A Numerical Study of the Trajectory of a Baseball
by
Javier Cisneros
An Engineering Project to the Graduate
Faculty of Rensselaer Polytechnic Institute
in Partial Fulfillment of the
Requirements for the degree of
MASTER OF ENGINEERING
Major Subject: MECHANICAL ENGINEERING
Approved:
_________________________________________
Ernesto Gutierrez-Miravete, Project Adviser
Rensselaer Polytechnic Institute
Hartford, CT
December, 2010
(For Graduation December 2010)
TABLE OF CONTENTS
A Numerical Study of the Trajectory of a Baseball ........................................................... i
LIST OF TABLES ............................................................................................................ iii
LIST OF FIGURES .......................................................................................................... iv
LIST OF SYMBOLS ......................................................................................................... v
ACKNOWLEDGMENT .................................................................................................. vi
ABSTRACT .................................................................................................................... vii
1. Introduction.................................................................................................................. 1
2. Theory .......................................................................................................................... 3
2.1
The Effect of Gravity on the Flight of a Baseball .............................................. 3
2.2
The Effect of Air Resistance .............................................................................. 5
2.2.1
The Effect of the Drag Crisis on the Drag Force ................................... 7
2.3
The Effect of the Lift or Magnus Force ........................................................... 10
2.4
Euler Method .................................................................................................... 12
2.5
Runge-Kutta-Fehlberg method used by Maple ................................................ 13
3. Results........................................................................................................................ 15
3.1
The Effect of Gravity on the Flight of a Baseball ............................................ 15
3.2
The Effect of Gravity and Air Resistance ........................................................ 17
3.3
Varying Air Resistance to account for the Drag Crisis .................................... 18
3.4
Adding the Lift or Magnus force ..................................................................... 20
4. Discussion and Conclusion ........................................................................................ 22
4.1
Suggestions for Further Research .................................................................... 23
5. References.................................................................................................................. 24
6. Appendix.................................................................................................................... 25
6.1
Spreadsheet screen shots and Maple work sheets ............................................ 25
ii
LIST OF TABLES
Table 1 Maximization of Distance by Launch Angle………….....................………16
Table 2 Comparison of Three Different Numerical Solutions………………………..16
Table 3 Modified Euler Compared to Maple worksheet for the case….………………18
Table 4 Comparison of 2 Dimensional Results……………………………………..…..19
Table 5 Comparison of 3 Dimensional Solutions………………………………………21
iii
LIST OF FIGURES
Figure 1 Coordinate axes and a projectile under the influence of gravity only ................ 3
Figure 2 Projectile under the influence of gravity and air resistance ............................... 6
Figure 3 Drag Force .......................................................................................................... 6
Figure 4 Laminar Boundary Layer and a Turbulent Boundary Layer .............................. 8
Figure 5 Variation of Reynolds number vs. golf ball and rough sphere drag. ................. 8
Figure 6 Velocity (mph) and Reynolds Number (Re) vs. Drag coefficient.................... 10
Figure 7 Bird’s eye view of a baseball field with coordinate axes. ................................ 11
Figure 8 Forces on a Baseball ......................................................................................... 11
Figure 9 Contact Angle (degrees) vs. Distance (ft.) ....................................................... 15
Figure 10 Baseball Trajectory with Cd = 0.5 ................................................................. 17
Figure 11 2D Trajectory with Adair’s model of air resistance. ...................................... 18
Figure 12 2D Trajectory with SHSs model of air resistance. ......................................... 19
Figure 13 Plots of three dimensional cases for both models of air resistance. ................ 20
Figure 14 Screenshot of a spreadsheet for equation 1.11 ................................................ 25
Figure 15 Screen shot of a spreadsheet using the modified Euler method to solve the 2D
constant drag case. ................................................................................................... 25
Figure 16 Screen shot of spreadsheet using modified Euler method for 2D case with
Adair’s air resistance model .................................................................................... 28
Figure 17 Screen shot of spreadsheet using modified Euler method for two dimensional
case with SHS air resistance model ......................................................................... 31
Figure 18 Screen shot of spreadsheet using modified Euler method for 3D case with
Adair’s air resistance model .................................................................................... 34
Figure 19 Screen shot of spreadsheet using modified Euler method for 3D case with
SHS air resistance model ......................................................................................... 38
iv
LIST OF SYMBOLS
m
mass of a baseball ( lbs )
rb
radius of a baseball ( in )
A
frontal area of a baseball ( in 2 )
r
position vector
t
time ( sec )
g
 ft 
gravitational constant 
2 
 sec 
v0
 ft 
baseball’s initial velocity after being hit by the bat 

 sec 
a0
 ft 
baseball’s initial acceleration after being hit by the bat 
2 
 sec 
θ
angle from the horizontal ( deg )
ρ
 kg 
atmospheric density  3 
m 
v
 kg 
kinematic viscosity of air  2 
m 
µ
 kg 
dynamic viscosity of air 

 m • sec 
Re
dimensionless Reynolds number
Cd
dimensionless drag coefficient
CL
dimensionless lift coefficient
v
ACKNOWLEDGMENT
I would like to thank my advisor, Professor Ernesto Gutierrez-Miravete for helping me
finally finish this project. I’d like to thank my family and friends who constantly
reminded me to get to work.
vi
ABSTRACT
This project describes results of numerical experiments to investigate typical
trajectories of baseballs. Homeruns have long been one of the most exciting facet in all
of sports.
Ever since Mickey Mantle’s tape measure shot in 1956 we have been
infatuated with finding out how long a feared hitter can hit the baseball. Baseball
trajectories were predicted considering the effects of gravity, wind resistance, and the
Magnus effect. Calculations were performed using both a simple Euler integration
method and the Runge-Kutta algorithm embedded in the Maple software. Reasonably
good agreement was obtained between the results of these two computation methods for
the gravity only case and the wind resistance with constant drag coefficient cases.
Significantly different trajectories were predicted by both methods when non constant
drag
coefficients
or
Magnus
vii
forces
were
considered.
1. Introduction
Ever since we started playing sports scientists have tried to gain a more thorough
understanding behind the trajectories of sports balls. Baseball was mostly ignored by
scientists and many of them even believed that the curveball was an optical illusion.
Today, however, there’s even an official physicist of the national league, Dr. Adair, who
wrote a very influential book on the physics of baseball. There’s no longer an argument
about whether or not the curveball curves but there is an argument over how far it will
go when hit by a bat. Sawicki, G.S., Hubbard, M., and Strong, E.J., predict that if a
batter hits a curveball for a homerun then it will travel farther than if he were to hit a
fastball.
Ever since Babe Ruth started hitting home runs baseball fans have been enamored
with finding out just how long a player can hit a homerun. Some sports casters even
began describing homeruns as “tape measure shots”. This term came around after a
famous Mickey Mantle homerun in which a publicity director got hold of a tape measure
so that he could measure exactly how long it traveled. As it turns out that publicity
director never actually used a tape measure, instead he walked it off with his size 11
shoes and estimated the distance. There’s just something majestic about seeing a ball hit
that long, some stadiums even have different colored seats that mark exactly where the
longest homerun was hit.
Just about every baseball video game has a homerun derby mode, where one can
play as their favorite players and try to hit nothing but homeruns. These games almost
always have a distance meter somewhere on the screen that tells you how long you hit it,
much like the real homerun derby during the all-star break. This paper is an attempt at a
sanity check for these distance numbers that seem to appear with every homerun hit
today.
The trajectory of a baseball will be modeled first in a very simple manner by only
taking into account gravity where according to Adair1 it would travel approximately
750ft. if this was hit at sea level. Next the equations of motion will take into account
wind resistance and these solutions will be plotted and optimized for maximum distance.
According to official baseball rules the pitchers mound is 60 feet 6 inches away
from home plate which has an effect in a baseball’s maximum velocity as it reaches
1
home plate. A baseball cannot weigh more than 5.25 ounces or less than 5 ounces and
its diameter must measure between 9 to 9.25 inches. It also has 108 raised red cotton
stitches which make the ball fly farther and are almost impossible to model. The rubber
that the pitcher pushes off from before every pitch is never more than 10 inches tall.
Some ”submarine” or underhand pitchers are known to throw the ball as low as six
inches off the ground. These models are assuming that the pitcher is throwing with a
more conventional motion that is most commonly referred to as a 3/4 type throw. The
fraction is referring to the arm angle, which is 3/4 of the 90 degrees between a side-arm
throw and an overhand throw.
2
2. Theory
Three different forces were accounted for in reaching a mathematical model.
The baseball will first modeled as if it were in a vacuum on earth and the only force
retarding its flight after being hit by bat is gravity. Air resistance will then be taken into
account followed by the Magnus force.
2.1 The Effect of Gravity on the Flight of a Baseball
The simplest model would be a baseball falling under its own weight. Under these
circumstances Newton’s second law becomes
m
d 2r
= − mg
dt 2
(1.1)
We can cancel the mass terms to get
d 2r
= −g ĵ
dt 2
(1.2)
where r is the position vector with respect to a fixed origin (home plate).
r
y
θ
ĵ
W
î
x
Figure 1 Coordinate axes and a projectile under the influence of gravity only
If at t = 0 the ball is traveling at speed v0 at an angle θ to the horizontal then the initial
velocity is
v = v 0 cos θ î + v 0 sin θ ĵ
(1.3)
d 2r
dr
∫ dt 2 dt = dt = v0 = v0 − gt ĵ
(1.4)
Integrating equation 1.2 yields.
3
Substituting for the velocity yields.
v 0 = v 0 cos θ î + v 0 sin θ ĵ − gt ĵ = (v 0 cos θ ) î + (v 0 sin θ − gt ) ĵ
(1.5)
Assuming r = 0 when t = 0 equation 1.4 can now be integrated with respect to time.
r = v 0t −
1 2
gt ĵ
2
(1.6)
Substituting for the velocity in this position vector we get.
r = (v 0t cos θ ) î + (v 0t sin θ −
1 2
gt ) ĵ
2
(1.7)
The respective components in the x and y directions are
x = v 0t cos θ
(1.8)
y = v0t sin θ −
gt 2
2
(1.9)
Solving equation 1.8 for t and plugging that into equation 1.9 yields
y = x tan θ −
gx 2
2v02 cos 2 θ
(1.10)
By setting y = 0 we have an equation for x that we can solve in order to find out when
the maximum range will occur.
 2v 2 cos 2 θ
x = tan θ  0
g

 2v 02 sin θ cos θ v 02 sin 2θ
=
=
g
g

(1.11)
The results from equation 1.11 can be verified using the equations of motion with
constant acceleration:
y = y0 + v 0 y t −
1 2
gt
2
(1.12)
Since y = 0 when the baseball hits the ground, and substituting for the vertical
component of the velocity, v0 y , equation 1.12 becomes:
0 = y0 + v 0t sin θ −
1 2
gt
2
(1.13)
This equation can be used to find the amount of time t that the ball was in the
air. Once this is known the distance that it traveled can be found using the
following equation:
x = x0 + v 0t cos θ
4
(1.14)
Yet another way to solve this equation is to divide up time into slices equal to
dt so that at time t + dt:
x(t + dt ) = x(t ) + v x (t ) ∗ dt
(1.15)
y (t + dt ) = y (t ) + v y (t ) ∗ dt
(1.16)
v x (t + dt ) = v x (t ) + a x (t ) ∗ dt
(1.17)
v y (t + dt ) = v y (t ) + a y (t ) ∗ dt
(1.18)
Given that at time t = 0 . x0 , y0 , v 0 x , v 0 y are known and
ax = 0
ay = − g
(1.19)
Then we can set dt to 0.1 or smaller and arrive at the range by subsequently
multiplying through until y is negative.
2.2 The Effect of Air Resistance
In this section I will ignore the role that the stitches play in the flight of a
baseball, and model it as a smooth sphere.
As air flows around a sphere it
accelerates and the surface pressure decreases until the air gets halfway around the
sphere. It is at this point where the velocity is maximized and the pressure is
minimized. Over the rear of the sphere the velocity decreases and the pressure
increases which creates what is known as an adverse pressure gradient.
As the sphere is in flight it carries with it an ultra thin region of air near the
surface, otherwise known as a boundary layer, which will begin to separate from
the surface because it cannot overcome the increased pressure over the rear of the
sphere. This boundary layer separation causes the pressure differential between the
front and rear of the sphere to become constant which in turn results in a drag force
that resists the sphere’s forward flight. Accounting for this force Figure 1 now
becomes:
5
dy*
dt *
r
y
ĵ
θ
x*
D
W
dx*
dt *
y*
x
î
Figure 2 Projectile under the influence of gravity and air resistance
The magnitude of the drag force is usually expressed as
FD = ½CD ρ Av 2
(1.20)
where ρ is air density, v is the baseball’s speed, CD is the non-dimensional drag
coefficient and A is the frontal area of the baseball which is equal to
A = π rb2
(1.21)
Taking into account this drag force the equations of emotion become:
d 2x
m 2 = − FDx
dt
(1.22)
d2y
m 2 = − FDy − mg
dt
(1.23)
The following figure illustrates the x and y components of the drag force.
y
v 
FDx = -FD  x 
 v 
 vy 
FDy = -FD  
 v 
vy
x
vx
Figure 3 Drag Force
Taking this into account equations 1.21 and 1.22 now become
6
max = -FDx = -½ C DρAvv x = -½ C DρAv x v 2x + v 2y
ma y = -FDy − mg = -½ C DρAvv y − mg = -½ CDρAv y v 2x + v 2y − mg
(1.24)
(1.25)
The previous two equations can be solved much like equations 1.15 to 1.18 by
dividing time into small slices, except that acceleration isn’t constant this time
around. The acceleration term is in each direction is arrived at by these equations:
-½ C DρAv x v 2x + v 2y
ax =
ay =
2.2.1
(1.26)
m
-½ CDρAv y v 2x + v 2y
m
−g
(1.27)
The Effect of the Drag Crisis on the Drag Force
The drag coefficient CD varies with the spheres velocity, the orientation of the
baseball stitches during flight, and the humidity of the air. The magnitude of this
coefficient depends on the state of the boundary layer which can be “laminar” or
“turbulent”. A laminar boundary layer consists of smooth tiers of air passing one of
top of the other and is usually very desirable because it reduces drag. Laminar
boundary layers, however, are also very fragile and separate from the surface of a
sphere very easily when they encounter an adverse pressure gradient. By delaying
or minimizing the separation of this boundary layer from the sphere the drag
experienced by it would be reduced.
Frohlic2 noted that a “drag crisis” occurs when the laminar boundary layer
begins to separate and become turbulent. As you can see from Figure 4 the turbulent
boundary layer reduces the size of the turbulent wake behind the ball which leads to
reduced drag.
7
Figure 4 Laminar Boundary Layer and a Turbulent Boundary Layer3
In this drag crisis the drag coefficient for a spherical object changes
drastically by a factor of 2 -5, due to a change from laminar to turbulent flow. This
effect is clearly seen in experimental data like that pictured below.
Figure 5 Variation of Reynolds number vs. golf ball and rough sphere drag.4
The previous diagram illustrates how the drag on a sphere varies with the Reynolds
number (Re). The Reynolds number is an important non-dimensional parameter
that is used to relate the size of an object, or sphere, to the flow conditions it
experiences and for a baseball it is defined by the following equation:
Re =
ρ | v | (2rb )
µ
8
(1.28)
In this equation v is the baseball’s velocity relative to the air, µ is the dynamic
viscosity of air, ρ is the density of the air and rb is the radius of the baseball. The
baseball velocities that this paper is concerned with lies in the Reynolds number
range of 1.1x10 5 (50mph) to 2.8 × 10 5 (127.3 mph), which as you can see from Figure
5
lies in the reduced drag region corresponding to a drag coefficient in the range of
0.1 to 0.35 for a rough sphere.
In his book professor Adair states that the
incremental drag on a spinning baseball will usually not be much larger than 5% of
the drag on a non-spinning baseball. In his freshmen-level course5 on the physics
of baseball Professor Adair defines an equation for the drag coefficient as:
CD = 0.24 +
0.26
1 + e(v-83.9)/10.3
(1.29)
This equation isn’t in exact agreement with the findings of SHS6 (Greg Sawicki,
Mont Hubbard and William Stronge), who extracted information on the drag
coefficient versus the Reynolds number from data taken at the 1996 Olympic
Games.
In their paper Mitiguy and Woo7 posted the following drag model
equations for the drag coefficient which were used by SHS:
CD min = 0.15 , CDrebound = 0.35 if Re < Re drop = 158, 000
CD = CD min
Re − Redrop

+ (0.5 − CD min ) ∗ 1 − e 10000





When Re drop ≤ Re ≤ Re rebound CD = CD min
(1.30)
And Re > Re rebound = 175, 000
Re rebound − Re


CD = CD min + (CDrebound − 0.5) ∗ 1 − e 30000 


I tried plotting equations 1.30, in order to recreate Figure 3 of SHS but ended up
using the following equation for a better fit for when Re > Re rebound = 175, 000 .
Re rebound − Re


CD = CD min + (0.21) ∗ 1 − e 30000 


(1.31)
The following figure is a comparison of equations 1.29, and 1.30 (with 1.31
modification), with the speed of the baseball in mph on the x-axis and the Reynolds
9
number on the other x axis. As you can see from this figure the Drag force is greatly
reduced when it reaches speeds above 60 mph. It’s also important to note that much of
baseball is played in this reduced drag regime of 60-90 mph.
Reynolds Number (Re)
0.0E+00
5.0E+04
1.0E+05
1.5E+05
2.0E+05
2.5E+05
3.0E+05
0.55
0.5
Drag Coefficient Cd
0.45
0.4
SHS
0.35
0.3
Adair
0.25
0.2
0.15
0.1
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
Speed of baseball (mph)
Figure 6 Velocity (mph) and Reynolds Number (Re) vs. Drag coefficient
According to this website http://www.hittrackeronline.com/index.php8 the fastest
known speeds for a baseball after being hit by a bat during a major league baseball
game in the last 4 years is in the range of 120 to 127.3 mph.
2.3 The Effect of the Lift or Magnus Force
This force is a reason why a curveball curves and it’s also why a fastball has the illusion
of rising as it gets to the plate. The magnitude of the lift or magnus force can be
expressed as
FL = 1 CL ρ Av 2u
2
(1.32)
CL is the lift coefficient and it’s a function of the spin parameter S, the Reynolds number
and the roughness of the ball. The spin parameter S is a dimensionless number equal to
the ratio of the tangential velocity U, to the baseball’s relative velocity. The tangential
velocity is equal to the product rω, where r is the radius of the ball and ω is the spin rate.
10
S=
rω
v
(1.33)
Equation (1.32) also has a variable u which is the following cross product
ωˆ × v̂
(1.34)
For this paper the lift coefficient will be approximated by the following equations
CL = 1.5 ∗ S
when
S ≤ 0.1
(1.35)
CL = 0.09 + 0.6 ∗ S
when
S > 0.1
(1.36)
Taking into account the spin on a baseball makes this a 3 dimensional problem where z
is up and pointing towards the sky. The y axis points directly to the pitcher and the x
direction is pointing to the batter’s right which is illustrated in the following figure.
Figure 7 Bird’s eye view of a baseball field with coordinate axes.
Re-visiting the forces on a baseball we now have:
F
V
Magnus
ω
F
Drag
F
W
Figure 8 Forces on a Baseball
11
As you can see the magnus force is perpendicular to both v and ω, and will change signs
depending on whether we have top spin or back spin. Our coordinate system is arranged
such that the lift force in each direction is:
FLx = − FL sin θ
(1.37)
FLy = FL cos θ
(1.38)
And our new equations of motion are:
a x = − KCD vv x − KCL vv y sin φ
a y = − KCD vv y + KCLv ( v x sin φ − v z cos φ )
(1.39)
a z = − KCD vvz + KCL vv y cos φ − g
K in the previous set of equations stands for a constant equal to
ρA
2m
. We can plug in the
value for the velocity and equation 1.39 is now:
(v
(v
(v
a x = − KCD v x
a y = − KCD v y
a z = − KCD vz
2
x
+ v y2 + vz2 ) − KCL v y sin φ
(v
2
x
+ v 2y + vz2 ) + KCL (v x sin φ − vz cos φ )
2
x
+ v y2 + vz2 ) + KCLv y cos φ
2
x
(v
2
x
+ v 2y + v z2 )
+ v 2y
(v + v
+v ) −g
2
x
2
y
+ vz2 )
(1.40)
2
z
2.4 Euler Method
The Euler Method is a first order numerical procedure for solving ordinary
differential equations with a given initial value. We know all of the initial conditions for
this problem so a spreadsheet calculating each solution with a small time step was used
as one numerical approach. For example in order to compute where the ball is after our
first time step of t = 0.001 the following equation was used.
y1 = y0 + v0 y idt + a0 y idt 2
(1.41)
In this last equation v0 y was given by setting our initial ball velocity to one of the
velocities from hittrackeronline, dt was set to be 0.001 and a0 y is given by
a1y = − KCD v0 y
(v
2
0x
)
+ v02y + v02z + KCL (v0x sin φ − v0z cos φ )
(v
2
0x
+ v02y + v02z
As you can see by using the initial values we can eventually find yn +1
12
)
(1.42)
yn +1 = yn + vn y idt
+  − KCD vn y

(v
2
nx
)
+ vn2y + vn2z + KCL ( vnx sin φ − vnz cos φ )
(v
2
nx
)
+ vn2y + vn2z idt 2

(1.43)
This approach led to rather large Excel files and somewhat inaccurate results.
2.5 Runge-Kutta-Fehlberg method used by Maple
A second numerical approach was done using the dsolve …numeric command in
Maple. These worksheets used this command in order to call a Runge-Kutta-Fehlberg
method that produced a fifth order accurate solution.
At each step two different
approximations for the solution are made and compared. If the two answers are in close
agreement, the approximation is accepted. If the two answers do not agree to a specified
accuracy, the step size is reduced. If the answers agree to more significant digits than
required, the step size is increased. Each Runge-Kutta-Fehlberg step requires the use of
the following six values
k1 = hf(t j , y j )
1
1 

k2 = hf  t j + h, y j + k1 
4
4 

3
3
9 

k3 = hf  t j + h, y j + k1 + k2 
8
32
32 

12
1932
7200
7296 

k4 = hf  t j + h, y j +
k1 −
k2 +
k3 
13
2197
2197
2197 

439
3680
845 

k5 = hf  t j + h, y j +
k1 − 8k2 +
k3 −
k4 
216
513
4104 

(1.44)
1
8
3544
1859
11 

k6 = hf  t j + h, y j − k1 + 2k2 −
k3 +
k 4 − k5 
2
27
2565
4104
40 

Then an approximation to the solution of the initial value problem is made using a
Runge-Kutta method of order 4:
y j +1 = y j +
25
1408
2197
1
k1 +
k3 +
k 4 − k5
216
2565
4104
5
(1.45)
A better value for the solution is determined using a Runge-Kutta of order 5:
z j +1 = y j +
16
6656
28561
9
2
k1 +
k3 +
k 4 − k5 + k 6
135
12825
56430
50
55
13
(1.46)
The optimal step size sh can be determined by multiplying the scalar s times the
current step size h . The scalar s is
1/4


∈h
s=
 2 | z − y | 
j +1
j +1 

Where ∈ is the specified error control tolerance.
14
(1.47)
3. Results
Following are the results of the previous equations on a case by case basis.
3.1 The Effect of Gravity on the Flight of a Baseball
Using equation 1.11 for a given initial speed v0 the maximum range will occur
when the sin term is equal to one and this will occur when α =
π
4
radians. Thus the
v02
maximum range will occurr at . This can be seen in the following figure where
g
equation 1.11 was plotted for several angles and several initial velocities. The deepest
outfield wall of any current major league ball park was plotted as the solid line at 436 ft.
Figure 9 Contact Angle (degrees) vs. Distance (ft.)
The 45° angle was expected as we are only taking into account gravity at this point. The
maximum velocity plotted was 127.3 mph, which corresponds to the fastest a baseball
has been hit off a bat in a major league ball game over the past four years. Plugging this
value in for v0 results in a distance of 1084.2 ft. which is close to 2.5 times as long as the
15
longest outfield wall in today’s stadiums. This goes to show that air drag or wind
resistance is extremely important.
Assuming that a baseball is hit at an initial height of 3 ft. off the ground, at a
speed of 100 mph (146.7 ft./sec.), and an angle of 45° equation 1.13 becomes
0 = 3 + 103.36t − 16.08t 2
(1.48)
which can be solved for t, the time that the ball spends in the air, by employing the use
of the quadratic equation. Using the t that was just found, and given that x0 = 0 we can
find the distance that the ball traveled using equation 1.14.
x = 0 + 104.05t
(1.49)
These two equations were input into an excel spreadsheet in order to find the best angle
that would correspond to the maximum distance and the following table shows these
results.
Table 1 Maximization of Distance by Launch Angle
Speed (mph)
Theta
Time in Air (sec)
Distance (ft.)
100
44.87
6.47
672
110
44.89
7.11
812.5
120
44.91
7.75
966.4
127.3
44.92
8.22
1087.2
The results correspond well with the previous method of finding the maximum distance
and the angle slightly increases as the initial velocity does. The equations of 1.15
through 1.19 were used to find the ranges of hit baseballs. This exact solution was then
compared with two other solution methods in the following table.
Table 2 Comparison of Three Different Numerical Solutions
Speed (mph)
Euler (Excel)
Maple
Exact
100
127.3
100
127.3
100
127.3
Time in Air (seconds)
6.476
8.231
6.47
8.2295
6.47
8.22
Distance (ft.)
671.62
1086.67
671.6
1086.5
672
1087.2
16
3.2 The Effect of Gravity and Air Resistance
Equations 1.23 and 1.24 were solved with a modified Euler approach in an Excel
spreadsheet that divided time into increments of 0.001, this approach assumes that over
very short time intervals velocity is constant.
The following constants and initial
conditions were used in this spreadsheet.
A = 0.046 ft 2
ρ = 1.23
kg
kg
= 0.3483 3
3
m
ft
x0 = 0 y0 = 3 CD = 0.5
This air density is taken at normal temperature and pressure. Like before the initial ball
exit velocities of 100, 110, 120 and 127.3 mph were used and the launch angle was
optimized in order to max out the horizontal distance.
110
100
90
Vertical Distance (ft.)
80
127.3 mph
θ = 37.1°
70
120 mph
θ = 37.6°
60
50
110 mph
θ = 38.2°
40
30
100 mph
θ = 38.7°
20
10
0
0
25
50
75
100
125
150
175
200
225
250
275
300
325
350
375
400
Horizontal Distance (ft.)
Figure 10 Baseball Trajectory with Cd = 0.5
The best angle to hit a baseball decreases as the velocity increases. As one can see from
Figure 10 the drag force significantly reduces the maximum horizontal distance and it
has also led to a smaller launch angle. These two equations were also solved using a
maple worksheet and compared to the modified Euler in the following table.
17
Table 3 Modified Euler Compared to Maple worksheet for the case with a constant drag coefficient
Speed (mph)
Excel Solution
Optimized Launch Angle
(Degrees)
Time in Air (seconds)
Distance (ft.)
Maple
100
127.3
100
127.3
38.7
37.1
38.7
37.1
4.56
5.14
4.549
5.1251
310.87
393.2
308.84
390.25
For the simple case of a constant drag coefficient the maple solution arrives at pretty
much the same values as the Excel spreadsheet, but this small difference is more
pronounced in the higher initial velocity case.
3.3 Varying Air Resistance to account for the Drag Crisis
The drag force was varied according the equations of Professor Adair (equation 1.29)
and SHS (equations 1.30 and 1.31) and plotted for the four varying initial ball velocities.
Figure 11 2D Trajectory with Adair’s model of air resistance.
18
Figure 12 2D Trajectory with SHSs model of air resistance.
As you can see from the previous two figures the reduced drag coefficient that we saw in
Figure 6 with the SHS model of air resistance leads to farther range. Both of these plots
were arrived at by optimizing the angle for maximum distance by using a modified Euler
in an xcel spreadsheet which can be seen in the appendix. These two models were also
solved using a maple worksheet and those results can be seen in Table 4.
Table 4 Comparison of 2 Dimensional Results
Modified Euler Approach
Adair
Optimized Launch
Angle (Degrees)
Time in Air
(seconds)
Distance (ft.)
SHS
100
127.3
100
127.3
(mph)
(mph)
(mph)
(mph)
37.9
34
34.7
32.1
4.716
5.379
4.725
5.182
339.537
476.49
389
511.93
19
Maple worksheets
Optimized Launch
Angle (Degrees)
Time in Air
(seconds)
Distance (ft.)
37
35
34.9
32.4
4.7054
5.4778
4.7385
5.22
337.350
473.171
386.436
507.785
3.4 Adding the Lift or Magnus force
The lift force takes into account spin and for these cases I will assume that the rotational
speed is 2000 rpm, which is approximately equal to what it would be if somebody hit a
curveball. I will also only take into account backspin since topspin would make the lift
force negative.
Figure 13 Plots of three dimensional cases for both models of air resistance.
20
The last figure plots all 8 cases for both models of air resistance and the SHS model is
still going the farthest. The optimal launch angle has also been steadily decreasing as
the initial launch speed has been increasing.
Table 5 Comparison of 3 Dimensional Solutions
Modified Euler Approach
Adair
Optimized Launch
Angle (Degrees)
Time in Air
(seconds)
SHS
100
127.3
100
127.3
(mph)
(mph)
(mph)
(mph)
28.12
23.31
23.6
19.9
5.303
6.454
5.4
6.218
445.756
581.418
Distance (ft.) 382.2364 544.9247
Maple worksheets
Optimized Launch
Angle (Degrees)
Time in Air
(seconds)
Distance (ft.)
38
34.9
34.8
32.3
4.716
5.4677
4.729
5.2095
337.56
473.46
386.623
508.096
Table 5 shows some significant differences between the modified Euler and the maple
solution. In all four cases the Maple solutions differed by at least half a second and 44
ft. which could very well be the difference between a long fly ball and a homerun.
21
4. Discussion and Conclusion
All numerical solutions match up pretty well for the simple case where we only
take into account gravity. When compared to the exact solution the maple solution was
off by 0.12% and the Euler solution was off by 0.13%. This serves as a good baseline
letting us know that our maple worksheet and the Euler spreadsheet are working well.
As you can see from Table 3 we also have good comparisons for the case with a
constant drag coefficient. There is no exact solution for this case, but the angles which
maximized both solutions were exactly the same. The Maple solution differed from the
Euler solution by at least 2 ft, and as only 0.015 seconds.
When comparing the SHS model vs. Adair’s version of air resistance it’s obvious
that the SHS model will always go farther because of its treatment of the drag
coefficient. It’s important to note that the Euler approach is starting to show higher
results, by as much as 4 ft. when compared to the Maple solution. Table 4 also shows
that the two models of air resistance differ by 49 ft. at the 100mph velocity and only 35
ft. at the higher velocity. These differences were about the same for both numerical
methods.
Table 5 tells a whole different story and it looks like the Euler spreadsheet is
showing unbelievable values. It’s hard to believe that a ball would travel 581 ft. (SHS
version with initial launch speed of 127.3 mph) without the help of a ferocious tail wind.
Based on the data from the website hittrackeronline.com the Maple approach seems to
be showing better real world results. This website tracks every homerun hit in the major
leagues for the last few years. The longest homerun I could find was Adam Dunn’s
homerun in 2008 which was hit 504 ft. This is showing a shorter distance than the SHS
Maple numerical method, but he also hit it with a slower initial velocity. This website
also tracks historical homeruns, like Ted Williams blast in 1946, which is now marked
with a famous red seat in a sea of green ones. According to hittrackeronline.com this
homerun went a total of 527 ft. with the help of a strong wind that day.
Based on this work I believe that a curveball would be hit farther than a fastball
and I really wish homerun derbys would incorporate some sort of curveball aspect to
them.
A fastball would travel a lot faster but its negative Magnus force would pull it
down and not allow it travel as far.
22
4.1 Suggestions for Further Research
Further research into this topic should include varying the air density to account
for the elevation of different major league stadiums. This would affect both the drag and
lift force. Another factor to take into account would be tailwinds. A strong tailwind
would also affect both forces. Tailwinds would be difficult to model since they would
be stronger as the baseball rises above and away from the seats of a stadium. A decay
factor for the spin factor should also be included in a complete modeling in the flight of
a baseball.
23
5. References
[1] Adair, R.K. 2002. The Physics of Baseball. New York: HarperCollins.
[2] Mestre, Neville De 1990 The Mathematics of Projectiles in Sport New York:
Cambridge University Press.
[3] Sawicki, G.S., Hubbard, M., and Stronge, E.J., “How to hit home runs: Optimum
baseball bat swing parameters for maximum range trajectories,” Am. J. of Phys.,
November 2003, Vol. 71, No. 11, pp. 1152-1162.
[4] Mitiguy, P., Woo, M., “A Controversial Study of the Aerodynamics of a baseball,”
Proceedings of IDETC/CIE September 2005, pp. 1901-1905.
[5] http://online.physics.uiuc.edu/courses/phys199bb/fall07/
[6] Frohlich, C., “Aerodynamic drag crisis and its possible effect on the flight of
baseballs,” Am. J. Phys. 1984, Vol. 52, No. 4, pp. 325-334
24
6. Appendix
6.1 Spreadsheet screen shots and Maple work sheets
Figure 14 Screenshot of a spreadsheet for equation 1.11
Figure 15 Screen shot of a spreadsheet using the modified Euler method to solve the 2D constant
drag case.
This is the maple worksheet for the case of constant air drag = 0.5.
>
>
>
>
>
25
>
>
>
>
>
>
>
>
26
>
>
27
Figure 16 Screen shot of spreadsheet using modified Euler method for 2D case with Adair’s air
resistance model
This is the maple worksheet for the 2 dimensional case with Adair's air resistance model
>
Cd is the drag coefficient
>
This plot is more of a sanity check just to make sure that the drag matches Figure 6 except that v is in ft/sec
>
>
>
28
>
>
>
>
>
>
>
>
29
>
>
>
30
Figure 17 Screen shot of spreadsheet using modified Euler method for two dimensional case with
SHS air resistance model
This is the maple worksheet for the 2 dimensional case with SHS' air resistance model
>
E stands for the reynolds number where E =
where p is the density of air, v is the baseball's velocity, r is the radius of the
baseball, and u is the dynamic viscosity of air.
>
Cd is the drag coefficient and this worksheet has SHS' model for drag.
>
This plot is more of a sanity check just to make sure that the drag matches Figure 6 except that v is in ft/sec
>
31
>
>
>
>
>
>
>
32
>
>
>
>
>
>
33
Figure 18 Screen shot of spreadsheet using modified Euler method for 3D case with Adair’s air
resistance model
This is the maple worksheet for the 3D case with Adair's model of air resistance.
>
Cd is the drag coefficient and this worksheet has Adair's model for drag.
>
This plot is more of a sanity check just to make sure that the drag matches Figure 6 except that v is in ft/sec
>
>
>
34
>
>
>
35
>
>
>
>
>
>
>
36
>
>
>
>
>
>
37
Figure 19 Screen shot of spreadsheet using modified Euler method for 3D case with SHS air
resistance model
This is the maple worksheet for the 3D case with SHS' air resistance model
>
E stands for the reynolds number where E =
where p is the density of air, v is the baseball's velocity, r is the radius of the
baseball, and u is the dynamic viscosity of air.
>
Cd is the drag coefficient and this worksheet has SHS' model for drag.
>
38
This plot is more of a sanity check just to make sure that the drag matches Figure 6 except that v is in ft/sec
>
>
>
L is the lift coefficient which is depends on the spin factor that's defined above.
>
>
39
>
>
40
>
>
>
>
>
>
41
>
>
>
>
>
>
42
1
Robert K. Adair, Physics of Baseball, 3rd Ed. (Addison-Wesley, New York, 2002).
2
Cliff Frohlich, “Aerodynamic drag crisis and its possible effect on the flight of baseballs,” Am. J. Phys.
52, 325-334 (1984)
3
http://www.aerospaceweb.org/question/aerodynamics/q0215.shtml
4
Bearman, P. W., Harvey, J. K. 1976. Golf ball aerodynamics. Aeronaut. Q. 27:112-22
5
http://online.physics.uiuc.edu/courses/phys199bb/fall07/
6
Sawicki, G.S., Hubbard, M., and Stronge, E.J., “How to hit home runs: Optimum baseball bat swing
parameters for maximum range trajectories,” Am. J. of Phys., November 2003, Vol. 71, No. 11, pp. 11521162.
7
Paul Mitiguy and Michael Woo, “A Controversial Study of the Aerodynamics of a Baseball,” ASME
2005 International Design Engineering Technical Conferences & Computers and Information in
Engineering Conference September 24-28, 2005, Long Beach, California USA
8
http://www.hittrackeronline.com/index.php (Reggie Abercrombie, Florida Marlins 04/19/06 Speed of bat
of 127.3 mph)
43