Recoil is not just an effect of the ignition in the chamber. Think about all the gasses and the projectile traveling down the barrel. Since they are moving away from you and you are holding a stationary object there are rearward linear forces imparted on the barrel that are then transferred to the recoil lug and action body.
Lug setback? What am I missing?
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Recoil is not just an effect of the ignition in the chamber. Think about all the gasses and the projectile traveling down the barrel. Since they are moving away from you and you are holding a stationary object there are rearward linear forces imparted on the barrel that are then transferred to the recoil lug and action body.
Johnny Jr.
Member
So....the force on the bolt is the same in both the 460Wby and the 30-378 Wby (per the f= area x PSI equation), but the extra 3700 ft lbs of energy of the 460 is transferred to the back end through the recoil lug and the action body. Thanks. That makes the most sense so far. I always thought the opposite reaction to the muzzle energy to the butt pad was going through the bolt face.
Johnny Jr.
Member
Yeah, I get the bullet and gases making felt recoil. The bottom line is that if you stay under around 60,000 psi loads the bolt and lugs can handle it.....even with an 8000 ft lb muzzle energy round. Thanks again.
fiftybmg
Well-Known Member
I'm coming late to the party, and this sort of summarises all the things said here that I don't understand.
Chamber pressure is one thing. Pressure on the bolt face is another. How do you take a gas pressure, which is what the cartridge psi rating is, and make it equivalent to a surface pressure.
The psi rating of the cartridge is not the force against the bolt face. Cartridge pressure and force on the bolt are not related that way.
If the force is transferred to the lugs, the bolt face area is not important, it's the lug area of contact that matters.
When you have larger lugs, or more of them, it spreads the force out more [ less contact pressure, not less chamber pressure ], which is why heavy recoil calibers have bigger lug area, and why Weatherby bolts have so many lugs.
This force on the lugs that is spread out more is all taken up by the action, but at the points of contact of the lugs, the pressure is less per unit area with larger lugs, so as not to overwhelm the tensile stress of the steel.
To refer to Kirby, "The larger the case head size, the lower the breaking point in chamber pressure will be for any receiver" cannot be real, because then our 24 inch battleship guns would have to run at 100 psi.
A .223 has a much higher ratio of lug area to bolt face than a .338 Lapua. That is why it can handle higher chamber pressure before overcoming the tensile stress of the steel.
It does not make the .223 receiver stronger than a .338 Lapua receiver, or the force on a .223 bolt more than on a .338 Lapua.
fiftybmg
Well-Known Member
Totally incorrect.
A pressure of 50,000 psi means 50,000 pounds of force over one square inch.
If you apply 50,000 pounds of force over 2 square inches, it is not 100,000 psi, it is 25,000 psi.
By that logic, if I put a .223 cartridge in the chamberof a tank barrel, and fire it, does the tank blow up ?
fiftybmg
Well-Known Member
62,000 psi against a .585 bolt face or on a .470 bolt face or on a 2" bolt face is still 62,000 psi. On it's own, it cannot be used for calculating bolt thrust.
62000 pounds of force over 0.1735 square inches is 357,348 psi.
62000 pounds over 0.2712 square inches is 228,613 psi.
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fiftybmg
Well-Known Member
Yes. But we are not starting with a force, then going to area, we have force per unit area already. It is 62000 psi. The size of the case is not relevant, a small case can have the same pressure as a large case, the small case does not need the bigger action.
fiftybmg
Well-Known Member
psi is a very specific measurement. pounds per square inch. its pressure. it is not pounds on the bolt face.
You can apply 62000 psi to one square yard, or a pinhead, it's still 62000 psi.
Cartridge psi does not mean force on the bolt head, because we know that the amount of steel on the action is there to deal with force, and there is much more steel on a 50BMG action than there is on a .223, but the .223 runs higher pressure, so pressure on it's own does not determine force on the bolt.
ie : you can't determine force on a bolt from a pressure number on it's own.
You can apply 62000 psi to one square yard, or a pinhead, it's still 62000 psi.
Cartridge psi does not mean force on the bolt head, because we know that the amount of steel on the action is there to deal with force, and there is much more steel on a 50BMG action than there is on a .223, but the .223 runs higher pressure, so pressure on it's own does not determine force on the bolt.
ie : you can't determine force on a bolt from a pressure number on it's own.
fiftybmg
Well-Known Member
I'm not questioning the equation. Math is fact. I'm questioning wether you can take a psi number and determine bolt thrust from it.
Maybe you can, I'm just questioning.
adam32
Well-Known Member
Fill up a 1"x1' long PVC pipe with 100psi of air, and another 1"x100' long pipe and tell me which makes the bigger boom...it's all about volume...the only thing the equations listed don't account for. It's not just a simple one step calculation.
