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

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I’ve been wanting to write this one for a while, but I had to wait for the right time to do it. Now that the semester is drawing to a close, I think that the temperature has finally fallen to an appropriate level for me to write this post.

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A quick note before we begin: the word ‘frozen’ is usually associated with coldness, but it’s actually synonymous with ‘solid’. Anything that is in the solid state of matter is ‘frozen’. Your phone is frozen. Your clothes are frozen. That hot coffee cup you burned your hands on? Also frozen. And as much as you may want it to be lava, the floor is also frozen.

Thermodynamics can be briefly described as “the flow of heat.” The first law of thermodynamics says that energy (heat) cannot be created or destroyed – it can only be redistributed to other parts of the universe.

Freezing is an exothermic process, which means that heat is removed from the object and released into the surroundings. That heat has to go somewhere. When Elsa froze the water around Arendelle, that should not have caused an eternal winter – the heat in the water should have been released into the atmosphere, increasing the temperature. It’s also possible that she absorbed the energy herself, but if she did, she would have been so hot as to melt the ice around her (and her clothes too).

Either way, this would require a massive amount of heat energy. This energy has actually been estimated at around 5.8*10^15 J of energy, or 100 times the energy released by the Hiroshima bomb! (Seems like measuring energy in terms of explosives is becoming a recurring theme on this blog.)

Another question is how the winter weather came about. Meteorology is not as simple as drastically reducing the temperature somewhere. Snow is caused by a wide range of factors including warm and cold fronts as well as temperature and pressure differentials. Perhaps Elsa was able to convert the heat energy into kinetic energy to create wind drafts. Elsa does appear to also be able to control kinetic energy, such as when she ‘froze’ the storm near the climax. But why would she continuously put in the effort to convert heat energy to make a blizzard?

What would happen if you were hit by Elsa’s freeze power? Water expands when it freezes, which is why pipes burst in winter. Well, blood is mostly water, and runs through blood vessels which are quite similar to pipes. You can see where I’m going – the blood would freeze and expand, destroying vital tissue around blood vessels. If that didn’t kill you outright, then you would die of deoxygenation because the frozen blood would not be able to transport oxygen from the heart. Being hit by Elsa’s freeze power would almost certainly be fatal, and you would probably die well before you suffered hypothermia or were frozen solid.

(Note: there are actually some animals, including certain species of frogs, that can survive freezing by replacing their internal water.)

(Note 2: Since Elsa can apparently create living organisms with perfect human vocal chords from inanimate snow, maybe this isn’t as a big an issue. The implications of that action would require far more analysis beyond the scope of this blog.)

Speaking of Olaf, how exactly does Olaf’s personal flurry work? Why would it be any less susceptible to summer than he is? If it does in fact regenerate itself, then A) autonomous infinite energy and B) why doesn’t Elsa just make Olaf out of that stuff?

At the end of the movie, Elsa is able to thaw out the entire country of Arendelle. Needless to say, this would require an equally massive amount of heat energy. If Elsa has the power to transfer this amount of energy at will, she should stop being queen, start generating infinite energy for Arendelle, and save the universe from heat death. (A true happy ending.)

Maybe if I love the leftovers in the fridge just a bit more I won’t need to use the microwave.

1. “Whatever floats your boat”

Contrary to this phrase’s meaning, there is exactly one thing that floats almost all boats: buoyancy. This is the force that a fluid exerts to counteract the weight of an object. It is proportional to the volume (and, by extension, the weight) of fluid that is displaced by the object. When a boat is submerged in water, the bottom of the boat is subject to a greater pressure than the top of the boat, which tends to push it upwards. The more water is displaced, the greater the pressure difference, so the greater the buoyant force.

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The only exception to this is hydroplaning. Hydroplanes use something very similar to an airplane wing to push water downwards at high speeds. In other words, they primarily use lift, instead of buoyancy, to float, as long as they are moving sufficiently fast. As their name suggests, they are pretty much like airplanes, but in water. However, when the boat is still or at low speeds, hydroplanes still rely on buoyancy to float.

So whenever you use the phrase “Whatever floats your boat,” you’re basically saying, “Buoyancy.”

“Hey, what are you writing about on your blog?”

“I’m writing about the physics of everyday phrases.”

“Uhh… okay. Doesn’t sound very interesting to me, but buoyancy, I guess.”

2. “Turn up the volume”

This one is quite confusing. Why do we refer to a measure of the amount of space an object takes up when we talk about loudness? There are in fact several ways to measure the physical strength of a sound, including amplitude, sound pressure (decibels), sound intensity, etc. However, when we talk about volume, we’re usually talking about how humans perceive sound. As an example, I’m going to bring up another phrase: “If a tree falls in the forest and no one hears it, does it make a sound?”

Physically, the tree falling sends out vibrations, and these vibrations have a certain amplitude, generate a certain sound pressure, etc. However, because no one hears it, it has zero loudness. Similarly, many animals can make sounds that are outside human hearing range. They definitely produce sound, but because we cannot hear them, they have zero loudness.

There is a possible correlation here: the volume, as in the space, of the room you are in will determine the volume, as in the loudness, of the sound you hear. This is particularly notable for firearms, as firing a gun indoors without hearing protection will cause permanent hearing damage, while firing the same gun outdoors will be painful, but probably won’t make you deaf. Interestingly, this is an inverse relationship. In other words, if you turn up the volume of the room, you’re turning down the volume of the sound!

3. “Rock star”

This is an oxymoron. Rocks are solid objects and stars are giant balls of superheated gas (plasma, to be more accurate).

Alright guys, I’ve got one more post on economics before normal service will resume. This time, we’re talking about the economics of choice. Decision making is a very important skill to learn in college and it’s one that many students, including myself, struggle with every day.

Every weekday, I am faced with a choice: should I attend class or not? By attending class, I gain the knowledge of whatever is taught in the class, plus other benefits like respect from the teacher or social engagement with classmates. The opportunity cost of attending class is time that could be used for sleeping, doing homework, etc. So when I am making a decision of whether or not to attend class, I am deciding whether or not my time is best spent learning in class or doing something else.

This is actually a much more complicated decision than you might think. There are so many factors in play that can affect the decision, it is bound to be different depending on the situation. For example, marginal benefit. Marginal benefit is the benefit gained from consuming one additional unit of a product. In this case, we’d be talking about one additional class period. The thing about marginal benefit is that it gets smaller and smaller as you add more and more units. You may know this as “diminishing returns.”

Here’s an example that my economics professor used in class. If you have a garbage truck, and no workers, it cannot function. If you add one worker, he can drive the truck, collect garbage, etc. and make the garbage truck functional. That’s a large marginal benefit. If you add another worker, you can now split the work and have one driver and one collector – still a large marginal benefit. But when you add a third worker, you don’t get as much benefit. What’s he going to do? As you add more and more workers, you get less and less benefit from them. Ten guys on a garbage truck really isn’t much more effective than two.

So how can we apply this to classes? Well, let’s take a look. If I don’t attend any of my Microeconomics lectures at all, I’m going to fail because I don’t know anything. As I attend more and more lectures, my grade will get higher and higher. But there’s a certain limit to how high my grades can go – they can’t go past an A. There will be a point at which attending a class doesn’t improve my grade as much as attending the last one did. Once I begin understanding the material, additional lectures aren’t going to help as much. Is it worth it to attend 100% of the lectures and get a 99, versus attending 90% of the lectures and getting a 94?

Of course, this will vary widely depending on the class in question. Economics is pretty easy to me, so I may be able to only attend 90% of the lectures and get an A. However, something like Differential Equations is a lot harder – I might need to attend 100% of the lectures, plus office hours, in order to understand the material and get an A. In other words, the point of diminishing returns comes later than I could possibly reach it.

However, there’s also a hidden cost here – the cost of tuition. By skipping class, I am forfeiting the money that was intended to be spent on classes. If there are 15 weeks in a semester and each week has 15 classes in total, then there’s 225 classes in a semester. An in-state student pays $16000 for a semester, so each class is worth about $70. That’s pretty scary! I could go on a tangent about how expensive college is and so on, but the reality is that every time I skip class I’m paying $70 to do something else like sleep or finish my homework.

It’s really tough having to consider all of these factors when making decisions, and in practice, we often don’t. We oversimplify and only consider a few obvious ones, and this leads to mistakes. We all have decisions we regret. Honestly, it would be impossible to perfectly process all of the information that goes into every decision every day, so it’s inevitable that we will make mistakes. Hopefully I helped you consider some factors that you might not have thought of before, and maybe this will help you next time you need to make a choice.

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.

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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.

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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. 

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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.

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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.”

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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:

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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.

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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.

Among his seemingly endless list of superpowers, super strength is probably one of Superman’s most recognizable and most commonly used superpowers. It was the one that was shown on the very first issue of Action Comics.

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We’ve seen things like this everywhere – Superman lifting cars, buildings, even planets with his bare hands. As an example, I’m going to use this image of Superman holding up a plane by its nose:

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Unfortunately, it doesn’t matter how strong Superman is – this feat is simply impossible. A Boeing 747 commercial airliner has a maximum take-off weight of almost 1 million pounds. In other words, in order to do what we see in the image, Superman has to exert a force equal to 1 million pounds. However, there’s something we’re not accounting for here: pressure.

Pressure is simply the amount of force applied over a certain area. Superman is applying a tremendous amount of force over a tiny area – the size of his hands, or approximately 16 square inches. Therefore, the plane has to sustain a pressure of more than 60,000 pounds per square inch – far greater than the pressure caused by being hit by a bullet. Imagine trying to stop a falling orange with a knife sticking straight up. If Superman had the strength to withstand the weight of a 747, he would simply pierce through the plane as it continued to fall.

Likewise, this is true for virtually anything that Superman is commonly depicted as lifting. Cars would buckle and bend under the pressure. Buildings would crumble and fall apart as soon as Superman tried to lift them. How does he even maintain balance while holding these massive things over his head?

Superman’s incredible strength is often attributed to the increased gravity of his home planet, Krypton. This article attempts to calculate how much larger Krypton would have to be in order for Superman to be strong enough to exert the amount of force he does. It’s not optimistic: Krypton would have to be “3,000 times the mass of the sun.” It would be nearly impossible just to send a space ship into orbit from Krypton, much less all the way to Earth.

Even further, Superman may have been born on Krypton, but he grew up entirely on Earth. His bones and muscles should have developed for Earth’s gravity, not Krypton’s. In fact, astronauts lose muscle and bone mass incredibly quickly in space, to the point that they often have trouble walking upon returning to Earth. They have to run on a special treadmill pretty much every day just to prevent too much bone loss. Superman, on the other hand, just flies around everywhere. That can’t be good for exercise.

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So Superman’s super strength is pretty much impossible in every way. Nobody will ever be able to lift a million pound plane. But actually… we can. Not only that, but we can also send it at 600 miles per hour through the air. Scientists and engineers have developed machines that are capable of both flight and super strength, and they have changed our lives by allowing us to travel across the globe. Sure, we may not be able to pack it into as small a package as Superman, but I don’t think that even Superman can claim to transport millions of people around the world every day, year after year. A commercial jet is a real life ‘superhero’ with real-life ‘superpowers.’

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To kick off this blog I will be taking a look at one of the oldest and most iconic superheroes – Superman! He also happens to be the one who probably breaks the most laws of physics, but let’s start small here. Today, we’re looking at Superman and flight.

The act of flying is something that eluded humanity for thousands of years. At its most basic, it can be described as “not falling” – in other words, counteracting the force of gravity. Take a look at this image of Superman flying:

http://cinematicjackass.files.wordpress.com/2010/09/s8.jpg

How exactly does he counteract the force of gravity? In order to fly, you must apply a force. Birds fly by pushing the air with their wings. Planes fly by pushing wings through the air, and using the wings to redirect air downwards. But Superman? He just stays still, holding a cool pose with his arms outstretched. He is doing absolutely nothing to oppose gravity. Logically, he should simply fall to the ground.

It is possible that Superman could simply exert a lot of force during takeoff, and then glide using the force of the initial push. After all, Superman was originally described as being able to “leap tall buildings with a single bound.” (For the record, flying was only added to Superman’s abilities after he first debuted, in order to make animating cartoons easier.)

Let’s assume that Superman can leap the tallest building when he debuted, the Empire State Building. How much force would that require? The Empire State Building is 1,454 feet tall at its tip. For my calculations, I will convert all units to metric. 1,454 feet is 443.2 meters. We can use the equation v^2 = 2gh to calculate the velocity that Superman would have to begin with in order to clear the building. Plugging in 443.2 m for h and 9.8 m/s for g (the acceleration of gravity of Earth), we find that Superman needs to achieve an initial velocity of 93.2 m/s, which is approximately 200 miles per hour. If we assume that Superman can achieve this speed in 1 second, and that he weighs 107 kg, then we can use Newton’s second law, F = ma, to find that the force required to leap the Empire State Building is 9,972 N.

http://i2.kym-cdn.com/photos/images/newsfeed/000/370/705/37c.gif

This figure is actually not that impressive. In fact, according to Wolfram Alpha, a weight lifter can achieve a maximum force of about 8000 N during a clean lift. Of course, we are not accounting for big factors such as air resistance, energy lost to the ground, or even how Superman would survive the impact of falling 1,454 feet to the ground. Still, with the right technology such as a jet turbine or a rocket, it’s not hard for humans to generate much greater thrust. After all, we have been able to leap to the moon in a single bound (actually, three rocket stages, but I digress).

However, there is still a problem with this. In most depictions of Superman, we are often treated to a scene where we see Superman flying or hovering, and then, out of sheer will, he accelerates himself even faster. For example:

http://media.giphy.com/media/Aon0ejxr6vdJu/giphy.gif

Sorry Superman, it doesn’t matter how strong Krypton’s gravity was, but you cannot accelerate without thrust. You cannot push off against nothing. (Also, why is his cape billowing? He’s IN SPACE.)

One more question. How high would Superman have to fly in order to be confused for a bird or a plane? When migrating, geese usually fly at an altitude of around 3,000 feet, which is about 1000 meters. Their wingspan is around 1.5 meters. Planes usually cruise around 12,000 m and have wingspans of around 60 m. If we simply take the ratio of wingspan to altitude, geese have a ratio of about 0.0015, and planes have a ratio of 0.005. On average, the arm span of a human is very similar to their height, so let’s assume Superman’s wingspan is 6’ 3”, or about 1.9 meters. Based on these figures, Superman would have to fly at an altitude of between 380 meters and 1270 meters to be confused for a bird or a plane. This would put him at between half and 1.5 times the height of the Burj Khalifa.

In the next part of this analysis, I will be taking a look at Superman’s strength. We’ll find that sheer strength is not enough for him to accomplish some of his deeds. Stay tuned!

I’m also looking for suggestions as to what I should cover next. Leave a comment and tell me what you’re interested in, and I may have some thoughts on it myself!