pressure of a .38 +p++ lswchp 2' barrel

[dgang]

In another shooting board I detailed how I worked up to a load consisting of a 158 gr lead swc hp /gc . By starting with a .38 load and increasing the load by 5% at a time until I reached a what I call a +p++ loading. Each "+" indicates a 10% increase of Power Pistol until I topped out at 7 gr, about 20% less than a .357 load. I am shooting this load out of a Ruger LCR 357 with a 2" barrel, using .38 spl brass and getting about 1020 fps. I am hoping that some of you experts can run this data through a ballistic software program and give me some idea of the pressure this load is generating. Thanks in advance and good shootin' to ya', dgang
__________________
I shoot, therefore I am.

[spacecoast]

Withdrawn, I'll be interested in the results. Seems like it would be easier and safer to load this in .357 brass, though.

[dgang]

spacecoast, here is the results from another poster: "QuickLOAD says that to get 1020 fps from a 2" revolver barrel you would be at about 26,600 psi, which is fine in your .357 magnum revolver, but would not be good for the half life of a light frame .38 Special revolver. Good practice would be to put these loads into .357 cases to prevent accidental chambering in a .38 Special. That's what the extra .357 brass length is for. "

[Brian Pfleuger]

QuickLoads estimates in this instance would be ballpark at best. First, QuickLoad is inherently less reliable for straight wall cartridges and, second, QuickLoad does not simulate the cylinder gap on a revolver.

[dgang]

peetzakilla, Good points, but I still figure I'm safely within .357 levels. Thanks, dgang PS. I have to wonder what kind of " non canister " powder BB uses to reach the same velocity with .38 +P pressures?

[Clark]

CAUTION: The following post includes loading data beyond currently published maximums for this cartridge. USE AT YOUR OWN RISK. Neither the writer, The Firing Line, nor the staff of TFL assume any liability for any damage or injury resulting from use of this information.


I have blown to pieces (4) 38 special revolvers and damaged (2) more.
I own so many 38 special revolvers I have no idea what the count is, and they are in boxes in storage.
I have overloaded many 38 special revolvers and 357 mag and 38 S&W and some so old they are just marked 38, but 38 special will fit.


I was so impressed with 1993 John Bercovitz's analysys of the 357 mag pressure effects on the brass, that I somehow recruited him in 2005 to help me with my "Prove the load books wrong about the CZ52 vs Tokarev" project.

A friend asked why steel cases aren't more common since they would
allow higher chamber pressures. I thought that as long as I had
written something up for him, I might as well post it here:

Material Properties
CDA 260 cartridge brass: barrel steels:
Young's modulus = 16*10^6 psi Young's modulus = 29*10^6 psi
Yield stress = 63,000 psi min. Yield stress: usually > 100,000 psi

I was going to get back to you and explain further why brass is a better
cartridge case material than steel or aluminum. Sorry I took so long. I
left you with the nebulous comment that brass was "stretchier" and would
spring back more so it was easier to extract from the chamber after firing.
Now I'll attempt to show why this is true given the basic material properties
listed above.

A synopsis would be that the propellant pressure expands the diameter of
the thin wall of the cartridge case until it contacts the interior wall
of the chamber and thereafter it expands the case and the chamber
together. The expansion of the cartridge case, however, is not elastic.
The case is enough smaller in diameter than the chamber that it has to
_yield_ to expand to chamber diameter. After the pressure is relieved by
the departure of the bullet, both the chamber and the cartridge case
contract elastically. It is highly desirable that the cartridge case
contract more than the chamber so that the case may be extracted with a
minimum of effort.

A quick review of the Young's modulus: this is sort of the "spring
constant" of a material; it is the inverse of how much a unit chunk of
material stretches under a unit load. Its units are stress / strain =
psi/(inch/inch). Here's a basic example of its use: If you have a 2
inch by 2 inch square bar of steel which is 10 inches long and you put a
10,000 pound load on it, how much does it stretch? First of all, the
stress on the steel is 10,000/(2*2) = 2500 psi. The strain per inch will
be 2500 psi/29*10^6 = 0.000086 inches/inch. So the stretch of a 10 inch
long bar under this load will be 10 * 0.000086 = 0.00086 inches or a
little less than 1/1000 inch.

Yield stress (aka yield strength) is the load per unit area at which a
material starts to yield or take a permanent set (git bint). It's not
an exact number because materials often start to yield slightly and then
go gradually into full-scale yield. But the transition is fast enough
to give us a useful number.

So how far can you stretch CDA 260 cartridge brass before it takes a
permanent set? That would be yield stress divided by Young's modulus:
63,000 psi/16*10^6 psi/(inch/inch) = .004 inches/inch.

How far can you stretch cheap steel? Try A36 structural steel:
36,000 psi/29*10^6 psi/(inch/inch) = .001 inches/inch.
How about good steel of modest cost such as C1118?
77,000 psi/29*10^6 psi/(inch/inch) = .003 inches/inch.
(Note that C1118 doesn't have anywhere near the formability of CDA 260.
Brass cases are made by the cheap forming process called "drawing"
while C1118 is a machinable steel, suitable for the more expensive machining
processes such as turning and milling.)

What about something that's expensive such as CDA 172 beryllium copper?
175,000 psi/19*10^6 psi/(inch/inch) = .009 inches/inch.
(This isn't serious because CDA 172 is pretty brittle when it's _this_
hard.)

Titanium Ti-6AL-4V
150,000 psi/16.5*10^6 psi/(inch/inch) = .009 inches/inch
(This is an excellent material though expensive and hard to work with.)

Really expensive aluminum, 7075-T6
73,000 psi/10.4*10^6 psi/(inch/inch) = .007 inches/inch
Cheap aluminum, 3003 H18
29,000 psi/10*10^6 psi/(inch/inch) = .003 inches/inch
(Aluminum isn't a really good material because it isn't strong and cheap
at the same time, it hasn't much fatigue strength, and it won't go over
its yield stress very often without breaking. So you can't reload it.
It makes a "one-shot" case at best. Also, 7075 is a machinable rather
than a formable aluminum, primarily.)

Magnesium, AZ80A-T5
50,000/6.5*10^6 = .0077
(Impact strength and ductility are low. Corrodes easily.)

+Here's the important part: Even if you stretch something until it
+yields, it still springs back some distance. In fact, the springback
+amount is the same as if you had just barely taken the thing up to its
+yield stress. This is because when you stretch it, you establish a new
length for it, and since you are holding it at the yield stress (at
least until you release the load) it will spring back the distance
associated with that yield stress. So the figures given above such as
.004 inches/inch are the figures that tell us how much a case springs
back after firing.

Changing subjects for a moment: How much does the steel chamber expand
and contract during a firing? Naturally this amount is partially
determined by the chamber wall's thickness. The outside diameter of a
rifle chamber is about 2 1/2 times the maximum inside diameter,
typically. The inside diameter is around .48 inches at its largest.
Actual chamber pressures of high pressure rounds will run 60,000 psi or
even 70,000 psi range if you're not careful.

One of the best reference books on the subject is "Formulas for Stress
and Strain" by Roark and Young, published by MacGraw-Hill. Everyone
just calls it "Roark's". In the 5th edition, example numbers 1a & 1b,
page 504, I find the following:

For an uncapped vessel:
Delta b = (q*b/E)*{[(a^2+b^2)/(a^2-b^2)] + Nu}

For a capped vessel:
Delta b = (q*b/E)*{[a^2(1+Nu)+b^2(1-2Nu)]/(a^2-b^2)}

Where:
a = the external radius of the vessel = 0.6 inch
b = the internal radius of the vessel = .24 inch
q = internal pressure of fluid in vessel = 70,000 psi
E = Young's modulus = 29 * 10^6 psi for barrel steel
Nu = Poisson's ratio = 0.3 for steel (and most other materials)

A rifle's chamber is capped at one end and open at the other but really
it's not too open at the other end because the case is usually bottle-
necked. You'd have to go back to basics instead of using cookbook
formulae if you wanted the exact picture, but if we compute the results
of both formulas, the truth must lie between them but closer to the
capped vessel.

For an uncapped vessel:
D b = (70000*.24/29*10^6)*{[(.6^2+.24^2)/(.6^2-.24^2)] + .3} = .00097

For a capped vessel:
D b = (70000*.24/29*10^6)*{[.6^2(1.3)+.24^2(.4)]/(.6^2-.24^2)} = .00094

There's not a whole heck of a lot of difference between the two results
so let's just say that the chamber's expansion is .001 inch radial or
.002 inch diametral.

The cartridge case's outside diameter is equal to about .48 inch after
the cartridge has been fired. So its springback, if made from CDA 260,
is .004 inches/inch (from above) * .48 inch = .002 inches diametral
which of course is just the amount the chamber contracted so we've just
barely got an extractable case when chamber pressures hit 70,000 psi in
this barrel. This is why the ease with which a case can be extracted
from a chamber is such a good clue as to when you are reaching maximum
allowable pressures. By the same token, you can see that if a chamber's
walls are particularly thin, it will be hard to extract cases (regardless
of whether or not these thin chamber walls are within their stress limits).
A really good illustration of this can be found when comparing the S&W
model 19 to the S&W model 27. Both guns are 357 magnum caliber and both
can take full-pressure loads without bursting. The model 27 has thick
chamber walls and the model 19 has thin chamber walls. Cartridge cases
which contained full-pressure loads are easily extracted from a model 27
but they have to be pounded out of a model 19. So manufacturers don't
manufacture full-pressure loads for the 357 magnums anymore. 8-(

We can see from the above calculations that a steel case wouldn't be a
good idea for a gun operating at 70,000 psi with the given 2.5:1 OD/ID
chamber wall ratio if reasonable extraction force is a criterion. Lower
pressures and/or thicker chamber walls could allow the use of steel cases.

I can get the pressure up to the point of the brass sticking in any 38 sp I have tested, if I use slow powder.

The max load for 357 mag is 18 gr LIL'GUN 158 gr, but I have shot 26 gr LIL'GUN 158 gr in a number of 38 specials, and the noise and recoil are horrific, but no damage.

But typically, with fast powder, at ~ 3 ~ 5 gr more than the max 357 mag load, a 38 special will split the cylinder and may break to top strap. This never seems to hurt me, but the pieces that fly to the sides put some awesome holes in the walls, and might kill by standers if I did this in public.

The SAAMI registered max pressure for the 38 sp is 17,000 psi.
The SAAMI registered max pressure for the 38 sp +P is 18,500 psi.
The SAAMI registered max pressure for the 357 mag is 35,000 psi.

I run all my 38 specials beyond 357 mag max published loads.
That would be 38 sp +P+++++ or something else like that with no meaning.

[sdt11670]

I enjoy having eyesight, two hands, and a beating heart! No way would i approach those loads.

[SL1]

Clark,

Can you explain to us how you managed to put 26 grains of Lil'Gun into a 38 Special case under a 158 grain bullet? QuickLOAD seems to think that the powder will need something like 58% more space than what is left. Even considering QuickLOAD's propensity for small default case volumes, that seems like a LOT of compression for a ball powder.

The max load for 357 mag is 18 gr LIL'GUN 158 gr, but I have shot 26 gr LIL'GUN 158 gr in a number of 38 specials, and the noise and recoil are horrific, but no damage.

Also, QuickLOAD seems to think that it will produce 878,000 psi. Of course, that assumes that it all ignites from the primer flash. I suspect that won't happen with the level of compression you seem to be using. What pressure does it take to actually blow-up your 38 Specials?

SL1

[dgang]

Clark, I appreciate the data you quoted, it will take me a few days to fully understand how it relates to the load I'm developing. In reflection , it seems like I am working well within the guidelines for a .357 revolver. I have never "blown up" a firearm and could live out my life quite happily without experiencing such a event. Like some of the others I wonder how or why you would load some cartridges to the extent that you knew would cause a catastrophic failure of cartridge and firearm?. Thanks for the input and may you shoot safely in the future. dgang

[Clark]

CAUTION: The following post includes loading data beyond currently published maximums for this cartridge. USE AT YOUR OWN RISK. Neither the writer, The Firing Line, nor the staff of TFL assume any liability for any damage or injury resulting from use of this information.
It is very hard to get 26 gr of LIL'GUN to fit under 158 gr.
And it kicks harder than a 44 mag.
Not a useful load.
Just and experiment.

In order to get it to fit, it must be compressed in several stages.
I put a pin gauge, typically smaller than the bullet, into a collet bullet puller die in another press.

I ream out the cylinder chambers of the 38 sp to 357 mag length with a few twists of the wrist on a 0.380" straight fluted chucking reamer.
I fill the 357 mag case with powder.
I compress the powder with the pin gauge.
I add more powder to the case.
I compress the powder with the pin gauge.
I add more powder to the case.
I compress the powder with the pin gauge.
I move the filled case to another press with a seating die and seat the bullet.
I resize the loaded ammo with a Lee factory crimp Carbide ring, so the bulged case will fit in the chamber.

[WESHOOT2]

Continued use of high-pressure ammunition will result in decreased service life and (in your case, radically) increased wear.
Guaranteed.

Power Pistol can act very progressively.