LugerForum Discussion Forums - View Single Post - Cartridge power

unspellable · 12-06-2002, 02:19 PM

Ok, we'll try this again.

Consider a closed system consisting of a tube with a spring inside that we will use to launch a ball bearing. You will note that there is no "chamber pressure" involved. Before launch, nothing is moving, the momentum of the system or any of its parts is zero, regardless of the mass. We will say the ball bearing has a mass of one ounce and the tube with spring has a mass of 10 ounces.

Now we launch the ball bearing at 100 fps. It is moving towards the target and has a momentum equal to its mass times its velocity or 100 oz*ft/s. (Non-standard unit, but we'll worry about that later.) Now the momentum of a closed system is constant and therefore the momentum of our system is still zero. Since the ball bearing, a part of the system, has a non-zero momentum, the tube with spring must have a momentum equal in magnitude but opposite in direction or sign. Now 100 oz*ft/s divided by 10 oz equals 10 ft/s, the velocity of the tube in the opposite direction.

Now if we knew the expected velocity of the ball bearing before we launched it, we would say that it had an impulse capable of imparting a momentum to the tube of 100 oz*ft/s.

Impulse and momentum are mathematically the same thing, for convenience we say momentum when talking about a moving object and we say impulse when we are talking about the ability of something to impart momentum. Momentum is expressed in units of mass x velocity. Impulse is expressed in units of force x time. A little algebraic manipulation will prove they are one and the same thing.

In the case of a cartridge's ability to impart momentum to the firearm, we speak of the cartridge impulse. It does not involve the weight of the firearm. (Recoil is another story, not to be confused with momentum, it takes into account the weight of the firearm.) In the case of the cartridge we must take into account not only the bullet, but any other ejecta. (Powder gas, birdshot, wads, the crud you didn't clean out last weekend, etc.) In the case of a Luger cartridge, we have to consider the powder gas. There will be no birdshot or wads, and we all know that everyone on this forum is faithful and fastidious about cleaning.

For the bullet, it is simply velocity times mass. At a given velocity, a bullet twice as heavy will produce twice the impulse. The powder gas is also part of the ejecta and must be taken into account. It's tricky to measure the velocity of the gas so we generally use one of several rules of thumb to get a decent approximation. The one I have been using for plain barreled closed breech pistols (Not revolvers, or pistols with compensators.) is the average gas velocity will be one and a half times the bullet velocity.

So impulse equals the bullet mass times the velocity plus the powder mass times one and a half times the velocity.

Or impulse = ( Mb + 1.5*Mp ) * v

Note that barrel length has a small effect on velocity and hence impulse. To compare cartridges and loads we must agree on a standard barrel. Also, the usual units are not very consistent so we have to throw in some constants to convert units.

For Impulse in pound force seconds, bullet and powder mass in grains, and velocity in feet per second we have:

Impulse = ( ( Mb + 1.5 * Mp ) * v ) / ( 7000 * 32.174 )

The numbers under the divisor are constants to convert to consistent units, 7000 gr to a lb, 32.174 lb to a slug.

For a hypothetical 7.65 Parabellum load with 5 gr powder, 93 gr bullet at 1220 fps.

Impulse = ( 93 + 1.5 * 5 ) * 1220 / ( 7000 * 32.174 ) or 0.5444 lbf*s

Now all this happens in less than a millisecond so the peak force on the cannon is in excess of 500 lbf. The cannon will weigh a around 14 oz so it will have a velocity of about 20 fps. In a perfect world this would be just enough to fully compress the recoil spring and no more.

You will note that the momentum of the cannon is largely transferred to the frame via the recoil spring so the action is delayed and spread over time. This is the reason that while a recoil operated pistol has the same true recoil as a single shot pistol or revolver, it seems much less as it is spread out over several milliseconds.

I hope this is all at least as clear as Mississippi River water.

12-06-2002, 02:19 PM	#12
unspellable User Join Date: Jun 2002 Location: Iowa Posts: 768 Thanks: 0 Thanked 19 Times in 11 Posts	Ok, we'll try this again. Consider a closed system consisting of a tube with a spring inside that we will use to launch a ball bearing. You will note that there is no "chamber pressure" involved. Before launch, nothing is moving, the momentum of the system or any of its parts is zero, regardless of the mass. We will say the ball bearing has a mass of one ounce and the tube with spring has a mass of 10 ounces. Now we launch the ball bearing at 100 fps. It is moving towards the target and has a momentum equal to its mass times its velocity or 100 ozft/s. (Non-standard unit, but we'll worry about that later.) Now the momentum of a closed system is constant and therefore the momentum of our system is still zero. Since the ball bearing, a part of the system, has a non-zero momentum, the tube with spring must have a momentum equal in magnitude but opposite in direction or sign. Now 100 ozft/s divided by 10 oz equals 10 ft/s, the velocity of the tube in the opposite direction. Now if we knew the expected velocity of the ball bearing before we launched it, we would say that it had an impulse capable of imparting a momentum to the tube of 100 ozft/s. Impulse and momentum are mathematically the same thing, for convenience we say momentum when talking about a moving object and we say impulse when we are talking about the ability of something to impart momentum. Momentum is expressed in units of mass x velocity. Impulse is expressed in units of force x time. A little algebraic manipulation will prove they are one and the same thing. In the case of a cartridge's ability to impart momentum to the firearm, we speak of the cartridge impulse. It does not involve the weight of the firearm. (Recoil is another story, not to be confused with momentum, it takes into account the weight of the firearm.) In the case of the cartridge we must take into account not only the bullet, but any other ejecta. (Powder gas, birdshot, wads, the crud you didn't clean out last weekend, etc.) In the case of a Luger cartridge, we have to consider the powder gas. There will be no birdshot or wads, and we all know that everyone on this forum is faithful and fastidious about cleaning. For the bullet, it is simply velocity times mass. At a given velocity, a bullet twice as heavy will produce twice the impulse. The powder gas is also part of the ejecta and must be taken into account. It's tricky to measure the velocity of the gas so we generally use one of several rules of thumb to get a decent approximation. The one I have been using for plain barreled closed breech pistols (Not revolvers, or pistols with compensators.) is the average gas velocity will be one and a half times the bullet velocity. So impulse equals the bullet mass times the velocity plus the powder mass times one and a half times the velocity. Or impulse = ( Mb + 1.5Mp ) * v Note that barrel length has a small effect on velocity and hence impulse. To compare cartridges and loads we must agree on a standard barrel. Also, the usual units are not very consistent so we have to throw in some constants to convert units. For Impulse in pound force seconds, bullet and powder mass in grains, and velocity in feet per second we have: Impulse = ( ( Mb + 1.5 * Mp ) * v ) / ( 7000 * 32.174 ) The numbers under the divisor are constants to convert to consistent units, 7000 gr to a lb, 32.174 lb to a slug. For a hypothetical 7.65 Parabellum load with 5 gr powder, 93 gr bullet at 1220 fps. Impulse = ( 93 + 1.5 * 5 ) * 1220 / ( 7000 * 32.174 ) or 0.5444 lbf*s Now all this happens in less than a millisecond so the peak force on the cannon is in excess of 500 lbf. The cannon will weigh a around 14 oz so it will have a velocity of about 20 fps. In a perfect world this would be just enough to fully compress the recoil spring and no more. You will note that the momentum of the cannon is largely transferred to the frame via the recoil spring so the action is delayed and spread over time. This is the reason that while a recoil operated pistol has the same true recoil as a single shot pistol or revolver, it seems much less as it is spread out over several milliseconds. I hope this is all at least as clear as Mississippi River water.