How many people are left?

I had time this evening to finally sit down and code up a simulator to answer the question I keep having when I watch this Last Man on Earth show (which I don't think is very good, since half the cast play horrible people, and because I can use the phrase "half the cast" on a show with that title).  Given that one person survived whatever killed everyone, how likely is it that there's another person alive as well.  This is basically a Bernoulli trial problem, where we don't know k or p.  I'm taking N = 300e6, which is close enough to the population of the US.  The results:

Normalized to the probability that everybody died, with the curves ranging for logarithmically spaced p values from 10**-8.4 to 10**-7.8.

If you think you're the only one alive, then you estimate that lp = -8.4.  However, once you find out that another person is alive, then lp >= -8.3 is the minimum (in this resolution), and those two cases have similar probabilities.  If you assume that k = 2 is the most likely probability, this moves lp to -8.1, and you have strong evidence that a third and fourth survivors are also likely.  Finding the third again bumps things up, and you can start expecting up to 8.  A fourth?  You've reached the other end of the simulation, and although the dynamic range of probability ratios increases, it's not excessive.

Basically, once the guy found Kristen Schaal, he's no longer the last man on Earth, and the likelihood of finding a lot more people jumps.  I will say that his strategy of travelling around writing the city he's living in seems like a good idea, assuming he wrote on a sign there that he's still driving around looking for people if they don't meet him.  That way, people will stay there and wait instead of assuming it's empty as well.  If he did meet a city with survivors, he could just change that sign to redirect to survivor-town.

I've spent way too much time on this stupid problem: Endor vs Death Star.

Today's main point is a continuation of thoughts on Ewoks.  Earlier this week, I saw this image in my RSS stuff:

And searching online for it again led me to this, which is an overly long discussion of a bunch of nonsense about how Endor would have been incinerated by the debris from the Death Star.  After thinking about it for a few minutes, I came to the conclusion that that didn't make much sense, as pre-explosion, the DS had to have been moving sufficiently to not crash, so after the explosion, only a small fraction should hit the planet.  Right?

So let's throw physics at the problem, and see.  According to the Star Wars wiki page, Endor has a radius of 2450km radius.  If it has the same average density as the Earth, this gives it a mass of 3.39e23 kg.  We're trying to slam stuff into it, however, so let's go with an alternate calculation that the surface gravity is the same as on Earth (the Earth density Endor has about half-Earth gravity).  This gives about twice the mass at 8.82e23 kg.

Now we know what the target looks like, but what about the DS?  It apparently has a radius of 80km.  What does that mean in terms of mass?  No idea.  It's made of "quadanium steel," so we have to make something up.  Is it steel?  That's 7.75 g/cm^3.  Maybe quadanium makes it lighter, so maybe something like aluminum at 2.70 g/cm^3?  Plus, a lot of the DS is empty, because otherwise you couldn't do stuff inside it.  So there's a fill factor to deal with.  Let's say all the rooms look like this one:

This one.

That's an imperial shuttle, and those are 20m long, and roughly square-ish.  Stamp that footprint around, and I come up with something like 120m * 120m * 60m = 864000 m^3.  If the walls are 5m thick, the fill factor for the DS made up of these rooms is 33%.  In the first movie, the walls tend to look thin like a regular wall.  This yields a fill factor of about 1%.  This results in less mass for the DS, which turns out to be super important.  The high end mass is steel with fill factor=1.0, or 1.66e19 kg.  The low end is aluminum with a 1% fill factor, at 5.8832e16 kg.

Why?  The DS blows up.  It's a good "kablooey" kind of explosion.  Assuming this is going to completely disperse all the mass to infinity, this means the explosion is comparable to the gravitational binding energy, U = 3/5 * GM^2 / R.  For our 80km radius, and the high and low mass estimates give binding energies of 1.38e23 and 1.73e18 MJ.  This has to be converted to kinetic energy of all the debris particles, so assuming equal mass, each particle gets, U = 1/2 (M/Np) * v^2, so v = sqrt(Np) * [129 7.67] km/s.

What's left to know?  Where the DS is located in association to Endor.  I remembered this scene:

Zap!

But that turns out to not help.  I had planned to measure the length of the chord on the arc of Endor, and use that to work out how far behind the DS it is.  However, since this is the CGI reconstruction of two models on a stage, this doesn't work.  Using that chord to work out the projected scale of Endor, given the known "real" radii of the two objects means that Endor is in front of the DS, as the model Endor wasn't big enough to create the correct arc.

This isn't a big deal, as I remembered that the DS is orbiting Endor because there's a shield generator that's protecting it.  This means the DS has to be in a geostationary orbit.  That wiki page for Endor claims it has a period of 18 hours.  So the DS orbits at 18433km, and has an orbital velocity of 1.79km/s (which was yesterday's plot).

Put this all together, and run it through the N-body simulator to see what happens, and you get:

It freaking explodes.

How much does it explode?  It explodes all the way.  For the randomly placed particles (they all start on the surface of the sphere at 80km), one triggered the "collision" flag in the simulation.  This makes sense, as at the distance to Endor, it subtends 0.13radians, which translates to 0.435% of the full sky close to the 1/200 simulation particles hitting.  This is still 2.94e14kg being dumped on the planet.  That's about a third the size of the Chicxulub impact, although it'd be spread out somewhat instead of being a single impact.

The other fun thing I tried was reducing the explosion strength by slowing the particle velocity:

Red is 100 times smaller and blue is a 1000.

In these cases, the particles continue to largely follow the original DS orbit.  In the lowest energy case, Endor is spared, and the extra energy goes to kicking particles out.

The Simpsons cast vs aging.

I came home, watched a Simpson's episode that was sufficiently bland that I sat down and reimplemented the thing I did for the Supreme Court for a generic data file.  Feeding the main voice actor cast for the Simpson's into it (and quantizing on years because working out partial years would have taken more time than I wanted to spend.

Individually.

Summed.

Based on this, The Simpson's is unlikely to lose any of the main cast until 2020 at the earliest.  Assuming they keep going after that, they can continue for about another twenty years before they lose half of the cast.  However, Harry Shearer is that first line, and he voices about a third of all characters on the show.  That would probably be hard to recover from.

Because some questions deserve answers.

I saw this episode of Bob's Burgers last night.  Basically, Bob is nearing his 100000 burger, and some jerk shows up with a cow to protest animal cruelty and asks "how many cows had to die for your burgers?"

First, let's take the National Cattlemen's Beef Association beef research guide, and accept their calculations for a standard 1300-pound steer.  Summing their numbers and then noting after-the-fact that they do it themselves on page 7, this results in a cow containing 638.2 pounds of meat, of which 265.1 pounds are ground beef.  This is where you have to decide if Bob can offset future ground beef with the other cuts contained in the cow or not (is he charged .41 cows or 1 cow per cow).  Assuming he's making standard quarter pound burgers, this results in either 39.173 cows (assuming he's allowed to offset) or 94.304 (if all cows that contribute count).

That's way lower than I would have guessed beforehand.