TITLE: Size of a sandpile.
QUESTION [7 upvotes]: Introduction
There is a structure called sandpile. Let us introduce it in the following way. Suppose we have a set of whole points $\mathbb{Z}^2$. Next, we have a function $g:\mathbb{Z}^2 \rightarrow\mathbb{N}$, which shows how many grains are in the point $(x,y).$ Also, there is a number of a maximum possible grains in a point which leaves point stable. We will denote this number through $T$ (threshold). Now execute the following algorithm:

if $g(x,y) > T$ then subtract 4 grains from $(x,y)$ and add one grain to each neighbor of $(x,y)$ i.e. $(x\pm 1, y)$ and $(x, y\pm 1)$.
if there is no points with $g(x,y) > T$ then terminate. Else, start with step 1.   

Simple example with $T=4$ and starting amount of grains $S$ (seed) at $(2,2)$ equals to 11 showed below. 
$$\begin{pmatrix}
0 & 0 & 0\\
0 & 11 & 0\\
0 & 0 & 0
\end{pmatrix} \rightarrow
\begin{pmatrix}
0 & 2 & 0\\
2 & 3 & 2\\
0 & 2 & 0
\end{pmatrix}$$
More information about this here
Question
Suppose we have sandpile with $T=t$ and $S=n_0$ at (0,0). Let us denote such sandpiles through $\Delta(n_0;t)$. My question is

Given a sandpile $\Delta(n_0;t)$ find the size of this sandpile, where $$\text{size} := |\Delta(n_0;t)| =\max_{g(x,y)>0}|x|=\max_{g(x,y)>0}|y|$$

Tries
Some reasons lead me to the answer
$$|\Delta(n_0;t)| \leq 3\log_{4}\frac{n_0}{t} + 1,$$
but empirical results say that for sufficiently large $n_0$ it is not true.
Pictures
Here are some pictures I made with Sage. The darker color the more grains in pixel. The first three pictures with threshold equals to 4, two last - 3.





EDIT: 1D sandpiles
Starting from the suggestion to look at 1-dim sandpile it is possible to say that there is nothing special (i.e. interesting) about that. It is almost obvious how they (1-dim SP's) look like (take a guess). More over, one can calculate the size of 1-dim SP:
$$\Delta_{1}(n_0; t) \sim \frac{n_0}{2t}$$

REPLY [2 votes]: This estimate for $T = 3$ works pretty good.
I asssumed the shape is a circle. Its area is $A = \pi r^2$.
Now, the number of grains in the entire pile doesn't change, so I assumed the average number of grains per cell is $\left \lceil T / 2 \right \rceil = 2$.
Solving for $r$ gives
$$r = \sqrt{ \frac{n_0}{2 \pi} }$$
This is just a rough estimate, but it seems to work quite well.
Here is a table of values I calculated with my formula and the exact values (for $T = 3$).

This seems to indicate that the average number of grains isn't quite two, but it is still quite good. This may be improved by multiplying with $0.9341$.