I am always suprised when someone reminds me that Pres. Obama smokes cigarettes or shows me a picture like the one below, wherein I realize, again, that he is left-handed. The White House released this 2012 picture to confirm that the president shoots skeet. I do a bit of that myself (with an Ithaca 20-guage double-barrel, modified over improved).
Maybe because I am a very inexperienced shooter, I don’t know what I’m seeing, but in the picture below there appear to be two smoke plumes coming from the president’s gun. The one coming from the muzzle is, obviously, the smoke that emerges right after a shot. But what’s the second plume, apparently at right angles to the first, that looks as though it is coming from some kind of vent in the side of the barrel? There’s certainly no opening or other potential source of such a plume on my Ithaca. If anyone reading this knows what it is, please let me know in the comments.
It’s ported to reduce muzzle rise and to reduce perceived recoil. Along with the bullet/shot being projected out the barrel, a lot of gas exits the muzzle at high velocity. The gas is moving so rapidly that, even though it has minimal mass, it represents a fair amount of momentum. The port redirects the momentum of the gas. This redirection exerts as force on the gun as the gas “turns the corner”. The exact vector of this force depends on the exact configuration of the ports, but one commonality is that they all have some component of their vector that points forwards in direct opposition to the recoil vector. This has the effect of “yanking” the gun forwards away from the shooter. So the felt recoil by the shooter becomes F_recoil minus F_port, reducing the felt recoil. If the ports are radially symmetric about the barrel, the net force is directly forwards, counteracting a significant portion of the recoil. If the porting is asymmetric, there’s a component of the vector that also pushes the gun in a direction opposite the porting. Because the barrel is typically above the point of contact with the shoulder of the shooter, the recoil impulse tends to cause the muzzle to rise as a torque reaction. If we port the top of the barrel, we get a resulting vector that points both forwards and up, counteracting both the recoil impulse and muzzle rise.
In a skeet situation, it means your followup shots can be quicker due to the gun staying on target better and knocking you around less.
Also, I came to your site because of the UV-3R testing, thanks for doing that.
Thanks, williaty. That’s a great explanation. Been a long time since I looked at the innards of a shell, but my recollection was that it is stacked (from the muzzle end down) with shot, wadding, and propellant. Doing a little basic physics with your explanation, and assuming no gas leaks past the wadding (which I think is probably a bad assumption, but let’s start with it), the momentum of the gas will be equal to the mass of the propellant (assuming the propellant burns completely into gas, and that it doesn’t change mass significantly as it burns by bonding with oxygen), times the muzzle velocity. (Now, the actual net momentum of the gas is probably half of that, since it forms an expanding cloud that is bounded at the shell; that is, some of it barely moves at all. However, we are only interested in the gas at the port, which is close to the muzzle in the picture, and that gas is probably moving at pretty close to muzzle velocity.)
The question then becomes: what part of the energy of the expanding gas are we willing to sacrifice from accelerating the shot and into correcting for the effects of the recoil? Put another way, how much are we willing to slow down the shot to vent the gun? Now, if I recall correctly, propellant in a typical shell has a lot less mass than the shot. To completely cancel recoil, the component of the F_port parallel to the barrel would have to equal F_recoil. Best case, if we could somehow vent 180 degrees (which we can’t, but that’s the absolute limit), that means the rearward momentum of the ported gas would have to equal the forward momentum of the shot and the forward gas. Since they all have equal velocity (again, that’s bull, for a bunch of reasons, but it’s a good limiting case to start with), we can cancel that out and say that Mshot + Mforward_gas = Mported_gas.
Now, that means that, for this to work at all, it seems that the mass of the propellant must at least be equal to the mass of the shot (although, interestingly, in that limiting case, the Mforward_gas = 0, in which case the muzzle velocity will also be zero, but that’s a typical seeming-anomaly in limiting case analyses). As we add propellant, we add more forward gas, until we get the muzzle velocity we want. However (and here I am mostly guessing, based solely on the fact that my shells all seem to have much more of their mass accruing to shot than to powder), I wouldn’t be surprised if the amount of propellant necessary to pump shot out of my gun at the velocity one needs, while simultaneously having enough to completely counteract its kick, also turned out to be enough to blow my little Ithaca apart.
The upward pull seems like something this method has a good chance of correcting. I can hold my gun in my hands without each shot flinging it into the sky, so I infer that means the right-angle component of the recoil vector is pretty small. But the sight-line component must be huge. Venting a bit of the gas in such a way as to counter that effectively seems outside the reach of Newtonian physics.
Of course, this whole exercise I’ve just cobbled up is based on a number of assumptions I admit could be wrong, and there may be other factors one could bring to bear (maybe the vent shape can accelerate that part of the gas, though I think conservation of energy laws say that can’t happen without giving up some of the mass you’d be accelerating… dunno about that one).
Anyway, thanks for cluing me in. Personally, the fewer design elements my gun has, the more I trust it. When two of the pigeons go up, I can see how shortening the recovery time (right phrase?) after the first shot would be a help, but I almost feel like a gun that copes with its own kick this way is, well… “inauthentic” somehow seems to apply. (I mean, heck, at some point, I could mount a webcam and servo-controlled gimbals on my gun, write a program to see moving targets, and get it to aim for me altogether; probably get a great score, but what fun would that be? 8-) )
Glad you liked the UV-3R test. Hardly scientific, but a useful comparison and, we thought, validation of the little Chinese cheapy as a bona fide option for actual ham radio service.
73!