Cartridge power - reduced loads

[Stu]

There's been a lot of talk about the Luger needing a stouter load than some other weapons. It seems this is particularly true of the 30 Luger with reference to modern loadings.

I was browsing and came acroos the following statements from the Swiss tests of 1899:

"For both pistols, the following tests were executed:

* Fire test with reduced loading (10%), 15 shots, followed by 47 shots with 20% load reduction, where both pistols worked correctly.
* 32 shots with no grease in the breechblok: both pistols ok.
* Tests with cut cases and limited in length, side-wise, and obliquely. The Borchardt-Luger worked correctly, while for the Mannlicher, two shots out of six where the breechblok does not open at all, one case split.
* Tests with sand and water: 16 shots are fired with heavy dust on the weapons, and 16 other shots are fired after the pistols have both been soaked with water. Both weapons function correctly."

I just thought it interesting that back then the gun worked fine with a 20% load reduction, how does that reflect on our modern ammo ?

The actual web site makes interesting reading for those of us without a decent library

http://www.regnier.ch/pages/BorchardtLugerHistory.htm

[unspellable]

I have run into this before. It doesn't really add up and is one of the reasons for starting a research project into the 7.65 mm Parabellum cartridge.

Just as a back of the envelope calculation, NO recoil operated pistol of any make or model should be expected to work with a stock recoil spring and a powder charge 20% below nominal. For the 7.65 Parabellum this would result in a velocity around 800 fps and an impulse around 63% of nominal. (Admittedly a very approximate calculation.)

No doubt there are a few pistols out there that will function with this sort of load, but they are the exception to prove the rule. A pistol could be made to work this way by choosing a stock recoil spring that is lighter at the expense of increased stress and wear with the nominal load.

Perhaps in time, further research will shed some light on this point.

I can make my Ruger with 40 S&W upper work with such a wimpy load by using the 7.65 Parabellum recoil spring, but a standard 40 S&W load with this spring might blow the slide off the back of the pistol. I also have a Luger in 7.65 that currently has a very wimpy recoil spring in it.

Note that the Borchardt had a recoil spring adjustment that allowed it to handle a wider range of loads. The original intention of this adjustment was to tune the individual pistol to the standard load and compensate for variations from one recoil spring to the next.

[Thor]

I think the only way the above vintage statement could have been true was if they were using VERY much higher loaded ammo to start with, then a reduced load might still be hot enough to work the action. Wasnt earlier 30 Luger Carbine Ammo loaded to 1400 fps +? All that said, I do know that unspellable is correct in that the standard 30 Luger you get today is pretty anemic. A load of 1 gr over max in some reloading DATA books are required to get some 30 Lugers to lock back the hold open. This is a 20% overload from that DATA based on volume of smokeless powder.

[Roadkill]

I'll throw in my .02 worth and give everyone another chance to talk about that dumba*s in Alabama again, but wouldn't the bullet weight make the difference? If they (and since its not stated what the bullet weight was nor do we know what composition the powder was in 1899)used a fairly heavy bullet and the case was full to begin with, wouldn't a 20% reduction in powder volume actually increase the recoil? I load 255g .45 with 5g of unique, 1911A1 goes big boom every time, plenty of recoil. I load 185g with 6.7g of the same, big boom, plenty of recoil, same with the 230g and 6.4g. Same gun, all work.

rk

[Stu]

The extract is supposedly from the official report, does anyone know if that's true ? The site referenced is a guy in Switzerland.

RK has an interesting point about bullet weight. When we come to the operation of a recoil based system, which side of the weight x velocity equation is more significant ? I could see that a heavier bullet might also be in the barrel a fraction longer, so increasing the duration of the impulse, hence total energy transfer ?

[Roadkill]

Also allows time to build up more chamber pressure?

rk

[AGE]

RK,

I really enjoy your posts and Alabama humor, especially when you include your dog and truck. In many ways we have a lot in common. I am an old country boy from PA who never really grew up. My toys just got more expensive (as in cars, guns and boats). I save a buck or two when I can and then go blow many $$ on a toy that I really need (I do?).

Anyway, I also find my 1911s work perfectly with fairly light as well as heavy reloads and factory loads. I never touch the recoil springs. I have not really messed with light bullets, I only use 230 grain roundnose because other shapes have malfunctioned in the past. Roundnose 230 grain bullets work great in both old and new guns with 4.6 grains of Bullseye powder and shoot one hole groups at 25 yards in my accurized guns.

I suspect the Luger would be more sensitive to light loads, but I can't be bothered reloading 9 mm when the Wallymart ammo is available.

[Edward Tinker]

I know that a heavier bullet takes a fraction of a second to leave the barrel, heavier bullet, longer time for pressure and it can cause problems.

So, RK is totally right on that one.

[unspellable]

The impulse resulting from launching the bullet is simply velocity times mass. Pressure and barrel time have no bearing on operating the action. (We are talking strictly recoil operated actions here.)

There is a minor complication in that the cartridge impulse includes both the bullet impulse and the powder impulse. It's difficult to get a handle on the average muzzle velocity of the powder gas so various rules of thumb have been proposed. The one I've been using for the Luger has the average gas muzzle velocity at 1.5 times the bullet velocity. Fortunately the bullet mass dominates so small errors in the gas velocity do not make a great difference.

So cartridge impulse is the bullet mass times muzzle velocity plus 1.5 times the powder mass times the muzzle velocity.

impulse = ( bullet mass + 1.5 * powder mass ) * muzzle velocity.

You have to be careful about using consistent units. The resulting number (around 0.5 lbf*s for the Luger) is the figure of merit for a particular load's ability to operate the action.

[Roadkill]

Unspellable, would you please dumb that down about 80% where I can understand it?

rk

[Edward Tinker]

[quote] The impulse resulting from launching the bullet is simply velocity times mass. Pressure and barrel time have no bearing on operating the action. (We are talking strictly recoil operated actions here.) <hr></blockquote>

Me too...

But you say the kick back (impulse) has little bearing. All my limited teachin' tells me that more mass in the barrel (heavier bullet) can bring pressure higher. More pressure, more kick?

Is that wrong. I was a cop and security consultant now, so I ain't no scineterst, so explain to RK and Ed's level [img]biggrin.gif[/img]

[unspellable]

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.

[AGE]

Unspellable,

That is a pretty technical presentation involving energy, momentum, impulse, etc. I even understand some of your discussion since I worked in that kind of stuff for years.

I suspect (from earlier discussions we have had) that you have made a lot of simplifying assumptions that you are comfortable with by now, but not everyone is. You could write a textbook on this stuff and you might have to in order that we (me too) can really understand and agree with it. We also need to be wide awake when thinking about such complicated stuff.

Anyway, nice try but it is still pretty mysterious stuff.

[Stu]

Unspellable's explanation that the key factor is momentum, (mass x velocity), is pretty reasonable. I think when the topic generally comes up most people ignore the charge component for simplicity's sake, as it does complicate the equation somewhat !

So we should be able to come up with a rule of thumb in terms of momentum to give reliable functioning, I assume ?

I know some people out there have chronos, could we get some figures on the loads that work well, (9mm Winchester 115gr, anyone have the velocity ?).

I'll volunteer one set of info for the 30; Western X 93gr @ 1220 fps for 307 ft. lbs.

[Mark A]

Master,
If I could but touch the hem of your garment perhaps I could balance my checkbook.

[Roadkill]

Thanks for the explanations, one of the good parts of working in a school is you can always find a smart kid to explain it to you. I do understand what you mean. Using my gun as an example, once I finally got it to shoot it handled the Walmart 115g well. Once I started reloading, it handled the 125g extremely well. If someone would have handed me the gun to shoot without my looking I would have sworn it wan't mine. Unbelievable difference.
Thanks again

rk

[unspellable]

Winchester claims their 30 Luger load with 93 grain hardball has a muzzle velocity of 1220 fps from a 4.5 inch test barrel in a test fixture. How ever I find that either this is optomistic or else the air over my chronograph is particularly thick stuff. I have to use at least six inches of barrel to get it near that velocity. Won't get close from a 4, 4.5, or 5 inch barrel.

Winchester's listed hand load for the 30 Luger is quite unlikley to cycle the action as Winchester calims it runs the same bullet from the same test barrel at around 1050 fps. Hopeless!

Right now I am sorting out recoil springs versus loads. Later on I'll try to sort out loads.

We have established 25 lbf as the absolute minumum for the recoil spring without regard to load. Below this the spring will not reliably close the action. It is possible for the action to stop almost closed but not locked and then fire from that position. So too light a recoil spring can be a safety hazard!