A new formulation for computing batted-ball speed

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14 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

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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 ball-before bat bat-before ball ball-aft
er bat bat-after
m v m v m v m v
  




(conservation of momentum
)

and the
experimentally determined
kinematic
coefficient of restitution (CoR)

ball-after bat-after
ball-before bat-before
v v
CoR
v v

 









(CoR)

to derive

the following equations

for the velocities of the ball and bat after the collision
:

ball-before ball bat bat-before bat
ball-after
ball bat
( ) (1 )
v m CoRm v m CoR
v
m m
  



Bahill

11/14/13

10

ball-before ball bat-before bat ball
bat-after
ball bat
(1 ) ( )
v m CoR v m m CoR
v
m m
  



or

bat-before ball-before bat
ball-after ball-before
ball bat
( ) (1 )
v v m CoR
v v
m m
 
 


bat-before ball-before ball
bat-after bat-before
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 ball-before bat bat-before
1 1
2 2
KE m v m v
 

where
bat bat-cm
v v


A collision at the center of mass will not make the bat spin.

2 2
after ball ball-after bat bat-after
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 bat-before ball-before
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 ball-before bat bat-before bat-before ba
ll-before ball ball-after bat bat-after
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

bat-cm-before
v

29.90

49

Outputs



ball-after
v

from
Eqs.

4, 5 and 6

41.693
2

93

ball-after
v

from
Eqs.

3, 4 and
6

41.693
2

93

ball-after
v

from Eqs. 3, 4 and
5

41.6932

93

bat-cm-after
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

knob-cm
d


distance from the center of the knob to the center
of mass

0.57 m

22.4 in

pivot-cm
d


distance

from the pivot point to the center of mass
(H)

0.42 m

16.5 in

spine-cm
d

Distance from the spine to the center of mass, an
experimentally measured value, (L
a
)

1.04 m

41 in

spine-ss
d

Distance from the spine to the sweet
spot of the
bat

(H+R+B)

1.17

m

46.2 in

ss-endOfBarrel
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

ball-before
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

ball-after
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.



cm-before
v

velocity of the center of mass of the bat before the
bat
-
ball collision.

21.9 m/s

49 mph

cm-after
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

ss-before
v

velocity of the sweet spot of the bat before the
collision.

24.6 m/s

55 mph

ss-after
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

bat-before


angular velocity of the bat about its center of mass
before

the collision

0.02 rad/s

0.2 rpm

bat-after


angular velocity of the bat about its center of mass
after the collision

-
31.8 rad/s

-
304
rpm

spine-before


angular velocity of the batter
’s arms

a湤⁴桥⁢a琠
a扯畴⁴桥⁳灩湥

㈱⁲2d


㈰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 ball-before bat bat-before ball ball-aft
er bat bat-after
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

ball-after ss-after cm-ss bat-after
ball-before ss-before cm-ss bat-before
v v d
CoR
v v d


 
 
 



(5)

A note on my notation:

bat-after


is the angular velocity of the bat
about its center of mass

after the collision.

ss-before

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 bat-after bat-before
( )
cm ss bat
d Ft I
 

  

Now this impulse

will also change

the linear
momentum

of the
ball
.

1 c ball ball-after ball-before
( )
Ft m v v
 

M
ultiply both sides of this equation by
cm cop
d

and

add these two equations to get

cm-ss ball ball-after ball-before bat bat-aft
er bat-before
( ) ( )
d m v v I
 
   




(6)

Equations 4, 5 & 6

[
Watts and Bahill
]

Definition of the variables

ball-before ball-after ss-before
ss-after bat-before bat-after
Inputs , and ,
Outputs , and .
v v v
v
 

First, I want to solve for
ball-after
v
.

Equations

Conservation of linear momentum

ball ball-before bat ss-before ball ball-afte
r bat ss-after
m v m v m v m v
  






(4)

Definition of kinematic CoR

ball-after ss-after cm-ss bat-after
ball-before ss-before cm-ss bat-before
v v d
CoR
v v d


 
 
 







(5)

An impulse torque during the collision changes angular momentum

cm-ss ball ball-after ball-before bat bat-aft
er bat-before
( ) ( )
d m v v I
 
   





(6)

For all baseball swings that I can think of
bat-before
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
ball-before bat bat ball bat cm-ss bat bat bat-b
efore cm-ss bat-before
ball-after
2
ball bat bat bat ball bat cm-ss
(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


ball-before
v

-
37.00

-
83

bat-ss-before
v

24.59

55

bat-before


0


ball-before


0


Outputs



ball-after
v

from Eqs. 4, 5 and 6

41.60

93

bat-ss-after
v

11.99

21

bat-after


-
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
cm-ss
d

from the center of mass

and
ss-end
d
from
the barrel end of the bat.

before after lost
KE KE KE
 

2 2 2
before ball ball-before bat ss-before 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 ball-after bat ss-after bat bat-aft
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 ball-before ss-before cm-ss bat-before
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 ball-before bat ss-before bat bat-before
ball-before ss-before cm-ss bat-before
2 2 2
ball ball-after bat ss-after bat bat-after
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 ss-before
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

ball-before ball-after ss-before
ss-after bat-before bat-after
Inputs , and ,
Outputs , and .
v v v
v
 

We want to solve for
ball-after ss-after cm-ss bat-after
, and
v v d

. First, I want to solve for
ball-after
v
.

The e
quations

are

C
onservation of energy





2
2 2 2 2
ball ball-before bat ss-before bat bat-before
ball-before ss-before cm-ss bat-before
2 2 2
ball ball-after bat ss-after bat bat-after
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 ball-before bat ss-before ball ball-afte
r bat ss-after
m v m v m v m v
  






(4)

Definition of
kinematic
CoR

ball-after ss-after cm-ss bat-after
ball-before ss-before cm-ss bat-before
v v d
CoR
v v d


 
 
 







(5)

An impulse

during

the collision
changes

momentum

cm-ss ball ball-after ball-before bat bat-aft
er bat-before
( ) ( )
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


ball-before
v

-
37.00

-
83

bat-ss-before
v

24.59

55

bat-before


0


ball-before


0


Outputs



ball-after
v

from Eqs. 4, 5 and 6

41.60

93

ball-after
v

from Eqs. 3, 5 and 6

41.19

92

Bahill

11/14/13

19

ball-after
v

from Eqs. 3, 4 and 5

42.67

95

bat-ss-after
v

11.99

21

bat-after


-
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 ball-before bat ss-before bat bat-before
ball-before ss-before cm-ss bat-before
2 2 2
ball ball-after bat ss-after bat bat-after
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 ball-before bat ss-before ball ball-afte
r bat ss-after
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

ball-after ss-after cm-ss bat-after
ball-before ss-before cm-ss bat-before
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

cm-ss ball ball-after ball-before bat bat-aft
er bat-before
( ) ( )
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
ball-after ss-after cm-ss bat-after
, and
v v d

.
First, I want to solve for
ball-after
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
ball-after
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
ball-after
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
ball-after ss-after cm-ss bat-after
, and
v v d

. First, I want to solve for
ball-after
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
ball-after
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
ball-after
v
that was derived from

equations 3, 4 and 5
.




Bahill

11/14/13

25

Abbreviations



ss-before ball-before cm-ss bat-after bat-be
fore
A (1 )( ) ( ) kg-m/s
m CoR v v d CoR
 
    



ss-before ball-before cm-ss bat-after bat-be
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
bat-cm
kg-m
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
ball-after ss-after bat-after
, and
v v

. First, I want to solve
for
ball-after
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
ball-before bat bat bat ball bat cm-ss bat-befo
re cm-ss bat-before bat bat
ball-after
2
ball bat bat bat ball bat cm-ss
(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

  
 
 





ball-before bat-before cm-ss bat-before bat b
at
ball-after ball-before
2
ball bat bat bat ball bat cm-ss
1
v v d m I CoR
v v
m I m I m m d

  
 
 

if

cm-ss
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
   
 

 
 


bat-before ball-before bat
ball-after ball-before
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

 
  
 
 





ball-before bat-before cm-ss bat-before ball
bat
bat-after bat-before
2
ball bat bat bat ball bat cm-ss
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