball (equation 3a) and
conservation of angular momentum (equation 7).
hosscebs
knob
pivot
cm
ss
d
knob

pivot
d
pivot

cm
d
cm

ss
d
pivot

ss
d
ss

end
planar
,
direct
,
eccentric impact at the sweet spot of the bat
Figure 3.
Configuration
s
2a
, b and c are
planar,
parallel

direct, eccentric impact
s at the sweet spot of the
bat.
Configuration 2a
, with no spin on the ball (for example
a knuckleball).
The impact is planar, because the equations are in the x

y plane. This collision can be drawn on a
flat piece of paper.
The impact is
parallel

direct, because the ball is moving along the x

axis and the bat’s motion is
parallel to the x

ax
is at the time of the impact.
However, t
he velocity of the bat’s sweet
spot is not collinear with the ball’s velocity.
The impact is eccentric (on the x

axis), because the line of impact does not pass through the
center of mass of the bat.
This type of
collision would produce a line drive back to the pitcher. The equations for this type of
impact are presented in the main body of this paper.
Bahill
11/14/13
6
Configuration 3 is a collision at the sweet spot of the bat, but above the horizontal axis of the bat. This is
the same as configuration 2c, but it adds that the bat hits the bottom part of the ball, equation (3b) for bat
twist and the equations of Brach [2007]. ssaha
knob
pivot
cm
ss
d
knob

pivot
d
pivot

cm
d
cm

ss
d
pivot

ss
d
ss

end
nonplanar
,
direct
,
eccentric impact at the sweet spot of the bat
Figure 4.
Configuration 3
is a
nonplanar,
parallel

direct
,
eccentric
impact
at the sweet spot but above
the
horizontal
axis of the bat, i. e. the bat
hits th
e bo
ttom part of the ball
.
The
impact
is not in the x

y plane: the bat and ball will both have z

axis motion after the
impact
.
The impact is parallel

direct, because t
he ball is moving along the x

axis and the bat’s motion is
parallel to the x

axis at the time of the impact.
The
impact
is eccentric
(on both x

and z

axes)
, because the line of impact does not pass through
the center of mass of the bat and
the ball.
This type of collision would
typically
produce a flyball to center field.
The equations for this type of
impact might be presented in the third section of this paper.
Bahill
11/14/13
7
Configuration 4 is an oblique collision at the sweet spot, but above the horizontal ax
is of the bat. This is
the same as configuration 3, but it adds that the bat is rotated short of (or beyond) the y

axis at the time
of the collision.
knob
pivot
cm
ss
d
knob

pivot
d
pivot

cm
d
cm

ss
d
pivot

ss
d
ss

end
nonplanar
,
oblique
,
eccentric impact at the sweet spot of the bat
Figure 5.
Configuration
4 is a
nonplanar,
oblique
,
eccentric
impact
at the sweet spot but above
the
horizontal axis of the bat,
This is the same as configuration 3, but it adds that the bat is rotated short of
(or beyond) the y

ax
is at the time of the collision
.
The
impact
is not in the x

y plane: the bat and ball will both have z

axis motion after the
impact
.
The
impact
is oblique
, because
the bat is
not
moving along the x

axis
at the time of the
impact
.
This means that
there will be
tangential (y

axis) velocity c
omponents
.
The
impact
is eccentric
(on both x

and z

axis)
, because the line of impact does not pass through
the center of mass of the bat and
the ball.
This type of collision would typically
produce a flyball to right (or left) field.
The equations for this type
o
f impact will be considered in a future paper.
Bahill
11/14/13
8
Configuration 5 is a collision at the center of mass of the bat, but above the horizontal axis of the bat.
This is the same as configuration 1, but it adds spin on the ball and a
vertical
offset between the b
at and
ball at the collision
.
cmaha
Configuration
5 is a
nonpl
anar, parallel

direct, central
impact at the center of mass
of the bat
but above
the horizontal axis of the bat.
The impact is not in the x

y plane: the bat and ball will
both have z

axis motion after the impact.
The impact is parallel

direct, because the ball is moving along the x

axis and the bat’s motion is
parallel to the x

axis at the time of the impact.
The impact is central, because the line of impact passes through
the center of mass of both the
ball and the bat.
This type of collision would typicall
y produce a flyball to center field, or maybe a popup
. The equations
for this type of impact will be considered in a future paper.
Bahill
11/14/13
9
Config 1
Config 2a
Config 2b
Config 2c
Config 3
Config 4
Config 5
planar or
nonplanar
planar
planar
planar
nonplanar
nonplanar
nonplanar
nonplanar
collinear

direct,
parallel

direct or
oblique
collinear

direct
parallel

direct
parallel

direct
parallel

direct
parallel

direct
oblique
parallel

direct
central or
eccentric
central
eccentric
along the
x

axis
eccentric
along the
x

axis
eccentric
along the
x

axis
eccentric
along the
x

and y

axes
eccentric
along the
x

and y

axes
central
spin on the
ball
no
no
no
yes
yes
yes
yes
Point o
f
contact
center of
mass
sweet spot
sweet spot
sweet spot
sweet spot
sweet spot
center of
mass
difference
from
previous
configurati
on
Moves
collision
to sweet
spot
Adds
conservati
on of
energy
Adds spin
and
conservati
on of
angular
momentu
m
Adds
vertical
offset
Bat is
not
parallel to
the y

axis
this is
simpler
than
configurati
ons 2, 3
and 4. It is
similar to
config 1
with the
addition of
a vertical
offset at
the
collision.
Head

on collision at the center of mass
,
configuration
1
We will now
derive the equations for a head

on collision at the cen
ter of mass of the bat, Figure 2
.
The
abbreviations
used in the following equations
are described in Table 1.
Many authors
,
for example
[Bahill and Karnavas, 1989, 1991; Brach, 2007]
,
have used
the Newtonian concepts of
conservation of
linear momentum
ball ballbefore bat batbefore ball ballaft
er bat batafter
m v m v m v m v
(conservation of momentum
)
and the
experimentally determined
kinematic
coefficient of restitution (CoR)
ballafter batafter
ballbefore batbefore
v v
CoR
v v
(CoR)
to derive
the following equations
for the velocities of the ball and bat after the collision
:
ballbefore ball bat batbefore bat
ballafter
ball bat
( ) (1 )
v m CoRm v m CoR
v
m m
Bahill
11/14/13
10
ballbefore ball batbefore bat ball
batafter
ball bat
(1 ) ( )
v m CoR v m m CoR
v
m m
or
batbefore ballbefore bat
ballafter ballbefore
ball bat
( ) (1 )
v v m CoR
v v
m m
batbefore ballbefore ball
batafter batbefore
ball bat
( ) (1 )
v v m CoR
v v
m m
These derivations start with a two

rotation model for the swing of a baseball or a softball bat
(f
igure 1)
and
linearize
the model by finding tangents to the circular motion, leaving a model with o
ne bat
translation and a collision
at the center of mass
(cm)
, as shown in figure 2
. Bahill and Karnavas
[1989]
expanded this model by measuring
the swing speeds of a few hundred baseball and softbal
l players and
u
sing this experimental data and m
odel, to derive
equation
s for the bat
ted

ball
speed for each indi
vidual
person.
This derivation
use
s
these assumptions:
1. Neglect the deformation of the bat, deformation of the ball and inertia of the batter’s arms.
2. Assume a head

on (direct, central) collision at the center of mass
of the bat
.
3. Ignore the
change
in the rotational kinetic energy of the
ball. The energy stored in the spin of the ball
is about ½% of the translational energy [Bahill and Baldwin, 2008]:
later, this paper shows that
for a curveball hitting the sweet spot of the bat
the total kinetic energy stored in the bat and the
ball is 373 J, of which 0.8 J is stored in the spin of the ball:
so neglecting it seems reasonable.
Furthermore, a direct
central
collision at the center of mass would not change the spin.
4. As
sume that
there are no tangential
forces during
the
collision.
Although an additional equation is not needed
to solve the equations for the batted

ball speed
, we will
now
present the conservation
energy equation as a consistency check.
There
is nothing
in the system
that
will release energy during the collision (loaded springs or explosives)
and the bat swing is level;
therefore,
there is no change in potential energy during the collision. Before the collision, there is kinetic
energy in the bat
and kinetic energy in
the ball.
2 2
before ball ballbefore bat batbefore
1 1
2 2
KE m v m v
where
bat batcm
v v
A collision at the center of mass will not make the bat spin.
2 2
after ball ballafter bat batafter
1 1
2 2
KE m v m v
before after lost
KE KE KE
Kinetic energy will be lost to heat in the
ball, vibrations in the bat or deformations
of the bat or ball. The
coefficient of restitution (CoR) models the energy that is lost in a frictionless head

on collision between
two objects [Dadouriam, 1913, Eq. (XI), p. 248;
Ferreira da Silva,
2007, Eq. 23;
Brach, 2007, p. 39].
Such a collision will have no tangential velocity components.
2
2
lost batbefore ballbefore
1
2
m
KE v v CoR
(2)
Bahill
11/14/13
11
where
1 2
1 2
mm
m
m m
Combining these
equations yields
2
2 2 2 2 2
ball ballbefore bat batbefore batbefore ba
llbefore ball ballafter bat batafter
1 1 1 1
1
2 2 2 2 2
m
m v m v v v CoR m v m v
These equations are consistent.
T
he particular
numbers
in
Table 1,
produce the following results.
Table 2. Typical values for
a
head

on
bat

ball
collision
at the center of mass
.
m/s
mph
Inputs
CoR
0.55
v
ball

before

37.00

83
batcmbefore
v
29.90
49
Outputs
ballafter
v
from
Eqs.
4, 5 and 6
41.693
2
93
ballafter
v
from
Eqs.
3, 4 and
6
41.693
2
93
ballafter
v
from Eqs. 3, 4 and
5
41.6932
93
batcmafter
v
9.30
21
Checks
before
KE
316
after lost
KE KE
316
In
the rest of
this
paper,
we will use
Newtonian mechanics
and derive equations for the speed of the ball
and bat after their collision, for collision
s
that do
not
occur at the center
of mass
of the bat
.
List of
parameters and variables
Table 1
Abbreviation
This table is
arranged
alphabetically
by the
abbreviations.
Description
specific values are for configuration 2b
Typical values for a
C243 wooden bat and a
professional major
league baseball player
SI
units
Baseball
units
CoR
Coefficient of Restitution of a bat

ball collision
0.55
0.55
bat
d
Length of the bat from the knob to the barrel end,
for a C243 Wooden Bat
0.864 m
34 inch
Bahill
11/14/13
12
cm ss
d
distance from the center of mass to the
sweet spot,
which we will define as the
center of percussion
0.134 m
5.3 in
knobcm
d
distance from the center of the knob to the center
of mass
0.57 m
22.4 in
pivotcm
d
distance
from the pivot point to the center of mass
(H)
0.42 m
16.5 in
spinecm
d
Distance from the spine to the center of mass, an
experimentally measured value, (L
a
)
1.04 m
41 in
spiness
d
Distance from the spine to the sweet
spot of the
bat
(H+R+B)
1.17
m
46.2 in
ssendOfBarrel
d
distance from the sweet spot to the barrel end of
the bat
0.16 m
6.3 in
g
earth’s gravitational constant
⡡琠t桥
啯rAF
㤮㜱㠠洯s
ball
I
moment of inertia of the ball
with respect to its
center of mass
0.000079
kg m
2
bat
I
moment of inertia of the bat with respect to its
center of mass
0.048 kg m
2
bat
I
moment of inertia of the bat with respect to the
knob
0.342 kg m
2
pivot
I
moment of inertia of the bat with respect to the
pivot point, this is colloquially called the swing
weight
0.208 kg m
2
before
KE
Kinetic energy of the bat and the ball before the
collision (typical outputs)
374
J
after
KE
Kinetic energy of the bat and the ball after the
collision (typical outputs)
215 J
lost
KE
Kinetic energy lost in the collision (typical
outputs)
165 J
ball
m
mass of the ball
0.145
kg
5.125 oz
bat
m
mass of the bat
0.905 kg
32 oz
m
1 2
1 2
mm
m
m m
0.125 kg
4.4 oz
f
dynamic coefficient of friction
0.5
ball
r
radius of the ball
0.037 m
1.45 in
bat
r
maximum radius of the bat
0.035 m
1.37 in
ballbefore
v
velocity of the ball before the collision, the pitch
speed.
When the pitcher releases the ball it is
going 10% faster, which is 91 mph.
Actually, it is
the velocity of the surface of the ball at the point
of contact.

37 m/s

83 mph
ballafter
v
velocity of the ball after the collision, often called
the launch speed or the batted

ball speed. The
value given is
not
an input
value;
it is
a
typical
output value.
41.6 m/s
93 mph
Bahill
11/14/13
13
bat
v
velocity of the bat. If a specific place or time is
intended then the subscript may contain cm
(center of mass), ss (sweet spot), before or after.
Actually, it is the velocity
of the surface of the bat
at the point of contact.
cmbefore
v
velocity of the center of mass of the bat before the
bat

ball collision.
21.9 m/s
49 mph
cmafter
v
velocity of the center of mass of the bat after the
collision. The value given is
not
an input
value;
it
is a
typical
output value.
9.3 m/s
21 mph
ssbefore
v
velocity of the sweet spot of the bat before the
collision.
24.6 m/s
55 mph
ssafter
v
velocity of the sweet spot of the bat after the
collision. The value given is
not
an input
value;
it
is a
typical
output value.
112 m/s
27 mph
batbefore
angular velocity of the bat about its center of mass
before
the collision
0.02 rad/s
0.2 rpm
batafter
angular velocity of the bat about its center of mass
after the collision

31.8 rad/s

304
rpm
spinebefore
angular velocity of the batter
’s arms
a湤⁴桥a琠
a扯畴⁴桥灩湥
㈱2d
⽳
㈰ㄠ2灭
Model for
collisions
at the sweet spot
,
configuration
2a
Now we want to consider collisions that are not at the center of mass of the bat.
This section is based on
Watts and Bahill [1990]
.
Our objective wa
s to derive an equation for the
velocity
of the ball after its
collision with the bat. W
e
will expand the above linear
model to the two

rotation model with the
ball

impact point off of the
center of mass
, at the sweet spot
.
There are
about a dozen definitions for the
sweet

spot of the ba
t [Bahill
, 2004
]. For the rest of this paper we wil
l
use the center of percussion
definition (cop).
We will u
se the symbols
defined in
Table 1.
knob
pivot
cm
ss
d
knob

pivot
d
pivot

cm
d
cm

ss
d
pivot

ss
d
ss

end
planar
,
direct
,
eccentric impact at the sweet spot of the bat
Coordinate System
We will use a right

handed coordinate system with the x

axis from home plate to the pitching rubber, the
y

axis from first base to third base, and the z

axis pointing straight up. A torque rotating from the x

axis
to the y

axis would be
positive
upward.
Ac
tually,
the ball comes down at a t
en degree angle and the bat
moves
up
ward
at about ten degrees so later the x

axis will be rotated back ten degrees.
Bahill
11/14/13
14
Assumptions
A1. Neglect the deformation of the bat
,
deformation of the ball
, and
inertia of the
batter’s
arms.
A2
. The swing of the bat is as modeled in the above figures.
A3
.
Initially w
e will ignore the kinetic energy stored in the rotation of the baseball.
A4.
The contact duration is short, for example one millisecond.
A5.
Because t
he contact duration is s
hor
t and the swing is level, we can
ignore the effects of gravity
during the collision
.
A6.
During impact the ball slid
es and does not roll on the bat, but the sliding halts before separation.
A7.
The
bat

ball
collision is a direct, eccentric impact in th
e x

y plane.
A8. The coefficient of restitution (
CoR
) for a baseball wooden

bat collision
at major

league speeds
is
0.55
.
A9.
Assume that the pitch is a knuckleball with no spin. Later we will consider
a curveball
.
A10.
The dynamic coefficient of friction has been measured by Bahill
at
0.5
f
.
This agrees with
measurements
by
Sawicki, Hubbard and Stronge [2003]
and Cross and Nathan
[2006]
.
A11. Collisions at the center of percussion will produce a
rotation but not a translation of the bat.
Conservation of linear momentum
The conservation momentum law states that
linear m
omentum will be conserved in
collisions
,
if there
are no external forces.
Although the bat is actually rotating, we will approximat
e its motion with the
tangent to the curve.
Every point on the bat has the same angular velocity, but the linear velocities
will
be different. For
a
collision anywhere on the bat
,
conservation of
momentum in the direction of the x

axis
states that the
momentum before plus the external impulse will equal the momentum after the
collision.
There are no external impulses
during the ball

bat collision
, therefore
ball ballbefore bat batbefore ball ballaft
er bat batafter
m v m v m v m v
(4)
Experimental data
The experimental data in Table 1 are based
on
the following assumptions. The batter is using a
Louisville Slugger C243 wooden bat and is hitting a regulation
major league
baseball. The pitch speed is

83
mph
. The
velocity
of th
e
sweet spot of the bat is 55
mph
: this is
the
average value of the
data
collected from
the San Francisco Giants meas
ured by Bahill and Karnavas
[1991]
.
These
velocities
produce a CoR of 0.55
and a batted

ball speed of
93
mph.
We assume an ideal
launch angle of 31
degrees. From Baldwin and Bahill
[2004]
we find a batted

ball spin of

2100 rpm. With these
values,
the
ball would travel
337
feet, which would produce a home run in all major league stadiums
.
D
efinition of the coefficient of restitution
The kinematic co
efficient of restitution (CoR) wa
s define
d
by Sir Isaac Newton
as the ratio of the
relative velocity of the two objects after the collision to the relative velocity before the collision.
In our
models, f
or a collision at the sweet spot
of the bat
we have
ballafter ssafter cmss batafter
ballbefore ssbefore cmss batbefore
v v d
CoR
v v d
(5)
A note on my notation:
batafter
is the angular velocity of the bat
about its center of mass
after the collision.
ssbefore
v
is the linear velocity
of the sweet spot
of the bat
in the x

direction
before the collision.
Bahill
11/14/13
15
An
impulse
changes momentum
According to Newton’s third
model
, w
hen a ball hits a
bat at
the sweet spot
there will be a force on the
bat
in the direction of the negative x

axis
, let us call this
1
F
, and an equal but opposite force on the ball,
called
1
F
.
This force will be applied during the duration of the collision, called
c
t
. When a force is
applied for a short period of
time,
it is called an impulse
.
According to Newton’s second
model
, an
impulse will change the momentum. Changing from linear motion to angular rotational, t
he force on the
bat will create a torque of
1
cm ss
d F
around the center of mass
of the bat
.
An impulse
torque
will produce
a change in angular momentum of the
bat
.
1 c batafter batbefore
( )
cm ss bat
d Ft I
Now this impulse
will also change
the linear
momentum
of the
ball
.
1 c ball ballafter ballbefore
( )
Ft m v v
M
ultiply both sides of this equation by
cm cop
d
and
add these two equations to get
cmss ball ballafter ballbefore bat bataft
er batbefore
( ) ( )
d m v v I
(6)
Equations 4, 5 & 6
[
Watts and Bahill
]
Definition of the variables
ballbefore ballafter ssbefore
ssafter batbefore batafter
Inputs , and ,
Outputs , and .
v v v
v
First, I want to solve for
ballafter
v
.
Equations
Conservation of linear momentum
ball ballbefore bat ssbefore ball ballafte
r bat ssafter
m v m v m v m v
(4)
Definition of kinematic CoR
ballafter ssafter cmss batafter
ballbefore ssbefore cmss batbefore
v v d
CoR
v v d
(5)
An impulse torque during the collision changes angular momentum
cmss ball ballafter ballbefore bat bataft
er batbefore
( ) ( )
d m v v I
(6)
For all baseball swings that I can think of
batbefore
0
These equations were derived for the bat

ball system. Therefore th
ere were no external impulses (I
f the
collision is at the sweet spot then the batters arms do not apply an impulse.)
Equations 4, 5 and 6
produce the following equa
tion for the batted

ball speed [Watts and Bahill, 1990 and 2000, page 140].
2
ballbefore bat bat ball bat cmss bat bat batb
efore cmss batbefore
ballafter
2
ball bat bat bat ball bat cmss
(1 )
ball
v m m CoR I m m d m I v d CoR
v
m I m I m m d
Table 2
. Typical values for bat

ball collisions at the
sweet spot
,
configuration
2a
SI units
(
m/s
, rad/s
)
mph
Bahill
11/14/13
16
Inputs
CoR
0.55
ballbefore
v

37.00

83
batssbefore
v
24.59
55
batbefore
0
ballbefore
0
Outputs
ballafter
v
from Eqs. 4, 5 and 6
41.60
93
batssafter
v
11.99
21
batafter

31.81
This is the end of the Watts and Bahill
[1990]
derivation
, called configuration 2a.
Configuration 2b
, adds conservation of energy
Now we will consider the conservation of energy equation for collisions that are not at the center of
mass of the bat. There are no springs in the system and the bat swing is level, therefore there is no
change in potential energy. Before the collision, th
ere is kinetic energy in the bat created by rotation of
the body and arms plus the translational kinetic energy of the ball. The following equations are for a
collision at the sweet spot (ss) of the bat, which is in general 16 cm from the barrel end of the
bat
[Bahill, 2004]
. In Figure
1,
the sweet spot is the distance
cmss
d
from the center of mass
and
ssend
d
from
the barrel end of the bat.
before after lost
KE KE KE
2 2 2
before ball ballbefore bat ssbefore bat bat
before
1 1 1
2 2 2
KE m v m v I
The collision will make the bat spin about its center of mass. If the collision is at the center of
percussion
for the piv
ot
point,
it will produce a rotation about the center of mass, but no translation.
2 2 2
after ball ballafter bat ssafter bat bataft
er
1 1 1
2 2 2
KE m v m v I
Kinetic energy will be
lost to
heat in the ball, vibrations in the bat or deformations of the bat or ball. The
kinetic coefficient of restitution (CoR) models the energy that is lost in a
frictionless
head

on collision
between two objects
[
Dadouriam, 1913, Eq. (XI), p. 248;
Ferr
eira da Silva,
2007,
Eq.
23; Brach, 2007,
p. 39
]
.
Such a collision will have no tangential velocity components.
2
2
lost ballbefore ssbefore cmss batbefore
1
2
m
KE v v d CoR
(2)
The following combination occurs often, so we will give it a name, m bar,
1 2
1 2
mm
m
m m
It is
sometimes
colloquially called the effective mass.
Combining these three equations yields
Bahill
11/14/13
17
2
2 2 2 2
ball ballbefore bat ssbefore bat batbefore
ballbefore ssbefore cmss batbefore
2 2 2
ball ballafter bat ssafter bat batafter
1 1 1
1
2 2 2 2
1 1 1
2 2 2
m
m v m v I v v d CoR
m v m v I
(3)
Reality check.
Putting numbers from Table 1 into
these equations
yields
2
bat ssbefore
1
273 J
2
m v
Using rotations instead of
translations
2 2
bat bat arms spine
1
( 1.04 ) 305 J
2
I m I
The a
verage
kinetic energy
in the swings
of 29
members of the
San Francisco Giants
baseball team was
292 J
[Bahill and Karnavas, 1991]. Given the different circumstances for the experi
ments,
these
numbers are in agre
ement.
Bahill
11/14/13
18
Equations
3, 4, 5 & 6
,
Summary
This is a summary of the next dozen pages.
I guess it should go after them.
Definition of the variables
ballbefore ballafter ssbefore
ssafter batbefore batafter
Inputs , and ,
Outputs , and .
v v v
v
We want to solve for
ballafter ssafter cmss batafter
, and
v v d
. First, I want to solve for
ballafter
v
.
The e
quations
are
C
onservation of energy
2
2 2 2 2
ball ballbefore bat ssbefore bat batbefore
ballbefore ssbefore cmss batbefore
2 2 2
ball ballafter bat ssafter bat batafter
1 1 1
1
2 2 2 2
1 1 1
2 2 2
m
m v m v I v v d CoR
m v m v I
(3)
Conservation of linear momentum
ball ballbefore bat ssbefore ball ballafte
r bat ssafter
m v m v m v m v
(4)
Definition of
kinematic
CoR
ballafter ssafter cmss batafter
ballbefore ssbefore cmss batbefore
v v d
CoR
v v d
(5)
An impulse
during
the collision
changes
momentum
cmss ball ballafter ballbefore bat bataft
er batbefore
( ) ( )
d m v v I
(6)
These equations
were derived for the bat

ball system. Therefore there were no external impulse
s (if the
collision is at the sweet spot
then the batters arms do not apply an impulse).
Values for the variables
The pitch is in the negative x

direction, so the pitch speed is a negative number.
Table 3
. Typical values for bat

ball collisions at the
sweet spot
,
configuration
2b
.
SI units (m/s,
rad/s, J)
mph
Inputs
CoR
0.55
ballbefore
v

37.00

83
batssbefore
v
24.59
55
batbefore
0
ballbefore
0
Outputs
ballafter
v
from Eqs. 4, 5 and 6
41.60
93
ballafter
v
from Eqs. 3, 5 and 6
41.19
92
Bahill
11/14/13
19
ballafter
v
from Eqs. 3, 4 and 5
42.67
95
batssafter
v
11.99
21
batafter

31.81
Checks
before
KE
373
after lost
KE KE
380
It seems that we should do
what the astronomers do
and introduce
some
dark matter
or a cosmological
constant
here.
There is a discrepancy in the equation 3. The kinetic energy
lost
is too large.
E
quation
(2)
used for
lost
KE
was derived for a
frictionless
(
no tangential
velocity
components)
,
head

on collision between two
objects, but this collision is eccentric.
Bahill
11/14/13
20
Simplifying the notation ala Szidar
Equations
3, 4, 5 & 6
Conservation of energy
2
2 2 2 2
ball ballbefore bat ssbefore bat batbefore
ballbefore ssbefore cmss batbefore
2 2 2
ball ballafter bat ssafter bat batafter
1 1 1
1
2 2 2 2
1 1 1
2 2 2
m
m v m v I v v d CoR
m v m v I
(3)
2
2 2 2 2 2 2 2
1 1b 2 2b b 1b 2b b 1 1a 2 2a a
1
mv m v I m v v d CoR mv m v I
Let
1b 2b b
C
v v d
2 2 2 2 2 2 2 2
1 1b 2 2b b 1 1a 2 2a a
C 1 0
mv m v I mv m v I m CoR
(3s)
Conservation of linear momentum
ball ballbefore bat ssbefore ball ballafte
r bat ssafter
m v m v m v m v
(4)
1 1b 2 2b 1 1a 2 2a
mv m v mv m v
(4s)
Definition of kinematic CoR
ballafter ssafter cmss batafter
ballbefore ssbefore cmss batbefore
v v d
CoR
v v d
(5)
1a 2a a
1b 2b b
v v d
CoR
v v d
Let
1b 2b b
C
v v d
1a 2a a
C
v v d
CoR
(5s)
An impulse
duri
ng the collision changes
momentum
cmss ball ballafter ballbefore bat bataft
er batbefore
( ) ( )
d m v v I
(6)
1 1a 1b a b
( ) ( )
dm v v I
(6s)
Bahill
11/14/13
21
Equations
3, 5 and 6
The equations are
2 2 2 2 2 2 2 2
1 1b 2 2b b 1 1a 2 2a a
C 1 0
mv m v I mv m v I m CoR
(3s)
1a 2a a
C
v v d
CoR
(5s)
1 1a 1b a b
( ) ( )
dm v v I
(6s)
We want to solve for
ballafter ssafter cmss batafter
, and
v v d
.
First, I want to solve for
ballafter
v
.
From (5s) solve for
2a
v
2a 1a a
C
v v d CoR
From (6s) solve for
a
1
a b 1a 1b
( )
dm
v v
I
Now substitute this
a
into the above equation for
2a
v
2
1
2 1 1 1
( ) C
a a b a b
d m
v v d v v CoR
I
To simplify define
2
1
D
d m
I
2a 1a b 1b
(1 ) +D C
v D v d v CoR

To check, let’s solve it another way.
From
(
5
s
)
solve for
a
1 2
1
C
a a a
v v CoR
d
Put this into
(
6
s
)
1 1 1 1 2
1
( ) ( C )
a b a a b
dm v v I v v CoR
d
rearrange
and divide by
I
1
1 1 1 2
1
( ) ( C )
a b b a a
dm
v v v v CoR
I d
Solve for
2a
v
2
1
2 1 1 1
( ) C
a a a b b
d m
v v v v d CoR
I
To simplify define
2
1
D
d m
I
2a 1a b 1b
(1 ) +D C
v D v d v CoR
Same as above, good.

Bahill
11/14/13
22
To
use (3s) we need
2 2
2a a
and
v
2
2 2 2 1 0
2 1 1 b 1b 1 b 1b
(1 D) 2(1+D) +D C D C
a a a a
v v v d v CoR v d v CoR
To simplify let
b 1b
E 2(1+D) +D C
d v CoR
2
b 1b
F= D C
d v CoR
So
2 2 2 1 0
2a 1a 1a 1a
(1 D) E F
v v v v
We also need to compute
2
a
1
a b 1a 1b
( )
dm
v v
I
Let
1
G
dm
I
a 1a b 1b
G G
v v
2
2 2 2 1 0
a 1a 1a b 1b 1a b 1b
G 2G +G G
v v v v v
Prepare to substitute these
2 2
2a a
and
v
into (3s)
2 2 2 1 0
2 2a 1a 2 1a 2 1a 2
(1 D) E F
m v v m v m v m
2
2 2 2 1 0
a 1a 1a b 1b 1a b 1b
G 2G +G G
I v I v I v v I v
Now substitute
2 2
2 2a a
and
m v I
into (3s)
2 2 2 2 2 2 2 2
1 1b 2 2b b 1 1a 2 2a a
C 1 0
mv m v I mv m v I m CoR
(3s)
Next,
we create a quadratic equation
for
ballafter
v
. In anticipation of using the
quadratic formula
, we
solve
for
the coefficients
a, b
and
c
.
2 2
1 2
2 b 1b
2
2 2 2 2 2
1 1 2 2 2 b 1b
(1 D) G
E 2G +G
F G C 1
b b b
a m m I
b m I v
c mv m v I m I v m CoR
These
coefficients come
from a quadratic equation in
ballafter
v
that was derived from equations 3, 5 and 6.
Bahill
11/14/13
23
Equations 3, 4 and 5
2 2 2 2 2 2 2 2
1 1b 2 2b b 1 1a 2 2a a
C 1 0
mv m v I mv m v I m CoR
(3s)
1 1b 2 2b 1 1a 2 2a
mv m v mv m v
(4s)
1a 2a a
C
v v d
CoR
(5s)
We want to solve for
ballafter ssafter cmss batafter
, and
v v d
. First, I want to solve for
ballafter
v
.
From (4
s
) solve for
2a
v
1 1a 1 1b 2 2b
2a
2
mv mv m v
v
m
From (5
s
) solve for
a
a 1a 2a
1
C
v v CoR
d
Put
2
a
v
into this equation
1 1 1 1 2 2
1
2
1
C
a b b
a a
mv mv m v
v CoR
d m
Let
1
2
H 1+
m
m
and
1 1b 2 2b
2
J
mv m v
m
1
1
H J C
a a
v CoR
d
To use (3s) we need
2 2
2a a
and
v
1 1a 1 1b 2 2b
2a
2
mv mv m v
v
m
1 1a
2a
2
J
mv
v
m
2
2 2 1 0 2
1 1
2a 1a 1a 1a
2
2 2
2 J J
m m
v v v v
m m
Now we need to compute
2
a
2
2 2 2 1 0
a 1a 1a 1a
2
1
H 2H J C J C
v v CoR v CoR
d
Prepare to substitute these
2 2
2a a
and
v
into (3s)
2
2
2 2 1 0
a 1a 1a 1a
2 2 2
H 2H
J C J C
I I I
I v v CoR v CoR
d d d
2
2 2 1 0 2
1
2 2a 1a 1a 1 1a 2
2
2 J J
m
m v v v m v m
m
Now substitute
2 2
2 2a a
and
m v I
into (3s)
Bahill
11/14/13
24
2 2 2 2 2 2 2 2
1 1b 2 2b b 1 1a 2 2a a
C 1 0
mv m v I mv m v I m CoR
(3s)
2 2
2
2 2 2 2 1 0 2 2 2 1 0 2 2
1
1 1b 2 2b b 1a 1a 1 1a 2 2 2a 1a 1a 1a
2 2 2
2
H 2H
2 J J J C J C C 1 0
m I I I
mv m v I v v m v m m v v v CoR v CoR m CoR
m d d d
Next,
we create a quadratic equation for
ballafter
v
. In anticipation of using the quadratic formula, we solve
for the coefficients
a, b
and
c
.
2 2
1
1
2
2
1
2
2
2 2 2 2 2 2
1 1b 2 2b b 2
2
H
2H
2 J J C
J J C C 1
m I
a m
m d
I
b m CoR
d
I
c mv m v I m CoR m CoR
d
These coefficients come from a quadratic equation in
ballafter
v
that was derived from
equations 3, 4 and 5
.
Bahill
11/14/13
25
Abbreviations
ssbefore ballbefore cmss batafter batbe
fore
A (1 )( ) ( ) kgm/s
m CoR v v d CoR
ssbefore ballbefore cmss batafter batbe
fore
B (1 )( ) ( )
m CoR v v d CoR
1b 2b b
C m/s
v v d
2
1
D unit less
md
I
b 1b
E 2(1+D) +D C m/s
d v CoR
2 2 2
b 1b
F= D ( C) m/s
d v CoR
1
G 1/m
md
I
1 1 2
2 2
H 1+ unit less
m m m
m m
2
batcm
kgm
I I
1 1b 2 2b
2
J m/s
mv m v
m
2
1 2 1 2
K mI m I mm d
2b b 2
1
L v d m I CoR
1 2
1 2
mm
m
m m
Bahill
11/14/13
26
Previously we had
Equations 4, 5 & 6
1 1b 2 2b 1 1a 2 2a
mv m v mv m v
(4s)
1a 2a a
C
v v d
CoR
(5s)
1 1a 1b a b
( ) ( )
dm v v I
(6s)
We want to solve for
ballafter ssafter batafter
, and
v v
. First, I want to solve
for
ballafter
v
.
From (6s) solve for
a
a b 1a 1b
( )
G v v
From (5s) solve for
2a
v
2a 1a a
C
v v d CoR
Substitute
a
into this
2a
v
equation
2a b 1a 1b 1a
C D( )+
v CoR d v v v
Prepare to substitute this
2a
v
into (4s)
by multiplying by
2
m
2 2 2 2 2 1 1 2 1
C D( )+m
a b a b a
m v m CoR m d m v v v
Now substitute this
2 2
a
m v
into (4s)
1 1b 2 2b 1 1a 2 2a
mv m v mv m v
(4s)
1 1b 2 2b 1 1a 2 2 2 1 1 2 1
C D( )+m
b a b a
mv m v mv m CoR m d m v v v
S
tart solving
this equa
tion for
1a
v
by putting all
1a
v
terms on the left.
1 1a 2 1 2 1 1 1b 2 2b 2 2 2 1
m D C D
a a b b
mv v m v mv m v m CoR m d m v
Replace the dummy variables C and D and
2 2
1 1
1 1 2 2 1 1b 2 2b 2 1b 2b b 2 2 1
+m
a b b
md md
v m m mv m v m CoR v v d m d m v
I I
Divide by the big parenthesis, rearrange and multiply top and bottom by
I
.
2
1 1b 1 2 1 2 2b 2 1b 2b b 2
1
2
1 2 1 2
b b
a
mv I mm d v m v I mCoRI v v d m d I
v
mI m I mm d
Multiply
both sides
by

1 and rearr
ange.
2
1 1b 2 1b 1 2 1b 2 2b 2 2b 2 b 2 b
1a
2
1 2 1 2
mv I mCoRIv mm d v m v I mCoRIv mCoRId m d I
v
mI m I mm d
Rearrange
2
1b 1 2 1 2 2b b 2
1a
2
1 2 1 2
1
v mI m ICoR mm d v d m I CoR
v
mI m I mm d
Expand the subscripts and we get
the following
equation for the batted

ball speed [Watts and Bahill,
1990 and 2000, page 140].
Bahill
11/14/13
27
2
ballbefore bat bat bat ball bat cmss batbefo
re cmss batbefore bat bat
ballafter
2
ball bat bat bat ball bat cmss
(1 )
ball
v m I m I CoR m m d v d m I CoR
v
m I m I m m d
We can rearrange this equation into
Brach’s form
.
2
1b 1 2 1 2 2b b 2
1a
2
1 2 1 2
2
1 2 1 2
2b b 2
2
1b 1 2 1 2
1a
1b
1b
2
1b 1 2 1 2 1b 1 2 1 2
1a 1b
1
Let K
L 1
L
K K
K
add to the right side
K
v mI m ICoR mm d v d m I CoR
v
mI m I mm d
mI m I mm d
v d m I CoR
v mI m ICoR mm d
v
v
v
v mI m ICoR mm d v mI m I mm d
v v
2
2 2
1b 1 2 1 2 1 2 1 2
1a 1b
1b 2 2
1a 1b
1b 2
1a 1b
L
K K
L
K K
L
K K
(1 ) L
K
v mI m ICoR mm d mI m I mm d
v v
v m I m ICoR
v v
v m I CoR
v v
1b 2b b 2
1a 1b
2
1 2 1 2
1
v v d m I CoR
v v
mI m I mm d
ballbefore batbefore cmss batbefore bat b
at
ballafter ballbefore
2
ball bat bat bat ball bat cmss
1
v v d m I CoR
v v
m I m I m m d
if
cmss
0
d
this
indeed
reduces to
1b 2 2b 2
1a 1b
1 2
2b 1b 2
1a 1b
1 2
(1 ) 1
1
v m I CoR v m I CoR
v v
mI m I
v v m CoR
v v
m m
batbefore ballbefore bat
ballafter ballbefore
ball bat
( ) (1 )
v v m CoR
v v
m m
which we derived on page 3.
Now we want to find
a
in terms of the input parameters. From (6s) solve for
a
1
a b 1a 1b
( )
md
v v
I
Bahill
11/14/13
28
2
1b 1 2 1 2 2b b 2
1
a b 1b
2
1 2 1 2
1
v mI m ICoR mm d v d m I CoR
md
v
I mI m I mm d
2 2
1b 1 2 1 2 2b b 2 1b 1 2 1 2
1
a b
2
1 2 1 2
1
v mI m ICoR mm d v d m I CoR v mI m I mm d
md
I
mI m I mm d
1b 2 2b b 2
1
a b
2
1 2 1 2
(1 ) 1
v m I CoR v d m I CoR
md
I
mI m I mm d
1b 2b b 1 2
a b
2
1 2 1 2
1
v v d mm d CoR
mI m I mm d
ballbefore batbefore cmss batbefore ball
bat
batafter batbefore
2
ball bat bat bat ball bat cmss
1
v v d m m d CoR
m I m I m m d
Let
’
s try the other form of
1a
v
From (6s) solve for
a
1
a b 1a 1b
( )
md
v v
I
Substitute for
1a
v
1b 2 2b b 2
1
a b 1b 1b
2
1 2 1 2
1b 2 2b b 2
1
a b
2
1 2 1 2
1b 2b b 1 2
a b
2
1 2 1 2
(1 ) 1
(1 ) 1
1
v m I CoR v d m I CoR
md
v v
I
mI m I mm d
v m I CoR v d m I CoR
md
I mI m I mm d
v v d mm d CoR
mI m I mm d
They a
re the same
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