How Fast Would Wonder Woman’s Lasso Need to Spin to Block Bullets?

Look. I know Wonder Woman is not real. I’m cool with that. It doesn’t matter that she does stuff that breaks the normal rules of physics. In fact, that’s what makes super hero movies like Wonder Woman 1984 so enjoyable.

However, this doesn’t mean that I can’t do both — appreciate the movie for what it is AND analyze the physics. You can’t stop me.

So, here we go. We have this scene in the trailer that shows Wonder Woman spinning her lasso to block a bunch of bullets. Here is the full trailer, but this is the part I’m interested in.

So, how fast would she have to rotate the lasso to block bullets? Let’s do this.

Direct Blocking Bullets

OK, here is my first approximation. Suppose the lasso is a cylinder with a length L and a diameter d. It is rotating around with some angular velocity, ω. Here is a rough diagram.

I will also assume that Wonder Woman rotates the lasso at a constant angular velocity (for now). That means that she’s not “aiming” to block the bullets, but rather spinning so fast that she automatically blocks the bullets.

This means that a bullet could be randomly (seemingly) at any point in this lasso circle. That means it will either hit the lasso, or miss it. The probability of hitting the lasso would be equal to the ratio of the cross sectional area of the lasso rope compared to the swept out circular area. It doesn’t matter how fast the lasso rotates.

I’m not even going to put in estimated values for the size of the lasso. It’s clearly not going to be very effective at bullet blocking. Also, it seems like spinning the lasso really fast should make a difference.

Deflecting Bullets

That leads to the second (and likely more effective) method to stop bullets from passing through her glowing circle of protection — at least, that’s what I’m calling the spinning lasso shield. What if the lasso hits the side of the bullet as it passes through the sweeping area? It’s possible this magical lasso could hit the bullet to change its trajectory and then miss its intended target.

So, let’s say the bullet is a cylinder (just a rough approximation) and traveling with some velocity, v. Like this.

I know bullets don’t really look like that. It’s OK though. But this bullet will pass through the diameter of the sweeping lasso area (this is a side view). How long will it take? Well, the bullet has to travel a total distance of (s + d) in order to complete clear the path of the lasso. Oh, in case it wasn’t clear the length of the bullet is s.

If the bullet is traveling with a constant velocity (v), then the time to pass through the whirling lasso shield would be:

This expression is just based on the definition of constant velocity (but I solved for the change in time). Now consider that the bullet enters the sweeping area RIGHT after the lasso passed by. In order for the rope to hit the bullet, it has to make almost a full revolution in the time it takes the bullet to pass through (and it would just hit the tail end of the bullet). From that, I can calculate the angular velocity (ω) of the lasso. Remember that the angular velocity is defined as the change in angular position divided by the change in time. In this case, the change in angle will be 2π radians (well, almost).

Now I just need some estimations to put in for these values. Honestly, I’m mostly just making up stuff here. This is what I have.

• Bullet velocity = 400 m/s (9 mm) to 960 m/s for a M855A1 round from an M16.
• Length of bullet (using the same bullets) = 1.3 cm to 2.1 cm (I mostly just estimated the length of the 9 mm.
• Lasso diameter = 1 cm (guess).
• Lasso length = 1.5 meters — of course this can change based on how much Wonder Woman choose to spin in her lasso shield.

Since I don’t have single value estimations, I’m going to put this calculation into python — that way you can change the values in whatever way you like. Here’s the code.

Umm… Damn. One million revolutions per minute. And that’s just for a lower estimation bullet. How about this? Let me calculate the speed of the end of the lasso as it travels around the edge of this sweeping area.

With the same values as above, I get a speed of 1.8 x 10⁵ m/s. Just for comparison, the speed of sound is 343 m/s meaning the tip is traveling at Mach 528. I assume this would make a heck of a sonic boom.

This lasso would spin at a frequency of 19 kHz. I assume it would make a “whooshing” sound at that same frequency. Technically, some humans can hear sounds up to 20 kHz — but that’s not me. I wouldn’t hear it.

Homework

There are too many unanswered questions. I’m assigning these for your homework. Note — some of these might not be so trivial. That’s what makes them fun.

• Suppose the lasso has the same density as a nylon rope (I’m just guessing here). What is the tension in the lasso as it spins. It’s not a constant tension since the outer parts of the rope have a different acceleration than the inner parts. You might have to do this one with a numerical calculation — I’m not sure.
• What if Wonder Woman wants to change the speed of the spinning lasso in order to perfectly hit a bullet? Let’s say that the lasso is spinning with a lower angular velocity (half of my calculated value). She see’s a bullet fired from a gun that is 10 meters away and it’s going to impact the shield at a location opposite the rope so that it has to travel π radians to block the bullet. What is the angular acceleration for this to happen?
• What is the rotational kinetic energy of this spinning lasso (using my original rotation rate). You can assume the rope has a density the same as nylon — or change this to some value that makes you happy.
• Estimate the power required for Wonder Woman to get this lasso spinning up to shield speed in 5 seconds.
• Imagine that the lasso hits a bullet on the side during it’s rotation. Estimate the angular deflection of the bullet.