Don’t Underestimate the Forces, Mandalorian

I’m really excited about this new Star Wars spin-off show, The Mandalorian. It’s not part of the Skywalker saga, so pretty much anything can happen. And there’s the whole mystery of the armored Mandalorian: Who is he? What’s his motivation? How does he eat with a helmet on?

But I just watched the trailer for the show, and right now I’m interested in this quick view of a fight with some other character in a cave. In the scene (near the beginning, at about 0:20), the Mandalorian uses some type of glowy energy thing on the end of his rifle to hit this other guy. The result shows the bad guy (he has to be bad, right?) flying backwards across the cave.

It not only looks cool, it also offers the opportunity to do some physics. You know that’s what I like to do anyway. So let’s get to it. Yes, there will be some video analysis included.

The Nature of Forces

In my initial viewing of the fight scene, I thought that the Mandalorian actually hit the bad guy. OK, that’s probably not true. Apparently, this is an Amban phase-pulse blaster. He didn’t hit him, but just shot at very close range. Oh, that’s fine. The physics still works the same.

There are two big physics ideas we need. One is the nature of forces. A force (not the Force) is a way to describe interactions between two objects, and forces always come in pairs. So when the Mandalorian hits or shoots this guy, he exerts a force on him. But then there has to be another force back on the Mandalorian.

That’s just how it works. If you push on a wall, the wall pushes back on you. If you drop a ball, a gravitational force from the Earth pulls down on the ball, but there is also a gravitational force from the ball pulling up on the Earth.

In general, if A and B are two interacting objects, and A is pushing on B with a force FA-B, then there will also be a force from B onto A (FB-A), which is of equal magnitude but opposite in direction. As an equation it looks like this:

Illustration: Rhett Allain

(The arrows show that these are vectors. A vector is a quantity for which direction matters along with magnitude. That’s all you really need to know about vectors here.)

Bottom line: The Mandalorian pushes on the dude, the dude pushes back. That’s how all forces work.

Gaining Momentum

The other big idea we need is the momentum principle. Momentum is the product of an object’s velocity and its mass, and it’s usually represented (for no evident reason) by the letter p:

Illustration: Rhett Allain

So the bigger something is or the faster it’s going, the more momentum it has. Now, the momentum principle says that when a net force acts on an object, it changes the object’s momentum. We can write that as an equation:

Illustration: Rhett Allain

Oh, the Greek letter Δ (delta) is often used to indicate “a change in.” So you can read this as saying the net force equals the change in momentum per unit of time (e.g,. per second). It’s not just the change in momentum that matters, but how fast it changes. Bonus: Using Greek letters makes you look cool.

The Forces Are All Around You

Now, let’s return to the fight and put these two ideas together. Here’s a force diagram for the Mandalorian’s short range shot.

Illustration: Rhett Allain

Yes, there are a bunch of forces there—but it’s not too bad. Let’s break it down.

First the attacker (I’ll call him Bob from now on): There are two forces acting on him, indicated by the two arrows. One is the gravitational force, mg. The other is the force of the Mandalorian’s blow, FM-B. This changes Bob’s momentum such that he is moving back from the impact.

For the Mandalorian, there are four forces shown in the diagram:

  • The force that Bob exerts on the Mandalorian, FB-M, which is the other side of the force-pair from the blow.

  • The downward-pushing gravitational force, mg. This depends on his mass, m, and the local gravitational field of the planet, g.

  • The upward-pushing normal force from the ground, FN. This is the paired force that goes with gravity—it keeps us from falling into the center of the planet we’re standing on. Which matters because of …

  • The sideways-pushing frictional force from the ground, Ff. This depends on the materials that are in contact (his shoes and the dirt) and the magnitude of FN. Yes, the harder the ground pushes up, the greater the frictional force.

Friction can get complicated, but we can make a simple model that works well in most cases. This says the maximum magnitude of the frictional force, Ff-max, can be found with the following equation:

Illustration: Rhett Allain

Here, 𝛍 (mu) is the coefficient of friction. This is a value, usually between 0 and 1, determined by the specific materials involved. Low values are slippery, high values are sticky. There are cool tables where you can look up different combinations of materials. Steel sled runners on ice? 0.05. Rubber tires on dry asphalt? Up to 0.9.

Here’s the deal: If the Mandalorian stays in place while delivering this haymaker, the net force acting on him must be zero. It has to be zero, since his momentum is unchanged. In the horizontal direction, this means the frictional force must be equal in magnitude to the opposing force from the hit, so as to offset it.

A Friction Prediction

Is that OK? Well … it’s pretty implausible, which is probably why it doesn’t feel right when you watch it. Remember, the force of the shot is so great it blasts the other guy through the air. But let’s actually get down to numbers.

First we’ll estimate the force that the Mandalorian lays on Bob (FM-B). According to the momentum principle, we can get that by looking at the change in Bob’s momentum and the time interval during which it changes.

To get that data I used a tool called Tracker . It lets you mark the position of an object in each frame of a video to track its position, and you can get the time from the frame rate (here 60 frames per second). That gives us a plot of Bob’s movement after he is hit: distance traveled (vertical axis) vs. time (horizontal axis):

Illustration: Rhett Allain

The slope of the fitted line is Bob’s velocity (coefficient A)—3.89 meters per second. Then, to get his momentum, we just need to estimate his mass. He looks burly, so let’s say 100 kg (220 lbs). Multiplying velocity and mass, we get a momentum of 389 kg ⨉ m/s. And since he started at rest, that’s also the change in momentum.

But what about the time of impact? Looking at the frames when the weapon’s blue blast is in contact with Bob, I get an interval of 0.167 seconds. This is just a rough estimate, since it’s difficult to tell when contact starts and ends.

Dividing the change in momentum by the time interval, we get the impact force:

Illustration: Rhett Allain

That’s a force of 2,329 newtons (equivalent to 523.6 pounds). Then we know the frictional force must also be 2,329 N, so we can use the friction equation above to compute the implicit coefficient of friction. We just need one more parameter: the upward-pushing normal force of the ground, FN.

Since there are only two forces in the vertical direction (the Mandalorian’s weight pulling down and the ground pushing up), these two must also be equal. Let’s say the Mandalorian is also 100 kg, and we’ll assume gravity is the same as on Earth. (Frankly, it looks suspiciously similar in the video). The gravitational field on Earth is 9.8 N/kg. Plugging the numbers in, we get an implied coefficient of friction:

Illustration: Rhett Allain

Oh. That’s not great. To deliver that blast while remaining stationary, it seems the Mandalorian would need a friction coefficient of 2.4 between his shoes and the ground. Problem is, that number is way off the chart. Not to mention the fact that he’s standing on slippery sand and gravel.

I can only think of two possible explanations. Either the Amban phase-pulse blaster pulls energy and momentum from another dimension so that it only appears to violate the fundamental nature of forces, or the Mandalorian has some really grippy shoes. Even Air Jordans wouldn’t come close.

But wait! Is there time for a pickup shot before the episode airs? There may be ways to fix this scene. Just spitballing here:

  • Instead of big Bob, what if the attacker is a pint-size Jawa? If he has a very low mass, we could knock him back with less force, and there would be less recoil on the Mandalorian. Of course, that wouldn’t seem very heroic …

  • What if the Mandalorian delivers the force in an upward direction, instead of straight forward? Then the interaction force would push back and downward on the Mandalorian. That, in turn, would increase the magnitude of the normal force pushing upward, thus increasing the forward-pointing frictional force to keep him from sliding backward. (This, in fact, is how all superpowered people should punch.)

Oh, there’s one other way to eliminate the problem: Declare it a feature, not a bug. After all, this is why we have fantasy and science fiction shows—to bend the rules, escape from reality, and enjoy a great yarn. But that needn’t stop us from having a little fun with physics.

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