Travis Atkinson Hey are you happy with your choice of cartridge (300RUM)? Have you considered 6.5 Creedmoor? What are your thoughts?

Travis Atkinson hard to read your comment because of no full stops.

Matt Hauk a lot has happened since this post have learned and made it to the top of the professional ranks shooting PRS. Do get the 6.5 creed!!!

Aaron wow.. you get an A, for good plagiarism ! so if u really wrote this, explain more about, pioberts law please. I would seriously like to know more on the issue!! ;)

narada314 that happens in automatic pistols. the springs are weaker than the gas propellant. thus the pistol cycles. the ar I built the gas system is towards the end of the barrel. t The gas escaping the bullet (gas leaving the muzzle before the bullet) is not enough to cycle until the bullet has past the gas system So when the weapon cycles the bullet is long gone.

It is just pressed in in most cases. Tight fit, and the bolt face holds it in while firing.

head space is the the distance between the head of the case (or the base, or the bolt face on the rifle) and the point on the case that contacts the chamber and stops the case's forward motion (which usually corresponds to the case datum). With a particular chamber the headspace is a constant. Free bore is the distance from the point on the bullet that will contact the lands (rifling) which is called the ogive. Free bore is the distance between the bullet ogive and the lands when the cartridge is chambered and is dependent on how deep you seat the bullet. You can seat the bullet far out of the case until it contacts the lands ( a common bench rest technique) or you can seat the bullet deeper in the case and your free-bore will increase. Caution... zero free bore is an advanced technique and can lead to over pressure!

You have a gross misunderstanding of physics here. You seem to grasp conservation of momentum just fine; my math for an M16 rifle moving back due to conservation of momentum with regards to the bullet+gases in one direction and the rifle in the opposite direction from a system initially at rest is close to your result: (m*v)_gas + (m*v)_bullet + (m*v)_rifle = 0. Assuming the COM_gas moves at the same velocity as the bullet (pretty close, actually), we can say (m*v)_gas+bullet = -(m*v)_rifle. I achieved a little over 5ft/s for the rifle, not too far off from your 6ft/s. This is based solely on conservation of momentum, which of course is applicable here. Using your number for velocity, the kinetic energy of the rifle would be about 4.81ft*lb or 6.52J. (*EDITED* due to unit error: Previously was 155ft*lb and 210J.) "If the rifle absorbed 2/3 of the energy from the 30 grains of powder, you'd have 2705 joules going into the rifle and much of that would slam into your shoulders." No, much of that would be in the form of a host of losses. About 90%, in fact. The problem is that you try and apply conservation of KINETIC energy in the incorrect manner instead of conservation of TOTAL energy. Kinetic energy is in fact NOT conserved; not even close. A rifle firing is a very inefficient form of inelastic collision except in reverse. Much energy from the combustion of the powder is transferred to the rifle in the form of deformation, aka compression waves, vibrations, harmonics, barrel flexing, as well as the energy used to cycle the action (in semi-auto firearms like the M16) where you have damping effects, spring losses and gas redirection. Some energy is also transferred to the rifle in the form of heat. Finally, your suggestion that the energy absorbed by the rifle of 1995 ft*lb is equivalent to dropping a 199lb weight 10 feet onto your shoulder is a gross misunderstanding. By your logic, the kinetic energy of the bullet of about 1330 ft*lb would be equivalent to dropping a 133lbm object (about 5 cinder blocks) 10 feet (1 story!) onto a target. This is enough force to crush in a car roof, yet in reality, the bullet barely makes a dent in steel plate. So what gives? You are trying to equate momentum and energy, which are completely different.

Mila McShanks Recoil calculation is entirely based on conservation of momentum. You can google it and find the formula for computing free recoil. Here's a calculator site that gives the numbers I give. www.shooterscalculator.com/recoil-calculator.php The gas velocity (average) of the exploded grain is higher than the velocity of the bullet. Most calculators use something around 4500 ft/sec to 5000 ft/sec. Also, I'll conceded I made an error in the second paragraph. The momentum of a bullet is much lower than a 199 lb weight with equivalent energy. Still, the rifle does not absorb anything close to 2/3 of the kinetic energy. The rifle absorbs about 1/340th of the energy of the bullet. That's a fact that anybody can look up.

Mila McShanks "Using your number for velocity, the kinetic energy of the rifle would be about 155 ft*lb or 210J." You made a unit mistake. You have to use 3.538 kg for the rifle weight and 1.92 m/s for velocity. Kinetic Energy is 0.5 * 3.538 * 1.92 * 1.92 = 6.52 joules. When you're using funky lbs and ft/sec units, you have to divide by 32 ft/sec^2 to get the correct ft*lb value. When you use kg and m/s, you only need to divide by 1 m/s^2 to get joules.

***** "Recoil calculation is entirely based on conservation of momentum." I know, I agreed with you on this. "The gas velocity (average) of the exploded grain is higher than the velocity of the bullet." While it is true that some gases are expelled at higher velocity than the bullet, you have to look at the center of mass of the entire mass of gas. Unless most of the gas mass is expelled BEFORE the bullet (not true), then the majority and hence the center of mass must be BEHIND the bullet, meaning the bulk of the gases and hence the COM of the gas a lower velocity than the bullet. You have to keep in mind that the gas is expanding in a piston (with some leak-by of course) with a non-moving wall and a moving wall, and the majority of the gas is between these walls. Meaning that the COM of the gas behind the bullet will be moving in the same direction but at approximately half of the velocity of the bullet. When you consider leak-by past the bullet, then the mass of gases is moving faster than the half the bullet velocity, but the only way for the COM of the ENTIRE mass of gas to move faster than the bullet is if most of the gas is expelled before the bullet leaves the barrel, and/or if the gas vacates the space near the chamber, moves in the same direction of the bullet, and compresses instead of expanding. This is not the case, obviously. Now if you take the bullet out of the equation, then yes, the gas will travel much faster than the bullet would. Draw a free body diagram of the gas expanding inside the piston/cylinder (this is all a bullet in a barrel is) and figure out where that COM for the gas mass goes. "The rifle absorbs about 1/340th of the energy of the bullet. That's a fact that anybody can look up." Correction: recoil energy is about 1/340th of bullet energy. The rifle absorbs a lot more energy than that and releases it in ways other than recoil, as I already listed above. There is a heck of a lot more than kinetic energy in the form of recoil involved here. There is about 3MJ/kg stored energy in gunpowder, so 30grains will hold about 5,832J of energy. The gas/bullet has about 1,974J kinetic energy, or about 1/3 of the energy stored in the powder. Where does the rest go? Only a small amount goes to recoil due to conservation of momentum, as we have already discussed and agreed. The rest gets transferred in the ways I listed previously: compression waves, vibrations, sound waves, heat, friction (more heat), action cycling (semi-autos), etc. Think of a head-on car accident: in a perfectly inelastic collision, 2 equal cars moving towards each other at equal speeds will hit each other and stop. Momentum (a vector) is conserved, kinetic energy (non-vector) is NOT conserved, it is transferred to deformation energies, breaking of bond energies, sound waves, vibration waves, harmonic waves and shock waves, etc. This is why crumple zones are so effective. "You made a unit mistake." You are absolutely right, I have edited with a note. I much prefer using metric units, and jumping around unit conventions is a great way to introduce mistakes in calculations.

CUP stands for Copper Units of Pressure and is a measure of how much a standardized copper crush gauge is deformed. That is... it's more of a unit of pressure, but yes it is --only-- a measure of the peak pressure.