Reaction of Luger barrel to torque of firing?

[stevetwo]

Originally Posted by Piet

I was thinking about the torque that will be applied on the hand(s) when a shot is fired.
Before I come to that - what is the effect of the rifling (turns per inch or cm)? I will argue that the more turns the rifling has per cm, the greater the force applied to the bullet in order to get the spinning started (the accelleration of the spinning or rotation is greater) and the bigger the torque on the barrel/pistol. I will also argue that a heavier bullet and or a hotter load wil have the same effect.
What will happen if we could "suspend" the pistol in the air and pull of a shot? Will the pistol rotate clockwise or anti-clockwise (seen from behind the pistol)? Anyone wants to deliberate?

I believe the rotational forces applied by a projectile on a gun would be minimal considering the weight of the gun itself. Also, the linear forces applied by the propellant on the projectile are going to be orders of magnitude greater than the rotational forces applied by the projectile on the bore; even small calibers can generate high pressures (e.g., a factory loaded .32ACP can be 21,000 psi [SAAMI]).

IMHO the high pressures (generating the linear force) driving the projectile down the barrel (linearly), along with the weight of the gun, are going to overcome any rotational force applied by the projectile itself.

Steve