Si Tacuisses, Philosophus Mansisses: Fun with numbers for November 22, 2019

How lucky can asteroid miners be?

So, I was speed-rereading Orson Scott Card's First Formic War books (as one does; the actual books, not the comics, BTW), and took issue with the luck involved in noticing the first formic attack ship.

Call it the "how lucky can you get?" issue.

Basically, the miner ship El Cavador (literally "The Digger" in Castilian) on the Kuiper belt had to be incredibly lucky to see the formic ship, since it wasn't in the plane of the ecliptic, and therefore could be anywhere in the space between 30 AU (4,487,936,130 km) and 55 AU (8,227,882,905 km) distance from the Sun.

The volume of space between $r_1$ and $r_2 $ for $r_2 < r_1$ is $4/3\, \pi (r_1 - r_2)^3$, so the volume between 30 and 55 AU is 219,121,440,383,835,000,000,000,000,000 cubic kilometers.

Let's say the formic ship is as big the area of Manhattan with 1 km height, i.e. 60 km$^3$. What the hay, let's add a few other boroughs and make it 200 km$^3$. Then, it occupies a fraction $9 \times 10^{-28}$ of that space.

To put that fraction into perspective, the odds of winning each of the various lotteries in the US are around 1 in 300 million or so; the probability of the formic ship being in a specific point of the volume is slightly lower than the probability of winning three lotteries and throwing a pair of dice and getting two sixes, all together.

What if the ship was as big as the Earth, or it could be detected within a ball of the radius of the Earth? Earth volume is close to 1 trillion cubic kilometers, so the fraction is 1/219,121,440,383,835,000, or $4.56 \times 10^{-18}$; much more likely: about as likely as winning two lotteries and drawing the king of hearts from a deck of cards, simultaneously.

Let us be a little more generous with the discoverability of the formic ship. Let's say it's discoverable within a light-minute; that is, all El Cavador has to do observe a ball with 1 light-minute radius that happens to contain the formic ship. In this case, the odds are significantly better: 1 in 8,969,717. Note that one light-minute is 1/3 the distance between the Sun and Mercury, so this is a very large ball.

If we make an even more generous assumption of discoverability within one light-hour, the odds are 1 in 42. But this is a huge ball: if centered on the Sun it would go past the orbit of Jupiter, with a radius about 1 1/3 times the distance between the Sun and Jupiter. And that's still just under a 2.5% chance of detecting the ship.

Okay, it's a suspension of disbelief thing. With most space opera there's a lot of things that need to happen so that the story isn't "alien ship detected, alien weapon deployed, human population terminated, aliens occupy the planet, the end." So, the miners on El Cavador got lucky and, consequently, a series of novels exploring sociology more than science or engineering can be written.

Still, the formic wars are pretty good space opera, so one forgives these things.

Using Tribonacci numbers to measure Rstats performance on the iPad

Fibonacci numbers are defined by $F(1) = F(2)= 1$ and $F(n) = F(n-1) + F(n-2)$ for $n>2$. A variation, "Tribonacci" numbers ("tri" for three) uses $T(1) = T(2) = T(3) = 1$ and $T(n) = T(n-1) + T(n-2) + T(n-3)$ for $n>3$. These are easy enough to compute with a cycle, or for that matter, a spreadsheet:

(Yes, the sequence gets very close to an exponential. There's a literature on it and everything.)

Because of the triple recursion, these numbers are also a simple way to test the speed of a given platform. (The triple recursion forces a large number of function calls and if-then-else decisions, which strains the interpreter; obviously an optimizing compiler might transcode the recursion into a for-loop.)

For example, to test the R front end on the iPad nano-reviewed in a previous FwN, we can use this code:

Since it runs remotely on a server, it wasn't quite as fast as on my programming rig, but at least it wasn't too bad.

Note that there's a combinatorial explosion of function calls, for example, these are the function calls for $T(7)$:

There's probably a smart mathematical formula for the total number of function calls in the full recursive formulation; being an engineer, I decided to let the computer do the counting for me, with this modified code:

And the results of this code (prettified on a spreadsheet, but computed by RStudio):

For $T(30)= 20,603,361$ there are 30,905,041 function calls. This program is a good test of function call execution speed.

Charlie's Angels and Rotten Tomatoes

Since the model is parameterized, all I need to compute one of these is to enter the audience and critic numbers and percentages. Interesting how the critics and the audience are in agreement in the 2019 remake, though the movie hasn't fared too well in the theaters. (I'll watch it when it comes to Netflix, Amazon Prime, or Apple TV+, so I can't comment on the movie itself; I liked the 2000 and 2003 movies, as comedies that they were.)

Late entry: more fun with Tesla

15-40 miles of range, using TSLA's 300 Wh/mile is 4.5 kWh to 12 kWh. Say 12 hours of sunlight, so we're talking 375 to 1000 W of solar panels. For typical solar panels mounted at appropriate angles (150 W/m2), that's 2.5 to 6.7 square meters of solar panels…

Yeah, right!