HELLBURN NERD PHYSICS

This document answers such questions as, How do hellburns work on a deeper
level? Why is the PTL putting out as much energy as a quasar? Why is the
thermoelectric generator still working with all the pipes burst? 

There are three critical things that make hellburns work:

 * Characteristic gas values (P/V/T/amt.) held in arbitrarily-sized variables
   (bignums)

 * Atmospheric/Chemistry system ignores standard enthalpy of reaction

 * Pipe (hot/cold loop) burst scheme works differently than how it is presented
   on-screen

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				    Bignums

SS13 was originally written as an atmospheric simulator by a physicist and is
impressively built. It uses 'bignums' to store the fundamentally important
values of gases around the station, specifically pressure, temperature, and
molarity (amount, volume, however you want to think about it). This has the
effect of making gasses infinitely divisible. In "real life", gasses are limited
in this regard by the fact they are constituted by amounts of atoms/molecules
which are at some point indivisible and finite. SS13 has no such limitations,
and can be compacted/divided to an infinite degree bounded by the server's
free memory (and not by bitness bounds exhibited by fixed-width floats, ints,
etc.)

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		     Enthalpy and Le Chatelier's Principle

Secondly, SS13 has no concept of certain factors relating to specific enthalpy
and standard enthalpies of reactions. This is a little difficult to describe
without using verbose & impenetrable language, so instead here's a though
experiment that demonstrates what I'm talking about:

Imagine you have a bic lighter filled with infinite fuel enclosed in a hollow
sphere made of handwavium, a material that perfectly insulates heat and emits
zero black body radiation. The bic lighter ignites and burns perpetually.

In real life, the interior temperature of the sphere would only reach a finite,
stable point. "Energy", here, is in the form of Brownian motion of gas
particles, which is also temperature. As temperature increases, the
differential (perhaps, potential) for the oxidation reaction to "push" against
the system atmosphere decreases, and eventually you reach an equilibrium where
oxidation products, oxidants, and fuel all coexist.

In SS13, exothermic oxidation reactions aren't as intricately modeled and
energy/temperature is just some scalar quantity which is incremented with a
constant amount when oxidation occurs. The interior of the sphere's temperature
would diverge towards infinity.


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				  Burst Pipes

Finally, the way burst pipes comprising the hot/cold loops work "under the hood"
isn't intuitive and the on-screen graphics misrepresent things: Burst pipes look
like they have large empty gaps in them that would completely break any kind of
flow. The reality (in SS13) is more akin to the pipes just forming a single
crack per tile that leaks only some of the gas, and never expands or forms more
cracks past the initial one. *The takeaway to this is that since there are a
finite number of pipe tiles, then there are a finite number of cracks, and a
finite "leakage" figure describing how much gas is leaving the system that is
achieved once all pipes are burst.*

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			    Putting It All Together

The second point concerning enthalpy allows gas in the combustion chamber to
reach infinite temperatures. As gas is pumped in and burns, the temperature
increases at a constant (albeit large) rate. In turn, this causes the hot loop
gas passing through the combustion chamber heat exchange pipes to increase in
temperature and pressure at non-constant rates.

Those rates are the absolute #1 most critical components of a hellburn.

There is an efficiency drop at higher pressures in the loops, as in the eyes of
SS13 higher pressures always means less quantities of gasses in the loops. Your
temperature increase rate has to out-compete this loss.

Your pipes will eventually burst. Every single pipe in the hot/cold loop will
burst. The aforementioned shows how this results in a fixed amount of gas
leaving both loops, and your temperature increase rate has to be so high as to
power-through this. Note that the amount of gas lost is initially very high as
there is still a lot of gas in the loops, but as that quickly approaches zero,
this effect diminishes.

Once the pipes burst, you will have an incredibly small amount of gas present in
the pipes. We're talking like, less than a trillionth of a trillionth of an
atom. Because of bignums this is OK, but your temperature increase rate has to
overcome the diminishing heat capacities of smaller and smaller amounts of gas
mixing in the TEG from both loops.

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I'll update this as time goes on, as I realize I've forgotten and/or misspoke
about things, and keep a log below.


									 kremlin