I love Batman. I really do. There's no way to precisely articulate how much of a fanboy I am over Batman. An introductory paragraph in a class blog certainly won't do it justice. Anyway, I was extremely excited to hear we were doing a comic book themed blog this week, and even though I love the Avengers and Marvel, I prefer my comic books and films with far less levity and more austerity. They don't call him The Dark Knight for nothing.
Keeping in line with the "film" part in the class title, I'm going to analyze some scenes throughout the newest Christopher Nolan movie trilogy.
- Chinese Take-Out
One of the most notable attributes about Batman is that he, well, dresses like a bat. In doing so, this gives him one of his prime abilities -- to glide around with his bat-cape. Now the question I'm not asking issn't whether or not it is possible for him to glide around with a material (I'm going to assume it's possible), but if it's possible or not that his arms would be able to take the force after an extended fall.
In the above scene, Batman attempts to retrieve the mob's accountant in Hong Kong and use him as a major witness to testify against the Joker in court. To do this, he needs to penetrate Lau's corporate building, which he does by gliding in from a taller, adjacent building to its side.
Bruce Wayne jumps off the building for 4 seconds, and using our old kinematic equations, we know that his final velocity at the point before spreading his wings was 39.2 m/s. After this, he does a circle maneuver, where go goes from completely vertical to completely horizontal. This circular motion makes it a centripetal force. To prevent his arms from giving way or buckling, Bruce Wayne would have to exert the same force back onto the air as it is pushing onto him, a la Newton's 3rd law.
We're assuming Batman's mass is 80 kg, and that the radius of the circular path to be about 20 m. With the knowledge that a centripetal force is involved, we can plug it into one of those equations.
F = mg + mv^2
r
m = 80kg
g = 9.8 m/s^2
v = 39.2 m/s
r = 20m
F = (80)(9.8) + (80)(39.2^2) = 6930.56 N
20
6930.56 newtons turns out to be 1558 pounds, so Batman would have to withstand 779 pounds on each of his arms. And since Batman doesn't have super strength, I don't think that can happen.
- Truck Derby
Another good scene in The Dark Knight is when Batman is on the Batpod chasing the Joker in a semi truck. To stop him, the Bat attaches some cables into the ground which halt the truck, making it flip in spectacular fashion. But the question is, can the cables Batman attached withstand the force its being put under?
F = (m)(Δv)
Δt
m = 15, 000 kg
Δv = 22 m/s
Δt = 0.1 s
F = (15, 000)(22) = 3, 300, 000 N
0.1
The truck experiences 3, 300, 000 newtons of force, and using Newton's 3rd law, the cable would have had to withstand just the same amount of force as well -- 3, 300, 000 newtons or 741,869 pounds of force. Since I love Batman so much, I'm going to let him slide this time and assume Luscious Fox gave him some super wire that can withstand that crazy amount of force.
- Conclusion
So I still love Batman. I always will. Forcing me prove that the physics in the movies aren't up to par with reality was soul crushing, so paging Dr. Fragile to save the day and come up with some cool-yet-accurate movie scenarios to make Batman credible once more.
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Wow, that photoshopping job at the end got a big chuckle! Very good job on calculating the force on Batman during his maneuver to level out from his freefall. The only thing I can say is that it looks like his bat wings may not be attached to his arms. If they are, instead, attached to a harness on his back, then he wouldn't need to have the incredible arm strength that you calculated. Nevertheless, he would experience about 8 g's of acceleration during that maneuver, which would make most mortals pass out.
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