==> physics/brick.s <==
You can create an infinite overhang.

Let us reverse the problem: how far can brick 1 be from brick 0?

Let us assume that the brick is of length 1.

To determine the place of the center of mass a(n):
a(1)=1/2
a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n)
Thus
      n   1        n  1
a(n)=Sum -- = 1/2 Sum - = 1/2 H(n)
     m=1 2m       m=1 m
Needless to say the limit for n->oo of half the Harmonic series is oo.