I guess the speeds here above are given in feet per second...
From that, just a matter of energy calculations: the energy of the ball (kinetic energy), compared to the energy of the energy of the hull (potential energy).
The simplest one is the energy of the ball: E1= m v² / 2, where m is the mass of the ball (pd/0.454),and v the impact speed of the ball (in meters per second)
The potential energy of the hull resistance is proportional to the cylinder to be dug in the hull. This cylinder is disformed by the impact angle, its length is equal to thick / cos(angle).
Its diameter is impacted by angle too. It is equal to the ball diameter divided by cos(angle).
The ball diameter can be calculated, assuming the density of steel (7.8), knowing that the volume of a sphere is equal to 0.75 x pi x r3, we get pd= 7.8 x 0.75 x pi x (d/2)^^3, thus d= (8 x pd / 7.8 / 0.75 / pi)^^(1/3).
So, we have
The ball energy = K1 x pd x (impact_speed)².
The hull energy = K2 x pound^^(1/3) x thick / cos²(angle).
K1 and K2 to be balanced, and maybe adding some randomisation (hull being weaken by previous impact that did penetrate or not, presence of structural wood at the impact location, etc)