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Lug setback? What am I missing?
- Thread starter Mram10us
- Start date
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Ha! Thanks for that
bigngreen
Well-Known Member
You might be able to look at recoil as a felt representation of the work that you can get done with the chamber pressure, you can change the tool (bullet) but the muscle behind it will remain the same.
MNbogboy
Well-Known Member
Imagine a 30 carbine round forced into a 223 chamber, bullet is pushed up against the neck area of the chamber....bang....bullet goes no where...primer is blown....gun locks up solid....bullet goes no where.....bolt thrust is sky high.............no perceptible RECOIL....
For every action there is an opposite and equal reaction....ie; recoil is the opposite reaction of a bullet moving.
Clarifying, the recoil is transferred through all the rifle parts into the shoulder through its construction including barrel, reciever threads, bolt LUGS, receiver ABUTMENTS and recoil lug. All rifle parts work together as a mass to transfer recoil to the shoulder.
For every action there is an opposite and equal reaction....ie; recoil is the opposite reaction of a bullet moving.
Clarifying, the recoil is transferred through all the rifle parts into the shoulder through its construction including barrel, reciever threads, bolt LUGS, receiver ABUTMENTS and recoil lug. All rifle parts work together as a mass to transfer recoil to the shoulder.
Agreed, but we are assuming (and so was Kirby Allen) that the cartridge is functioning properly. Assuming this, why Wouldn't recoil be a good determinant of the lug forces experienced?
Im enjoying this discussion because sooooo much has come out of it for many other issues
adam32
Well-Known Member
Except it was stated earlier (I forget by who?) that recoil does NOT transfer thru the bolt lugs and it has ZERO effect on bolt lug setback...
I've been saying since the beginning that it is ALL things working together that contribute to bolt lug set back, not just one simple equation that converts chamber PSI to SQ IN on the bolt face.
adam32
Well-Known Member
Elaborate a bit more...so if recoil is a felt representation of chamber pressure then recoil should be the same across the board with equal chamber pressure? The bullet and case will have zero effect on recoil?
jpndave
Well-Known Member
For the forces we are dealing with on the lugs the weight of the bullet does not matter. Think of a balloon. The pressure at any given point on the inside surface is the same as any other. A gas under pressure eqalizes against all the walls. If anyone doesn't understand this then they need to go back to physics 101. That's beyond what we are going to be able to cover here. I and I suspect many others here have made an assumption that we're all at a minimum level here and have a basic understanding of rifles and physics. If I have assumed incorrectly I apologize.
Force on the lugs is the pressure (we've been talking peak - so at that instant in time the pressure is highest) times the rear inside case area minus any other factors such as case adhering to the chamber wall. That's it. All that calculation does is tell us the peak instantaneous load on the lugs. The total force to the lugs is a combination of all the forces and the duration (time). The bullet weight would impact the time because it will take longer at a give force to move the bullet to the end of the barrel. For that the weight of the bullet would impact the cause we are using but not the calculations directly.
Recoil is the combined energy delivered rearward from all the forces related to firing the bullet. The force transferred to the lugs is one component of that. For that calculation the weight of the bullet does matter. As does the gas blast, weight of the rifle, length of the barrel, brake or not, etc.
Sorry for the wordy response and I hope I'm not being offensive here. Again if so I apologize.
jpndave
Well-Known Member
Are you really wanting a discussion here or just to antagonize and argue? If the former I'm in for the duration. If the later then I'm out on what could be a valuable discussion.
MNbogboy
Well-Known Member
One more example; 338 Lapua...225 grain @ 3000....13# gun...approx 27 pounds recoil energy....
Now turn the barrel down and flute it...switch the stock to lightweight carbon fiber...switch the scope to a lightweight....now the same gun weighs 6-1/2#.....shoot the same load 225 grain at 3000 fps....approx recoil energy is now 54 pounds....
The calculated bolt thrust is the same in both cases. The only thing that has changed is the calculated barrel strength (which would "burst" much easier).
The point here is static strengths of materials are irrelavent to dynamic recoil calculations.
A pressure VALUE within a chamber/barrel is required to calculate the static strength of the bolt lugs and abutments.
The ACTUAL pressure within the chamber/barrel will propel the bullet forward and the rifle/components backwards.
Now turn the barrel down and flute it...switch the stock to lightweight carbon fiber...switch the scope to a lightweight....now the same gun weighs 6-1/2#.....shoot the same load 225 grain at 3000 fps....approx recoil energy is now 54 pounds....
The calculated bolt thrust is the same in both cases. The only thing that has changed is the calculated barrel strength (which would "burst" much easier).
The point here is static strengths of materials are irrelavent to dynamic recoil calculations.
A pressure VALUE within a chamber/barrel is required to calculate the static strength of the bolt lugs and abutments.
The ACTUAL pressure within the chamber/barrel will propel the bullet forward and the rifle/components backwards.
jpndave
Well-Known Member
This is incorrect.
First the forces against the lugs contribute to some of the recoil. There are also other forces going on that push on the lugs not directly realized as recoil. For example, in taper walled shouldered cases chamber pressure is pushing forward on both the case taper and the shoulder. Those forces would cancel out a portion of the forces to the lugs as it pertains to recoil. However as it relates to lug pressure and potential setback (remember the original question for this thread?) the lugs take all of that pressure.
Second,
The force on lugs is the total of all the pressure exerted from firing the round with its relating time. See my previous post. It's the gas pressure and time less any factors taking any of that pressure such as case sticking to the chamber wall. Nothing except the pressure applies against the lugs. It's not all things just the pressure adding to but there are factors that may take away.
jpndave
Well-Known Member
Great question. Yes, more lugs would spread the force out. Force/3 is less than force/2. More lug area spreads the load out better to hopefully get it below the yield strength of the lug/receiver material.After taking a few days to get through this, it has been interesting.
idcwby
