From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Re: Colt GM dynamics - conclusion Keywords: physics Date: 19 Sep 90 20:14:42 GMT In article <8496@orca.wv.tek.com> bercov@bevb.bev.lbl.gov (John Bercovitz) writes: >In article <8474@orca.wv.tek.com> cbl@uihepa.hep.uiuc.edu (Chris Luchini) writes: >>................ for some portion of the pressure curve >>the brass shell is 'welded' to the chamber ............ > > Many in licentious youth have fabricated odd little guns with smooth >barrels, no breech block, and with ignition energy provided by rubber band. >These folks would attest that the "welding" is somewhat incomplete. ..... Chris Luchini's article in which he mentioned cartridge cases 'welding' to chambers intrigued me. We all know it happens or else we'd never see case stretching. But we also know that some cartridge cases, like that of the 45 ACP, don't stretch. So what's the difference? I think I've come up with a possible answer and I'd like to run it past you folks for comment. In short I propose that some cases have enough body surface area to grip the chamber walls and some don't. Since we already know that a 270 will exhibit case stretching if you set its shoulder back, let's look at some lower-pressure cartridges: 30-30: The body of a 30-30 averages about 0.4" diameter and is about 1.5" in length. A fired cartridge shows that the case expands from about 0.3" forward of the case head on to the case mouth. This leaves 1.2" for gripping the chamber (I'm ignoring the shoulder and neck). The area of this surface is (pi)*D*L= 1.5 in^2. So the force resisting axial motion of the case in the chamber when propellant pressure is high is: Faxr = (coefficient of friction) * (1.5 in^2) * (pressure of the propellant) The inside diameter of the case at its largest point is about 0.35" so the area on which the pressure acts is: (pi/4)*0.35^2 = 0.096 in^2 Axial force on the interior of the case head is then: Fax = (pressure of the propellant)*( 0.096 in^2) I think we'll all agree that if the force on the case head trying to ex- tract the case from the chamber (Fax) is larger than the gripping force of the case body (Faxr), the case will move (unless the case stretches or the head separates). Since the pressure of the propellant is the same on the case head and in the case body, and Fax = Faxr to see some incipient motion, it is fair to combine the two equations above and find: (coefficient of friction)= (0.096 in^2)/( 1.5 in^2) = 0.064 So as long as the coefficient of friction of the 30-30 case in its chamber is greater than 0.064, the case will 'weld' to the chamber. Note that it would be unusual for the coefficient of brass against steel to be quite so low; therefore the 30-30 case must be a 'welder'. Hey - here's a crazy and dangerous thought: to fireform cases, why not reduce the powder charge to 2/3 so as to reduce the pressure to about 1/2 and then put a thin layer of high pressure grease on the case? It might could lower the coefficient of friction enough to prevent case stretching. But I digress. 45 ACP: The same process applies mutatis mutandis to the 45 ACP: Pressure ring is 0.4 forward of case head leaving 0.49" for grip length; outside diameter is about 0.48 for a grip area of about 0.74 in^2. Cross sectional area of the internal diameter of 0.45" is 0.159 in^2. So minimum coefficient of friction for case welding is: 0.159 / 0.74 = 0.21. Since it would be unusual for a smooth chamber to have that high of a coefficient of friction against a brass case, the 45 ACP must be a 'slipper' rather than a 'welder'. All bets are off if you have a rough chamber; then you have gross deformation of the brass into grooves which, in the case of really decent rubber tires against asphalt, can give a pseudo-coefficient of friction higher than one. Be assured that I checked a few other things such as what pressure it takes to expand the case walls into contact with the chamber (low - about 7000 psi for the 30-30 and about 2700 psi for the 45 ACP) and what pressure it would take to stretch a 30-30 case in a rough, overlong chamber (about 30,000 psi give or take). Well, what do you think; will this fly or do I get to go down in flames? JHBercovitz@lbl.gov

From: bercov@bevsun.bev.lbl.gov (John Bercovitz) Subject: Re: handgun calibers in rifles Organization: Lawrence Berkeley Laboratory, California In article <54351@mimsy.umd.edu> bercov@bevsun.bev.lbl.gov (John Bercovitz) writes: Referring to the 45 ACP Marlin Camp Carbine: #Just for kicks I measured the bullet's barrel travel and found it was 16 #inches (the barrel length is 16 5/8). Then I weighed the bolt and found #it was 14.5 ounces. So the case backs out of the gun .58 inches by the #time the bullet exits the muzzle. This means that if you have a short #case, about .3 inches of it is left in the chamber. That leaves an awful #lot of thin brass supporting pressure. My friend says the cases don't #bulge appreciably so I guess pressures drop very quickly in his loadings. #I don't know if I'd want to try this with a compressed charge of Blue Dot. #The chamber is pretty smooth so I don't think we have an accidental #retardation lock here. Bzzzt!! Wrong! You know the old saying about keeping things as simple as possible but no simpler? Well, Henry Schaffer points out that I've violated the "but no simpler" part of the rule. Henry is always kind enough to allow me to correct my own mistakes unlike SOME people around here. 8-) 8-) Here's the error: The grip of the cartridge case on the interior of the gun's chamber considerably reduces the amount of force pushing back on the breechblock. Since the above analysis was based on a pure impulse- momentum analysis it is incorrect. The impulse is merely integral F(t)dt. The t function doesn't change; it is all the little times that the bullet is at all the various little places along the barrel. 8-) But the F does change and if you think about it, it just gets a proportional reduction. It's proportional because the frictional force and the backthrust are both proportional to the gas pressure. To find the remaining force, we need the force of the gas pressure on the case head and the amount of the frictional force. The force of the gas pressure is merely the pressure times the cross-sectional area or Pg*Ag = Pg*.159 in^2. The frictional force is not much more difficult to calculate: The area the friction is acting over is the grip length of the case, about .49 inches, times the circumference of case, .48*pi inches, or: .74 in^2. The pressure, Pg, on this area gives the normal force and all we have to do is to multiply this normal force by the coefficient of friction to find the total frictional force. The coefficient of friction of brass against steel ranges from .1 to .2 according to what kind of junk you have in the interface. So let's use .15. Then the frictional force is .74*.15*Pg = .111*Pg. The remaining back force is .159Pg - .111Pg = .048Pg. .048Pg is a lot less than the .159Pg that the too-simple pure impulse-momentum approach uses. In fact, it's only .048/.159 = .3 times as much. So the case doesn't back out .58 inches. It has only backed out .3*.58 or .175 inches at the time the bullet exits the muzzle. Remembering that the bullet and breechblock travel are proportional to each other through the ratio of their masses, we can say that the breechblock travels .175 inches while the bullet travels 16 inches so that the breechblock travels .011 inches for each inch the bullet travels. Since the bullet doesn't travel many inches before the pressure gets really low, it doesn't look like there's much of a problem here at all. It is left as an exercise for the student to prove that the case head doesn't get ripped off during its backward motion by the retarding force of friction. John Bercovitz (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz) Subject: Re: handgun calibers in rifles Organization: Lawrence Berkeley Laboratory, California For those who think this a discussion of range fees on the left and right coasts as well as the flyover, sorry for the ol' switcheroo. 8-) At one point there was some consideration of what lube might do in a purely blowback gun. The following is a continuation of that discussion. I did a couple of experiments with my friend's 45 ACP Camp Carbine this weekend. The first experiment was to determine the effect of lubricant in the chamber on the brass. I started with a dry chamber, fired three rounds and measured the largest diameter of the brass. It was .476 inch. Next, I liberally oiled three rounds with Break Free and fired them. The maximum diameters of these rounds were also .476 inch but they ejected a little more rapidly, I think (from the way they bounced around) and one of them had a lube dent. In successsion, I tried Lubriplate, Sears wheel bearing grease, and lithium/graphite chassis lube. In no instance was the maximum diameter of the ejected case anything other than .476 inch. Many lube and mouth dents were noted. The cases were still well-lubed after firing in all instances except for two of the Break Free cases. For reference, the load was 7.0 gn of Unique, Federal #150 primer, and Hornady #4515 200 gn JSWC. Also for reference, the chamber of the Marlin was not pretty as you'd expect on an expensive gun, but it was reasonable. Some circumferential tool marks were noticeable in it. The second experiment was to compare the muzzle velocity of a round in the Colt Gold Cup to the muzzle velocity of the same round in the Marlin Camp Carbine. Using this same round, I found a muzzle velocity of 900 fps in the Colt and 1160 fps in the Marlin. For those who would prefer KE, that's 360 ft-lb in the Colt and 597 ft-lb in the Marlin. I also tried a second load, similar to the first but with a #68 H&G 200 gn bullet and 6.2 gn of Unique. This round had a muzzle velocity of 890 fps in the Colt and 1150 in the Marlin. John Bercovitz (JHBercovitz@lbl.gov)

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