Would that mean recoil would be less with the smaller boltface?
Lug setback? What am I missing?
- Thread starter Mram10us
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Johnny Jr.
Member
So I read the wiki page on bolt forces
ive got another example
460 Weatherby 600 gr putting out 8000 ft lbs at 62000 psi
30-378 Weatherby 150 gr putting out 4300 ft lbs 62000 psi
identical interior case head diameter (stipulated in the wiki reference) and same psi. According to the formula, psi and head diameter are the only variables that dictate bolt forces. How does that extra 3700 ft lbs of energy bypass the lugs and get to my shoulder?
birdiemc
Well-Known Member
No idea.
But if every action has an equal and opposite reaction, the bullet leaving the muzzle might be where the recoil is coming from? So equal sized bullets at equal speeds would have equal recoil, whereas increasing the mass of the bullet would increase the recoil...have I got it right here?
But if every action has an equal and opposite reaction, the bullet leaving the muzzle might be where the recoil is coming from? So equal sized bullets at equal speeds would have equal recoil, whereas increasing the mass of the bullet would increase the recoil...have I got it right here?
bigngreen
Well-Known Member
I think I just got dumber reading this thread, unbelievable some of you are actually putting guns together and can't grasp this!!
MNbogboy
Well-Known Member
Pressure is equal in all directions. At what point peak happens is variable. All objects are considered "immovable" because of their construction. The push against the case head/lugs is where the "weak" link is in question. The actual weak link is the bullet, which travels down the barrel where pressure is relieved.
Getting back to the "immovable" case held by the chamber which is designed to hold the "max" pressure. The immovabable objects include the case walls and shoulder.
The pressure puts an equal force rearward where the "whole" case can actually move backwards because of headspace. This motion backwards closes the headspace with a "slam" against the boltface. The area that the pressure is pushing backwards against is the inside diameter where the pressure cup ends and thins out to form the case wall.
It all boils down that this area in the larger diameter cases is more so the force backwards is more.
If I wasn't retired I could mock up all of this in a finite element analysis (FEA) program and show the actual pressures in the case/barrel/action. This would show why lower lug setback is more.
In actuality most actions are normally weaker on the bottom because of the magazine cutouts. Less support behind the lower lug. The action actually flexes back on the bottom driving the top lug into its abutment. I believe most wear usually shows on the top lug/abutment.
drakehammer
Well-Known Member
Yep. I've never seen dumb progress at such a rapid pace. Impressive! Carry on...my 3rd bag of popcorn is almost done.
My rifles I build are very accurate, thank you
I'm betting there are many theories we don't have a grasp on. It's good to debate and learn
That is very informative! I am so glad you added such wisdom to the thread .... wait ... finding an eye roll face....
MNbogboy
Well-Known Member
Recoil is independent from bolt thrust.
Bolt thrust is mainly a STATIC force on the bolt face & lugs and is caused by chamber pressure.
Recoil is a DYNAMIC force (the equal and opposite reaction of the bullet motion) which is also initiated by chamber pressure. Muzzle energy directly correlates to recoil. In actuality recoil begins when the bullet starts moving at ignition.
The term bolt thrust is very misleading to some. Would make you believe the bolt is moving. But rather it is a measure of the actual forces placed upon the bolt.
While recoil is a measure of the forces placed upon the whole rifle as one unit.
Bolt thrust is mainly a STATIC force on the bolt face & lugs and is caused by chamber pressure.
Recoil is a DYNAMIC force (the equal and opposite reaction of the bullet motion) which is also initiated by chamber pressure. Muzzle energy directly correlates to recoil. In actuality recoil begins when the bullet starts moving at ignition.
The term bolt thrust is very misleading to some. Would make you believe the bolt is moving. But rather it is a measure of the actual forces placed upon the bolt.
While recoil is a measure of the forces placed upon the whole rifle as one unit.
Johnny Jr.
Member
Yes, it is kind of hard to grasp an equation that calculates forces placed on a bolt and lugs (and ultimately my sore shoulder) that doesnt take into account bullet weight and speed. Lets stipulate that the equation is right and the PSI and case head diameter are the same......the force on the bolt is the same. The kinetic energy of 8000 ft lbs and 4300 ft lbs is NOT the same ....so explain how the extra energy of recoil is NOT transferred through the bolt and lugs.
MNbogboy
Well-Known Member
The calculation for bolt thrust is for bolt thrust only and unrelated to recoil.
The calculation for recoil is based on bullet weight, rifle weight, charge weight & impulse, and muzzle velocity. You can find many recoil calculators on line. They do not use chamber pressure unlike the calculation we used here for relative bolt thrust.
If you accidently lodged a bullet in your barrel and fired another round in the chamber and neither one exited the barrel, chamber pressure and bolt thrust would be sky high.
Ironically, if the gun failed to blow up there would be no recoil either.
The calculation for recoil is based on bullet weight, rifle weight, charge weight & impulse, and muzzle velocity. You can find many recoil calculators on line. They do not use chamber pressure unlike the calculation we used here for relative bolt thrust.
If you accidently lodged a bullet in your barrel and fired another round in the chamber and neither one exited the barrel, chamber pressure and bolt thrust would be sky high.
Ironically, if the gun failed to blow up there would be no recoil either.
drakehammer
Well-Known Member
You see, there really isn't much to add that hasn't already been said and repeated ad nauseam for your benefit...wait...microwave just beeped.
Johnny Jr.
Member
So the force of dynamic recoil is not delivered to me and my stock through the bolt and lugs? Or is it possible that the "bolt force=area x psi" equation is just part of the energy imparted on the bolt? I'm not purposely being hard headed, but I can't imagine that the equal and opposite reaction of the kinetic muzzle energy does not impinge upon and transfer through the bolt face and lugs, somehow. I mean, that punishing pain of 8000 ft lbs of energy has to get back there by some mechanism. If it's not through the bolt, I'm at a loss. Sorry. Maybe I AM as dumb as my wife thinks.
Without going through and reading every post I can already see that there is some miscommunication going on here. @cohunt has got it correct so let me see if I can help fuse the disconnect. Let's star with firing a round out of the rifle. You have certain primary and secondary forces.
Pull the trigger and ignite the powder sending the projectile forward. At this point the brass swells and seals of the chamber as part of the primary forces. When it does this it creates a piston in which the secondary forces are going rearward.
At this point the internal pressure is pushing on the case head which is making contact with the blot face. This is where a lot of the miscommunication is happening. You have to convert this now to force which has been applied to the bolt face. That force is then transferred from the bolt face to the bolt body and then on to the lugs.
Here is where it can get confusing. Without going through all of the calculations we will just skip to the important parts. 1psi + 1 lbf/sqin. Using that calculation you can easily see that if you keep the pressure the same but increase the surface area in which the pressure is applied then the force imparted upon that surface area is greater then the smaller surface area.
Someone mentioned 1000lbs on a 4' table is the same as on a 1' table and that is absolutely true but when you figure forces they are vastly different. Using pi*r^2 you get a surface area of just under 1,810sqin for a 4' round table which means that a force of approximately 0.55 lbf/sqin is being imparted upon the table. Make that table a 1' round one and you get a surface area of just over 113sqin. That same 1000lbs is now exerting approximately 8.85 lbf/sqin
Pull the trigger and ignite the powder sending the projectile forward. At this point the brass swells and seals of the chamber as part of the primary forces. When it does this it creates a piston in which the secondary forces are going rearward.
At this point the internal pressure is pushing on the case head which is making contact with the blot face. This is where a lot of the miscommunication is happening. You have to convert this now to force which has been applied to the bolt face. That force is then transferred from the bolt face to the bolt body and then on to the lugs.
Here is where it can get confusing. Without going through all of the calculations we will just skip to the important parts. 1psi + 1 lbf/sqin. Using that calculation you can easily see that if you keep the pressure the same but increase the surface area in which the pressure is applied then the force imparted upon that surface area is greater then the smaller surface area.
Someone mentioned 1000lbs on a 4' table is the same as on a 1' table and that is absolutely true but when you figure forces they are vastly different. Using pi*r^2 you get a surface area of just under 1,810sqin for a 4' round table which means that a force of approximately 0.55 lbf/sqin is being imparted upon the table. Make that table a 1' round one and you get a surface area of just over 113sqin. That same 1000lbs is now exerting approximately 8.85 lbf/sqin
