Originally Posted by mcguyver
Remember the gyroscopic effect of hop-up on a BB. The heavier it is, the more energy it has, hence the less affected by gravity. It will "appear" to the naked eye to fall to the ground slower, all the while travelling straighter.
It's not gryoscopic. It's the Magnus effect which explains why hop-up works.
Gravity is constant not matter what mass object; it does not have less of an effect because a mass is greater or lesser than another. You calibrate hop-up for a straight trajectory where in effect, you are adding a force with an acceleration of 9.81m/s^2 to counteract that of gravity's.
The only reason why a heavier BB "could" retain it's hop-up effect better than a lighter BB, and therefore counteract gravity better, is due to it's mass being greater such that the force of friction from the surrounding air is not able to impart enough force to decelerate its spin when compared to a lighter BB.
I.E. Backspin is reduced faster on a BB with lighter mass when compared to a BB with heavier mass.
This explanation itself holds true to why a heavier BB retains more kinetic energy than a lighter BB when both energy outputs are equal; air resistance has less effect on an object with more mass, therefore it decelerates less than an object with less mass.
Less deceleration means it retains more of it's velocity in a given period of times than a lighter BB, and so it will retain more energy, and that dictates impact power.
It's not due to the momentum or anything like that, those are only relative to the reason and definition as to why more mass retains more energy. It's due to air resistance that a heavier projectile will
retain more energy than a lighter projectile.
Another thing to note is that a sphere traveling faster experiences more drag (air resistance) than a slower traveling sphere; that would also be indicative of why a heavier BB will keep its velocity, therefore its kinetic energy, better.
An equation to relate why velocity and mass dictate air resistance, and how that relates to the retention of energy:
a = (Cd * r * V^2 * A / 2) / m
a = deceleration due to air resistance
Cd = coefficient of drag for a smooth sphere, being 0.5
r = density of air, being 1.229 kg/m^3 at sea level
V = velocity, being variable
A = cross-sectional area of the ball, being 2.826e-5 for a 6mm BB
m = mass of BB, being variable
Simplifying the equation yields:
a = (8.68e-6 * V^2) / m (at sea level)
This shows the relationship of velocity and mass to air resistance, and that an increase in velocity or a decrease in mass will increase air resistance. That causes more deceleration, and more loss of energy.
(This equation is for an instantaneous calc. of deceleration with a given velocity, as the velocity decreases for the next calc. of deceleration since air resistance is continuous and therefore velocity changes continuously)
So even if two different mass BB's were fired with the same Joules and with their corresponding velocities, the heavier BB will retain more energy and have more "impact power".