I considered what would happen if you shaved two opposite corners off a d8 dice (Octahedral)
The result of which would be as so
How much should be removed from the corner for it to be fair?
I think there are probably quite a lot of factors going on. An initial guess would be to make the areas of each face equal. But it would probably be better to ensure each face has the same equivalent area when projected onto a circle from the centre.
Are there any good models that include some mechanics for things like center of mass, intertia etc.
disclaimer: I no nothing about dice mechanics....
Not only this is very difficult question, it's also ill-defined. I would argue, that if you managed to make this die fair for a specific material of the die/surface, it's possible to alter one of those to make the die unfair again.
In other words, the answer will be different for different physical models of the dice throwing.
Having said that, there is a particular setup which allows for easy fairness calculation. That is when the die is not bouncy at all (inelastic collision). In that case, when it lands on a vertex or an edge, the face that intersects with a gravity ray from its center of mass will be the face on which it will fall after. In other words, we want for all faces to have the same solid angles (if seen from center).
Take initial octahedron with side 1. If the square face of a new die has side $d$, then the distance from center is $s=\frac{1-d}{\sqrt2}$ and solid angle: $$ \Omega_1(d)=\int\limits_{-d}^{d}\int\limits_{-d}^{d}\frac{s\ dx\,dy}{(s^2+x^2+y^2)^{3/2}} $$
For a trapezoid face the math is a little harder. With $h=1/\sqrt6$ the distance from the center to the face in octahedron: $$ \Omega_2(d) = \int\limits_{-\frac{1}{2\sqrt3}}^{\frac{2-3d}{2\sqrt{3}}}\int\limits_{-\frac13+\frac{x}{\sqrt3}}^{\frac13-\frac{x}{\sqrt3}} \frac{h}{(h^2+x^2+y^2)}dy\, dx $$
Numerical calculation of $\Omega_1(d)=\Omega_2(d)$ gives $d=0.343866$