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Old 03-11-2006 | 09:10 PM
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Pglasgow
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Default RE: Pressures, .50 caliber rifles, strength of materials, PART 1

ORIGINAL: Wolfhound76


Force = Pressure (lb/sq in.) * Area of bore (sq. in.)

So at peak pressure, the force is:

Force = 51000 * (PI*.25^2)/4 = 2503 lbs. force

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Pressure (psi)= Force (lbs force) / bore area (sq in.)

Pressure = 2503 / ((PI*.5^2)/4)= 12,750 psi
With the top formula are you assuming a 100 gr bullet weight? If that's so since you used the 2503 then you have a 50 cal shooting a 100 gr bullet for that pressure. As you increase bullet weight the pressures go up.
Re-read it Wolfhound. Take some time and it will all be clear to you. I don't have a 100 grain bullet in the .50 cal. The force on the bullet is completely independent of the bullet weight. All we calculated was how much pressure it takes to impress a given force onANY bullet. For a given force, it takes 4 times the pressure in .250 cal as it does in .500 cal. It is indisputable.

It is also indisputable that loads with peak pressures in the neibhorhood of 13000 psi can produce muzzle energies of 2300 ft. lbs. In .500 cal. that is.

Now that said. Given a particular charge. Will a heavier bullet cause higher pressures to build in the breech? You bet it will. But thats not the topic. The actual topic is this. One doesn't need 3 pyrodex pellets for a service load.

Happy Hunting, Phil


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