[avg_freeper]

The video at the site says there's no recoil. That must mean that mass leaves in at least two directions. If the recoil is very small then there is very little mass projected, the exit velocity is small, the weapon mass is great, or some combination of the above.

Nothing in the centrifugal concept would allow them to violate f=ma.

[MeanWestTexan]

"Nothing in the centrifugal concept would allow them to violate f=ma."

I would think the more accurrate discription would be the recoil would be more of a steady "push" vs. bang!(recoil); bang!(recoil).

Helps with post-first-shot aiming, I suppose, as you are really firing a stream of rounds.

[Steely Tom]

Well, there is recoil, it's true, but it happens when the projectile is dropped into the center of the spinning mechanism and slides out along whatever guide or tube takes it to the outer radius. There would be no recoil at the moment of release.

The recoil would, I imagine, show up as vibration, as the rotor mechanism would intermittently be out of balance. Of course, this could be compensated by some sort of movable counterweight.

My point is, the recoil would not necessarily be in a direction opposite to that of the projectile. A fine point.

(steely)

[SlowBoat407]

The video at the site says there's no recoil. That must mean that mass leaves in at least two directions. If the recoil is very small then there is very little mass projected, the exit velocity is small, the weapon mass is great, or some combination of the above.

I'm not a physics whiz, but it seems all you'd need is something to absorb the counter-rotational reaction. A flywheel maybe?

[mnehring]

If this is a centripetal weapon, doesn’t this mean theoretically the opposing force of the exit volicty would be distributed along the entire centripetal loop (or path), thus giving a great force with little true recoil as it is spread evenly around the circle?

[c-five]

The "recoil" is angular, a torque impulse, in the opposite direction of the spin. Also, as Tom says, it occurs at load or spin-up, separated in time from the release. There would be little or no recoil at the moment of release.

There could be considerable vibration when loading, but if the projectiles distributed evenly in chambers around the circumference, there need not be a lot of vibration when loaded but not firing.

[Mark Felton]

"The video at the site says there's no recoil. That must mean that mass leaves in at least two directions."

The mass is already accelerated before it is released. When released there is no acceleration of the balls. They are already in full speed motion.

Before being released, the centrifugal force is equally and oppositely counteracted by a centripetal force which holds the ball bearing in place (centripetal prevents the ball bearing from flying out)

When the balls are released the centripetal force goes to zero and the opposing centrifugal force goes to zero but the ball bearing continues its motion tangentially without any real loss of energy, or change in total inertia (ideally).

Thus when the bearings are released there is no loss of energy in the system (or gain of energy as in a shell explosion) so there is no recoil.

[Note: It may be difficult to hold the aim on target since the angular torsion will vary as the mass is unequally released from the circumferance of the spinning device.]

[IronJack]

There is no recoil because there is no force imparted along the exit line save for the centrifugal force generated by the spin. The reaction force would come when the ball bearings were spun up to speed, and that would come in the form of counter-torque.

In other words, at the time the projectile is released, the propelling force has already been imparted to it. So there would be no recoil.