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Newtonian motion/guns/pressure/materials


YT2095

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here`s a problem I can`t quite figure about Bullets/ammo.

they fall into 2 catagories, center and rim fire.

both of which need a "Hammer" to strike the primer to start the reaction.

 

now the power a bullet comes out at is incredible! but Newton states that for every action, there is an equal and opposite reaction.

so how come, the primer case (being only thin metal) doesn`t rupture and blow off the firing pin?

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Obviously the metal is thick enough that it doesn't rupture; the gas expansion forces the bullet out, and that's the path of least resistance for the gas. The section for the primer is much smaller, so even for a large pressure, the actual force on that section is smaller than the force on the bullet by the ratio of the areas.

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well for center loaded types (most things above .22cal) the primer is loaded from the Outside (on a re-load), we tap them out with a tool from the inside and they "pop out".

the new primer is then loaded back into place by gentle presure from the outside to push it in, now it doesn`t take alot of pressure to remove these on a re-load at all!, so why not when firing?

 

it still doesn`t make sense even if you consider surface area, think .223, the width of the back of that round against that of the primer, what is it? maybe 3 perhaps 4 times the area thereabout?

 

so 1/4`th of that pressure is pushing back against it!, now a .223 is a Supersonic round (in most cases) and even a 100`th of that pressure would blow the primer out and ruin the firing pin.

 

so what`s going on exactly?

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I belive this is to do with, like swansont said, the path of least resistance.

 

When the primer blows there's a big pressure, it's so easy for that pressure to force the bullet out there's a small recoil relative to the force on the bullet.

 

As for "equal and opposite reaction" that doesn't apply to the gun and the bullet as the gun itself isn't applying a force on the bullet. What you have is the explosion pushes the bullet with a large force and so the bullet pushes the explosion with a large force, by explosion I'd be referring to the expanding gas.

 

As the gun itself isn't exerting a force on the bullet the bullet isn't exerting a force on the gun.

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  • 3 weeks later...

The chamber that surrounds the cartridge actually fits the casing to very close tolerances. When the round goes off the entire case is presses very hard against the "breach", this surrounding chamber. This includes the thin skin of the primer, except for the rather small hole where the pin passes through the breach to strike the primer. It is the breach that resists the pressure, not the casing. So what about the firing pin hole?

 

Well, as I said it's small, .050" or maybe a little larger. And it's filled by a firing pin that is moving toward the primer at the moment of detonation. So the momentum/inertia of the firing pin itself prevents the primer from rupturing at the point of impact.

 

And by the way, equal and opposite reaction absolutely applies. We call that the recoil of the firearm. Ever fire a 12 guage shotgun? ;)

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So the momentum/inertia of the firing pin itself prevents the primer from rupturing at the point of impact.

 

And by the way' date=' equal and opposite reaction absolutely applies. We call that the recoil of the firearm. Ever fire a 12 guage shotgun? ;)[/quote']

 

I understand how the gun(s) work and yes I`ve fired 12 guage and many other different rounds quite regularly.

it`s only the part about the firing pin and weight/momentum of the hammer keeping the cap in place that I find hard to understand, there just doesn`t seem to be enough stopping "Energy" in that mechanism to prevent it blowing back?

 

thnx though, Good answer :)

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The 12 guage remark was for 5614, who seems to be a little confused.

 

As to the question of how the inertia/momentum of the firing pin can resist the explosion, think of it this way.

 

Suppose you have 2 hydralic cylinders, A and B, connected to each other by a line that lets oil flow back and forth between the two. The piston of B has 100 times the area of A. You press in on B with 100 lb. To counter that force, you only have to push 1 lb. on A.

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