I saw the answer to airspeed and rate of turn but this I think isn't what I want. I've seen it many times that aircraft will pitch up much slower when at higher speeds particularly fighter aircraft. What is the mathematical relation of these two characteristics if any?
I am trying to make a game about fighter aircraft and want to find a realistic solution to this problem.
A constant speed pull-up is basically a circular motion, albeit in vertical plane:
$$q=\frac{(n_z-1)g}{V}$$
where $q$ is pitch rate, $V$ is true airspeed assuming no wind, $n_z$ is the load factor (lift to weight) and $g$ is gravitational acceleration.
The amount of load factor is limited by the airplane and its flight condition. As you can see, the faster the speed, the smaller is the obtainable pitch rate for a constant load factor.
Note: a constant speed pull-up is a theoretical idea that only approximates a pull-up at the very beginning, before the airspeed begins to decay. But the conclusion is valid.
If the aeroplane flies straight and level, and the pilot pulls on the stick in a step input, the elevator deflects => increased Angle of Attack => increased lift. The aeroplane climbs while continuing to rotate backwards, trading kinetic energy for potential energy.
In the situation depicted above, the centripetal acceleration is $V \cdot q$, and hence: $$N - W = m \cdot V \cdot q$$. So for a given elevator deflection and therefore increase in N, at higher V the achieved q is lower.
As a guideline for maximum n = N/W, from this wiki page:
