How to predict the cramped frequency of a digital filter based on an analogue frequency? I have a second order analogue high pass transfer function (unity gain at infinity). It's magnitude response hits the -20 decibel line at a frequency of 5706 Hz (the corner frequency is 18000 and sample rate is 44100). When I convert this analogue filter to a digital IIR via the BLT method, the digital filter's magnitude response hits the -20 line at 11149 Hz. This will happen with the standard RBJ audio EQ cookbook formula i.e. it's not a problem with my maths. I know that digital filters suffer from cramping and that's what I'm interested in. Is there a formula to predict what an analogue frequency will be cramped to when the digital filter is realized given the analogue filter's frequency? i.e. something that will tell me 11149 given 5706. Just to be super clear I'm not asking how to do pre-warping using: $ tan(\frac{w_c}{2}) $.
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:

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