Happy Star Wars Day (May the 4th).

Now for some physics. Here is the setup. The Imperial forces are assaulting the rebel base on the icy planet of Hoth using the impressive looking AT-AT walkers. After being shot down, Luke Skywalker proceeds to gain access to the underside of a AT-AT and destroys it with some type of bomb device. I hope that doesn't spoil the movie too much in case you haven't seen it. It wasn't a major spoiler though, so you should be fine. At least I didn't say something about the part where Luke finds out that Darth Vader is his father, right? That would have been a major spoiler.

Falling Luke ————

There are actually two things to look at here. The first is after Luke throws the bomb and drops back down to the snow. Here is a diagram during that fall.

If he starts from rest in his fall, then I can write the following kinematic equation (where -g is the vertical acceleration).

If I know the height of the fall, I can estimate the time of fall. Here is a big assumption: I will assume that g is 9.8 N/kg just like on Earth. Why? If you look at other scenes on Hoth (like inside the rebel base), falling stuff appears to fall the same as on Earth.

NEWS FLASH: Hello blogger. You are dense. Stuff in the rebel base appears to fall like on Earth, true. Do you know why? BECAUSE IT WAS FILMED IN A STUDIO ON EARTH! Is this what it takes to be a blogger? Do you have to avoid the obvious to be a physics professor? You should be fired. Wow.

Who said that? Oh well, let me continue. Wookieepedia lists the AT-AT with a height of 22.5 meters. If I go with this value, then Luke would fall from a height of about 12 meters. Using the equation above, this would be a free fall time of 1.56 seconds.

What does it show in the movie? Using Tracker Video (my favorite video analysis tool), I get from the time Luke let go until hitting the ground was a time of 1.2 seconds. Not bad. Not bad at all. This time is still off - but the movie doesn't show the entire fall, so I will just count this as an editing error.

Actually, there seems to be enough footage to get a plot of Luke at the end of his fall. Using a scale based on the height of Luke at 1.75 meters, I get the following plot of his vertical position.

This isn't enough data to get the acceleration, but I can get an estimate of the final velocity at 7.98 m/s. If he fell for 1.2 seconds, he should have a final speed of 11.76 m/s. Either Luke is already using the force to slow himself down or the gravitational field on Hoth is lower than 9.8 N/kg. However, if g was lower, he would have taken longer to fall. I am going to stick with the idea of him using the force.

But really, this falling Luke thing was just a warmup.

Falling AT-AT ————-

When something tips over instead of just dropping, it will take longer to hit the ground. This is actually a more advanced problem, so I will skip some of the details. Let me start with a model of a mass on the end of a stick and the stick is set in the ground so that it doesn't slip as it tips over. Here is a diagram.

If the base doesn't slip, this falling thing can only increase its angular position. It's what we call constrained motion. Really, the best way to deal with this would be with Lagrangian mechanics, but we can set up it as a torque problem also. The torque on this AT-AT is just due to the gravitational force. I am assuming that most of the mass is up top and the mass of the legs is negligible. This gives a torque of (I am writing torque as a scalar since the axis of rotation is fixed):

The angular momentum principle says that the torque on an object change its angular momentum. For a point object (like the top of the AT-AT), that would look like this:

Sure, this can be simplified some. However, the point is that the angular velocity (ω) changes and the rate of change depends on the angle. Since the angular velocity is the derivative of the angular position, I can write this as:

This is your basic second order differential equation. If you are saying "Hey. That looks a lot like the equation for a pendulum!" - you are right. The only difference is that there is a negative sign in there so that the mass oscillates back and forth. Now to solve this, there are several ways to do this but a numerical solution will be the most practical.

In the numerical solution, I will use python with the following strategy:

Break the motion into small time steps. During each time step, do the following.

Based on the current angle, calculate the sin(&theta) and use this to calculate the second derivative of θ (from the above equation). Let me call the second derivative of θ the angular acceleration (α).

With the angular acceleration, calculate the new angular velocity at the end of this time interval as though the acceleration was constant.

With the angular velocity, calculate the new angular position as though the angular velocity was constant.

Repeat until you get where you want to get to.

There are other numerical recipes, but I like this one because it is the most straightforward. Ok, there is one problem. If I want to find out how long this thing takes to fall over, it is VERY dependent on the starting angle. Look, if the object starts at θ = 0 then the torque will be zero also. It will never fall over.

With that in mind, let me make a plot of the angle as a function of time for an object that starts tilted 5 degrees from vertical.

From this, you can see it takes 4.9 seconds to fall over. What if I change the starting angle? With the power of python, this is pretty easy to do. Here is a plot of the total time it takes the object to tip over as a function of the starting angle.

First, you can see that as the starting angle gets closer to zero, the total time starts to explode. Second, even at a starting angle of something like 30°, the object would still take about 2.5 seconds to tip over.

Analysis of the Actual Falling AT-AT ————————————

Now let me look at the video from Empire Strikes Back. Here is the plot of the angular position of the falling AT-AT.

This shows that it took about 3.5 seconds for the AT-AT to fall over if I start counting time at a 5° tip angle which is a bit quicker than my estimated 4.9 seconds. Of course, the key is that this falling over time is length dependent. Let me go back to my model and plot the tip over time for different length AT-ATs. Remember, that I am making the assumption that all the mass is concentrated at the top portion of the walker.

According to this, how tall would the center of mass have to be in order to just take 3.5 seconds to fall over? It would just be about 9 meters tall. So, here are my options.

The gravitational field on Hoth is not like Earth. I crunched the numbers (re-ran the calculation) and you would need g to be about twice the value of Earth's in order to get a tip over time of 3.5 seconds (starting from 5 degrees). However, this would not agree with the falling Luke.

The center of mass of the AT-AT is not where you think it is. This could be the case if the legs were super massive. Why would they be so massive? Who knows? (well, maybe George Lucas would know)

The AT-AT is not 22.5 meters tall but instead like half that height. Of course, this wouldn't agree with Luke's fall time.

The AT-AT didn't actually tip over. Instead, it was an inside sabotage job by some disgruntled Storm Troopers. Wait, this wouldn't explain the fall time.

So, you see there are some problems with this scene. I guess the only reasonable thing to do is to make a new version of The Empire Strikes Back. In this new version, the AT-AT would take another second to fall over. Sure, this might be a lot of work to redo the whole movie for just one scene - but think of all the new Star Wars Blu-ray sales.

I'm just kidding about the Blu-ray sales. I don't even have a Blu-ray player anyway.

Update: Comparing Data and Models ———————————

Why didn't I include this when I first wrote this? I have no idea. Here is further evidence to support my claim that the AT-AT is much shorter than they claim. This plot shows the angle vs. time data from the actual movie along with the times for three different length numerical models.

Here you can see that the 12 meter tall model fits quite well. The other lengths don't work quite so nicely - especially the 18 meter model.