# Speed of a Subway: Batman Begins

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## Abstract

The purpose of our experiment is to determine the speed of a subway car in the movie Batman Begins.

## Introduction

The movie that the calculations were based on was Batman Begins directed by Christopher Nolan and starring Christian Bale as Batman, Michael Caine as Alfred Pennyworth, and Gary Oldman as James Gordon). Having witnessed the death of his millionaire parents, young Wayne travels to the East after fate diminished his only chance of revenge. There, he seeks counsel of the leader of an honorable ninja cult, Ra's Al-Ghul. Years later, he returns to Gotham to find it full of crime. With his company at stake, he turns to the cave under his mansion and becomes Batman, a shadow that strikes fear into the hearts of criminals. With the help of James Gordon, Batman sets out to take down various individuals, such as Falcone, the "scarecrow" and Ra's Al-Ghul, who were notorious for creating the biggest crime streaks and dangers of Gotham.

For this experiment, we analyzed chapter 36. This chapter involved Batman fighting the villain Ra's Al-Ghul on a speeding elevated train without a conductor, just as part of the track was blown apart. Just as the train was approaching the gap in the track, Batman subdued his adversary and let him die on the falling train. For this experiment, we are going to use projectile motion to perform the calculations mentioned in the introduction.

Projectile motion is the motion of a body in two dimensions. As the body is launched (or thrown), it travels with both a horizontal velocity and a vertical velocity. Due to our neglecting of drag, it will not have any horizontal acceleration. Thus, the horizontal velocity stays constant throughout the entire free fall. The body will, however, have a downward vertical acceleration (due to gravity) of 9.8 m/s2.

The equations commonly used in projectile motion include:

vx = x/t

where vx is the horizontal velocity, x is the horizontal displacement, and t is the time it takes for the body to hit the ground.

vy = v0y + gt

where vy is the vertical velocity after some time t, v0y is the initial vertical velocity(which is our case is zero), g is the constant for the acceleration due to gravity at the surface of the earth, and t is the time it takes for the body to hit the ground.

s = s0 + v0yt + ½gt2

where s is the vertical displacement after some time t, so is the initial vertical displacement (which is most cases, as well as in our case, is zero), v0y is the initial vertical velocity (which in our case is zero), g is the constant for the acceleration due to gravity at the surface of the earth, and t is the time it takes for the body to hit the ground.

In the fighting scene from chapter 36, as the train travels off the tracks (due to the gap), it resembles a projectile in its fall (from the moment the first car leaves the tracks to the moment that it hits the floor of a garage).

## Procedure

To determine the quantities mentioned in the introduction, we will use a laptop (with a good media player), the actual chapter 36 (from the movie), Wikipedia, the Internet Movie Database, and a calculator.

• Mr. Elert cut chapter 36 out of the movie and installed a frame count on it.
• We played the chapter, and started our frame count at the scene where the first car is just about to go off the tracks. We counted the frames, and concluded that there were 85 of them until the first car collided with the floor of the garage.
• Using the frame rate (given to us by quick player), we calculated how many seconds it took for the first car to hit the ground.
• We then assumed that the train in Gotham is the same size as a subway in NYC and that the approximate distance that the train fell was (before the first car hit garage floor) was equal to the length of two Gotham train cars. The New York City subway system has many classes of cars. For our purpose, we used the R160B class.

## Analysis

To find the horizontal speed of the cars as they fly off the track, we had to first determine the duration of the fall. To do so, we used the frame rate (mentioned in the procedure) of 23.72 frames per second. Thus, since we had 85 frames, we found that:

t = # of frames/23.72 frames/s

t = 85 frames/23.72 frames/s = 3.583s

Based upon the movie, the train cars took 3.583 seconds to fall. Since it is clear by watching the movie, the producers slowed it down by a factor of two, the actual fall time of the train is half of 3.583 seconds or 1.7915 seconds. Knowing this, we are able to calculate the height of the bridge, since:

s = s0 + v0y t + ½gt2

Where

g - Acceleration due to gravity (9.8 m/s2)
v0y - Initial vertical velocity of the first car.

But since the vertical component of the first car as it drives off the track is zero and the initial vertical displacement equals zero, the height (the final vertical displacement) only depends on the time and the acceleration due to gravity. Thus:

s = ½gt2

s = ½(9.8 m/s2)(1.7915 s)2

s = 15.72 m = 51.57 ft.

As already mentioned, Gotham cars are roughly the same size as a standard NYC subway car such as the R160B class. In our scene, the trains fell in such a way that they formed a right triangle with the ground being one leg, the height of the bridge being the other leg, and the displacement of the train cars being the hypotenuse. The displacement was equivalent to about the length of two Gotham cars, and the hypotenuse is therefore 36.68 m. We can use this quantity to find the horizontal displacement of the train cars by using the Pythagorean Theorem:

a2 + b2 = c2

1. height of the bridge
2. horizontal displacement of the cars
3. the displacement of the first car from the beginning of its fall to the end of its fall

(15.72m)2 + b2 = (36.68m)2

b = 33.14 m

Now that we have the horizontal displacement of the train, we can find its horizontal velocity:

vx = x/t

1. Horizontal displacement

vx = (33.14 m)/(1.7915 s) = 18.50 m/s = 41.39 mph

As our last calculation, we decided to find vertical component of the velocity just before the first cart hit the ground.

vy = v0y + gt

Where vy is the vertical component of the velocity of the car at some point.

vy = 0 + (9.8 m/s2)(1.7915 s)

vy = 17.56 m/s = 39.29 mph

## Conclusion

We found the height of the elevated track to be 15.72 m (51.57 ft), the initial horizontal velocity of the first car just as it left the track to be 18.50 m/s (41.39 mph), and the vertical component of the velocity with which the first car collided with the ground to be 17.56 m/s (39.29 mph).

## Sources of Error

1. We neglected any drag that might have acted on the falling train cars. Air resistance would decrease the acceleration of the train and would affect all our calculations. For example, the horizontal velocity would not stay constant.
2. The train collided with the top of a garage, which ordinarily would reduce the speed with which it falls to the bottom. The collision also impacted the time, since the train would fall much faster if its whole trip down was in free fall.

Gennadiy Rozentsvayg and Dasha Mulyukova -- 2007