You don't actually remove the forcing cone, but you knew that. Some folks refer to it as "removing the forcing cone". It is a misnomer. You only remove enough to reduce the angle making the transition from chamber to bore less abrupt which (a) reduces the tendency for the shot to spill and become deformed, which results in blown patterns, and (B), because the transition is smooth and less dramatic, less energy is needed to make the move, and we all know that less expended energy equals less recoil. Which is certainly not "snake oil". And, look at any modern trap gun, their forcing cones are what?
All of you that think like Malm should try this thought experiment. The laws of physics require that the recoil is determined by 3 numbers: Mass of the gun, mass of the ejecta, muzzle velocity of the the ejecta. At the risk of beating a dead horse here's the equation:
Gun velocity( recoil) = [Mass( ejecta) X Velocity (ejecta)]/ Mass of gun.
This is the relation that determines the operation of a ballistic pendulum. The gun is suspended horizontally on 2 wires and fired remotely. The height that the gun rises to in recoil is proportional to the recoil energy. Nothing that happens inside the gun before the ejecta leaves the muzzle has any effect because the gun is a free body and there are no fores acting on it externally. In essence, the gun is like a bomb until the ejecta is blown out of the muzzle.
I say that if the two cases of a standard barrel and a lengthened forcing cone barrel are fired with identical muzzle velocities and ejecta masses, the gun will recoil exactly the same with either barrel. If you say not, what is the exact reason for your contention.
RAN