Hmmm still not convinced.
Every 0.1s there is a bullet leaving the barrel. So that means there is one bullet passing any point every 0.1 seconds too. I don't see how this interval is being shortened by bullet speed at all. (If the muzzle velocity would shorten this interval, the bullets would catch up with each other if the muzzle velocity were high enough - sounds illogical to me)
So every 0.1s there is a hit on point X.
An extreme example would be a light flashing once per second. "Muzzle velocity" = speed of light. But someone 100ft away would still have 1 second to cross a certain point, and not less because the light is so freaking fast.
Okay, let me see if I get it. Every .1 seconds a round travels down range. That .1 seconds determines the time between each bullet. The velocity of the bullet determines the physical separation between each round.
Derivation:
t=.1 vsub1=500m/s vsub2=300m/s
d=vt
d=(500m/s)(.1)=50m between each round
d=(300m/s)(.1)=30m between each round
So, we have a smaller space between each round. However, what matters is frequency. How many bullets pass through a single area in any given time. Because the frequency at the barrel is equal to the frequency down-range (the velocity is constant, therefore the spacing is constant) the amount of time between each round firing is equal to the amount of time between each round passing a single point in space. Because ROF is the same, frequency is the same, and therefore you have an equal amount of time between each round regardless of velocity.
HOWEVER, Sourdaukar nailed the true problem unintentionally. It is our ability to perceive the placement of the round downrange. Consider the extremes of a laser versus a slingshot. The laser is the 500m/s round, the slingshot is the 300m/s round. The longer the time between the bullet leaving the gun (point A) and the bullet arriving at the intersection of bullet path and target (point B) determines how much we must lead the aircraft.
As we all know, t=d/v, therefore, assuming the target is 300m away from you (Reasonable snapshot distance in AH), we get the following equations:
t=v/d
t=(300m)/(500m/s)= .6 seconds between the leaving the gun and arriving at the target.
t=(300m)/(300m/s)= 1 second between leaving the gun, and arriving at the target.
We have two factors here, lead, and pilot reaction. The 500m/s round gives the target only .6 seconds to move out of the line of fire. The 300m/s round gives the target a full second to leave the point of impact. So, the 500m/s round must lead the target by .6 seconds of flight time for the target; The 300m/s round must lead by 1 second. Assuming flight time for the targets are equal, and d=vt so, derivations applied assuming the aircraft is moving at, say 200m/s (Not realistic, just a random number),
d=vt
d=(200m/s)(.6)= 120m lead
d=(200m/s)(1)= 200m lead
So, you must put your line of fire 120m in front of the target with a 500m/s round, and 200m for the 300m/s round. As we know, the farther we must lead, the more time we must lead by, the harder it is to predict where the plane will be. The shorter the lead, the easier the snapshot, and the less thinking necessary for the shot, therefore, less lead=easier kill.
So, long story short, while the 500m/s round is not MATHEMATICALLY more effective, PRACTICALLY it is much easier to put the round on a moving target than it is to use a 300m/s round.