Originally posted by William:
Look at it like this:
a guass rifle masses aprox. 4 kg
a guass rifle magazine includes 40 projectiles and the power source for a total of 300 g. Without subtracting that battery, the most the 4mm projectile could mass is 7.5 g. Most likely signifcantly less. There is no chemical explosive. Only an electromagetic pulse pushing a sliver of metal.
Yes Newton is still there. But do you see why I say that the recoil is insignificant? I doubt that anyone with the strength to be able to pick up a guass rifle in the first place will be able to notice the recoil; from a purely game mechanics level that's not the same as an AK-74. Hence, IMTU, I would assign 0 to the guass rifle for recoil.
All right, now the
REAL math begins.
Lets assume the mass of the bullet is 5g and the gun is 4000g.
Now, we need a muzzle velocity. I don't have handy a number for the expected muzzle velocity for a Gauss rifle - all of the comments in TA#1 and T20 Lite just say High velocity. I'm also not a gun-expert, so I don't know what high vs. low velocity means. Let's assume a muzzle velocity of equal to Speed of sound, which in dry air is 331.5m/s.).
Thus firing a round generates 1657.5 g-m/s or 1.658 kg-m/s of momentum (mass x velocity).
The 4kg rifle sees the same amount of momentum in the opposite direction. The recoil force on the shooter is determined by how long the transfer of momementum takes. Again, not being an expert, I don't know how quick this is. So lets look at a range of values.
The bullet is accelerated from the gun in 0.0045 seconds ((d=1/2 vt OR t=2d/v - constant acceleration and barrel length of 0.75m)), so the recoil on the gun can't be any faster than that.
Ft=mv so F=mv/t or 1.658 kg-m/s / time (sec) with the answer in kg-m/s^2 or Newtons (4.445 N=1 pound force)
So:
for t=0.0045 sec, f=368.3 N, or 82.9 lbs-force
for t=0.01 sec, F=165.8 N or 37.3 lbs-force
for t=0.1 sec, F=16.58 N or 3.7 lbs-force
But its not an issue of force - its momentum. So the same momentum value gets transferred to the shooter - lets assume 100kg basic mass and another 20kg of stuff (armor, teddy bear, etc.). That means that the shooter will see a net velocity of 0.014 m/s or ~0.54 inches/sec.
Doesn't seem like much, but if you are in zero-g, it will cause you problems if not properly braced (hence the Zero-g Familiarity Feat). This only gets worse if you use auto-fire - a 4-round burst generates another 2 inches/sec.
This number may seem small, but take out a ruler, and move your finger across 2 inches in a second - now imagine your whole body moving like that.
And this is a simplistic point of view - the force of the rifle isn't directly through the shooters center of mass - it at one corner, so the momentum becomes an angular momentum as it twists the shooters frame. The first mass accellerated is the shoulder, which has much less mass than the entire body, so the actual backwards velocity of the shoulder would be higher, more like 8-12 inches per second - now that's a kick.
The above are very simplistic calculations - probably over-simplified given the complex mechanics of the human body in terms of how the momentum is transferred. Also the assumptions on muzzle velocity obviously drive this train.
The point is not to pin a real value on the momentum, but rather to show qualitatively that the recoil is non-trivial in a zero-g environment.
