Is there a point at which spaghettification is highest? I've read anything getting close to a regular black hole would experience spaghettification but not when you get close to super-massive black hole.
Is there a point of "peak spaghettification" where the mass of the black hole exerts the greatest tidal force? Or, have I misunderstood something along the way? 
 A: A tidal force happens because different parts of the infalling object are trying to move along different geodesics. Suppose we take the 3-vector $\eta$ to be the distance between two points, then we can calculate how $\eta$ changes as the object falls inwards. If $\eta$ is constant then the distance between the points isn't changing and there is no tidal force. If $\eta$ is increasing then the points are being stretched apart while if $\eta$ is decreasing the points are being compressed together.
Anyhow, after some frenzied pen scratching (the details can be found in any GR textbook) we get:
$$ \frac{D^2\eta^r}{d\tau^2} = \frac{r_s}{r^3}\eta^r $$
$$ \frac{D^2\eta^\theta}{d\tau^2} = -\frac{r_s}{2r^3}\eta^\theta $$
$$ \frac{D^2\eta^\phi}{d\tau^2} = -\frac{r_s}{2r^3}\eta^\phi $$
where $r$, $\theta$ and $\phi$ are the spatial Schwarzschild coordinates and $r_s$ is the Schwarzschild radius $r_s = 2GM/c^2$. $D$ is the covariant derivative. The quantities on the left hand side are effectively an acceleration, so they are a force per unit mass that we can interpret as a tidal force.
The point of all this is that the tidal forces are proportional to $1/r^3$. This means they are greatest at $r \rightarrow 0$ i.e. they reach a maximum as the infalling object reaches the singularity.
Note that the tidal forces are finite at any value of $r \gt 0$ so they are finite at the event horizon. In fact at the horizon, i.e. when $r = r_s$, we get (I'll show just the radial equation):
$$ \frac{D^2\eta^r}{d\tau^2} = \frac{1}{r_s^2}\eta^r = \frac{c^4}{4G^2M^2}\eta^r $$
So the tidal force at the horizon decreases as the inverse square of the black hole mass.
