I have a question which I cannot quite figure out. Was going to post on /sci, but that place is dark and especially angry, so I'm giving /k a shot. Feel free to speak in math, though please explain your variables so that a clueless hack can figure them out. Disclaimer: I am not savvy about firearms or physics... it will be obvious soon. Feel free to correct numbers and terms which I have mistaken. Numbers are LOOSE, just trying to get ballpark answer.
.30-06 rifle cartridge has about 2700 Nm of energy at muzzle. Assume gun barrel is 0.6 m long. This number is staggering to me, about ten times the power of most handguns? Regardless, this equates to an avg. force of 450 kg (1000 lb) accelerating the bullet down the barrel for its full length. Assume 4.5 kg or 10 lb rifle. I looked up chamber pressure for .30-06 and the numbers all work out, at least 450 kg of force behind the bullet.
Every action has equal and opposite reaction. This concept makes me incorrectly think that an assault rifle would have to be mounted VERY securely. Calculated muzzle energy would accelerate 15 lb gun backwards at 60 mph (?), which would seriously injure an arm. Yet they are shot from the shoulder with very manageable recoil.
What am I missing?
Thanks-
>>29190333
Kinetic energy has nothing to do with recoil impulse.
It's all about momentum.
The easiest way to explain it is conservation of momentum. At the start, the bullet and rifle can be considered one combined unit. When you pull the trigger, you are causing a separation of the bullet rifle unit. Initially the momentum was zero, so after this action the total momentum, ie the sum of the bullet and rifle momentum, must be zero. Momentum is a vector, which means it has magnitude and direction, so momentum of the bullet and momentum of the rifle point in opposite directions, and must have the same magnitude in order for our conservation statement to hold true. But, just because the rifle and bullet have the same magnitude of momentum, it is easy to see that they will have different velocities when we look at the equation for momentum: p=mv
Equating the rifle and bullet magnitudes we see
m_r * v_r = m_b * v_b
m_b / m_r = v_r / v_b
This says the ratio of masses is the same as the ratio of the velocities. We know that the mass of the bullet is much less than the mass of the rifle, so the term on the left is small. This indicates thay the term on the right must also be small, or that the velocity of the bullet must be much greater than the velocity of the rifle.
>>29190333
at no point has there been an assault rifle chambered in 30-06, that would be ridiculous and impractical especially in full auto
>>29190680
>das it mayne
>>29190731
Assault rifles are chambered in intermediate cartridges, so of course there hasn't been any produced in .30-06.
>>29190731
>BAR
Yeah, intermediate, shut-up.
>>2919068
Thank you for detailed response. I didn't put the numbers into your equation yet, but I will work on it. If someone feels like applying some real world numbers, please do, and post results!
also I apologize for duplicate post...
>>29190831
The BAR is a light machine gun, not an assault rifle. Also hinging upon the whole full sized cartridge thing...
Assault rifle misnomer aside... though feel free to do the math for one too...
Putting the numbers into
>>29190680
the equation, it says the rifle will be accelerated to 35 m/s. 180 g bullet fired at 820 m/s and a 4535 g rifle.
35 m/s is seventy something miles per hour. If a 10 lb rifle was coming that fast at my shoulder, it would not be pretty. What's the missing variable?
>>29190937
32.5 m/s, mistyped
>>29190937
I'm just curious what happens when you add the mass of the shooter to the rifle? I'm assuming a 10 lb rifle and 180 lb shooter would drop the velocity to 1.5 m/s or so
>>29190937
doesnt gas block help stop this shit
>>29190937
For the bullet weight, are you using grams or grains? They're two totally different units of measurement.
>>29190937
This missing variable is that you cannot use non-SI units when using basic kinematic equations.
Using the ratio equality, I solve for the velocity of the rifle:
m_b * v_b / m_r = v_r
(0.011kg)(850 m/s) / (5.3kg) = 1.8 m/s
Or ~4mph.
>>29190978
Weight of shooter would be relevant in space, but what I'm trying to figure out is the interaction between rifle and shooter. The shooter is the damper, the decelerator between rifle and ground. They are the object to absorb the energy.
Recoil might be .1 meters, in shoulder movement. According to math so far, it's absorbing a LOT more energy than it does in practice.
>>29191020
GOOD CATCH
works out to 2.2 m/s or 5 mph
>>29191039
I did use SI units, except I pulled the wrong number (gr) off the chart (should have used g), grams or kilograms doesn't make a difference as long as they're used consistently.
We've cracked the case. Thanks for help-
>>29191095
As a physicist, it matters very much if you do or don't use SI units...
>>29191040
http://www.wolframalpha.com/input/?i=150gr+*+2910ft%2Fs+%2F+9.5+lbs
recoil isn't that bad, you fucked up your math somewhere
>>29190937
The missing variable is the length and duration of the momentum absorption.
