It’s finally time to analyze the latest Mythbusters episode again. This one has Grant, Tory, and Jessie testing another myth about that bus from the movie Speed. According to the Mythbusters, in the movie the bus, traveling at 70 miles per hour, was able to jump over a 50 foot gap in the highway, land safely, and continue on its way.

There’s an obvious physics question in here: could the bus even make the jump? Well, while it’s in the air, the bus is basically just a projectile, and projectile motion is one of the most basic topics in physics. This shouldn’t be hard to calculate. The equation for uniformly accelerated motion in one dimension is

In a two-dimensional system, like a flying bus (up and forward: two dimensions, assuming it doesn’t move sideways), we use two of these equations, one for each dimensions. And if we ignore icky things like air resistance, it’s easy to determine each of the individual factors in the equation:

• Let’s choose coordinates such that \(x_0 = 0\) and \(y_0 = 0\), setting the edge of the road where the bus jumps off as the origin.
• We know the speed of the bus is \(v_0 = \unit{70}{\mileperhour}\). The Mythbusters said that in the movie, the highway was angled upwards by \(\theta = \unit{1.2}{\degree}\) (don’t ask me how they know this) We’ll need the horizontal and vertical components, \(v_0 \cos\theta\) and \(v_0 \sin\theta\).
• The only force acting on the bus while it’s in the air is gravity, and that’s vertical, so \(a_x = 0\) and \(a_y = -g = -\unit{9.8}{\frac{\meter}{\second^2}}\).

Putting it all together,

These equations tell us everything we want to know about the bus’s motion through the air. For instance, what does its trajectory look like? For this, we can solve the first equation for time and plug it in to the second, getting

This equation tells you how high the bus is, relative to its starting point, after it’s traveled a given horizontal distance \(x\). Fortunately, this is exactly what we need to figure out whether the bus makes it across the gap: we just figure out how high the bus is after it’s traveled \(\unit{50}{\foot}\) horizontally, and compare that to the actual elevation of the landing ramp, which the Mythbusters reported as \(\unit{2}{\foot}\) below the takeoff ramp. Plugging in the values from above gives an answer of

So it looks like the bus is not going to make it, but that’s pretty darn close. One has to wonder how much faster they would have had to be going to get across the gap successfully, according to this equation. We can do that too: just solve the equation for \(v_0\), and you get

This time we plug in all the same values (except for \(v_0\)), along with \(y = -\unit{2}{\foot}\), and get

Again, that’s pretty close. Maybe a little more gas just might have done the trick… you think.

But if you’ve seen this episode of Mythbusters, you’ve probably noticed that something seems fishy. When the Mythbusters tested their scale model of a bus on the highway, it didn’t just barely miss the road — it failed spectacularly, crashing down about 2 feet below its target. What’s up with that?

First, let’s make sure it wasn’t because of the reduced scale of the model. Thankfully, the equation above applies equally well to the model as it does to the real world. In the model, all the distances were scaled down by a factor of 12, which is simple enough: just plug in \(\unit{\frac{50}{12}}{\foot}\) for \(x\) and \(-\unit{\frac{2}{12}}{\foot}\) for \(y\). But as Grant pointed out on the show, the acceleration of gravity doesn’t change, so they needed to make an adjustment for that. The only thing left to change is velocity, so how can we adjust velocity to compensate for the effect of full-scale gravity?

Take a look at what happens to the trajectory equation

when you scale down the distances \(y\) and \(x\). The left side becomes smaller by a factor of 12, and so does the first term on the right. So if we could make the second term smaller by 12 as well, we’d have a perfect scale model of the equation, so to speak. Well, we can do that by making sure the ratio \(\bigl(\frac{x}{v_0}\bigr)^2\) is a factor of 12 smaller than in the real world, like so:

Combine that with \(x_\text{scale} = \frac{1}{12}x_\text{real}\), and you get

With the full-size velocity \(\unit{70}{\mileperhour}\), this gives the proper scale velocity as \(\unit{20.2}{\mileperhour}\), which is just what Grant calculated in the show. I didn’t catch exactly what he wrote on the side of the bus, but I’ll bet it was pretty similar to the equations I’ve got here.

So now we see that yes, the Mythbusters did run their scale-model bus at the right velocity for it to be an accurate \(\frac{1}{12}\)-scale representation of the real situation. It should have missed the landing by less than an inch. Clearly, that didn’t happen. Why not?

If you were watching the show closely, you might have noticed that not only did the bus fall while it was in the gap, but it also rotated downward as it fell. This happens because a bus is not really a simple projectile. When physicists talk about a projectile, they mean something small and often roundish, like a baseball — basically an object that acts more or less like a point. This is definitely not true of a bus, which is much longer than it is wide or tall, and has all sorts of complicated substructures like wheels, seats, and an engine. So things are going to happen to a bus that the equations above don’t account for.

The most obvious (relatively speaking) of these corrections comes from the fact that the bus is a long object resting on two sets of wheels. It begins to fall as soon as the front wheels leave the road. But even once that happens, the bus is not yet falling freely (as a projectile would), because the rear wheels are still on the highway, and the road is still pushing up on them, applying a torque which acts to spin the rear of the bus upward as it travels forward.

We can figure out what that torque is by first estimating the moment of inertia of the bus. Since we don’t know just how much of the bus’s weight was supported by the rear wheels, we’ll use the rear axle as our axis so we can ignore the reaction force from the road on those rear wheels. The moment of inertia is thus

I’ll assume for simplicity that the mass is distributed uniformly along the length of the bus, even though this is not at all true in real life. (Hmm, a paper about automatic bus steering?) In this case we can do the integral to get

Now, if the mass is uniformly distributed, that puts the center of mass at the middle of the bus, a distance of \(\frac{l}{2} - r_\text{axle}\) from the rear axle. The torque resulting from the bus’s weight acting at that distance from the axis of rotation is

and the resulting angular acceleration is

In the show, they said the bus was \(\unit{35}{\foot}\) long and weighed \(\unit{25000}{\pound}\). I’ll just estimate that the rear axle is about \(\unit{5}{\foot}\) from the back of the bus; plugging in these numbers gives

This angular acceleration is going to be in effect for as long as the rear wheels of the bus are on the road and the front wheels are not, which is just the amount of time it takes to traverse the wheelbase of the bus — let’s estimate \(\unit{27}{\foot}\) — going at \(\unit{70}{\mileperhour}\):

What we’ve got here is an instance of accelerated motion for a specified time — just like we started out with, but now it’s rotation instead of just linear motion. Still, the equation looks similar:

And now we’ve got all the numbers needed to figure out how much the bus rotates while its rear wheels are on the pavement: \(\theta_0 = \unit{1.2}{\degree}\), \(\omega_0 = 0\), \(\alpha = -\unit{89}{\degree s^{-2}}\), and \(t = \unit{0.26}{\second}\), which gives

But of course the bus continues to rotate at the rate

for the time it takes to cross the gap, or more precisely, from the time that its rear wheels leave the road to the time that its front wheels hit the next section of the road (assuming that they are somehow going to do so):

so it rotates an additional

for a final orientation of about \(\unit{10}{\degree}\) below the horizontal. That puts the front of the bus lower than what we previously predicted by \(\frac{\unit{35}{\foot}}{2}\sin\unit{10}{\degree}\), or \(\unit{3.1}{\foot}\). Still not enough to account for what we saw in the show, but at least it’s a pretty clear miss. (And before you ask: yes, this should transfer appropriately to the scale model as \(\unit{3.1}{\foot}/12\); you can verify that by going through the preceding calculation and scaling the lengths down by a factor of 12 and the velocity down by \(\sqrt{12}\) as we found before.)

What else could be causing the bus to miss the jump the Mythbusters tested it on by the wide margin we saw in the show? There’s always air resistance, although a bus is pretty huge and you wouldn’t think the air would have much of an effect on it. But perhaps that effect was exaggerated in the scale model, since smaller things do tend to be influenced more by drag forces. As I’m writing this there’s an active, or at least nearly active, discussion at PhysicsForums regarding the Mythbusters’ experiment. Perhaps new ideas will show up there. (I must acknowledge the contributors on that thread for their development of some of the material in this post.)