A safe load in a safe rifle produces a certain, safe amount of pressure (for that rifle) thus, a safe amount of stress on the combination bolt-receiver-barrel and a certain amount of recoil.
If one compares two cartridges in a given caliber, let's say a 150 gr. bullet and a 200 gr. bullet, in a 30-06, both propelled by different (but safe) powder charges that produce equal pressure, would it be justified to assume that more recoil will be felt with the 200 gr. bullet and therefor firing the 200 gr. bullet will create more stress on the rifle due to the greater bullet mass that has to be moved?
Raindeer, As a general rule, the pressure is regulated to stay within acceptable parameters by using a powder that burns slower or by reducing the charge of the same powder. If you increase the weight of the projectile without changing the powder charge, the pressure will go up. That's a bad thing.
Does this babble answer your question?
Posts: 234 | Location: 40 miles east of Dallas | Registered: 21 December 2002
Yes. No. Maybe. Sort of. Finding 2 loads with different weight bullets, the same pressure curve, and the same charge weight would be difficult!!! Assuming we can do so, for arguments sake...
The stress on the stock is more, obviously, as you feel the kick harder, but the internal stress on the bolt and action is pressure determined. The amount of time the stress is applied will change...more for the heavier bullet, as the lower velocity would leave it in the bore longer...but as long as the loads are in the safe range, it is a pretty good assumption that the difference would not affect the useful life of the action.
Posts: 1780 | Location: South Texas, U. S. A. | Registered: 22 January 2004
there will be more recoil with the heavier bullet but since you say the pressure is the same or very close there shouldn't be increase in stress to worry about
Posts: 110 | Location: Minden , Nebraska | Registered: 23 July 2004
The free recoil of your firearm can be found using the formula below:
((weight of the bullet in pounds x the velocity of the bullet in feet per second)+(4,700 x the weight of the powder in pounds))squared / (64.348 x the weight of the gun in pounds)
Nowhere in the formula does the pressure come into consideration...
Posts: 3282 | Location: Saint Marie, Montana | Registered: 22 May 2002
ALF, I believe the gas jet is figured in the equation Steve gives. If you look close at the equaton it shows the bullet wgt X vel. PLUS the powder Wgt(gas) X velocity (4700fps is the figure for the velocity of excaping gases "jetting" from a barrel) The recoil is reduced with a brake by redirecting the gas at a angle to the bore, reducing the amount exiting the barrel (wgt x vel) and causing a "second jet effect" (wgt x vel)to counter act the recoil. JMHO...
Posts: 2535 | Location: Michigan | Registered: 20 January 2001
