Normalized collision energy calculation for Q Exactive

I get this question all the time.  And I got the answer when I was in Bremen 2 summers ago, and always mean to put it up here:

Q:  What is the formula the Q Exactive uses for normalizing the collision energy?  Or how do we go from a NCE setting to the actual electron volts used for each fragmentation?

I'm going to give you the equation and explain it as I understand it (I'm a biologist, so chill out):

The formula:

Absolute energy (eV) = (settling NCE) x (Isolation center) / (500 m/z) x (charge factor)

Charge factors are as so:
Charge/charge factor
2/0.9
3/0.85
4/0.8
5/0.75
>5/0.75

So.... if you set the QE to do a NCE of 40 and you hit a 1,000 m/z ion with a charge state of 1
(40) x (1,000)/500 x 1 = 80 eV

Now, this was the equation for the original QE and good for the software version that was out 2 summers ago (August,2012).  I can't speak to how this equation has been adjusted or changed for new software versions or for the QE Plus, but it at least gives you a good idea, right?