Thank you, Johnny-Come-Lately
If I understand the links (and I think I do), BC could be 'dumbed down' to this: if a .308 has a BC of .435, that's about 125% of the bullet diameter (and 125% of the weight).
Prior to it's MRT, (half-way to its MRT) a .308 has about the same energy as though it was a .435 weighing 200 gr., if the proj. is 150 gr. (bullet wt X 125%).
At the equal distance PAST the MRT the factor for energy would be a corresponding .75% of .308 / 150 gr or the energy of a .210 weighing 118 gr.
I would guess, that for a .308, that it would mean it would have that energy at about 500 yds. (I don't have the chart in front of me - I DO, but I would have to go to another window - THR would probably time-out).
I don't understand the significance of barrel length on a traveling projectile once it leaves the gun. The bullet has no memory: if a .223 leaves the barrel at 3,000fps., given its twist, the barrel length no longer affects the behavior of the round. It affected it in giving it it's muzzle velocity; if it took 16" or 23" that's history.
I don't know, but if a given bullet wt reaches a speed due to barrel length, would the round, at it's given speed vary from a longer barrel that produced the same speed? If the shorter barrel had a tighter twist, wouldn't it produce more stability - hence, better accuracy at a greater distance?
My 'dumbed-down' explanation seems to crunch the numbers. If it is way off as a definition of BC, please (you who seem to understand this equation) tell me why I'm wrong in the simplest terms you can better use.