How much pressure is required to "compress the boolit"

[TCLouis]

Bound to be a thread on this but I did not find it . . .
Maybe just the fact that my google foo is weak

Anyway how much pressure is required to cause a boolit to be compressed in length.

Will any practical load do that, and if so I would guess that we wold get some expansion in diameter?

[Newboy]

Search for the term obturation.

It will depend on the hardness of your bullet.


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[Land Owner]

Metallurgy and your gut should suggests that ANY positive pressure (force over area) will "compress" the chambered boolit as its inertia is overcome by the expanding gas spanking it from behind, into forward movement, before slamming it into the lands and grooves at the forcing cone.

In fact, so long as the internal stress (EX: extremely low force over time) in the lead remains in that material's elastic range, zero deformation will be present after the pressure wave has subsided (the boolit has not moved). Instantaneously, at very low pressure, the boolit will "compress" over a tiny and imperceptible length. With greater pressure, its sides will attempt to expand as this picture depicts to not less than the diameter of the side walls of the chamber.



A "pinch" more pressure and the boolit overcomes its inertia and MOVES. As it moves forward it meets the forcing cone and with increasing pressure is subsequently deformed into receiving the lands and grooves. With the force from a primer detonation the compressed boolit moves somewhat down the barrel. More pressure pushes the boolit barely out of the barrel. A whole lot of pressure pushes the boolit to the target.

Inertia, alloy hardness, lubricity of lead against steel (friction), boolit diameter, chamber diameter, powder "speed", are a few factors in "compression" of the boolit in its stress and strain relationship with pressure.

I am not sure at this point what the question ought to be?

[JonB_in_Glencoe]

compress means something different than what I think you are asking.
I would use the term "slump" when referring to a boolit becoming shorter in length, which happens to the unsupported nose when a condition of too much pressure occurs.

But honestly, the amount of Pressure needed to obturate the boolit base is likely the info you are looking for and there is a formula for that midway down the page, at this linked website.

http://www.lasc.us/CastBulletNotes.htm

Bullet BHN / "Minimum" Chamber Pressure For Lead Alloys (PSI)

The formula (from the pages of HandLoader Magazine) to determine at what pressure an alloy of given BHN will obturate the base of the bullet and seal the bore. If the bullet is too hard to obturate, gas cutting usually occurs on the base band on the non-driving side of the rifling and barrel leading is likely. Simply multiply the alloy BHN by 1,422.
Example: Alloy BHN of 12 multiplied by 1422 = 17,064. An alloy of 12 BHN should be used with a load that develops a "minimum" of 17,000 psi. Need more info on minimum / maximum alloy BHN? These Glen E. Fryxell articles explain alloy BHN in easy to understand language.

[Larry Gibson]

Bound to be a thread on this but I did not find it . . .
Maybe just the fact that my google foo is weak

Anyway how much pressure is required to cause a boolit to be compressed in length.

Will any practical load do that, and if so I would guess that we wold get some expansion in diameter?

Yes, there are numerous threads over the years that discuss this topic. However, to answer it here;

Consider, since you mention "any practical load", that you mean when fired(?). Well that also varies additionally on the type of firearm and how well the bullet "fits" along with how well the case neck fits the chamber neck. Assuming you're not talking undersized bullets (?) the bullets and they fit the throat and are over groove diameter then they will be swaged down when inside the barrel. They will also be swaged down, if the bullets are lubed, an additional amount as they ride over a layer of lube. Given that the barrel bore and groove diameter + the layer of lube determine the diameter size of the bullet. The bullet can not be any larger than that.

Another variable not considered in the formula is the "rise" of the time/pressure curve (acceleration rate). Given the same peak pressure a fast burning powder will "punch the bullet harder" so to speak than will a slower burning powder. Punching a door to close it or pushing the door to close it takes the same amount of force....one is easier on the hand and the door.

Older bullet designs with large lube grooves and small diameter column of bullet in the lube grooves, when cast of softer alloy can collapse into the lube groove as the lube is expunged or if the lube groove is not filled if the acceleration rate is fast enough. Older bullets with large 'scraper grooves" in front the front driving band are also more susceptible to collapse, bending or "sloughing" as mentioned.

Additionally if the bullet does not collapse and is swaged down it will be longer on exit from the muzzle than before firing. If the bullet is swaged down the alloy must go somewhere, if the bullet does not collapse into the lube or scraper grooves. and it is lengthened.

The above mentioned formula is valid only if you are using pure lead as the constant of "1422" is based on the yield or tensile strength of pure lead. Once you mix the tead with tin for a binary alloy or mix the lead with antimony and tin the yield and tensile strengths become very different.

As an example of how the formula fails; I shoot a modern designed 30 XCB bullet (lubed with 2500+) cast of #2 alloy that are WQ'd from the mould. The BHN runs 20 - 22. Given the 22 BHN the formula says the bullet should "fail" at 32,000 psi. Yet I shoot that bullet at 2900+ fps with 50,000 psi (measured in the rifle via an Oehler m43 PBL) and have maintained 10 shot moa accuracy to 600 yards (numerous targets and data posted on this forum and such has been witnessed by several members). If the bullet was to fail at 32,000 psi how could that be? Answer is it can be because the formula doesn't take numerous variables into account, uses and incorrect constant and is therefore unreliable if not wrong.

[bmortell]

Ive never seen it do anything usefull, whenever i had sufiecient softness and psi that it should obturate i immediatly get gas cutting even 5 ten thousandths undersized.

[JonB_in_Glencoe]

Larry,
You are the last person I would ever try to argue with, and I won't.

With that said, the formula is not for calculating a Maximum or Fail Pressure, it is for figuring Minimum pressure an alloy of given BHN will obturate the base of the bullet and seal the bore.

I surely agree with you that a single formula, with a constant of 1422, will not take into account the many variables and is surely not precise enough for laboratory type testing, but I believe it is close enough to get a boolit caster into the Ballpark of what alloy might work with a desired load.
That's my 2¢

[Larry Gibson]

JonB

That's what Fryxell says as further delineated in http://www.lasc.us/FryxellCBAlloyObturation.htm. However, (and I don't consider this an "argument" by any stretch but simply further discussion) Fryxell is talking about cast bullets in a revolver. Even there the constant is based on pure lead. An alloy with a 12BHN is much harder and can withstand higher psi before obturation. If the formula was correct our COWW cast bullets with a BHN of 12 +/- when fired at magnum levels in revolvers [Keith's 22 gr 2400 under a 429421 as example] with psi's of 35,000 should then collapse inside the cylinder/barrel into a solid glob of alloy. Yet we recover them with little or no sign of deformation from obturation or setback/collapse.....

I know Fryxell and the LASC article are a sacred cow to many but reading his article you will note his "formula" came from reading "Handloader" articles with no further vetting as to accuracy or correctness. I'm simple pointing out there is too much evidence to the contrary for the formula to make sense. Lee on the other hand attempts to quantify his conclusion as to why the formula works. His friend "John" ran a simple test with a M1903 30-06 using a 200 gr cast bullet. The point of accuracy loss seems to coincide with what the formula says. However, Lee failed to understand the adverse effect of RPM at the same coincidental point. Thus too many of us are now shooting cast bullets at higher velocity with higher psi than the formula says should happen. Again, too much contradictory evidence for the formula to be correct.

The OP doesn't state what cartridge or type of firearm he's asking about. Further confusion as most these days who ask the question aren't reading Fryxell but are reading Lee who uses the formula for all cast bullets in rifles and handguns. Now, to give both Fryxell and Lee their due I have found the formula is pretty close if pure lead or 40-1 alloy is used with attendant BHNs of 5 - 6 maybe 7.

[kens]

it doesn't take a whole lot to obturate, if your bullet is right
Muzzle loaders (black powder) will obturate soft lead bullets even with light loads.
Black powder cartridges all obturate with a medium hard alloy
as long as you match your alloy with your powder/pressure/gun, it will obturate

[bmortell]

If the formula was correct our COWW cast bullets with a BHN of 12 +/- when fired at magnum levels in revolvers [Keith's 22 gr 2400 under a 429421 as example] with psi's of 35,000 should then collapse inside the cylinder/barrel into a solid glob of alloy. Yet we recover them with little or no sign of deformation from obturation or setback/collapse.....

actually when I starting powder coating and no lube I was using keith 44 loads 3.5%sb 2.5%sn air cooled, so about 12bhn at full charge and the lube grove fully obturates out flat. so I do see signs of obturation but it also had signs of gas cutting since it started a hair undersized. that's why I order molds with small or tumble lube groves now because obturation does happen. but it didn't seem to help I still get gas cutting and worse looking recovered boolits if there not big enough to start with. it dont seem like obturation grows them fast enough to prevent problems for me.

[TCLouis]

Thanx Guys.

This will give me enough to ruminate on for a while.

In one way or another I have the the answer I sought and some additional info to consider.

[JonB_in_Glencoe]

Larry,
I can't see anything in Glen E. Fryxell's article "Cast Bullet Alloys and Obturation" about calculating/predicting when failure occurs, it is all about when Obturation starts. Why are you trying to point out that the formula fails at calculating a pressure that will create a condition of failure? When there is nothing said about that.

[Land Owner]

Obturation, or pressure induced expansion of a bullet's diameter for optimized sealing against the bore, is caused by 1.) searing hot gas melting the lead base, plus 2.) the explosively expanding pressure spike smashing into and expanding THE WHOLE BULLET against the smooth walls of the chamber, as 3.) inertia, 4.) friction, and 5.) rearward compression of the forcing cone restricts the bullet's forward movement.

You cannot "spank" (by explosive pressure wave) the rear end of a slightly oversized and SMOOTH SIDED bullet "that hard" while simultaneously smashing its forward edge into the restriction of lands and grooves and NOT cause the bullet's entire body to deform (i.e. SHORTEN instantaneously and EXPAND to the chamber wall diameter), as it is ultimately swaged (squeezed) back down to bore diameter...UNLESS the bullet is "infinitely hard" such that only the hot gas has a melting effect.

An "infinitely hard" bullet would immediately obstruct the bore, not take the lands or grooves, and potentially blow the chamber apart if the pressure spike is sufficiently large. Coming back from the Land of Make Believe, we don't have "infinitely hard" bullets, so we create other ways to blow up guns. That's food for another discussion.

It is assured that pure lead is "soft" by comparison to steel or even compared to alloy mixtures of lead. So, in a lead bullet's relative "softness" (compared to the steel barrel) we can think about the incremental nano-seconds in the chamber surrounding powder ignition and surmise what is happening to a lead alloy bullet (as above). Obturation then is a phenomenon of heat, pressure, and compression (resistance to change) EXPANDING the WHOLE BULLET, not just the base.



"Swage" is the term to describe lengthening of the bullet and lead redistribution from its obturated (expanded) chamber diameter as it is press fit into the barrel's lands and grooves. A fundamental function of "press fitting" (swaging) a bullet into its bore diameter is the redistribution of lead into the lube grooves.