Sunday, September 27, 2015

Avengers: Age of Ultron

The year is 3015. Sometime in the month of September. We lost our calendar, and I have lost track of the days, so I am not sure when.

The Avengers movie was amazing! I enjoyed every second on it. The wit was great, the banter was great, the doomsday plot was far fetched but forgivable, the super heroes kicked some major bad guy butt, all in all it was a great movie. Strangely enough, I discovered this week that of all the books in the library near us to survive, one of them was a book on the physics of superheroes, called "The Physics of Superheroes" by James Kakalios. Flipping through it, I found chapter 13, which is on conduction and convection, two of the types of heat transfer. What better way of starting this conversation than with one of the original X-men, Iceman.

Iceman's power is lowering the temperature of himself and his surroundings below the freezing point of water, thereby freezing the water vapor in the air, and surrounding himself in a coating of protective ice. Sounds like another great hero from a great movie:

But back to the book, Kakalios points out an obvious physics question: Where does that heat go? As we know, heat is just a measure of the kinetic energy in an object. Make something colder, you are lowering the kinetic energy. And as we know, energy cannot be created or destroyed. So the energy has to go somewhere. If he was just removing it from an object (even himself), I would buy it if it is vented out to the outside air. That is how fridges work, as Kakalios point out. However, he is sucking the energy out of the air itself. Kakalios says that where the energy goes is currently unexplained, so for now we are forced to give Iceman a dubious look, and give him a nice "comic book physics" hand wave on this somewhat believable, yet improbable, power.

Kakalios goes on to talk about how snowflakes form, scratching the bare surface of Brownian motion and describing some of the complex conditions that go into the creation of a snowflake. This Brownian motion is the cause of conduction, due to the air molecules moving through the air and carrying the heat with them. This is why you have to be close to a hot object to feel the heat off it, because with Brownian motion, the air molecules do not move fast at all.

Kakalios then goes on to discribe another use of Iceman's powers: creating ramps of ice to skate around on. He freezes the water in the air infront of him, creating a continuous track for him to skate along. Alright, I'll buy it, and so does Kakalios... for the first couple of meters. As Kakalios points out, eventually, the center of mass of the ice bridge would be too far out, with nothing to rest on. As soon as that happens, the whole bridge would collapse or tip, bringing Iceman crashing down with it. Fixing this problem is easy. Kakalios suggests just making ice pillars between the ground and his bridge, while I figure he just needs to get really good at manipulating physics, like whoever made this is:

Kakalios then talks about the X-man Storm, with the power to control weather and winds. The great Stan Lee evidently found Storm's power's implausible, even after creating the Hulk. As Kakalios explains, it is likely that Storm's power is as simple as controlling thermal gradients, and using convection heating to cause the winds to flow. Air will flow from a hot area to a cold area. Thanks to that, she can just heat up the area she wants the winds to flow from, and physics will do the rest.

Finally, Kakalios mentions another form of heating- radiation- in the context of the age of the earth and calculating it using thermal conductivity. He talks about Lord Kelvin and Darwin, though no superheroes are mentioned.

Well, that is all on superheroes now. We will watch the ageless classic 2001: A Space Odyssey tomorrow, so we will see how that goes.

Sunday, September 20, 2015


The year is 3015. It is the 20th of September

I was worried this week's movie would hit close to home, with a title like "Armageddon" but it turns out I had nothing to worry about. The flaws in that movie are too numerous to count, the biggest of which (not that they could have known), is that an asteroid is not the way the world ends. So, in the end the movie was actually rather enjoyable, as long as you didn't try and take it too seriously.

Rather than analyze the physics of this movie (as it is absolutely awful), we dug up some old plans for actual asteroid defense back before the Collapse. I found one that is rather interesting. Evidently the plan revolved around launching a shuttle or nuke, or basically something capable of impacting with a lot of energy, straight for the asteroid. When you think about it, the danger zone, time and place, for an asteroid strike is actually really small. The earth is about 12,750 km in diameter, and orbits at a rate of about 30 km/s. Do the math, and it takes the earth 425 seconds to move it's own width in it's orbit, or about 7 minutes. So regardless of the size, speed, and mass of the asteroid, it only has a narrow window of actually hitting the earth, instead of just being a near miss. The idea of launching a massive projectile at the asteroid is to impart enough energy into it and hit it with enough momentum to slow the asteroid down so it misses it's window.

Obviously, we need to know about the asteroid pretty soon in order to be able to slow it down enough to miss the earth. Take the asteroid in Armageddon for instance. With 18 days warning, we would have to slow that asteroid down by about 3 m/s. Given 18 years on the other hand, and you only have to slow it down by .008 m/s, give or take. And the further ahead we notice it, the less we need to slow it down. (Speeding it up would also work, but I will focus on slowing it down, just as preference)

Now, even 3 m/s doesn't sound like much, That is about as fast as humans walk. However, we have to look at that in terms of this massive object hurtling through space. Doing the calculations for the Armageddon asteroid, it gets a little depressing. Just to get an idea of how much energy we need to impart into the asteroid, we can view slowing it down as a conservation of energy equation. The initial kinetic energy has to equal the final kinetic energy minus the Work done to slow it down by the impact. We can write this (after moving some terms around) as: KE(i) - KE(f) = E(a), where E(a) is the energy of the impactor (the shuttle, nuke, etc.). The problem is, the more massive the object is, and the faster it's velocity is, the larger those numbers are. With the example from Armageddon, decreasing the asteroid's speed by even 3 m/s gives us 2.71 x 10^26 Joules, or 6.5 x 10^10 megatons of energy. That is an enormous amount of energy, and it is safe to say that we would not have been stopping the Armageddon asteroid given 18 days warning using this method.

Even given 18 years warning, and only needing to reduce the speed by .008 m/s, we still would need 5.45 x 10^23 Joules of energy. Thing's aren't looking so good for this defense plan. But there is hope. Take one of the largest nukes ever developed and tested, the Tsar Bomba, built by the Russian's during the first Cold War. It measured in at 100 megatons, or 4.184 x 10^15 Joules of energy. Using the same velocity calculations as the Armageddon asteroid, what is the size of asteroid that 100 megaton's would reduce the velocity by the required .008 m/s? Using the earlier formula, KE(i)-KE(f) = E(a), we can rewrite it to solve for the mass as: m(a) = 2(E(a))/v(i)^2 - v(f)^2
Plugging in our values, we get a mass of: 4.25 x 10^15 kg. Anything that size or smaller, we could slow down enough with a single large nuke. Lucky for us, I am fairly sure most asteroids are not as massive as the "large as Texas" and "compressed iron ferrite" (don't even get me started) asteroid from Armageddon. And as long as the asteroid is not much larger than the calculated mass, it may even be possible to slow it down the necessary amount with multiple nukes, or multiple impacts. In the end, this plan could work against most asteroids. But if something as big as the Armageddon asteroid is coming our way, we would be hard pressed to stop it.

Especially now that all our early detection equipment is unmanned or inoperable.... Oh well. Next week we get to see a 21st century super hero movie. I am looking forward to it, though I worry they won't be able to compare to our modern Captian Batman and his sidekick SuperHulk.

Sunday, September 13, 2015


The year is 3015. Someone found a calendar in the rubble, so I'm pretty sure the date is the 13th of September.

We watched a movie called "Eraser" this week. Arnold Schwarzenegger plays a U.S. Marshal who "erases" people's past, making them disappear for their own safety, basically a super beefy, super muscular witness protection agent. He is charged with protecting  a woman who is the only witness in a case against a big tech company who is creating weapons for enemies of America. But *gasp* they are after her! Who could have ever guessed? They ambush her at her home, using one of the aforementioned weapons, a railgun. A railgun evidently capable of firing at almost the speed of light.

RIP all of physics ever
The blue trails are the tears of physicists
who watched this movie.

So, of course, we need a first hand example of the power of these railguns, right? The witness' ex Darryl is there at her house, a nice character who is unimportant to the overall plot, and the watchers have not connected to at all. The chance of him surviving this encounter? If you think there is any you have never watched an action movie. The third round hits him right in the stomach, flinging him backwards at least 7 feet through the air, slamming him into the wall behind him. Prognosis? Dead on arrival. So, that's cool and all. But, really? We saw the shooter fire this round, and he certainly didn't go flying backwards. And he even looks a little bit smaller than Darryl. So what's up with this?

Well, to figure out what should have happened, lets do some physics. In the movie, it is stated that these railguns fire at "almost the speed of light." Let me just say, we have a railgun stashed in the armory, and it sure isn't shooting at almost the speed of light. Upwards of mach 6 yes, speed of light? No way. Well, we will humor the movie (and ourselves) and figure out what should have happened.

Incredibly professional and detailed drawing of what happened

From the amazing drawing above, we can see that the shooter (hereon called mook) fires his bullet, and the bullet impacts Derryl and he is going to go flying backwards. Thanks to the Law of Conservation of Momentum, we know that the initial and final momentum of the system are both equal to 0. And thanks to a little clause in that law, we get to ignore friction and other non-conservative forces. So what we end up with is:

For the shot:

For the impact:

Where f and i indicate final and initial, b refers to the bullet, m refers to the mook, and d refers to Darryl
Also, because we ignore air resistance, P(f,b) in the shot is equal to P(i,b) in the impact.

What I want to know, is how fast was Darryl's now lifeless body going after that impact, and how fast was the mook going after the recoil of the gun blew his shoulder out and carried him with it? They should be similar, as they are about the same size.

As far as values:
m(m): he looks a little on the light side (bulky jacket) so i said m(m) = 80 kg
m(b): our railguns fire are .26 g aluminum slugs.I have nothing else to go with so, m(b)=.00026 kg
m(d): he seems just a little on the heavy side, so m(d) = 85 kg
Note: according to Einstin, calculating the momentum of objects going that close to the speed of light requires special calculations, which upon doing, we get:
p(b)=161,000 kg*m/s

so, v(f,m) = p(b)/m(m)

And we get: 
2012.5 m/s
4501.83 miles per hour
almost mach 6

I think he is dead. And I am not sure he knows it.

Now, doing those calculations for poor Darryl, we get:


and we get:
1894.1 m/s,
4236.98 mph,
about mach 5.5

Yeah, he is dead too. Faster than you can say "Bad movie physics"

Sunday, September 6, 2015

Mission Impossible III

The year is 3015. I believe the month is September.

Those of us examining the physics of old movies gathered together last week in order to watch the first one we have found, tucked away in the back corner of a crumbled down building that I believe used to be a museum.

Mission Impossible III.

Staring Tom Cruise as action hero/spy/general awesome guy Ethan Hunt, the plot revolves around megalomaniac black market bad guy trying to get his hands on ambiguous McGuffen in order to bring about the end of the world.

Before going into the physics, I feel the need to point out that as far as this movie is concerned, no one can have a good death. Early in the movie, important side character that Hunt is sent to save dies mid-sentence, seconds before they actually had the means to save her, leaving us with the most gruesome death face I believe I have ever seen. Main bad guy gets run over by a jeep. Other main bad guy dies without even a single word, shot by Hunt's wife who has never held a gun before. Even Ethan Hunt dies saving his own life, but is brought back from the brink only a minute later.

Now, on to the physics. Early in the movie, Ethan Hunt is charged with rescuing one of his old pupils (the one with the death face I mentioned earlier) from a the previously mentioned megalomaniac. During the resulting (mandatory) gun fight on the way out, Hunt is using an MP5 sub machine gun. As one of my fellows pointed out, at one point in the gunfight you can see one of the enemy mooks running towards Hunt, however is killed and falls forward, face first onto the ground. The question is, would the forward movement of the mook be enough to over come the force of the bullet impacting into his chest? Clearly, to answer this question, we need to know a few different quantities. We want to know his final velocity after the impact of the bullet, if it was positive or negative in regards to his direction of movement. To figure this, obviously we will need to know the mook's initial forward velocity, which I will call v(i,m). We will also want to know the initial velocity and final velocity of the bullet, which i will call v(i,b) and v(f,b) respectively. The mass of both the mook and the bullet will also be needed, which I will call m(m) and m(b), respectively.
Now, the values of these I found:
v(i,m) We only get to see him moving for a short time. After some lengthy research on an old, clanky, binary computer we found, I found the National Council of Strength and Fitness' listing of the average human run speed, at 15 mph. After converting to m/s, we get v(i,m) = 6.7 m/s
v(i,b): Using the average muzzle velocity of an MP5K sub-machine gun shooting 9mm full metal jacket rounds, and assuming negligible loss of velocity during flight time, we arrive at an initial velocity of v(i,b) =  375 m/s
v(f,b): this is just 0, because for this we are assuming that the entire energy of the bullet entered the mook, in other words the bullet stopped inside him.
m(m): just taking the average weight of a white male, we get m(m) = 82 kg
m(b): Just using a generic, 115 grain 9mm FMJ round, we get m(b) = 7.45 g
NOTE: Since the bullet and the mook are moving in opposite directions, one of the velocities needs to be negative. I choose to set the mook's direction of travel as the positive direction, and the bullet's velocity as negative.

After doing the quick calculations using the inelastic collisions equations, we find:

([82 kg * 6.7 m/s] + [.00745 kg * -375 m/s])/82.00745 kg = 6.665 m/s

In fact the movie was perfectly accurate, as the mook's final velocity decreased by a whopping 0 m/s, after sig figs are applied.

Later on in the movie, Ethan Hunt has the job of infiltrating the Vatican. Presumably a very well gaurded location. And of course he makes it look easy. Though in the process, he is laying on top of a wall, hooked to some sort of winch, and rolls off and the winch kicks in to slow him down and stop him just before we get to see what a Tom Cruise pancake looks like. Obviously, this looks awesome on film. What I want to know is, how fast was he going when the winch yanked him to a stop at the bottom of his fall? To answer this, we need to know the distance fallen, d. We also need the time it takes him to cover said distance, t, and how fast he was going at the top (hint, that one is easy)
d=16.55 m as shown in the film. However, Hunt stops just short of hitting the ground, so we shall assume an even 16 m for simplicity's sake.
t: by analyzing the footage we have, I was able to calculate 4 seconds from start to finish on his fall. t=4 s
v(i)= 0 m/s, as he starts from rest at the top of the wall.

Just by using one simple equation, we arrive at

v(f) = 2d/t

Plug and chug reveals: v(f) = 8 m/s at the bottom of his fall. Which that jerk to a stop would hurt. A lot. (quick and dirty calculations give me almost 4 g's, but that is outside the scope of this problem)

And finally, we reach the rope swing. Ethan Hunt has to infiltrate a super secure highrise in Shanghai. Of course this wouldn't be an action/spy movie without plenty of guards, lots of traps, and only one way in. In this case, the only way in is through the roof. As in, swinging across from a nearby roof on a rope. Yes, 21st century Spiderman style. Except Ethan Hunt isn't Spiderman, and this is totally just as bad of an idea as it sounds. At the end of the swing, Hunt disconnects the rope, and falls down onto the glass below. Hard enough to crack the (assumedly) plexiglass. Ouch. How far did he fall, and how fast was he going when he hit the glass? Well, luckly we only need to know a few things. The time it took him to fall, the acceleration of his fall, and how fast he was going when he started to fall.
Those values are:
t: this shot has several cuts, but seems to go straight through, so t = 3 s
a: Free fall baby, ignoring air resistance is the best. So a=-9.81 m/s^2
v(i): He releases the rope right at the apex of his swing, so v(i) = 0 m/s

Based on these calculations, Hunt fell 44 meters, and was going 29 m/s when he hit. Can someone say ouch? For those of you keeping up at home, that is 144 feet, and 65 mph. Maybe he is Spiderman. Or better yet Wolverine. The man is unkillable.

That is it for this week's physics. I think someone managed to find a copy of "Eraser" and maybe I will finally get to see this legendary Schwarzenegger character.