I'm attempting to make a game similar to Ace Combat, and I'm still confused about banked turns, namely the speed of bank turning depending on the aircraft's roll.
Can I assume, for simplicity's sake, that it's something like $Rgy = K * Rz$, where $Rgy$ is the rate of turning on the global Y axis (the banked turn's angular speed), $Rz$ is the roll angle of the aircraft, and $K$ is a constant? And if I can do so, how do I calculate $K$?
It's computed with centripetal force:
$$F_C = m \cdot V \cdot \omega $$
Centripetal force is a function of bank angle $\Phi$ and lift, or mass and flight path. For a coordinated, level turn:
Vertical lift $L \cdot cos(\Phi)= m \cdot g$ and turning lift $L \cdot sin(\Phi) = F_C$, so in a steady turn
$$ F_C = m \cdot g \cdot tan(\Phi)$$
$$\omega = \frac{g}{V} \cdot tan (\Phi) \tag{Steady Turn}$$
A more general equation, also valid for 90 degree bank angles and not concerned with the aircraft staying up in the air, can be found by considering horizontal forces only:
$$ \omega = \frac {L \cdot sin(\Phi)}{m \cdot V} \tag{always} $$
And then if you assume that lift L is still equal to weight $m \cdot g$ which means the aircraft is losing altitude in this turn, you're almost back at the Steady Turn equation but now with a sin instead of a tan.
The above is for SI units. For aircraft units with feet and knots, please check this answer, and here is one that introduces lift coefficient into the equation.
