Dihedral angle for icosahedron with many faces I would like to make a rough sphere by icosahedron-ish shape which has many faces, N, and It will go to infinity. 
My question is what is the functionality and behavior of dihedral angle between two adjacent faces respect to N. does it increase as a $\delta\propto \pi- c/N^2$ or $\pi -c/N$?
Can we say something about higher dimension and $S^d$ spheres as well?
e.g. In the one-dimensional case, $S^1$, we can make a circle with  a polygon with many edges and the dihedral angle is simply proportional to $\pi/N$.
 A: I'll be assuming a subdivision of the sphere into triangles, as discussed in this Stack Overflow post. The angular defect of the whole sphere will remain $4\pi$, but it gets distributed over $V$ vertices, with $O(V)=O(N)$ or more precisely $V=\frac{3N+12}6\approx\frac N2$. So on average the angular defect per vertex will be $\frac{4\pi}V\approx\frac{8\pi}N$ with the approximation getting better as $N$ increases.
So how does this translate into dihedral angles? As I didn't find a formula just now, let's derive this relationship. Consider a regular hexagon, inscribed in the unit circle. Connect its vertices to the point $(0,0,z)$ to form six triangles. Choose $z$ such that the angular defect becomes $\frac{8\pi}N$ then compute the dihedral angle.
The base $a=1$ of each triangle is an edge of the regular hexagon. The opposite angle is $\alpha=\frac16(2\pi-\frac{8\pi}N)=\frac\pi3(1-\frac4N)$. The two base angles are $\beta=\frac{\pi-\alpha}2=\frac\pi3(1+\frac2N)$ as the triangle is isosceles. Which means the distance between hexagon vertex and common apex is $b=\frac{\sin\beta}{\sin\alpha}a=\frac{N+2}{N-4}$. That gives $z=\sqrt{b^2-1}=\frac{2}{N-4}\sqrt{3(N-1)}$
or more usefully $z^2=\frac{12(N-1)}{(N-4)^2}$.
The vector from the vertex $(-1,0,0)$ to $(0,0,z)$ is $(1,0,z)$. Likewise, the vector from the vertex $(-\frac12,-\frac{\sqrt3}2,0)$ to $(0,0,z)$ is $(\frac12,\frac{\sqrt3}2,z)$. Their cross product,
$n_1=(-\frac{\sqrt3}2z,-\frac12z,\frac{\sqrt3}2)$, is a normal of that triangle. Swapping the $y$ coordinate gives the normal of one of the adjacent triangles as $n_2=(-\frac{\sqrt3}2z,\frac12z,\frac{\sqrt3}2)$. So the dihedral angle between these is calculated as
$$ \delta
=\arccos-\frac{n_1\cdot n_2}{\lvert n_1\rvert\,\lvert n_2\rvert}
=\arccos-\frac{2z^2+3}{4z^2+3}
=\arccos-\frac{N^2+8}{N^2+8N}
$$
So what happens for large $N$? Using some trial and error diagram drawing and function guessing I found out that
$$\lim_{N\to\infty}\frac{\mathrm d}{\mathrm dN}
\frac{1}{(\pi-\delta)^2}=\frac1{16}$$
so
$$
\frac{1}{(\pi-\delta)^2}\approx\frac N{4^2}
\quad\text{and}\quad
\delta\approx\pi-\frac4{\sqrt N}
$$
for large $N$. This approximation is actually a lower bound on $\delta$. I would assume that one could prove that
$$\pi-\frac4{\sqrt N}\le\delta\le\pi-\frac c{\sqrt N}$$
is satisfied for $c=3$ for any $N\ge 20$ (i.e. the regular icosahedron or any subdivision thereof) and is also satisfied for any $c<4$ if $N$ is sufficiently large. But I don't have a proof for this claim, just strong experimental evidence.
Notice that the frequency $\nu$ of the subdivision is proportional to $\sqrt N$. Actually $N=12\nu^2$ for a class I geodesic sphere. So you might want to think about this as $\delta\approx\pi-\frac2{\sqrt3\nu}$.
