I assume every z-level is infinitely high, but every creature is infinitely tall. Lying down makes them infinity minus one in height, allowing an infinite number of lying down creatures to occupy a space two standing ones could not.
This doesn't work. Infinite minus one is still infinite. What does work is if they are finitely wide and erm... "deep", but infinitely tall. When they lie down, an infinite amount of them can get crammed in one tile, since now they each occupy a certain height in an infinitely high distance. When a creature standing up enters the tile, the total height of the creatures in that tile equals n*d+inf=inf where n is the amount of prone creatures, d is the depth of a creature and inf is, obviously, infinity. Technically, you should also account for a portion of the prone creatures to be on their side, but since that just gives the same result, it can be ignored.
This gives a good result when describing the height problem issued, but when we take a loo at the wiki and raws, you'll find that the size of a creature is finite. If one measurement goes to infinity while the other two are real numbers, the volume of all creatures would be infinity: V=d*w*h=lim[h->inf](d*w*h)=d*w*lim[h->inf](h)=inf This is obviously not the case. Therefore, one of the other two measurements has to be a function of h tat goes to zero as h goes to infinity, in such a matter that lim[h->inf](d(h)*h)=c a real number. One such function is sin(1/h)*h or in a more recognizable form: sin(x)/x with h*x=1 This function, also called the sinc function, goes to 1 for x going to 0. There will probably be an infinite amount of functions that comply with this equation.
The fun really starts when you try to find the depth of your creatures. Since the product of d and h has to go to a finite number for h going to infinite, d has to go to 0: d*h=c <=>h=c/d <=> lim[h->inf](h)=lim[h->inf](c/d(h))<=>c/lim[d->0](d) <=> h->inf=>d->0 This implies that a creature has a real, non-zero width, an infinite height and a zero depth. What this means is that all creatures are planes with a certain width. If we assume that all creatures have the same function for d, then we know that the limiting factor of the creatures volume is it's width. If we were to assume that all creatures are infinitely flexible, this would mean that it would be able to fold itself in such a manner that it would occupy a finite volume, and thus tiles could have a finite volume as well. However, this explanation would only allow a finite number of finite creatures in a finite tile, which is against our basic assumption that an infinite amount of creatures fits in a single tile, whether it be finite or infinite.
Another explanation would be a quantum mechanical one. As known, quantum mechanics describes particles (dwarves) as waves, which can be represented in Euler's notation: D(
r,t)=A*e
(i*(r*k-w*t) With
r=x
ex+y
ey+z
ez the location vector,
k the
wavenumber vector, w the angular frequency, t time and i=(-1)
1/2 the imaginary unit. In general, something written in Euler's notation can be rewritten in function of a sine and a cosine: A*e
(i(x+f))=A*cos(x+f)+i*A*sin(x+f) It is obvious that we can only perceive the real part, so what we observe of A*e
(i(x+f)) is Re[A*e
(i(x+f))]=Re[A*cos(x+f)+A*i*sin(x+f)]=A*cos(x+f) If we now assume that the wavefunction D(
r,t) is related to the dwarf's volume, it is still so that we can only observe the real part of D. This is also the solution of a particle trapped in an infinitely deep potential well, which can be a good approximation of a dwarf in a tile. This leads to the conclusion that we know that the dwarf is in the tile, but not /where/ exactly in the tile. Introducing a second dwarf in the tile adds an element of perturbation to the potential energy function, on which the Schrödinger equation and hence the wavefunction relies. The exact solution depends on how the particles interact with each other, but it's possible that the wavefunction of a dwarf in a tile with n total dwarfs in it is augmented by two functions f and g of n so D(
r,t)=A*f*cos(
r*
k-w*t)+A*g*sin(
r*
k-w*t) Seeing <D|D>=f²+g²=1, it is obvious that one should increase and the other should decrease with larger n. If we assume that f decreases with larger n, then what we observe is still the real part of D, but since it decreases with increasing n, the observed part of the wave function will be smaller when more dwarfs are in the same tile simultaneously. When n goes to infinity, the particles still interact with each other, but the observed part of the wavefunction will go to zero, allowing an infinite amount of dwarfs in a single tile.
colossi are not liquid and hence compressible 
A solid steel bar isn't a liquid and hence compressible? Technically true, you could compress it a bit, but not significantly. If your bronze colossus is a gas, I'd agree. But that would mean they'd have a temperature of about 2836K (assuming the current composition of 88% copper and 12% tin and a linear relation between the amount of each material and it's boiling point. 50% of each would result in a temperature of 2855K). This would ignite the surrounding grass. This is not the case, hence the bronze colossus is not a gas, and not compressible. The reason why a gas can be compressed is that in a gas, the atoms move freely with rather large spaces in between. In a liquid or a solid, they have almost no space in between them, making it harder to move them even closer.
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Topic: Trying to visualize z-levels... (Read 3120 times)