Recent content by Johnny Jr.

Mram10 us.... thanks for starting this thread. At first, I was dumbfounded that people were trying to say recoil has NOTHING to do with forces on the bolt. To me it was obvious that recoil from muzzle energy had to get to my shoulder mostly through the bolt lugs. I understood the pressure and...

I think some confusion can be attributed to a quote attributed to Kirby Allen, as stated in the very first post that started this thread ”2. the lugs feel MORE energy due to the bigger boltface” if you read the wiki reference on Bolt Forces it clearly says it’s the inside case diameter that...

Q: 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...

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.

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...

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...

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...

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...

I always assumed the forces on the boltface (and lug surfaces) was the ft lbs of energy derived from the bullet weight and it’s speed out the barrel. I also presume expansion of the case and it’s grip on the chamber wall would soak up some (maybe a lot?) of the energy reaching the bolt face...

I didn’t think so. This whole thread started with questioning the statement that “2. the lugs feel MORE energy due to the bigger boltface”. So lug forces are NOT a function of the boltface (or head diameter)? seems to me it’s a function of how big a bomb you are setting off in the chamber.

I am now really confused on the physics involved. So, I drastically reduce lug forces by rebating my 338LM to .473??Seems to me it would still kick like a mule, but be easy on the lugs. How can it recoil the same forces without going through the lug faces first?