jpeanuts

Gnuplot has an output mode - called "dumb" - that prints an ASCII plot to the terminal. See e.g. http://pastebin.com/WHcLZHPr This is very handy when running scientific codes on remote machines, and works in any terminal (not just iTerm2 like the submission).

I've been looking for a matplotlib extension to do the same thing, without success.

Not the example you requested, but the threat of it, from: https://en.wikipedia.org/wiki/Aaron_Swartz#Arrest_and_prosec...

> federal prosecutors filed a superseding indictment adding nine more felony counts, which increased Swartz's maximum criminal exposure to 50 years of imprisonment and $1 million in fines.[13][101][102] During plea negotiations with Swartz's attorneys, the prosecutors offered to recommend a sentence of six months in a low-security prison, if Swartz would plead guilty to 13 federal crimes.

The discussion is quite abstract - perhaps that's why it's difficult to grasp. As I understand it, the climate can be regarded as a dynamical system currently existing at a stable equilibrium. I.e. small perturbations to the state tend to be damped. If there's a massive volcanic event, we will have a few cold years, but the climate will return to something familiar.

It's stable, as when you add up all the positive and negative feedback effects, the result is negative. The most important negative comes from the thermal radiation of the Earth into space, which increases with increasing temperature, resulting in more cooling. Methane in Siberian ice is a hypothesised positive feedback effect: higher temperature => more methane => more warming.

If all these effects ever add up to more than 0 (the "tipping point"), the system will no longer be stable, and will travel around until it finds another stable equilibrium. The actual consequences of this are nowhere stated because they are completely unknown. But we're not talking about measuring sea-levels, we're talking a completely different climate (e.g. Venus).

This is a question I've asked myself before too. Here's a back-of-envelope calculation which seems to suggest that diffusion alone will never be enough to keep someone alive inside a box. I'd be really interested to hear if someone can confirm or refute this.

Let's assume CO2 transfer is entirely diffusion-limited, i.e. the worst-case of completely still air. The diffusive flux of some gas-component (per unit area per unit time) J is governed by concentration-gradient (dn/dx) multiplied by a magic diffusion coefficient D, which depends on the molecular properties of the gas. For CO2 in air D = 1.6e-5 m^2/s. Let's assume the gradient is linear so dn/dx = (n_{inside} - n_{outside}) / length. Now n_{outside) = 0.04% * 1 kg/m^3 = 4e-4 kg/m^3. We should decide what level of CO2 we can tolerate, 0.5% should be on the safe side, so n_{inside} = 5e-3 kg/m^3.

Length is a bit trickier. If (a) we assume completely stagnant air inside and outside, length should be chosen as the size of the box ~1 m say. However if (b) we assume that inside and outside are pretty well-mixed individually, due by breathing, wind etc., then length should be about the depth of the hole, or thickness of the box-material, say 1 cm = 0.01 m. For case (a) CO2 leaves our box at a rate of about 7e-8 kg/s/m^2, for (b) 7e-6 kg/s/m^2.

From the original article, a human breaths .84 kg of O2/day, let's estimate a production of 1kg CO2/day (the extra carbon atom can't be all that significant), or 1e-5 kg/s.

To compute the area of hole needed to maintain 0.5% CO2, we just divide 1e-5 by J, to get the area. In case (a) we need an enormous area of 135 m^2, in (b) "only" 1.35 m^2.

This seems to suggest that "air-holes" must work (if indeed they work at all) with air-flow. This doesn't necessary require an over- or under-pressure in the box - a suction can be generated by wind passing over a box with holes on both sides - this process would be far more efficient at restoring atmosphere than diffusion. But would be less reliable.