### Lecture 10 powerpoint

```Dynamic Events
parachute simulations
free fall in vacuum – exact
v(t) = -gt + v0
y(t) = -½gt2 + v0t + y0
2000 meters, zero velocity
0 = -½gt2 + 2000 (g = 9.81 ms-2)
t2 = 2000*2/9.81
t = 20.2
free fall – Euler’s method
see http://spiff.rit.edu/classes/phys317/lectures/heun/heun.html
ti, ti+1 vi, vi+1 ai
so,
vi+1 = vi + ai*( ti+1 - ti)
yi+1 = yi + vi*( ti+1 - ti)
euler’s method
%free fall in vacuum
dt = 1; tmax = 30;
clear y1 t t1 v y;
t = 0:dt:tmax;
y0 = 2000;
v(1:length(t)) = 0;
y(1:length(t)) = 0;
v(1) = 0;
y(1) = y0;
for i = 2:length(t)
y(i) = y(i-1)+v(i-1)*dt;
v(i) = v(i-1) + -9.81*dt;
end
comparison with exact
% formula follows below
texact = 0:tmax/100:tmax;
yexact = y0 +.5*(-9.81)*texact.^2;
vexact = -9.81*texact;
above = yexact > 0;
yexact = yexact.*above;
subplot(1,2,1)
plot(texact,yexact,'r',t,y,'b'),grid
vexact = vexact.*above;
subplot(1,2,2);
plot(texact,vexact,'r',t,v,'b'),grid
results
air resistance
If you do a free fall through air the acceleration is not constant. There
are terms which change as your velocity changes.
term proportional to v , A1
a measure of viscosity
term proportional to v2 , A2
drag
In air, the viscosity can be ignored ( A1 << A2). The drag means in
mathematical terms is that we do not have an analytical solution to
our problem. What it means in practical terms is that sooner or later
we will hit a maximum velocity, or terminal velocity at the point when
the deceleration due to drag just equals the acceleration of gravity.
FD = ½CρAv2
FD = ½CρAv2
Drag coefficients: C
Can of soup
Flat dinner plate
Parachute
.49
.88
1.11
1.34
“The drag coefficient C is 0.5 for a spherical object and can
reach 2 for irregularly shaped objects …”
(Physics for Scientists and Engineers, Volume 1
By Raymond A. Serway, John W. Jewett)
acceleration with drag
ma = ½CρAv2 - mg
a = ½CρAv2/m - mg
You can drop a mouse down a thousand-yard
mine shaft; and, on arriving at the bottom it gets
a slight shock and walks away, provided that the
ground is fairly soft. A rat is killed, a man is
broken, a horse splashes.
On Being the Right Size
by J. B. S. Haldane
Air density and altitude
Michael Richmond, Rochester Institute of
Technology:
density = (1.21 kg/m^3) * exp(-height/8000 m)
Ignore temperature dependence.
For much more detail, see:
http://spiff.rit.edu/classes/phys317/lectures/mu
ltiple_funcs/temp_profile.html
free fall in air
Cd = .5;
Area = .8;
mass = 85;
v(1:length(t)) = 0;
y(1:length(t)) = 0;
v(1) = 0;
y(1) = y0;
for i = 2:length(t)
acc = -9.81 + .5*Cd*Area*v(i-1)^2*airDensity(y(i-1))/mass;
y(i) = y(i-1)+v(i-1)*dt;
v(i) = v(i-1) + acc*dt;
end
Free fall in air
2000
0
-20
1500
-40
-60
1000
-80
-100
500
-120
-140
0
-160
-180
-500
0
10
20
30
40
-200
0
10
20
30
40
parachutes
Expert skydivers use parachutes that range in
size from 80 square feet to 200 square feet
[ about 7.5 to 18.5 square meters]
(http://www.livestrong.com/article/417149-thesize-of-skydiving-parachutes/)
With parachute
2000
0
1800
-20
1600
-40
1400
-60
1200
-80
1000
-100
800
-120
600
-140
400
-160
200
-180
0
0
50
100
150
200
-200
0
50
100
150
200
Heun’s method
while y(i) > 0
% current acceleration
a = calc_a(y(i),area,v(i),m);
vfuture = v(i) + a*dt;
%future acc based on future v
afuture = calc_a(y(i),area,vfuture,m);
%average acc
a = (a+afuture)/2;
% future velocity based on average acc
v(i+1) = v(i) + a*dt;
% future position based on average velocity
y(i+1) = y(i)+(v(i+1)+v(i))*dt/2;
t(i+1) =t(i) + dt;
i = i+1;
end
HALO
High Altitude, Low Opening: the trooper exits
the aircraft at 10,000 metres, except the chute
is deployed at around 760 metres.
(http://www.sasspecialairservice.com/sasbritish-air-troop.html)
```