Mechanical Systems
Professor Flint O. Thomas
Department of Aerospace & Mechanical Engineering
Hessert Center for Aerospace Research
Selection of Material on the Subject of
Projectile Motion given by Professor Thomas
for EG 111 - Fall 2000
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Mechanical
Systems


Today’s Lecture: “Flight Dynamics”
Goal: Develop a model of the flight of
the ball. Our analysis commences at the
instant the ball leaves the pouch.
y
V0 y
Trajectory will depend on the initial speed V0 and launch
angle q imparted by the launcher.
V0
q0
V0 x
y0
"Initial Conditions" at the
instant the ball leaves the
pouch: time t=0.
x
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

Consider “snapshots” of the ball taken at equal time
intervals Dt = ti+1-ti. At any instant of time the ball
will be acted on by two forces: weight and aerodynamic
drag.
y
t3
ti
ti+1
t2
t0
V
t1
yo
q
Dx
Dy
D
W
x
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

The motion of the ball will be governed by Newton’s
second law:


F  m a
Remember: force and acceleration are vectors which
possess both magnitude and direction. Vector addition
is by the parallelogram law.

Fy
F
Fx

Newton’s 2nd Law may be written for each component:
 Fx  m a x
Introduction to Engineering Systems
 Fy  m a y
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame


Let’s apply Newton’s Second Law to the flight of the
ball:
V(t)
q(t)
Dx
q(t)
Dy
D
W
Note that the drag force D always opposes the flight
direction. Also we see that:
Dx  D cosq
Introduction to Engineering Systems
D y  D sin q
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

Newton’s Law for the x-component forces:
 Fx  D cosq  m a x

Newton’s Law for the y-component forces:
 Fy  D sin( q)  mg  m a y

These equations relate the instantaneous x- and
y-component accelerations of the ball to the
instantaneous forces acting on the ball during the flight.
V(t)
Dy
Introduction to Engineering Systems
Dx
q(t)
D
q(t)
W
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame


Before analyzing the trajectory of the ball, let’s look
at the simpler problem of its motion in a vacuum.
In this case there is no air resistance and the
equations governing the motion can be obtained
by setting D = 0:
 Fx  0  m a x
ax  0
 Fy  m g  m a y
a y  g
In this case there is no x-component acceleration so
Vx will be constant. The vertical acceleration is constant.
The motion of the ball is a superposition of uniformly
accelerated motion in y and constant speed in x.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
We will divide the flight of the ball into many small
sub-flights, each of duration Dt. We start with known
initial conditions from launch at time t = 0:

Vx 0  Vo cos q0 

Vy 0  Vo sin q0 
x0
y  yo
The acceleration is the time rate of change of velocity.
Over any of the short time intervals Dt we have:
Vx ( t  Dt )  Vx ( t )
ax 
0
Dt
ay 
Vy ( t  Dt )  Vy ( t )
Dt
 g
Vx (t  Dt)  Vx (t )
Vy ( t  Dt )  Vy ( t )  g Dt
“Recipe” for finding the speed of the ball at t+Dt from the known speed at t.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
The average x and y-component speeds of the ball over the
time interval Dt are given by:

V ( t  Dt )  Vx ( t )
Vx  x
2

Vy 
Vy ( t  Dt )  Vy ( t )
2
The new x, y position of the ball is approximated as,
x(t  Dt )  x(t )  Vx Dt

y(t  Dt )  y(t )  Vy Dt
To illustrate consider the following example:
Ball Launched with Initial Conditions (time t = 0):
V0 = 40 m/s,
Introduction to Engineering Systems
q  35 degrees, x0 = 0, y0 = 5 m
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Vx 0  V0 cosq0   32.77 m / s = constant
Vy0  V0 sin q0   22.94 m / s
Suppose we divide the flight into Dt = 0.1 second
intervals. The component speeds of the ball at
the next instant of time t = 0.1 second are,
Vx (0.1 sec)  Vx (0)  32.77 m / s
Vy (0.1 sec)  Vy (0)  g Dt  21.96 m / s
The average speed over this interval is,
Vx  32.77 m / s
Introduction to Engineering Systems
Vy 
Vy (Dt )  Vy (0)
Module 1: Mechanical Systems
2
 22.45 m / s
Copyright ©2000, University of Notre Dame
The position of the ball at t=0.1 second is obtained from,
x (0.1 sec)  x(0)  Vx Dt  3.27 m
y(0.1 sec)  y(0)  Vy Dt  7.25 m
The procedure is now repeated sequentially for t=0.2,
0.3, 0.4 seconds etc. Microsoft Excel is an ideal tool
for performing these calculations. A sample spreadsheet
Vx (m/s) Vy (m/s) deltat (sec) x (m)
y (m)
time (sec)
is shown below.
Initial
32.76711 22.94158
0.1
0
5
0
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
32.76711
Introduction to Engineering Systems
21.96158
20.98158
20.00158
19.02158
18.04158
17.06158
16.08158
15.10158
14.12158
13.14158
12.16158
11.18158
Module 1: Mechanical Systems
3.276711
6.553423
9.830134
13.10685
16.38356
19.66027
22.93698
26.21369
29.4904
32.76711
36.04383
39.32054
7.245158
9.392317
11.44147
13.39263
15.24579
17.00095
18.65811
20.21727
21.67842
23.04158
24.30674
25.4739
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
condition
time
Copyright ©2000, University of Notre Dame

The trajectory (and range) of the ball is shown below:
35
30
25
20
15
10
5
Range
0
0

20
40
60
100
80
X (meters)
120
140
160
180
How do we know if this is correct?
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
!! Whenever possible numerical solution methods should be
checked against exact analytical solutions in order to
insure the accuracy of the technique. We will apply that
approach here.

Recall from your calculus course that:
Vx ( t  Dt )  Vx ( t ) dVx
a x  lim

Dt 0
Dt
dt
a y  lim
Dt 0

Vy ( t  Dt )  Vy ( t )
Dt

dVy
dt
Acceleration is the time derivative of velocity
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Newton’s
Second Law becomes:
dVx
0
dt
and
dVy
dt
 g
These are called differential equations because they involve
derivatives of the velocity of the ball. Note that in a
vacuum the x and y motions are independent, as noted
before.
 In order to obtain the velocity Vx(t) and Vy(t)
of the ball from the above equations we need to take
anti-derivatives (i.e. integrate) of both sides of the
equations.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
We will integrate with respect to time over the interval
from t = 0 to some arbitrary time t during the flight.
Vx
 dVx  0
and
Vx 0
Vy
t
Vy 0
0
 dVy  g  dt
So the velocity as a function of time is,
Vy  Vy 0  g t
Vx  Vx 0
Now recall:
Introduction to Engineering Systems
dx
Vx 
dt
and
dy
Vy 
dt
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

The differential equations for the trajectory of the
ball are,
dx
dy
 Vx 0 and
 Vy 0  g t
dt
dt
As before, we integrate with respect to time to obtain
the trajectory of the ball x(t) and y(t). Remembering
that at t = 0, x=0 and y = y0 we get,
x  Vx 0 t
1 2
y  y 0  Vy 0 t  g t
2
This is the analytical solution for the trajectory of the ball.
How does it compare to our numerical solution ?
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Excellent agreement!
35
30
25
20
15
analytical
numerical
10
5
0
0
Introduction to Engineering Systems
20
40
60
80
100
X (meters)
120
Module 1: Mechanical Systems
140
160
180
Copyright ©2000, University of Notre Dame

With the model validated we can put it to work.
Example: For a fixed initial velocity what launch angle gives
maximum range?
Vacuum (No Drag) , y0 = 5m
Launch Velocity Fixed at 40 m/s
180
160
B
B
140
For projectile motion
in a vacuum, maximum
range occurs at a
launch angle of 45 degrees.
B B B
B
B
B
120
B
Maximum Range
for 45 degree
launch angle
B
100
B
B
B
80
B
60
B
B
40 B
B
20
0
0
Introduction to Engineering Systems
10
20
30
40
50
60
70
Launch Angle (degrees)
Module 1: Mechanical Systems
80
B
90
Copyright ©2000, University of Notre Dame
Example: What combinations of V0 and q0 yield a desired range of
100 meters? Solution... construct a ballistics chart:
250
45 m / s
40 m / s
200
35 m / s
30 m/s
150
Desired range
of 100 meters
100
50
0
0
10
20
30
40
50
60
Launch Angle (degrees)
70
80
Numerous possibilities indicated…impossible for V0 = 30 m / s.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

We do not live in a vacuum. We are at the bottom of an
ocean of air! What is the effect of the air on the
flight of the ball?
Photograph by F. N. M. Brown, Notre Dame
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

Visualization of flow over a cylinder:
Photo by Thomas C. Corke
Where does the energy to put the air in motion come from?
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Conclude that the passage of the ball through air
will give rise to a loss in the kinetic energy of the
ball that must be accounted for in order to
realistically model the trajectory. In other words,
we must account for aerodynamic drag!
Back to the original form of the equations of motion:
 D cosq
ax 
m
D sin( q)
a y  g 
m
These equations don’t help us unless we know how
to determine D !!
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

Goal : Determine the aerodynamic drag on the ball and
incorporate it into our numerical simulation of the
flight.
This requires some basic fluid mechanics (the branch of
mechanics which deals with the dynamic behavior of gases
and liquids). The objective of fluid mechanics is most
often the determination of flow-induced forces.

Example: Wing lift and drag:
V
LIFT
DRAG
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Select a reference frame:
observer sees
ball moving at
speed V through
air at rest.
V
Two ways of
looking at the same
problem. The drag
would be identical
in both cases since
it is the relative
motion of the air
and ball that
determine drag.
This equivalence is
one basis of wind
tunnel testing.
ball
observer sees the
ball at rest in a
flow of air at
speed V.
V
ball
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame

How is drag created?
density = r
V
ball
"Cylinder of Air"
A
L  V Dt

The momentum of the “cylinder of air” is the product
of its mass times its velocity.
Mass of air = density X volume = r A L  r A V Dt
2
r
A
V
Dt
Momentum of air = Mass X velocity =
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
As the cylinder of air and ball collide the momentum
of the air is imparted to the ball.

density = r
ball
"Cylinder of Air"
A
V
L  V Dt

Another way of writing Newton’s second law is in
terms of the time rate of change of momentum.
 

 d mV

ma
F 
dt
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
If we assume all the momentum of the cylinder of air
is imparted to the ball during the time interval Dt then
the force on the ball D is given by,
Momentum of air
Dm V ball r A V 2 Dt
2
D

rAV
Dt
Dt
Transferred during time Dt
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Is this result correct? Not necessarily.... Our
calculation assumes that all of the momentum of
the cylinder of air is transferred to the ball. This
may not be the case!
What we can say with some certainty is,
DrA V
Remember:
2
r
= air density
A
= area of the body “seen” by
the oncoming flow. [m2]
[ kg/m3]
V = fluid velocity = flight speed [m/s]
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
To make this an equality the convention is to
incorporate a factor CD/2 (typically determined
from experiment) such that,

CD
D
r A V2
2
CD is called the
“drag coefficient”
The
drag coefficient CD depends on the geometry of
the body and also varies in a complicated way with flow
speed.
 Fortunately, for flow over a sphere at the velocities
that we’re likely to encounter during the launch project,
to good approximation experiments show that:
CD  0.5
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
We will model the drag on the ball by the relation,

D  0.25 r A V2

Now that we have a model for the drag force we can
perform a numerical simulation of the flight of the
ball including the effect of drag.
As before, we will divide the flight of the ball into many
small sub-flights, each of duration Dt. We start with
the known initial conditions from launch at time t = 0:

x0
Introduction to Engineering Systems
y  yo
Vx 0  Vo cos q0 
Module 1: Mechanical Systems
Vy 0  Vo sin q0 
Copyright ©2000, University of Notre Dame

The acceleration is the time rate of change of velocity.
Over any of the short time intervals Dt we have:
Vx ( x  Dt )  Vx ( t )  D( t ) cosq
ax 

Dt
m
ay 
Vy ( t  Dt )  Vy ( t )
Dt
D( t ) sin q
 g 
m
These provide a “recipe” for advancing the speed of the ball
in time…If I know Vx(t) and Vy(t) I can use these equations
to find Vx and Vy at time t+Dt later.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
The recipe...
D( t ) cosq
Vx ( x  Dt )  Vx ( t ) 
Dt
m
D( t ) sin q 

Vy ( t  Dt )  Vy ( t )   g 
 Dt
m


with drag given by our model,
D( t )  0.25 r A


Vx2 ( t )  Vy2 ( t )
Vy
V
Vx
q
Introduction to Engineering Systems
Module 1: Mechanical Systems
What about q ? !!
Copyright ©2000, University of Notre Dame

The angle q depends on the the relative size of the xand y-components of the velocity of the ball (as shown
below) and is therefore a function of time.
35
q
30
q
V
25
Vy
q
q
15
q
‘looks like simple trig
to me…”
20
Vx
10
Vy ( t ) 
1


q(t )  tan 

V
(
t
)
 x 
5
0
0
20
40
Introduction to Engineering Systems
60
80
100
X (meters)
120
140
Module 1: Mechanical Systems
160
Copyright ©2000, University of Notre Dame
Now I see the procedure!
1. Using Vx(ti) and Vy(ti) compute the drag force
and angle theta.
Repeat 1-4
2. Use “the recipe” provided courtesy of Sir
Isaac Newton to find the new velocity Vx(tiDt
and Vy(tiDt a time Dt later.
3. Calculate the average x- and y-component
velocity for the time interval Dt.
4. Use the average velocity and knowledge of the
position of the ball x (ti), y (ti) to find the
of the ball x(tiDt), y (tiDt).
NO
Introduction to Engineering Systems
Has the ball hit the ground?
Module 1: Mechanical Systems
YES
Copyright ©2000, University of Notre Dame
To
illustrate, reconsider the example we looked at before:
Ball Launched with Initial Conditions (time t = 0):
V0 = 40 m/s,
q0  35 degrees, x0 = 0, y0 = 5 m
With the additional information that:
m = 300 grams = 0.3 kg
Diameter = 0.051 meters
A = 2.043 X 10-3 m2
Initial x- and y-component velocities are:
Vx 0  V0 cosq0   32.77 m / s Vy0  V0 sin q0   22.94 m / s
As before we will divide the flight into Dt = 0.1 second
intervals.

Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
Compute the drag force:
D  0.25 r A V 2



 0.25 1.225 kg / m3 2.043 103 m 2 1600 m 2 / s 2

 1.00 N
Compute the new x- and y-component velocities:
D cosq
Vx (0.1 sec)  Vx (0) 
Dt
m

1.00 N  cos35 
0.1sec 
 32.77 m / s 
0.3 kg
 32.50 m / s
Introduction to Engineering Systems
D sin q 

Vy (0.1sec)  Vy (0)   g 
 Dt
m 

 0.1sec


1
.
00
N
sin
35
2
 22.94 m / s   9.8 m / s 

0.3 kg

 21.77 m / s
Module 1: Mechanical Systems


Copyright ©2000, University of Notre Dame
Compute
the average speed for the interval,
Vx (0.1sec)  Vx (0)
Vx 
 32.64 m / s
2
Vy (0.1sec)  Vy (0)
Vy 
 22.36 m / s
2
The
position of the ball at t=0.1 second is obtained from,
x(0.1 sec)  x(0)  Vx Dt  3.26 m
y(0.1 sec)  y(0)  Vy Dt  7.24 m
The
procedure is now repeated using Microsoft Excel.
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame
You will develop your own spreadsheet in the learning center.
ti
q x(ti) y(ti)
Vy
Vx
D
Initial Velocity(m/s) Launch angle (degrees) time step (sec) mass (kg)
diameter (m) Area (m^2)
40
35
0.1
0.3
0.051 0.00204282
time (sec)
Vx (m/s)
Vy (m/s)
Drag (N)
theta
initial conditions:
0 32.76608177 22.94305745 1.000982105
35
0.1 32.49276293 21.77167754 0.957054997 33.8239428
0.2 32.22773783 20.61409827 0.915628445 32.6045423
0.3 31.97062641 19.46963994 0.876602466 31.3407983
0.4 31.72106099 18.33765828 0.839883683 30.0318182
0.5 31.47868523 17.21754305 0.805384829 28.6768412
0.6 31.24315307 16.10871666
0.7730243 27.2752652
0.7 31.01412783 15.01063311 0.742725739 25.8266756
0.8 30.79128138 13.9227769 0.714417648 24.3308753
0.9 30.57429336 12.84466225 0.688033035 22.7879157
1 30.36285052 11.77583233 0.663509071
21.198128
1.1 30.15664614 10.71585867 0.640786786 19.5621541
1.2 29.95537953 9.664340625 0.619810774 17.8809747
1.3 29.75875564 8.62090493 0.600528916 16.1559355
1.4
29.5664847 7.585205375 0.582892124 14.3887684
1.5 29.37828207 6.556922475 0.566854095 12.5816068
1.6 29.19386804 5.53576321 0.552371083 10.7369934
1.7 29.01296785 4.521460781 0.539401674 8.85788029
1.8 28.83531172 3.513774359 0.527906584 6.94761813
1.9 28.66063497 2.512488837 0.517848456 5.00993572
2 28.48867828 1.517414528 0.509191675 3.04890836
x
0
3.262942
6.498967
9.708885
12.89347
16.05346
19.18955
22.30241
25.39268
28.46096
31.50782
34.53379
37.5394
40.5251
43.49136
46.4386
49.36721
52.27755
55.16997
58.04476
60.90223
How significant is the influence of drag?
Introduction to Engineering Systems
Module 1: Mechanical Systems
Initial conditions
y
5
7.2357367
9.3550255
11.359212
13.249577
15.027337
16.69365
18.249618
19.696288
21.03466
22.265685
23.39027
24.40928
25.323542
26.133847
26.840954
27.445588
27.948449
28.350211
28.651524
28.853019
time
Copyright ©2000, University of Notre Dame
A very significant effect on range!
35
30
25
20
15
10
5
0
B with drag
JJJJJJJ
JJ
JJ
J
J
J
J
J vacuum
J
J BBBBBBBBBB
J
JB
BB
J
H diameter * 2
JBB
J
B
B
JB
J
B
JB
J
B
H
H
H
H
H
B
H
H
J H
H
J
B
HH
BHH
J
B
J
H
BH
H
B
H
J
J
H
B
H
B
H
J
J
BH
H
B
JBH
H
J
B
H
JBH
J
H
B
H
J
H
B
J
H
B
H
JH
B
J
B
H
J
B
H
B
H
J
H
B
JB
H
J
H
B
JB
H
J
H
B
range = 130 meters
H
J
B
H
0
20
40
60
80
100
120
140
160
The trajectory is
no longer
parabolic!
180
X (meters)
Introduction to Engineering Systems
Module 1: Mechanical Systems
Copyright ©2000, University of Notre Dame