How to predict the cramped frequency of a digital filter based on an analogue frequency?
I've come up with a solution.

Let $ f_a = $ analogue frequency, $ f_d $ = digital frequency, $ f_s = $ sampling rate and $ f_c = $ corner frequency with $ \omega_c = \frac{\pi f_c}{f_s} $ and $ \omega_a = \frac{\pi f_a}{f_s} $ then:

$$ c = \frac{\omega_c}{\tan(\omega_c)} $$

$$ \omega_d = \arctan\left(\frac{\omega_a}{c}\right) $$

Giving:

$$ f_d = \frac{\omega_d f_s}{\pi} $$