professor flint o. thomas department of aerospace & mechanical engineering

Introduction to Engineering Systems

Copyright ©2000, University of Notre Dame

Module 1: Mechanical Systems

Professor Flint O. ThomasDepartment of Aerospace & Mechanical Engineering

Hessert Center for Aerospace Research

Mechanical Systems

Selection of Material on the Subject of Projectile Motion given by Professor Thomas

for EG 111 - Fall 2000




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.

x

y

V 0 y V 0

y0

V 0 x

"Initial Conditions" a t theinstant the ball leaves the

pouch: time t=0 .

Trajectory will depend on the initial speed V0 and launch

angle imparted by the launcher.




Consider “snapshots” of the ball taken at equal time intervals t = ti+1-ti. At any instant of time the ball will be acted on by two forces: weight and aerodynamic drag.

x

y

yo

t1

t2

t3

t0

ti ti+1

W

V

D

D x

D y




The motion of the ball will be governed by Newton’s second law:

amF

Remember: force and acceleration are vectors which possess both magnitude and direction. Vector addition is by the parallelogram law.

F

Fx

Fy

Newton’s 2nd Law may be written for each component:

xx amF yy amF




Let’s apply Newton’s Second Law to the flight of the ball:

W

V(t)( t)

D

D x

D y

( t)

Note that the drag force D always opposes the flight direction. Also we see that:

cosDDx sinDDy




Newton’s Law for the x-component forces:

xx amcosDF Newton’s Law for the y-component forces:

yy ammg)sin(DF These equations relate the instantaneous x- and y-component accelerations of the ball to the instantaneous forces acting on the ball during the flight.

W

V(t)( t)

D

D x

D y

( t)




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:

xx am0F 0ax

yy amgmF ga y

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.




We will divide the flight of the ball into many small sub-flights, each of duration t. We start with known initial conditions from launch at time t = 0:

0o0x cosVV 0o0y sinVV oyy0x

The acceleration is the time rate of change of velocity. Over any of the short time intervals t we have:

0t

)t(V)tt(Va xxx

gt

)t(V)tt(Va yyy

)t(V)tt(V xx

tg)t(V)tt(V yy

“Recipe” for finding the speed of the ball at t+t from the known speed at t.




The average x and y-component speeds of the ball over the time interval t are given by:

2

)t(V)tt(VV yyy

2

)t(V)tt(VV xxx

The new x, y position of the ball is approximated as,

tV)t(x)tt(x x tV)t(y)tt(y y

To illustrate consider the following example:

V0 = 40 m/s, 35 degrees, x0 = 0, y0 = 5 m

Ball Launched with Initial Conditions (time t = 0):




s/m77.32cosVV 000x s/m94.22sinVV 000y

Suppose we divide the flight into t = 0.1 second intervals. The component speeds of the ball at the next instant of time t = 0.1 second are,

= constant

s/m77.32)0(Vsec)1.0(V xx

s/m96.21tg)0(Vsec)1.0(V yy

The average speed over this interval is,

s/m45.222

)0(V)t(VV yyy

s/m77.32Vx




The position of the ball at t=0.1 second is obtained from,

m27.3tV)0(xsec)1.0(x x

m25.7tV)0(ysec)1.0(y y

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 is shown below. Vx (m/s) Vy (m/s) deltat (sec) x (m) y (m) time (sec)

32.76711 22.94158 0.1 0 5 032.76711 21.96158 3.276711 7.245158 0.132.76711 20.98158 6.553423 9.392317 0.232.76711 20.00158 9.830134 11.44147 0.332.76711 19.02158 13.10685 13.39263 0.432.76711 18.04158 16.38356 15.24579 0.532.76711 17.06158 19.66027 17.00095 0.632.76711 16.08158 22.93698 18.65811 0.732.76711 15.10158 26.21369 20.21727 0.832.76711 14.12158 29.4904 21.67842 0.932.76711 13.14158 32.76711 23.04158 132.76711 12.16158 36.04383 24.30674 1.132.76711 11.18158 39.32054 25.4739 1.2

time

Initialcondition




The trajectory (and range) of the ball is shown below:

0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140 160 180X (meters)

Range

How do we know if this is correct?




!! 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:

dt

dV

t

)t(V)tt(Vlima xxx

0tx

dt

dV

t

)t(V)tt(Vlima yyy

0ty

Acceleration is the time derivative of velocity




Newton’s Second Law becomes:

0dt

dVx and gdt

dVy

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.




We will integrate with respect to time over the interval from t = 0 to some arbitrary time t during the flight.

t

0

V

Vy dtgdV

y

0y

and0dVx

0x

V

Vx

So the velocity as a function of time is,

0xx VV tgVV 0yy

Now recall:dt

dxVx and

dt

dyVy




The differential equations for the trajectory of the ball are,

0xVdt

dx and tgV

dt

dy0y

As before, we integrate with respect to time to obtainthe trajectory of the ball x(t) and y(t). Rememberingthat at t = 0, x=0 and y = y0 we get,

tVx 0x 20y0 tg

2

1tVyy

This is the analytical solution for the trajectory of the ball.How does it compare to our numerical solution ?




0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140 160 180X (meters)

analytical

numerical

Excellent agreement!




With the model validated we can put it to work.Example: For a fixed initial velocity what launch angle givesmaximum range?

For projectile motionin a vacuum, maximumrange occurs at a launch angle of 45 degrees.

B

B

B

B

B

B

BB

B B BB

B

B

B

B

B

B

B0

20

40

60

80

100

120

140

160

180

0 10 20 30 40 50 60 70 80 90Launch Angle (degrees)

Vacuum (No Drag) , y0 = 5mLaunch Velocity Fixed at 40 m/s

Maximum Rangefor 45 degreelaunch angle




Example: What combinations of V0 and yield a desired range of 100 meters? Solution... construct a ballistics chart:

Numerous possibilities indicated…impossible for V0 = 30 m / s.

Desired rangeof 100 meters

0

50

100

150

200

250

0 10 20 30 40 50 60 70 80Launch Angle (degrees)

45 m / s

40 m / s

35 m / s

30 m/s




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




Photo by Thomas C. Corke

Visualization of flow over a cylinder:

Where does the energy to put the air in motion come from?




Conclude that the passage of the ball through airwill give rise to a loss in the kinetic energy of theball 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:

m

cosDax

m

)sin(Dga y

These equations don’t help us unless we know howto determine D !!




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

D R AG




observer seesball m oving at

speed V throughair a t rest.

ball

V

Select a reference frame:

observer sees theball a t rest in a

flow of a ir a tspeed V .

ball

V

Two ways of looking at the sameproblem. The drag would be identicalin both cases since it is the relativemotion of the air and ball that determine drag. This equivalence is one basis of windtunnel testing.




How is drag created?

A"C ylinder of A ir"V

density =

tVL

ball

The momentum of the “cylinder of air” is the product of its mass times its velocity.

Mass of air = density X volume = tVALA

Momentum of air = Mass X velocity = tVA 2




A"Cylinder of Air"

V

density =

tVL

ball

As the cylinder of air and ball collide the momentum of the air is imparted to the ball.

Another way of writing Newton’s second law is in terms of the time rate of change of momentum.

am

dt

VmdF




22

ball VAt

tVA

t

VmD

If we assume all the momentum of the cylinder of air is imparted to the ball during the time interval t then the force on the ball D is given by,

Momentum of air

Transferred during time t




Is this result correct? Not necessarily.... Our calculation assumes that all of the momentum of the cylinder of air is transferred to the ball. Thismay not be the case!

What we can say with some certainty is,

2VAD

Remember: = air density [ kg/m3]

A= area of the body “seen” bythe oncoming flow. [m2]

V= fluid velocity = flight speed [m/s]




To make this an equality the convention is to incorporate a factor CD/2 (typically determinedfrom experiment) such that,

2D VA2

CD 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:

5.0CD




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.

We will model the drag on the ball by the relation,

2VA25.0D

As before, we will divide the flight of the ball into many small sub-flights, each of duration t. We start with the known initial conditions from launch at time t = 0:

0x oyy 0o0x cosVV 0o0y sinVV




The acceleration is the time rate of change of velocity. Over any of the short time intervals t we have:

m

cos)t(D

t

)t(V)tx(Va xxx

m

sin)t(Dg

t

)t(V)tt(Va yyy

These provide a “recipe” for advancing the speed of the ballin time…If I know Vx(t) and Vy(t) I can use these equationsto find Vx and Vy at time t+t later.




The recipe...

t

m

cos)t(D)t(V)tx(V xx

t

m

sin)t(Dg)t(V)tt(V yy

with drag given by our model,

)t(V)t(VA25.0)t(D 2y

2x

Vx

Vy V

What about ? !!




The angle depends on the the relative size of the x- and y-components of the velocity of the ball (as shown below) and is therefore a function of time.

0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140 160X (meters)

xV

yVV

)t(V

)t(Vtan)t(

x

y1

‘looks like simple trigto me…”




Now I see the procedure!1. Using Vx(ti) and Vy(ti) compute the drag force and angle theta.

2. Use “the recipe” provided courtesy of Sir Isaac Newton to find the new velocity Vx(tit and Vy(tit a time t later.

3. Calculate the average x- and y-component velocity for the time interval t. 4. Use the average velocity and knowledge of the position of the ball x (ti), y (ti) to find the of the ball x(tit), y (tit).

Has the ball hit the ground?NO YES

Repeat 1-4




To illustrate, reconsider the example we looked at before:

Ball Launched with Initial Conditions (time t = 0):

V0 = 40 m/s, 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

s/m77.32cosVV 000x

As before we will divide the flight into t = 0.1 second intervals.

Initial x- and y-component velocities are:

s/m94.22sinVV 000y




N00.1

s/m1600m10043.2m/kg225.125.0

VA25.0D22233

2

Compute the drag force:

s/m50.32

sec1.0kg3.0

35cosN00.1s/m77.32

tm

cosD)0(Vsec)1.0(V xx

Compute the new x- and y-component velocities:

s/m77.21

sec1.0kg3.0

35sinN00.1s/m8.9s/m94.22

tm

sinDg)0(Vsec)1.0(V

2

yy




Compute the average speed for the interval,

s/m36.222

)0(Vsec)1.0(VV yyy

s/m64.322

)0(Vsec)1.0(VV xxx

The position of the ball at t=0.1 second is obtained from,

m26.3tV)0(xsec)1.0(x x

m24.7tV)0(ysec)1.0(y y

The procedure is now repeated using Microsoft Excel.




time

D x(ti) y(ti)ti Vx Vy

You will develop your own spreadsheet in the learning center.

How significant is the influence of drag?

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 x yinitial conditions: 0 32.76608177 22.94305745 1.000982105 35 0 5

0.1 32.49276293 21.77167754 0.957054997 33.8239428 3.262942 7.23573670.2 32.22773783 20.61409827 0.915628445 32.6045423 6.498967 9.35502550.3 31.97062641 19.46963994 0.876602466 31.3407983 9.708885 11.3592120.4 31.72106099 18.33765828 0.839883683 30.0318182 12.89347 13.2495770.5 31.47868523 17.21754305 0.805384829 28.6768412 16.05346 15.0273370.6 31.24315307 16.10871666 0.7730243 27.2752652 19.18955 16.693650.7 31.01412783 15.01063311 0.742725739 25.8266756 22.30241 18.2496180.8 30.79128138 13.9227769 0.714417648 24.3308753 25.39268 19.6962880.9 30.57429336 12.84466225 0.688033035 22.7879157 28.46096 21.03466

1 30.36285052 11.77583233 0.663509071 21.198128 31.50782 22.2656851.1 30.15664614 10.71585867 0.640786786 19.5621541 34.53379 23.390271.2 29.95537953 9.664340625 0.619810774 17.8809747 37.5394 24.409281.3 29.75875564 8.62090493 0.600528916 16.1559355 40.5251 25.3235421.4 29.5664847 7.585205375 0.582892124 14.3887684 43.49136 26.1338471.5 29.37828207 6.556922475 0.566854095 12.5816068 46.4386 26.8409541.6 29.19386804 5.53576321 0.552371083 10.7369934 49.36721 27.4455881.7 29.01296785 4.521460781 0.539401674 8.85788029 52.27755 27.9484491.8 28.83531172 3.513774359 0.527906584 6.94761813 55.16997 28.3502111.9 28.66063497 2.512488837 0.517848456 5.00993572 58.04476 28.651524

2 28.48867828 1.517414528 0.509191675 3.04890836 60.90223 28.853019

Initial conditions




A very significant effect on range!

The trajectory is no longerparabolic!

B

B

B

B

BB

BB

BBBBBBBBBBBBBBBBBBBBBB

BBBBBBBBBBBBB

B

B

B

B

J

J

J

J

JJ

JJ

JJ

JJ

JJ

JJ

J J J J J J J J J J J J J J J JJ

JJ

JJ

JJ

JJ

JJ

J

J

J

J

J

J

H

H

HHHHHHHHHHHHHHH

HHHHHHHHHHHHHHHHHHHHHHHHH0

5

10

15

20

25

30

35

0 20 40 60 80 100 120 140 160 180

X (meters)

B with drag

J vacuum

H diameter * 2

range = 130 meters

professor flint o. thomas department of aerospace & mechanical engineering

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