Simple math on it - or rather simple PHYSICS on it: 3000fps, 200yrds = 600ft, neglecting any velocity loss, 600ft/3000ft/sec = 0.2 sec ToF. Drop due to gravity is -32.17 feet per second per second, and the formula to calculate non-propelled gravitational drop is = 1/2 * G * ToF^2 (where G is -32.17ft/s/s and ToF = time of fall) = .5 * -32.17 ft/sec/sec * 0.04 sec*sec * 12"/ft = -7.7" total drop at 200yrds. 220yrds = 660ft, 660/3000 = 0.22 sec ToF, .5*-32.17*0.0484 * 12 = -9.3" total drop at 220yrds. So between 200 to 220yrds, that's about an inch and a half of drop. A guy can make a small correction that for 100 or 200yrd zeroes, you'll peak somewhere between 75-125yrds, so call it 100yrds, so the bullet only starts dropping there. So if you cut off the "rise" part of the trajectory, then you get the drop from 100 to 200 and 100 to 220yrds = 0.1sec vs. 0.12sec, yielding -1.9" vs. -2.8" for 200 and 220yrds respectively. Good for about an inch of drop over those 20yrds - obviously, real world bullets slow down in flight, so those ToF's are too short, and the drop would be slightly more. This math doesn't care about zero'd range, only about Time of Fall.

Seems to be pretty much exactly one inch for a 300wm from 200-220yrds, whether you have a 100yrd zero or a 200yrd zero...