3.x - How Many Mirror Mephits Can A Mirror Mephit Sim If A Mirror Mephit Can Sim...

[unseenmage]

Can someone more math-minded figure out how fast a single Mirror Mephit spamming Sim for more Mirror Mephits who also spam etc create a Mirror Mephit singularity?

Or how fast the ones at the bottom of the pile get so crushed they cant even SLA anymore?

Like how fast does the outer edges of the pile even grow?

How dense do the Mephits get before they cannot even cause their Sims to come into being because there's no room?

I'm imagining that the pile of Mephits grows to cover the planet all the way to the outer atmosphere and they asphyxiate before they can gather black hole levels of mass.

However, private demiplanes could still provide an environment for Mirror Mephit singularity farming.

[unseenmage]

So a friend JUST reminded me that Sims could only pile up to the point that those on the bottom expire and evaporate.

Still though, depending on the weight, Str, and hp of the average mirror mephit they might pile up quite a bit.

It was suggested to use the dragon horde pile math from Draconomicon plus the ceiling collapse rules to help figure out how large a pile of mephits is lethal.

[Hellpyre]

Depends on whether a 2HD Mirror Mephit can still use Simulacrum or not.

[newguydude1]

Simulacrum takes 12 hours even as SLA. So it grows real slowly.

Thank Hellpyre for reminding us.

[PoeticallyPsyco]

Okay, let's take this from the top. We'll assume the mephits reproduce in a perfect hemisphere; this makes the math a lot easier.

Terms:
M = total number of mephits
m = number of mephits with an empty space that they can spawn a new mephit into.
x = the number of 'cycles' of reproduction; since the SLA takes 12 hours, x = hours/12 (rounded down to the nearest whole number)
R = the radius of the hemisphere

First off, you can get a rough approximation of the growth of the sphere with the simple equation M = 2x
This ignores that mephits in the center of the pile can't cast the SLA, because they have no open space for a new mephit to appear.

The volume of a sphere is V = 4/3*pi*R3. Since we're assume this pile is forming on the ground and is thus a hemisphere:
M = 2/3*pi*R3

However, only the ones on the outside have an open space to produce new mephits into. We can use the surface area of a sphere (halved) to represent this, so:
m = 2*pi*R2

R = (3M/2/pi)1/3
m = 2*pi*(3M/2/pi)2/3 = (18*pi*M2)1/3

Now we move into calculus territory. The rate of change of M at any given time is equal to the number of producing mephits (m), because the total will increase by that amount each cycle as they clone.
dM/dx = m
dM/dx = (18*pi*M2)1/3
M3/2*dM = (18*pi)1/3*dx
Then we integrate:
(2/5)M5/2 = (18*pi)1/3x + C
At x = 0, M = 1, therefore C = 2/5
M = [(5/2)(18*pi)1/3x + 1]2/5

M = [(5/2)(18*pi)1/3x + 1]2/5

M = (9.59579x + 1)2/5 (now in decimals because they look prettier)