Could Thor Punch The Hulk Without Knocking Himself Over?
Summer is blockbuster movie season. That means entertaining, flashy action and superhero feats. But do real physics ever come into play? This year’s slate of high-octane superhero movies got us thinking about whether our favorite characters’ signature moves are rooted in reality.
Could Wonder Woman act fast enough to stop a bullet with her bracelet? Could Thor punch the Hulk without knocking himself over? Rhett Allain, an associate professor of physics at Southeastern Louisiana University, joins Ira to test the physics of these superhero stunts.
Rhett Allain is the author of Geek Physics and an associate physics professor at Southeastern Louisiana University. He also writes WIRED’s Dot Physics blog.
IRA FLATOW: This is Science Friday. I’m Ira Flatow.
Summer is blockbuster movie season. Entertaining, flashy action, superhero stunts, but real physics? This year’s slate of high-octane superhero movies got us thinking about whether anything our favorite fictional heroes do has a shot at being real. Could Wonder Woman act fast enough to stop a bullet with her bracelets? Could Thor punch the Hulk without knocking himself over?
So this week, we’re bringing the mighty hammer of mathematics to bear on these impossible-looking stunts. They’re geeky, I know, but stay with us on this. Because joining us on our test of phony physics is superhero movie and physics expert Rhett Allain, Associate Professor of Physics at Southeastern Louisiana University and author of WIRED’s Dot Physics blog.
Welcome back to Science Friday, Rhett.
RHETT ALLAIN: Thank you for having me.
IRA FLATOW: Let’s begin our journey through the physics of superheroes with who else but Wonder Woman.
SPEAKER 1: Be careful of mankind, Diana.
IRA FLATOW: Whoa. Bullets flying everywhere in this one because, of course, it takes place during World War II, and that got us thinking about the physics of blocking bullets. Now, wouldn’t all these bullets, wouldn’t they knock Wonder Woman down. How do you deal with that, Rhett?
RHETT ALLAIN: So this is a great example of your classic introductory physics problem. You have a bullet coming in, and it deflects off. We can calculate the momentum of that bullet and looking at how, let’s say, it bounces straight back. And you can, from that, say the momentum has to be conserved, and Wonder Woman would have to be pushed back also.
But if you look at the mass, momentum is mass times velocity. And so the mass of the bullet times its velocity compared to the mass of Wonder Woman, bullets actually don’t have that much momentum. They don’t really knock you back that much even if you’re wearing a bulletproof vest. So she can deflect these bullets and not get knocked back, and that’s not really a big problem with Wonder Woman blocking bullets.
IRA FLATOW: And could a normal person ever bring up their arm fast enough, though, right? These bullets are coming pretty fast. She has to bring her bracelet up to block the bullet.
RHETT ALLAIN: Right. Now that’s a much more superhuman feat. So if you just take some basic guesses about the bullet speed and how far away you can see it– let’s say it’s maybe 30 meters before you can even see the bullet to know where to block it– that gives you just a small reaction time that you have to take your hand and move it all the way up to wherever the bullet is.
So if I calculate that, it’s going to be around 43 milliseconds. So that’s very, very short amount of time. And so during that time, she would have to take her arm. She would have to move it up, increase the speed, and then decrease the speed over a distance of– I’m just going approximate– about one meter.
So when you’re increasing the speed of this hand, that’s increasing the kinetic energy. So that takes energy. And if you do that in a short amount of time, then we can calculate the power. Power is change in energy over change in time. So it’s a pretty fun problem. And doing that, I get a power that she would need to have 76,000 watts. And you can convert that to horsepower like a car, and that’s about 100 horsepower. So just to block, the energy, the power she would need to move her hands up in order to get that bullet in time, we’re talking like a car.
IRA FLATOW: That’s why they call her Wonder Woman.
RHETT ALLAIN: That’s exactly why they call her Wonder Woman. That’s right.
IRA FLATOW: What about all those different bullets? I mean, the clip we played had a bunch of machine gun bullets coming at her. Could she actually hit them all?
RHETT ALLAIN: Well then, you would have to move your hand for each one. So it would just be continuous 100 horsepower that you would have to exert. And just for comparison, a normal average human if you’re riding your bike at a good speed, that’s about 100 watts. A professional athlete can do that for maybe 300 watts. And for short amount of times, you could possibly– very, very short amount of time– do maybe 1,000 watts, but not 76,000.
So she is definitely a superhero. That’s why they call her Wonder Woman, I guess.
IRA FLATOW: I’m not going to argue with that. Let’s move on to our arachnid hero, Spiderman.
PETER PARKER: Finally, here we go.
SPEAKER 2: Good evening, Peter.
PETER PARKER: Wow.
SPEAKER 2: You have 576 possible web shooter combinations.
PETER PARKER: That is awesome.
IRA FLATOW: Oh. Let’s talk about that web because it’s Spiderman’s main mode of transit. As we all know, he swings from building to building, right? It certainly looks cool, but how do we know that that’s actually faster than running on the ground is?
RHETT ALLAIN: So again, it’s another great introductory physics problem because if you have Spiderman on the end of a web, it’s just like a pendulum. It’s just like a mass on the end of the string, and it swings down. So as he swings down, he increases in kinetic energy until he gets to the lowest point, and then he slows down going back up. So that’s your basic pendulum.
But if you look at from where he started to where he ended and you measure the time it takes, you can get his average speed. Now, this I’d like to point out is what I call a Tarzan swing. So Tarzan, Lord of Apes, and he grabs one vine and he swings to the next vine. So he goes vine to vine. And that’s the easiest way to calculate because it’s just a normal pendulum.
Then there’s the Spiderman swing, which he swings and then let’s go and flies through the air and then shoots another web. So it’s a little bit more complicated. But the average speed for a Tarzan-type swing depends on the length of the swing, the length of the spider web, and the angle that he starts with. And so you can calculate this for best case scenario of maybe 30 meter web with an angle of about 45 degrees, and it’s around 8 meters per second or 18 miles per hour.
And so that’s a speed that you could probably run that fast. So you may not be gaining too much. If you do a much larger swing with a Spiderman jump at the end, you can go quite a bit faster, up to about 16 meters per second. It’s still in the running speed range.
But then you think about other things. Maybe I don’t want to be down on the ground. Maybe I want to be swinging so everyone can see me because I’m Spiderman and I want to be seen, or maybe I don’t want to interact with the people on the ground, or it could be a number of reasons. It just looks cool. I mean, I would not want to be watching a movie with just a running Spiderman. I want Spiderman to swing. So even if it’s not necessarily that big of an advantage, it just looks so cool.
IRA FLATOW: Yeah. Of course, if you want running, we could always call in the Flash to come in–
RHETT ALLAIN: That’s right.
IRA FLATOW: And to do his thing. Now, we always see Spidey, he’s shooting out spider webs all over the place. I mean, there’s got to be a limit to what he can carry in his arm, mustn’t there?
RHETT ALLAIN: Yeah. In one of the movies a while ago, there was no limit because the webs came out of his hand as a biological thing. But in the original comics and in their latest movie, he builds web shooters that use a chemical. So we can estimate the volume of chemical that he would need for his webs based on the volume of one shot.
So if you shoot it let’s say five stories high and it’s a thin stream only a millimeter radius, you can calculate the volume of that one shot. And for that one shot, the volume is really small, but it’s the same as a three-meter-long pencil. That’s what I like to think of. That one shot is like a three-meter-long pencil.
So if you want 50 shots– I’m just picking 50 because you have got to have a number–
IRA FLATOW: Sure.
RHETT ALLAIN: Then you have some serious volume of the webs that he would have to have on his wrist. That’s like 50 three-meter-long pencils he’s going have to carry along with him to have all of those webs, and it would look a lot different on his actual body.
IRA FLATOW: Do we know if there’s anything in the real world that comes close to having the properties of Spiderman’s web?
RHETT ALLAIN: Yeah, there could be. So if you look at what he’s done with the webs, he doesn’t just swing from them, but he’s stopped falling cars and things like that. And based on a falling car that gets stopped, it’s much larger force than just to hold a car. And I get about 40,000 newtons it would have to support. And the amount of force that a cable can support depends on its thickness.
So if I look at all of the same one-millimeter-radius cables, steel would just be 6,000 newtons, so it’s not enough. Nylon rope is not anywhere near. Spider silk, because he’s Spiderman, that would be about 3,000 newtons. But the one that could work is carbon nanotubes. Those are around 200,000 newtons that it could support.
So it’s a science thing, and Peter Parker is a scientist. So carbon nanotube webs would feasibly work, and I think that would be pretty cool. In my mind, he’s shooting a carbon nanotube webs, so that’s what I’m going to stick with.
IRA FLATOW: Are they thick enough, though? Can he get all that stuff?
RHETT ALLAIN: No, he can’t. It won’t be thick enough, but I’m going to let it slide. I’m still going to be a Spider Man fan and just let that one slide.
IRA FLATOW: OK. Yeah. If can hold a space elevator, it can hold Spiderman.
All right. Let’s move on to our next clip, for the upcoming movie Thor, Ragnarok. It’s setting up what is essentially a gladiator-style duel between two of our favorite superheroes. We have Thor and the Hulk.
THOR: Yes! We know each other. He’s a friend from work.
HULK: The Hulk
THOR: Come on.
IRA FLATOW: Hehehehe. That was the trailer. It was kind of funny, wasn’t it?
RHETT ALLAIN: Yeah. And that’s all we have to go on right now. The movie is not out.
IRA FLATOW: Yeah. We’ll have to wait for that to come out. But let’s look at a preview of what to expect. This got us thinking about what would happen if Thor punched the Hulk in mid-air, if they met mid-air, and the physics about superhero punching in general.
RHETT ALLAIN: Right. I think the most important physics idea here is that if Thor punches the Hulk, then he’s exerting a force in the Hulk but forces come in pairs. They always come in pairs. So no matter what the Thor pushes on the Hulk, the Hulk pushes back. It’s not that he has a choice. It’s just the way forces work.
And this force does something. Forces change the momentum, and momentum is mass times velocity. So there’s a big difference between the Thor and the Hulk, and that’s the mass. The Hulk is much bigger, so he most likely has a larger mass. So they jump up in the air, Thor punches, and those forces do different things. Because now that Thor has a smaller mass, in order to have the same change in momentum, his velocity is going to have to change more.
So Thor’s going to jump up. He’s going to hit him, and he’s the one that’s going to get pushed back way more than the Hulk even though they probably want to depict this as Thor really hitting the Hulk and sending them back. But it just can’t– it can’t work that way. So yes, he may hit them hard, but Thor is one that’s going to get thrown back, not the Hulk.
IRA FLATOW: This seems to be like the simplest form of physics calculation you could make of all the ones we’re talking about.
RHETT ALLAIN: Right. And it really helps. In the clip, what got me excited is that he punches him in the air. So you don’t worry about forces from who’s pushing on the ground more or anything. They’re not touching the ground, so it’s a great collision in the air that you could easily calculate what’s going to happen.
IRA FLATOW: You know, these sound like great teaching moments for a teacher to say go out into the movie. We’ll talk about the physics when you get back.
RHETT ALLAIN: That’s right.
IRA FLATOW: I’m Ira Flatow. This is Science Friday from PRI, Public Radio International.
OK. Leaving the superhero battlefield for a second, what about Transformers, a blockbuster, popular blockbuster movie franchise? We’ve all know. We’ve seen the Transformers, but do they square with the laws of physics? Could we actually have something like a Transformer in the real world?
RHETT ALLAIN: So I think we do have something like a Transformer. We have planes that turn into helicopters and cars that can fly. They’re not great, but we can do that. But I think the interesting thing about the Transformers is how would you build a large robot because they’re much bigger than a normal human? And I think that’s the interesting part.
This is the age-old story of scale in physics, and it’s that small things aren’t the same as big things. And that may seem obvious, but if you think about it, the proportions of things would have to be different. So I like to use an example of a box, a cubicle box. It’s one meter on a side. So it’s one by one by one meter. I can easily calculate the volume. It would be 1 times 1 times 1, which is 1 cubic meter.
And I can also calculate the surface area if it’s sitting on the ground, and it’s 1 times 1, so one square meter. So it has a volume which is related to its mass, and it has a surface area. So that’s all fine.
Now I’m going to double the box. I’m going to make it twice as big, so it’s two by two by two. So it has a volume of eight, and the surface area is going to be two by two with a surface area of four. So by doubling the size, the length of one side, I’ve increased the mass by eight, but I’ve increased the surface area by four.
So if it’s sitting on the ground, it’s going to push down a lot more, and it’s going to compress into the ground more because the surface area compared to the mass is much smaller.
So if you build a robot, a big robot, their legs would have been much thicker than a normal human. You can’t just say, oh, it’s a big human and it’s just big. It would look a lot different because its legs have to be much thicker to support its much, much, much larger mass.
And we see this in real life. Look at the chicken. Look at the legs of a chicken compared to the legs of a dinosaur, a Tyrannosaurus rex. They have very, very thick legs because they’re so much bigger. So big things are not the same as small things. You can’t just make a giant chicken. That wouldn’t work.
IRA FLATOW: And I guess this is something any math or superhero nerd can do at home, right?
RHETT ALLAIN: Yeah. They can and they should. And the great thing about these kind of problems is that they are at all different levels. You could just take, let’s say, who’s faster, the Flash or Quicksilver? They both run fast and they are in different universes. And you can find how far did they go, and how long did it take, and what’s the speed. That’s a very simple calculation. And you can compare it and have a nice argument with your friends about who’s faster.
Or you could do something very complicated like Spiderman swinging and jumping. That’s not so easy. So there’s a wide range of problems that can satisfy your interest in comic books and physics, and everyone can do it.
IRA FLATOW: Well, Rhett, thank you for taking time to be with us today. As one former famous movie reviewer said, we’ll see you at the movies.
RHETT ALLAIN: Well, thanks for having me.
IRA FLATOW: Rhett Allain is associate professor of physics at Southeastern Louisiana University and author of WIRED’s Dot Physics blog.
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Have a great weekend. We’ll see you next week. I’m Ira Flatow in New York.