OK, somethingis wrong with the math here!

Status
Not open for further replies.
B_Scott,

Yes, 6 Newtons is the force the bullet exerts on the rifle. But then the rifle, once accelerated by the bullet, begins to exert a different force on the shooter's shoulder. I think you have to treat the rifle and the shooter as two independent things here, unless of course the rifle is welded into the shooters skeleton. My guess is that the rifle decelerates on the shoulder over a much longer period of time than the bullet accelerates out the barrel, so the force will be smaller than you might expect.

Sumpnz,

Good point, I just assumed that the only significant source of mass was the bullet, but in many cases the powder can be as much as 30% of the total mass exiting the muzzle, which makes a big difference. The magnums kick harder not only because the bullet is faster, but because there is additional mass, disguised as burned powder. I suppose this is why muzzle brakes work so well on magnum rifles, because you can turn all that gas momentum into your advantage by bouncing it off plates attached ot the front of the muzzle.
 
B_Scott, Powder mass most certianly affects the recoil of a rifle. If you made a magnum rifle heavy enough, it would recoil like a normal .223. It would take a lot more than 1/2 pound though.

Think about it this way. The powder is being ejected at high speed from the muzzel. So is the bullet. I think we can all agree on that. So why would the bullet's mass contribute to recoil, but not the powder?

Magnums kick harder mostly becuase of the increased velocity and/or mass of the bullet being fired. But the extra powder most certainly affects that recoil too.

Are you a reloader? If so, try using a few different powders that have significantly different weights required to produce the same velocity for the same bullet from the same rifle. Measure the recoil from each load and then tell me that powder weight makes no difference.

FYI, some reloading manuals contain formulas for calculating recoil. Guess what those formulas include: Powder mass and rifle mass.
 
A .243 bullet fired from 300 meters can push a 1 lb object for a distance of 1000 feet. Remington Catalog lists stats for a .243 bullet: 243 - 100 gr bullet 1,945 ft - lbs energy .30-06 - 150 gr bullet 2,820 ft - lbs energy .300 Win mag 150 gr bullet 3,605 ft - lbs energy
 
OK - I learned something new today.
I though that all the powder would be burnt up by the time the bullet exited the barrel.
My mistake.
Carry on.
Another thought - you might want to divide the force into 2 vectors due to the kick of the rifle. ???? just a thought. Or would the energy already be transfered at this point???
 
B_Scott, In many cases the powder is completly burned by the time the bullet exits the muzzel. However, just becuase something is burned does not mean its mass goes to zero. Burning is simply a chemical reaction in which a lot of heat is generated by oxygen combining with, in this case, smokeless powder. If you could somehow capture and measure the mass of the propellant gasses you would find that the mass of the powder, combined with the mass of air consumed in the combustion process, remains equal to the propellant gas mass. Hope that made sense.

Not exactly sure what you are getting at with your second point.
 
This is the coolest thread I think I've ever seen! Physics, shooting, AND Monty Python!

Thanks to you guys that gave those links, I'll be reading that hyperphysics site for days, thanks WT!

oh yeah, and "Look, look! I'm bein repressed!"
 
This is the coolest thread I think I've ever seen! Physics, shooting, AND Monty Python!

...

oh yeah, and "Look, look! I'm bein repressed!"
I would have thought you'd rather "go back and face the peril."
 
Status
Not open for further replies.
Back
Top