Well basically (physics student mode on):
damage = energy transfered from the round to the target (more or less)
Now assuming if a hit occurs then the same proportion of the projectile's energy wiill be transfered for every type of bullet (which isn't of course the case in real life).
Damage = Energy transfered = some fraction of total energy
and
Total energy = (Kinetic Energy + Explosive Energy)
(ignoring secondary effects like fire from incendiary rounds)
Only cannon shells have explosive energy (obviously)
so for machine guns
Damage is proportional to kinetic energy
and Newton will tell you
Kinetic energy = 1/2 * mass of bullet *
velocity of impact squared The first point to note is that aparently speed (or in fact velocity) has a much higher effect on damage than mass. .50 cal rounds have a much higher muzzle velocity than 7.7mm rounds and also
.50 cal rounds are much larger (greater volume) for a similar density (assume both bullets are mostly lead) and so have greater mass as well. Therefore .50 cal rounds are likely do do a lot more damage.
Once you get to cannon you have to add in the explosive energy (though they tend to have lower muzzle velocities) and you also have to consider that they are much less likely to overpenetrate (i.e. travel right through the airframe) and so are likely to transfer more of their energy and thus do more damage.
There will be a test later
