Category Archives: Physics

Views of Rolling Clouds

I bring fresh showers for the thirsting flowers,
From the seas and the streams;
I bear light shade for the leaves when laid
In their noonday dreams.
From my wings are shaken the dews that waken
The sweet buds every one,
When rocked to rest on their mother’s breast,
As she dances about the sun.
I wield the flail of the lashing hail,
And whiten the green plains under,
And then again I dissolve it in rain,
And laugh as I pass in thunder.

The Cloud, Percy Bysshe Shelley

As a child I spent what seemed like hours at a time watching clouds move across the sky, shifting shapes as they went. Seeing dragons, devils, ships, and castles moving and morphing across a blue canvas. I can’t be the only one. Rain clouds rolled in this morning and I found myself watching as a few low-lying, dark gray ones trundled along beneath the overcast sky.

Lower clouds appear to be moving faster than higher ones, but this is an illusion. In reality, wind speed increases with altitude. But when a low cloud bears down in its dark and shadow and immensity, it’s nearly impossible not to tremble at one’s own insignificance.

So, what creates the illusion of faster movement? The answer lies in the changing angle of an observer’s eye as it tracks a cloud. The observation angle changes faster when a cloud moves faster or when it’s closer to the observer. Closer can mean altitude–the cloud is lower in the sky, or it can mean distance over the ground–the cloud is closer to being directly above the observer.

So how much difference does it make?

Start with the sky, and a cloud, and it’s a sunny day, and there’s a guy standing on the ground looking at the cloud. The cloud’s altitude is a, and the distance over ground is d. Take a line straight into the sky and another that goes from the guy’s eyes to the cloud. Those two lines make an angle, θ. A breeze blows on the cloud, pushing it horizontally with velocity H, and vertically with velocity V.

So now for the nerdy stuff. When the cloud’s to the right of the dude, d and θ are positive, and to the left they’re negative. a is always positive. H is positive when going right and negative when going left. V is positive when the cloud moves up and negative when it moves down. The tangent of θ is d divided by a, and can be calculated if their lengths are known.

The total change of the angle θ with time is found by adding the change in angle due to horizontal movement to the change due to vertical movement:

Gad, that’s ugly to work with. It basically says when the cloud flies left, the angle changes in the negative direction. When the cloud is to the right of the observer, the angle changes in the positive direction when the cloud moves down, and in the negative direction when it moves up. And when the cloud is to the left, vertical movement causes changes in the opposite direction. How does it look when calculated?

I started with altitude and horizontal distances of 200 feet, and since they’re equal, the angle is 45 degrees. The cloud flies by at 10 feet per second, and the observers eyes track it across the sky. Here’s what the angle, θ looks like over time. It starts out at postive 45 degrees, reaches zero when the cloud is directly overhead, and goes negative as it flies to the right of our guy on the ground.

So what happens if the cloud is now 20 feet off the ground instead of 200, and still whizzing by at 10 feet per second? Well, at first it’s just a cloud on the horizon, getting bigger and bigger, and our guy’s head doesn’t even have to move. It takes 15 seconds for the angle of observation to go from 85 to 70 degrees. Then the cloud flies over in a tear, going to an angle of -70 degrees in only 13 seconds, before shrinking into the horizon.

This reminds me of something:
“How did you go bankrupt,” Bill asked.
“Two ways,” Mike said. “Gradually and then suddenly.”
Ernest Hemingway, The Sun Also Rises

A shot of the excel sheet and the formulas are below. Happy cloud watching.

B3 = A2 + E2 and copy down
C3 = C2 + D2 and copy down
F2 = DEGREES(ATAN(C2/B2)) and copy down
G2 = F2 – F3

Baumgartner’s Jump

This was originally posted at The Cameron Hoppe Project.  It’s popularity inspired this site.  It’s been updated since.

          “And He will raise you up on eagle’s wings,

           bear you on the breath of dawn,

           make you to shine like the sun,

           and hold you in the palm of His hand.”

                        — Josh Grobman “On Eagle’s Wings”

At 128+ thousand feet, Baumgartner looks down on the blue sky.

Felix Baumgartner completed his much pre-hyped jump from 128,100 feet, achieving a top speed 1.24 times the speed of sound.  It must have been an amazing ride, with the bill footed by Red Bull and everything.  I am neon green with envy.  Sure, it could be death to try, but it would be worth it just to see the blue-glowing world one time.  Besides, the life insurance is paid.

Gravity creates constant downward acceleration.  Friction with air produces drag that runs the opposite direction in which one is moving, and is proportional to velocity squared.  So the sum of all the forces on Felix, as with any skydiver, during his fall were equal to his mass times his actual acceleration.  Since drag is proportional to the square of velocity, the drag force and the gravitational pull become equal, and acceleration reaches zero.  This is known as terminal velocity.  Not nearly as exciting as the name sounds.  I know; I was disappointed, too.  In equation form it looks like this:

This is a pretty standard force balance on an object moving through a liquid or a gas.

The drag coefficient includes half the surface area of the guy in the suit and a proportionality constant.  Based on his reported time in free fall and an area of 4.3354 square meters I found this to be 1.15, which is actually quite reasonable.

The next issue to deal with in the problem is the air density.  Anybody who’s ever been up a tall mountain or a plane ride knows air gets thinner with elevation.  As a result, drag forces are higher near the ground than at the elevation Baumgartner jumped from. Without that included, we’re only left with the beautiful sky, without any interesting math.  So we can expect that Felix’s speed went way up, and then it actually decreased until he pulled his parachute at 270 seconds.  Then it really dropped.  The air density equation is described below:

Air Density Eqn

Air density is a function of initial pressure, temperature, and height off the ground.

There’s a ton more I just didn’t have time to do.  For example the gravitational force also decreases as elevation increases and the temperature gradient is probably not a constant from the ground to the stratosphere, etc, etc.  But time is short.  Well, we’ve got two coupled differential equations.  There’s only one thing to do–code it and find out how it looks!  I punched this one out in Matlab:

Baumgart Posistion

Baumgartner’s position with time into the jump. The green circle marks the point at which his descent slows and the red line is the time where the chute was deployed.

The scale of this is in tens of thousands of feet, so you can see where he pulls the cord at 270 seconds and about 8400 feet.  At that point, his descent slows way down.  Now, the velocity of the fall:

Baumgartner's Velocity

Baumgartner’s Velocity over the time period of the jump. Maximum velocity predicted by the model is too low by about 4.1%.

Two important spots here.  On the right, you can see where his chute deploys.  On the left, you can see where he reached terminal velocity in the upper atmosphere about 50 seconds in.  This was his maximum velocity.  After that, the rising density of the atmosphere continually slowed him down.  Kinda like being married.

You can see the model is a little off; the maximum velocity should be 1223.75 feet per second, while I’ve got him maxing out around 1175.  It’s probably due to my crude modeling of gravitational and temperature gradients at high elevation.  What can I say?  There’s only so many hours in a Sunday afternoon. The model is named for Baumgartner but it can be used generically for any object falling through a gas.  It works out everything in SI units, then converts to feet at the end before plotting.  Changing the plot command to small letters puts it in  SI.  The rest is pretty straightforward.  The function call is:

Baumgart(r,L,CD,p0,u0,a0,rhoR,rho0,T0,Tf)

r = Radius of object         L = Length of object        CD = Drag Coefficient

p0 = initial position       u0 = initial velocity         a0 = initial acceleration

rhoR = density of object        rho0 = density of air at sea level

T0 = Temperature at sea level in Celsius         Tf = Temperature at p0

The call I used was:

Baumgart(0.65,1.9,1.15,39045,0,0,1062,1.48,-25,25)

Cheers!

PS–Are you following @CameronHoppe on Twitter yet?  If you follow me, I’ll follow you.  Plus you should leave a comment.  Just sayin’.

Code:

function Baumgart(r,L,CD,p0,u0,a0,rhoR,rho0,T0,Tf)
tstep = 0.00001;
 Tgrad = (Tf - T0)/30000000;
for n=1:30000001
 t(n) = (n-1)*tstep;
 u(n) = 0;
 p(n) = 0;
 end
n = 0;
for n=1:27000001
 a(n) = 0;
 rho(n) = 0;
 T(n) = 273.15 + T0 + (n-1)*(Tgrad);
 end
a(1) = a0;
 u(1) = u0;
 p(1) = p0;
 rho(1) = rho0*exp(-0.0284877*9.8*p0/(T(1)*8.3144621));
n = 0;
V = pi*L*r^2;
 A = 2*pi*r*(r+L);
 g = -9.8;
 k = 0.5*CD*A;
massR = rhoR*V;
for n=2:27000001
 u(n) = u(n-1) + tstep*a(n-1);
 p(n) = p(n-1) + tstep*u(n);
 rho(n) = rho0*exp(-0.0284877*9.8*p(n)/(T(n)*8.3144621));
 a(n) = (massR^-1)*(V*(rhoR-rho(n))*g + k*rho(n)*(u(n)^2));
 end
uS = u(27000001)*.25;
for n = 27000002:30000001
 u(n) = uS;
 p(n) = p(n-1) + tstep*u(n);
 end
for n = 1:30000001
 P(n) = p(n)*3.28084;
 U(n) = u(n)*3.28084;
 end
 A
 k
 m = plot(t,P);
 xlabel('Time (seconds)')
 ylabel('Position (feet)')
 set(m,'LineWidth',2)
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