May The Mass Times Acceleration Be With You Deconstructing Pop Culture with Physics

Hi everyone, I know that this blog is supposed to be about physics, but I’m currently taking a class in microeconomics, and I’ve had some thoughts on the subject. I think that the they are still consistent with the theme of my blog, so I’m going to talk a little bit about sports economics.

Why do sports athletes get paid so much? Many athletes easily make millions of dollars a year, and that’s not considering the advertisement and sponsorship deals they get. Stars like LeBron James and Miguel Cabrera easily make tens of millions, but even no-name benchwarmers get paid more than most people will make in their lifetimes. It would be tough to argue that athletes contribute to society more than scientists, engineers, or teachers. So why are athletes being paid so much more?

Well, the truth of the matter is that athletes generate revenue. I’m going to use baseball as my example, because there are a lot of statistics on this topic. As I write this, the Giants have just won the World Series.

A player’s job is to generate wins for his team, and through statistics such as WAR (Wins Above Replacement), we can estimate with startling accuracy just how many wins a player generated. And what happens when a team starts winning more games? More fans show up at the stadium. More fans watch on TV. More fans buy merchandise. Players generate wins, and wins generate money.

Without access to the checkbooks of MLB teams, we won’t be able to figure out exactly how much money is generated from one additional win. But we can estimate this based on how much teams are willing to pay free agents to generate wins. According to FanGraphs, this is about $5 to $7 million per win. Why is this so high? And why don’t professors get paid $5 to $7 million per research paper?

Well, it’s because of you. And by you, I mean the thousands of fans that are willing to spend $30 on an average ticket or or willing to pay to watch their favorite team on TV. Without people like you spending so much money on sports, teams would not have enough money to pay their players millions of dollars per win. Professors don’t get paid millions of dollars per research paper because there aren’t thousands of die-hard fans who pay to read their papers and root for their favorite professors. If our society valued professors the way we value athletes, we’d be able to see professors on TV every night, there’d be magazines and websites and blogs about professors, and you bet they would make millions of dollars.

http://i.imgur.com/GAK4BvJ.jpg

You can’t really blame the athletes or the football coaches here. They simply generate more money than scientists, engineers, and professors do. And it’s not really surprising that people value the emotion and drama in sports over reading research papers. That’s just the reality of the society we live in. But if you’re a sports fan and you complain about how much money players make, just remember that you’re the one who is giving them that money.

And for the record, I really enjoyed the MLB playoffs this year.

http://fc05.deviantart.net/fs71/f/2013/221/c/4/pacific_rim_by_kclub-d6hbl8m.jpg

For a robot built in the future, powered by a nuclear reactor, and controlled using some sort of neural interface, a Jaeger’s combat tactics are surprisingly primitive. Yes, they have plasma casters and missile launchers, but most of the time, the Jaegers fight with old-fashioned punching and brawling. They talk about the Kaijus’ thick skin and how conventional weapons cannot penetrate it, yet the most effective weapon in the movie is a sword that can cut through entire limbs with ease. There’s a reason why modern soldiers don’t fight with swords. Why fight up close with a sword when you can stand far away and propel a sword at high velocity towards your enemy?

 A rocket punch from a Jaeger has similar kinetic energy to a Boeing 747 at 60 mph, at 125 MJ of energy. This is approximately equivalent to 30 kg of TNT. This may seem like a lot, but the MOAB, the largest non-nuclear bomb in the U.S.’s arsenal, has a blast yield of 11 tons of TNT, or almost 10000 kg. A MOAB would do damage similar to 300 rocket punches, all at once. 

http://i.imgur.com/eD5i4.jpg

We don’t need expensive, physically impossible robots to fight giant monsters. Popular Mechanics laid out a strategy of using current day helicopters and bombers to counter the Kaiju threat. These bombers could deploy payloads such as the MOAB, a bunker buster bomb, or a continuous hail of carpet bombs. A MOP is a bunker buster which is claimed to be able to penetrate 60 m of reinforced concrete. This would appear to be a great weapon to use against the Kaijus’ thick skin.

Here’s another idea. We know exactly where the kaiju are coming from, so why not surround the location with tons of naval mines? Sure, a single naval mine would probably not be very effective, but a minefield of them should at least slow a kaiju down as it swims toward the shore. You might even be able to pack a MOAB into a naval mine, which could potentially cripple a kaiju before it was able to do any damage.

If you extend the time frame slightly into the future, there are more weapons available that could be potentially even more effective. One idea is to use a railgun. Yeah, the deus ex machina in that other giant robot movie.

http://upload.wikimedia.org/wikipedia/commons/d/d3/Railgun_usnavy_2008.jpg

These weapons use electromagnetic propulsion to send a projectile at over 10 times the speed of sound. Current railguns have kinetic energies in the 10-60 MJ range. Even if this is not as much as a rocket punch or a MOAB, the railgun has many potential advantages. First, given correct projectile design, a railgun would almost certainly be able to penetrate a Kaiju’s skin and deal damage to internal organs. Second, railguns have tremendous range and accuracy. With a range of 100 miles, they could engage Kaiju well clear of any potential danger. With the money spent on Jaegers, you could build a large battery of hundreds of small railguns, or develop a massive railgun with even greater kinetic energy.

 Finally, there is my favorite idea: kinetic orbital bombardment, also known as “Rods from God.”

http://img1.wikia.nocookie.net/__cb20100924143021/endwar/images/4/41/Rods_from_God.jpg

Under this concept, massive projectiles would be housed in orbital satellites. When a kaiju is detected, a projectile would be released from orbit, and as it falls it would accelerate to velocities as high as 10,000 m/s. This would result in energies of over 100 tons of TNT – 10 times the yield of the MOAB! That’s 3000 rocket punches delivered at once. Moreover, since it is deployed from space, it could easily target the kaiju and eliminate it as soon it is detected, far away from the coast.

If these weapons were to succeed, the consequences would be clear: If they’re so effective on kaiju, they’d be even more effective on human armies! Whoever was in control of an orbital bombardment satellite would be able to quickly deploy 100 tons of TNT to major targets all around the globe. Sometimes, it’s scary to see just how good we have become at using physics to create ever greater weapons of destruction. But at least we know we’d be safe from giant monsters traveling through an interdimensional portal. Probably.

I’m beginning a new multi-part series in which I tackle Pacific Rim and all of its giant robot goodness. Although I really enjoyed the movie, it’s not hard to find all of the scientific inconsistencies. In fact, Pacific Rim got so much science wrong I could probably write about them for months. But honestly, it’s no fun to analyze exactly why giant robots are physically impossible. When you’re talking about giant robots punching giant monsters in the face, realism has to take a backseat to sheer awesomeness. If you’re interested, you can read about the square-cube law or this detailed rundown of many errors.

So instead of looking at the biggest, most glaring mistakes, I’m going to look at a brief, inconsequential scene that is used purely for comedic effect. I’m talking about the Newton’s cradle scene.

The action of the balls is correct, of course – you’ve probably seen it before. The problem is how the action is initiated. In order to produce the distinctive action seen in the movie, you need to lift up exactly one ball and let it fall. On impact, the energy and momentum of the first ball is transferred through the middle balls and into the ball on the other side. The physics behind this is surprisingly complex, but the short of it is that both momentum and energy are conserved in these collisions, and that each collision passes them to the next ball. Even though the balls in the middle don’t move at all, they play an important role in transferring energy. You can see this energy visualized in this Mythbusters video. As a result, lifting up two balls results in two balls bouncing up on the other side, lifting three results in three, and so on. This YouTube video does a good job in explaining the phenomenon intuitively.

If you want, you can play with an interactive version of Newton’s Cradle here.

However, in the Pacific Rim scene, there is no action that lifts up only one ball. It appears that the massive arm of the Jaeger gently bumps into a chair, which in turn transfers the energy and momentum into the desk. Therefore, the desk, and the arms of the cradle, should move toward the left. The balls, being subject to Newton’s First Law (inertia) would be shifted by the sudden movement of the desk. And now here’s the problem: All of the balls would be set in motion. They would all swing back and forth parallel to each other. There is absolutely no way that what we see in the movie could possibly produce the single ball reaction. The scene was pretty funny, but it completely ignored all of the physics behind Newton’s cradle.

If you were dismayed by the relatively small amount of giant robot in this post, fear not. Next week, I’m going to be looking at the weapons of Pacific Rim. What did the Jaegers use against their massive enemies, and what could have been used instead? Stay tuned!

By the way, did you know that pretty much all of the big fights in Pacific Rim are available on YouTube, in glorious HD? The internet is a wonderful thing.

These days, video games are becoming closer and closer to looking like real life. Battlefield 4 is one such example, with vehicles and weapons modeled down to the very last detail. In many ways, Battlefield 4 attempts to create a realistic and immersive environment. And then there’s this.

Yes, that is a tank being launched into the air with C4 and shooting a jet out of the sky. For all of its realistic graphics and audio, Battlefield 4 has to compromise on gameplay, and one of those compromises is friendly fire. In real life, the tank would just have been blown to smithereens, but in Battlefield, it takes no damage because the C4 was detonated by a teammate. Instead, all of the force of the C4 is transferred to the tank, propelling it into the air.

But would this really be possible? What if all of the force of a C4 explosion could be used to launch a tank? How much C4 would it take to do this? In order to calculate this, we need a very basic understanding of kinetic and potential energy. Then, we can C 4 ourselves. (ba dum tss)

First, I need to make a very rough estimate of the vertical distance that the tank traveled. In the third person view of the video (just over a minute in), you can see the distance marker on the tank which tells us how far it is from the viewpoint. It appears that the tank begins at 63 m away from the viewpoint and reaches its maximum height about 90 m away from the viewpoint. I’m going to assume that the tank is horizontal with the viewpoint at the beginning and that it travels completely vertically (both of these assumptions aren’t true if you look at the video, but bear with me here). Then, we can draw a simple right triangle:

The ’93’ should be a ’90’.

The Pythagorean theorem can tell us h, the vertical distance traveled by the tank. It’s about 64 m. When the tank is 64 m in the air, it has a certain amount of gravitational potential energy, and no kinetic energy. As it falls to the ground, this potential energy is converted to kinetic energy. Just before it hits the ground and explodes, all of the potential energy has been converted to kinetic energy. The amount of kinetic energy right before it hits the ground is the same as the amount of potential energy at its maximum height in the air.

In order to launch the tank into the air, we need to do the opposite: create kinetic energy which is then converted to potential energy as the tank rises. Therefore, to calculate the kinetic energy to launch the tank all we need to do is calculate the potential energy of the tank in the air.

Potential energy is equal to mass * gravitational acceleration * height. The mass of a Type 99 main battle tank, according to Wikipedia, is 55 metric tons, which is equivalent to 55,000 kg. The gravitational acceleration of Earth is 9.8 m/s^2. So we can plug in the numbers and find that the potential energy is 34.5 megajoules. Therefore, this is the same amount of energy required to launch the tank 64 m into the air.

According to WolframAlpha, 1 kg of TNT can produce an explosive yield of 4.2 megajoules. This table on Wikipedia tells us that C4 is 1.34 times as effective as TNT, so 1 kg of C4 produces 5.6 MJ. It would require 6.2 kg of C4 to launch a tank 64 m into the air. Of course, this is assuming that all of the energy of a C4 explosion is converted to kinetic energy, and none of it is converted to heat or sound energy, and that energy is somehow perfectly transferred into pushing the tank instead of dissipating into the air or ground.

In Battlefield 4, C4 is used in small packs, which look like this:

http://bf4central.com/wp-content/uploads/2013/08/battlefield-4-c4-explosives.jpg

In order calculate how many packs of C4 are required to generate 34.5 MJ, I need to know the mass of 1 pack of C4. I wasn’t able to find this online, but I was able to find the texture sheet used by the C4 model. Credit to elementofprogress for extracting this texture from the game.

According to this, 1 pack of C4 weighs 1.14 lb, which is about half a kilogram. Therefore, it would take just over 12 packs of C4 to launch a Type 99 tank 64 m into the air. Unfortunately, I’m not sure how many packs of C4 they used in the video, so I can’t tell you how accurate the game is. But sometimes, it’s just way more fun to be unrealistic!

For this week’s installment in my blog, I’m sticking with the superhero theme but switching over to Marvel. This post was inspired by the many T-shirts I’ve seen around campus bearing a distinctive logo:

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While Captain America may not have superpowers like Superman, his iconic shield certainly does. It is supposedly indestructible, able to completely absorb the kinetic energy of any impact. In addition, Captain America is able to use it as a powerful offensive weapon. He can bludgeon enemies with it or even throw it and have it ricochet off multiple targets.

Now, I’d like you to think about this for a minute: what everyday objects do you know are very, very good at absorbing blunt impacts? It’s not going to be a steel shield, or anything else made of metal for that matter. I’ll give you a hint: in a car crash, what is used to absorb the energy of your head jerking forward? That’s right, an airbag. Similarly, what do many action movie heroes use to survive jumps from multiple-story buildings? It’s always something soft like trash bags. Absorbing kinetic energy is a property of soft materials, not rigid ones.

If we take the core assumption that the shield can absorb all kinetic energy, then many of its other abilities fall apart. For instance, it would not be an effective weapon. Hitting someone with the shield would be like hitting them with a pillow: the energy used to hit them with the shield would be absorbed, causing no damage. The shield’s absorption abilities work both ways; if it does not allow energy to be transferred to the wielder, then it cannot transfer energy to the enemy either.

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Similarly, the shield would not be able to ricochet at all. In any collision, some energy is lost, which is why a ball will never bounce back to the same height it was dropped from. However, if all that energy is lost (absorbed), then the ball will not bounce at all. Pillows do not bounce. We often see Captain America use his shield to deflect bullets – in the most recent movie, he even uses it to deflect bullets toward enemies and kill them. Clearly, if the shield truly absorbs energy, then this is wrong – the bullets should just fall to the ground.

http://www.americansansoo.com/captain-america_full-size.jpg

In the movie, Captain America takes fire from a minigun. For this example, let’s assume that the energy of each bullet is being properly absorbed. A M134 minigun fires a 7.62x51mm cartridge which generates about 4000 joules of energy at the muzzle. It fires these rounds at a rate of 2000 to 6000 rounds per minute. Let’s assume that it fires at 2400 RPM – this is equivalent to 40 rounds a second. If the minigun is fired at that rate for 10 seconds, the total energy of all of the rounds is 4000*40*10 = 1,600,000 joules. The first law of thermodynamics says that energy cannot be created or destroyed, so where does it go?

Let’s assume that it is completely converted to heat. Now we’re going to use the specific heat equation to find out how hot the shield would get by absorbing that energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/shta.gif

We don’t know what the specific heat of Captain America’s shield is, but let’s use the specific heat of steel to get an idea: 420 J·kg−1·K−1. The mass of the shield has been estimated at 20 kg. So now we can plug in the numbers and solve for ∆T, the change in temperature. It turns out that the temperature of the shield would rise by 190°C, or 343°F. That would pretty much instantly result in third degree burns.

So what do we learn from this? Even Captain America’s shield cannot defeat physics. If you could perfectly absorb the kinetic energy from a minigun’s bullets, the heat energy would still be more than enough to do significant damage.

Update: Some readers may find fault with the estimate of the mass of the shield that I linked to, seeing as how it is based on an incorrect assumption (if the shield truly absorbed all energy, Captain America would not have had been pushed back from the impact of the shield). The official Marvel wiki lists its weight as 12 lbs, or 5.4 kg. This would actually make the results even scarier: the temperature would rise by 705°C, or 1300°F! That’s hot enough for Captain America’s uniform to literally melt and fuse to his skin. I think I’d rather just be shot to pieces by the minigun than face such a terrible end.