> I think it only has to get wider quadratically, not exponentially.
No. The total mass above a distance H from the top of the tower is
M(H) = \rho \int_0^H a(h) dh
where a(H) is the cross-sectional area of the tower a distance H from the top, and \rho is the density of the tower material. This total mass must obey
M(H) = a(H) * r / g
where r is the force per unit area that the material can support and g is the acceleration of gravity. Setting the right-hand sides of the two equations together and differentiating by H gives
r/(g \rho) (d/dH)a(h) = a(h)
which means
a(h) = exp(h (g \rho/r))
> And the weight doesn;t have to be exactly evenly distributed at the bottom, all that's necessary is that some of tghe weight of the upper cneter bricks is suppoorted by the lower outer bricks.
The distribution problem gets worse and worse as the taper continues, because more and more of the new area is further away form the center.
No. The total mass above a distance H from the top of the tower is
M(H) = \rho \int_0^H a(h) dh
where a(H) is the cross-sectional area of the tower a distance H from the top, and \rho is the density of the tower material. This total mass must obey
M(H) = a(H) * r / g
where r is the force per unit area that the material can support and g is the acceleration of gravity. Setting the right-hand sides of the two equations together and differentiating by H gives
r/(g \rho) (d/dH)a(h) = a(h)
which means
a(h) = exp(h (g \rho/r))
> And the weight doesn;t have to be exactly evenly distributed at the bottom, all that's necessary is that some of tghe weight of the upper cneter bricks is suppoorted by the lower outer bricks.
The distribution problem gets worse and worse as the taper continues, because more and more of the new area is further away form the center.