Example a pistol can that is rated for 300blk on a 8.5 barrel but not for a 223 with a long barrel......

300blk with 8.5" barrel 10kpsi

223 on a 24" barrel has 7.3kpsi

I just dont how the rating works or is it just a marketing thing.....

- Thread starter jcwarrior87
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Example a pistol can that is rated for 300blk on a 8.5 barrel but not for a 223 with a long barrel......

300blk with 8.5" barrel 10kpsi

223 on a 24" barrel has 7.3kpsi

I just dont how the rating works or is it just a marketing thing.....

300 Win Mag - 22 inch minimum

308 18 inch minimum

300 BLk Out 10 inch minimum

Just pulling numbers out of my ass, I stayed with 30 cal cartridges.

Can you explain how the rating work a little more?

We do a full failure and strength analysis for the most common intended cartridges, including high volume field testing, and then develop a set of conservative rules we can extend to other cartridges and barrel lengths.

But my question is bigger than just this one instance. I am wanting to learn more about suppressors and how their limits are set

You build a few and test to point of failure. Then either accept the limitations it has, or redesign or use better materials.

Zak can easily tell me I am wrong but everything has trade offs. For example Titanium suppressors save weight but can't handle the heat of stainless steel.

I thought I read (another example) that the Harvester is glued together. Cheap, but not made for large amount of abuse.

I've personally seen and shot many .375 Cheytac rounds through a Crux .45 acp pistol suppressors. Their suppressors are tough.

Now now now, they have had issues. They were taken care of ASAP. And I have no complaints on mine (even though mine were affected).

I've personally seen and shot many .375 Cheytac rounds through a Crux .45 acp pistol suppressors. Their suppressors are tough.

I just like to make sure the kool-aid doesn't get to sweet.

That was back when they were first starting out as Templar Tactical - correct? They certainly had their teething issues, like all companies.Now now now, they have had issues. They were taken care of ASAP. And I have no complaints on mine (even though mine were affected).

I just like to make sure the kool-aid doesn't get to sweet.

Its done using a mathematical calculation called hoop stress.

https://www.engineersedge.com/calculators/hoop-stress.htm

A quick and dirty example:

We will use 2 calculations. One, to find the pressure inside the suppressor and two, to find the maximum allowable hoop stress for the suppressor material. In this 'dirty' example im going to cut alot of corners fore the sake of explanation.

We will use the simple algebraic p=f/a and work it backwards and forwards a couple times.

A 6.5 creedmoor has a saami maximum chamber pressure of 62000 psi. So first we will set p to 62000.

So, now were at 62000=f/a

Now we will calculate the inner sirface area of a 6.5 creedmoor loaded round. (Im gonna cut corners). (.473×pi x1.75=2.4025"^2

So, we go with 62000=f/2.4025, then solve for f. F=148955

So now we will rework it for p, using the new calculations for f and a. But this time we recalculate a to include the area of the barrel. We will assume a 24" barrel. So for a, we use 2.4025+(.264)(pi)(24)=22.308

So, back to p=f/a we use p=148955/22.308 and solve for p. For p we have 6677.201 psi at the muzzle.

Now we work it over again, but include the area of the silencer. Again, dirty and cutting corners. We will assume a 1.5"od'8"long silencer. So, using the previous calculations for the area of the cartridge and bore, but now adding the silencer we get 22.308+(1.5)(pi)x8= 60.007 inches squared

So now, back to p=f/a to solve for p again.

P=148955/60.007 and we get 2482.294 psi inside the silencer.

So, now that we know the pressure inside the silencer, we can use the hoop stress for thin wall pressure vessels calculation to check to find the minimum wall thickness for the silencer.

This part gets complicated, so im going to use engineers edge hoop stress calculator linked above:

For a 4130 chromoly silencer, using a tensile rating of 63100 psi and assuming perfect surface finishes, welds, and assuming a bunch of other shit you should never assume, i get a minimum wall thickness of .030". Using the above numbers, below .030ish wall thickness, it blows up in your face.

So, obviously youd want lots of margine for error.

As the barrel length is reduced, pressure goes way up.

As the silencer inner area, aka volume, goes down, pressure goes way up.

Welds have heat affected zones that can create spots weaker than the steels max tensile load.

Poor surface finishes give sharp microtroughs that give cracks a place to start.

Baffles inside the suppressor reduce volume and increase pressure.

The algebraic formulas above can be increased in accuracy and reduced in complexity by integrating the increase of area with respect to the bullets movement down the bore with the change in pressure, but i figured a calculus function would be harder to follow and visualize.

For me anyway, using the algebraic solutions, while not as accurate, lets me see whats going on in my head.

I used to manufacture suppressors, and for a typical .308 suppressor that i intended to be used on and "rated for" use on barrels as short as 16", i used 4130 steel at a wall thickness of .062", with tig fusion welded end caps.

As an example of how barrel length changes pressure at the muzzle, the above example had the 24" creedmoor putting 6700ish psi at the muzzle.

The same gun with a 16" barrel produces 9500 psi at the muzzle.

Please please,! dont use any of the mathy shit above to make a silencer. I rushed through it, didnt double check or verify anything. Its just a dirty example of how its done! Overbuild it

Another random thought, but cool as shit nonetheless.

You can use the above to calculate the force applied to the bullet at any given point as it moves down the bore, and using force mass acceleration method, f=m×a calculate the rate of change of velocity for the bullet. For example, in the throat the bullet has 3394 lbs of force pushing against the base.

So, the bullet has 3394lbs of force, so 3394=m×a

So now we find mass. A 140 grain bullet weighs .02lbs. So the mass is .0062 slugs.

So now 3394=.0062×a and we solve for accelleration. A=547419 feet per second squared.

So, at accelerating at 547,419 feet per second per second, how long does it take to get to 2700fps?

0.0027 seconds. not very long.

But, the formula above is an oversimplification too. But it wont be far off.

People dont like us nerds much, till theyre looking out the winder of their 747-8, looking down at a cloud layer 3 miles below them as its climbing through 35,000 feet. Then that shits a big deal. 900,000 lbs and 100,000 horsepower going 500mph

Seems like a miracle, but its engineering!

Time to go clean my pocket protector anf get another juice box.

https://www.engineersedge.com/calculators/hoop-stress.htm

A quick and dirty example:

We will use 2 calculations. One, to find the pressure inside the suppressor and two, to find the maximum allowable hoop stress for the suppressor material. In this 'dirty' example im going to cut alot of corners fore the sake of explanation.

We will use the simple algebraic p=f/a and work it backwards and forwards a couple times.

A 6.5 creedmoor has a saami maximum chamber pressure of 62000 psi. So first we will set p to 62000.

So, now were at 62000=f/a

Now we will calculate the inner sirface area of a 6.5 creedmoor loaded round. (Im gonna cut corners). (.473×pi x1.75=2.4025"^2

So, we go with 62000=f/2.4025, then solve for f. F=148955

So now we will rework it for p, using the new calculations for f and a. But this time we recalculate a to include the area of the barrel. We will assume a 24" barrel. So for a, we use 2.4025+(.264)(pi)(24)=22.308

So, back to p=f/a we use p=148955/22.308 and solve for p. For p we have 6677.201 psi at the muzzle.

Now we work it over again, but include the area of the silencer. Again, dirty and cutting corners. We will assume a 1.5"od'8"long silencer. So, using the previous calculations for the area of the cartridge and bore, but now adding the silencer we get 22.308+(1.5)(pi)x8= 60.007 inches squared

So now, back to p=f/a to solve for p again.

P=148955/60.007 and we get 2482.294 psi inside the silencer.

So, now that we know the pressure inside the silencer, we can use the hoop stress for thin wall pressure vessels calculation to check to find the minimum wall thickness for the silencer.

This part gets complicated, so im going to use engineers edge hoop stress calculator linked above:

For a 4130 chromoly silencer, using a tensile rating of 63100 psi and assuming perfect surface finishes, welds, and assuming a bunch of other shit you should never assume, i get a minimum wall thickness of .030". Using the above numbers, below .030ish wall thickness, it blows up in your face.

So, obviously youd want lots of margine for error.

As the barrel length is reduced, pressure goes way up.

As the silencer inner area, aka volume, goes down, pressure goes way up.

Welds have heat affected zones that can create spots weaker than the steels max tensile load.

Poor surface finishes give sharp microtroughs that give cracks a place to start.

Baffles inside the suppressor reduce volume and increase pressure.

The algebraic formulas above can be increased in accuracy and reduced in complexity by integrating the increase of area with respect to the bullets movement down the bore with the change in pressure, but i figured a calculus function would be harder to follow and visualize.

For me anyway, using the algebraic solutions, while not as accurate, lets me see whats going on in my head.

I used to manufacture suppressors, and for a typical .308 suppressor that i intended to be used on and "rated for" use on barrels as short as 16", i used 4130 steel at a wall thickness of .062", with tig fusion welded end caps.

As an example of how barrel length changes pressure at the muzzle, the above example had the 24" creedmoor putting 6700ish psi at the muzzle.

The same gun with a 16" barrel produces 9500 psi at the muzzle.

Please please,! dont use any of the mathy shit above to make a silencer. I rushed through it, didnt double check or verify anything. Its just a dirty example of how its done! Overbuild it

Another random thought, but cool as shit nonetheless.

You can use the above to calculate the force applied to the bullet at any given point as it moves down the bore, and using force mass acceleration method, f=m×a calculate the rate of change of velocity for the bullet. For example, in the throat the bullet has 3394 lbs of force pushing against the base.

So, the bullet has 3394lbs of force, so 3394=m×a

So now we find mass. A 140 grain bullet weighs .02lbs. So the mass is .0062 slugs.

So now 3394=.0062×a and we solve for accelleration. A=547419 feet per second squared.

So, at accelerating at 547,419 feet per second per second, how long does it take to get to 2700fps?

0.0027 seconds. not very long.

But, the formula above is an oversimplification too. But it wont be far off.

People dont like us nerds much, till theyre looking out the winder of their 747-8, looking down at a cloud layer 3 miles below them as its climbing through 35,000 feet. Then that shits a big deal. 900,000 lbs and 100,000 horsepower going 500mph

Seems like a miracle, but its engineering!

Time to go clean my pocket protector anf get another juice box.

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Yep, and as I said. I ment nothing derogatory, just try and keep it real. Everyone has their bad days.That was back when they were first starting out as Templar Tactical - correct? They certainly had their teething issues, like all companies.

And when you were in high school you probably thought, "I will never use this shit".Its done using a mathematical calculation called hoop stress.

https://www.engineersedge.com/calculators/hoop-stress.htm

A quick and dirty example:

We will use 2 calculations. One, to find the pressure inside the suppressor and two, to find the maximum allowable hoop stress for the suppressor material. In this 'dirty' example im going to cut alot of corners fore the sake of explanation.

We will use the simple algebraic p=f/a and work it backwards and forwards a couple times.

A 6.5 creedmoor has a saami maximum chamber pressure of 62000 psi. So first we will set p to 62000.

So, now were at 62000=f/a

Now we will calculate the inner sirface area of a 6.5 creedmoor loaded round. (Im gonna cut corners). (.473/2)^2×pi (1.75)=.3075"^2

So, we go with 62000=p/.3075, then solve for p. P=19065

So now we will rework it for p, using the new calculations for f and a. But this time we recalculate a to include the area of the barrel. We will assume a 24" barrel. So for a, we use .3075+(.264/2)^2x24=1.621

So, back to p=f/a we use p=19065/1.621 and solve for p. For p we have 11761.26 psi at the muzzle.

Now we work it over again, but include the area of the silencer. Again, dirty and cutting corners. We will assume a 1.5"od'8"long silencer. Sousing the previous calculations for the area of the cartridge and bore, but now adding the silencer we get 1.621+(1.5/2)^2x8=15.758 inches squared

So now, back to p=f/a to solve for p again.

P=19065/15.758 and we get 1209.861 psi inside the silencer.

So, now that we know the pressure inside the silencer, we can use the hoop stress for thin wall pressure vessels calculation to check to find the minimum wall thickness for the silencer.

This part gets complicated, so im going to use engineers edge hoop stress calculator linked above:

For a 4130 chromoly silencer, using a tensile rating of 63100 psi and assuming perfect surface finishes, welds, and assuming a bunch of other shit you should never assume, i get a minimum wall thickness of .014". Using the above numbers, below .014ish wall thickness, it blows up in your face.

So, obviously youd want lots of margine for error.

As the barrel length is reduced, pressure goes way up.

As the silencer inner area, aka volume, goes down, pressure goes way up.

Welds have heat affected zones that can create spots weaker than the steels max tensile load.

Poor surface finishes give sharp microtroughs that give cracks a place to start.

Baffles inside the suppressor reduce volume and increase pressure.

The algebraic formulas above can be increased in accuracy and reduced in complexity by integrating the increase of area with respect to the bullets movement down the bore with the change in pressure, but i figured a calculus function would be harder to follow and visualize.

For me anyway, using the algebraic solutions, while not as accurate, lets me see whats going on in my head.

I used to manufacture suppressors, and for a typical .308 suppressor that i intended to be used on and "rated for" use on barrels as short as 16", i used 4130 steel at a wall thickness of .062", with tig fusion welded end caps.

As an example of how barrel length changes pressure at the muzzle, the above example had the 24" creedmoor putting 11000ish psi at the muzzle.

The same gun with a 16" barrel produces 16156.78 psi at the muzzle.

Please please,! dont use any of the mathy shit above to make a silencer. I rushed through it, didnt double check or verify anything. Its just a dirty example of how its done! Overbuild it

Hypersonic burning powder granules impacting a 6061 aluminum blast baffle vs an inconel, titanium or chromoly blast baffle. Gas velocity, erosion, etc.

A suppressor may be manufactured to take the pressure, but of materials that wont stand up to particle impacts, etc.

I got self conscious and thought, what if some day somebody might come across this thread while researching their form 1 silencer build. I went back through and cleaned up the math just in case.

For future reference, if anyone blows themselves up, its on you. Leave me out of it. Using data you randomly come across on a website is a shitty idea.

But, like you would tell a teenage kid. I dont want u having sex, but if you do it anyway, use a condom.

I dont want anyone using my info to build a silencer, but if you do, the math was corrected with an edit in my original post. When i originally posted it i was bullshitting with family over thanksgiving, liquored up on Makers Mark, and seriously wasn't thinking about the possibility that some dumbass could actually build a silencer with a thin wall and get himself or someone else assholed!

For any future form 1 builder, do your own research and be careful. A few thousand psi in a thin wall pressure vessel, a couple feet away from your face is nothing to play with.

For future reference, if anyone blows themselves up, its on you. Leave me out of it. Using data you randomly come across on a website is a shitty idea.

But, like you would tell a teenage kid. I dont want u having sex, but if you do it anyway, use a condom.

I dont want anyone using my info to build a silencer, but if you do, the math was corrected with an edit in my original post. When i originally posted it i was bullshitting with family over thanksgiving, liquored up on Makers Mark, and seriously wasn't thinking about the possibility that some dumbass could actually build a silencer with a thin wall and get himself or someone else assholed!

For any future form 1 builder, do your own research and be careful. A few thousand psi in a thin wall pressure vessel, a couple feet away from your face is nothing to play with.

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Basic method: compute thin wall cylinder hoop stress and which includes knowing the uncorking pressure at the muzzle and within an existing suppressor. This has already been published using mil-spec 5.56 ammo, as the example here, for 14.5 and 10.5 inch barrels. Rounding up to the nearest 100 psi is about 2100 and 3000 psi respectively (as I recall and pretty close). Using 3000 psi, the hoop stress for 17-4PH stainless with a mean diameter of 1.305" and wall thickness of 0.10" is about 19,000 psi. Then we compare to yield strength of the 17-4PH under certain conditions, such as ambient temperature and also realistic temperature increases, depending on design parameters, like UCIW test, full auto, semiauto, or other cyclic rate. Ambient tensile for 17-4 is about 190 ksi. Increasing temperature to about 850F, the tensile and yield strength drop to just over 60%. Assuming we are looking for breaching 2% yield at 850F, we are down to about 115 ksi, as I recall (someone can verify the 17-4PH properties per spec). We need a safety factor. Assume our hoop stress was 20 ksi and yield is 100 ksi at 850F (duration is also a factor). Here we have a 5 to 1 safety factor. At a 1 to 1 factor, there is a 50% chance of failure, which can be a bulge or rupture. 2:1 is minimum safety factor. Aerospace metallurgy is 2.5:1 minimum with redundancies. Say full auto fire is sustained at maximum barrel cyclic tolerance, and the can can heat even more such that the safety factor can get much lower. So design and marketing parameters have an impact on final product. This is all math. This is, or should be, followed by destructive testing to actually prove the failure rate and is not a cheap process. Consider titanium rods are over $200 per foot for huge, bulk orders, and before machining starts. A good machine is well over $1 million. So, trying to make cans to sell as a going concern is difficult from one's garage that has a car in it and household stuff. Then add other designs for different calibers and barrel lengths and the variables increase.

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