What works for me
For me the following works pretty good in my head (though mathematically probably not exactly correct):
Based on the nautical miles per minute e.g. 540 kt equals 9 nm/min (or mach number at higher levels).
1 degree change of flightpath times 9 equals 900 ft/min descent /climb rate.
To calculate the flightpath-angle: Altitude to loose in FL (e.g. 20000ft 200FL)
divided by distance to go e.g. 50nm equals 4 degrees.
So if you want to loose 20000ft in 50nm at a speed of 540kt (or .9mach) you need 4 degrees nose down which gives 3600 ft/min descent rate.
Cheers