Center of rotation (rigid body)

Thought experiment -

Given a free body $B$ that is being acted on by forces, find its instantaneous center of rotation $P1$ and its instantaneous center of zero acceleration $P2$.

Does this let us solve for the center of curvature for any point $P$ on the body?

To do so, we need the acceleration vector at $P$ and the velocity vector at $P$. That lets us define the osculating circle to the path of $P$ at that instant.

Given the coordinates of $P$, we know the direction of the velocity vector is perpendicular to the segment $P$ to $P1$, but what is it's magnitude? We need the magnitude of the velocity at some other point (not $P$) to calculate it. The only other point is $P2$. But we know only it's velocity direction also, not it's magnitude. Similarly, we know the direction of the acceleration vector is perpendicular to the segment $P$ to $P2$, but not it's magnitude, and we don't know the acceleration magnitude of $P1$ either, so we can't derive it.

So knowing the ICR and center of zero acceleration does not provide sufficient constraints to determine the ICC. We know the vector towards the ICC will be co-linear with segment $overline{PP1}$. But we don't know the distance.