26.9 C
New York
Wednesday, May 22, 2024

How Long Would It Take for a 747 to Stop, Like in 'Tenet'?

It's not uncommon for people to email me and ask me questions. If it's a question about a perpetual motion machine, I probably will just ignore it. But there was one email question that I didn't ignore. It went something like this—oh, it was from someone at Warner Bros. So clearly, this wasn't just a normal email.

Hello Rhett. We are working on a movie and we need your help. There is going to be a stunt in which we take a stripped-down 747 and get it up to 20 mph on a flat runway. Then we are going to stop it. So, the question: What is the minimum braking distance for this aircraft?

The email included some details like the estimated mass (200,000 lbs) and the fact that it has 8 of its 16 brakes installed. Oh yeah, I was interested in this challenging question. Game on. Little did I know that this was for a scene in the movie Tenet. It wasn't until I saw the trailer for the movie that I realized that the calculation I had performed was for this particular film.

OK, but how do you estimate the stopping distance for this 747? You can't just do an internet search for "stopping distance of a 747"—although, if you do you might find this page describing the physics of brakes heating up on 747 test stop (yes, I wrote it). But this calculation shouldn't be too difficult, right? Isn't it something that you would cover in an introductory physics class? Well, that's a good place to start.

The key idea here is that of acceleration. Acceleration is defined as the rate of change of velocity. As an equation, it looks like this (in one dimension).

This acceleration is for any change in velocity. It doesn't matter if the object is increasing or decreasing in speed—it's still an acceleration. If you know the acceleration for something, you can find the stopping distance using the following kinematic equation (here is a derivation if you want it).

In this expression, v1 is the starting velocity (20 mph in this calculation) and v2 would be the final velocity—hopefully zero since it will stop. So, with a known acceleration the stopping distance (Δx) would be:

Now I just need to get a value for the acceleration of a stopping Boeing 747. Ah ha! That's not so easy. Sure, large aircraft stop all the time—this is normally called "landing." However, the normal method during a landing won't work here. Usually a large aircraft like the Boeing 747 will use two things to slow down. It not only uses the wheels, which have brakes, but it also has reverse thrusters. The reverse thrusters are essentially the force from the engines directed backwards (thus the "reverse" part). This backward-pushing thrust force, along with the brakes, slows down the aircraft.

For this stunt in Tenet, the 747 will only have brakes since it's not a fully working aircraft. So, what would be the acceleration if an aircraft didn't use the reverse thrusters? Well, we are in luck. Here is this thing called a rejected take off test (RTO). For this maneuver, an aircraft starts off and gets up to take off speed. At that point, the pilot slams on its brakes (no reverse thrusters) and comes to a stop. It's a worst-case scenario test to make sure the plane's brakes can handle extreme cases.

Here is a nice video of a rejected takeoff test.

The 747 goes from its approximate takeoff speed of 200 mph (89.4 meters per second) to 0 mph in 27 seconds. Using the definition of acceleration, this means the brake-only stopping 747 has acceleration magnitude of 3.31 m/s2. So, let's assume that the aircraft starts at 20 mph (8.94 m/s). Using the kinematic equation above, I get a stopping distance of 12.1 meters (39.7 feet). That at least seems plausible. It's fine for a first estimate, but we can do better.

Notice that this estimation assumes that the mass of the plane doesn't matter. It also doesn't take into account the fact that only half the brakes are working. So, how can we get a better estimate? How about the following assumption: Each wheel can exert some maximum braking force. So, if the plane has fewer wheels that brake AND a lower aircraft mass (because it's stripped down without any real engines) it could have a different stopping distance.

Let's go back to the RTO example. In that case, the 747 used 16 braking wheels and had a mass of 443,000 kg (975,000 pounds). There is relationship between force and acceleration, it's called Newton's Second Law. In one dimension it says that the net force is equal to the product of mass and acceleration.

If each wheel provides an equal braking force, then for the RTO 747 example we have the following.

Now we can use this braking force for the stripped down 747 from the movie stunt. In this case, there are only 8 brakes and the mass is lower since it doesn't have engines and stuff—the value would be 90,718 kg (200,000 pounds). From this, the stopping acceleration would be:

Wait. Why does this aircraft with half the number of brakes stop with a greater acceleration? So the force is lower, but the decrease in mass is more significant to give it a greater acceleration. Now we have one more thing. If the stripped-down 747 starts with a speed of 20 mph, how much distance would it take to stop? Using the same kinematic equation above but with the new acceleration, I get a distance of 4.9 meters (16.2 feet).

If you don't like my numbers, here are all my estimates and calculations in a Python program (so you can change them and recalculate if it makes you happy).

OK, so what does this say about the crashing 747 stunt? My first estimate was a stopping distance of 12 meters (about 40 feet). Using a modified 747, and this calculation stops shorter than that. The key here is to set some maximum stopping distance that you are absolutely sure the aircraft won't go past. If you put this value at 100 feet (30 meters), it's pretty difficult to imagine it going past that. You should be good.

In the end, I never heard back from the crew on the exact stopping distance. Maybe one day I will find out how accurate my calculations were.

Related Articles

Latest Articles