OK, let's try recoil.
Mouse Hunter said:
Likewise, since there is an equal and opposite reaction, my rifle ought to recoil with 3,000 FPE.
You are correct about the equal and opposite reactions part, although that applies to forces, NOT Kinetic energy as you stated. The FPE that you are referring to is the KINETIC ENERGY of the bullet. You're not going to understand it at a basic level until you understand the difference between FORCE and ENERGY.
To use equal and opposite reactions you'll have to calculate the FORCE which is pushing the bullet down the barrel, which is equal to the FORCE pushing the rifle backwards. Once you know the force on the rifle, you can calculate it's acceleration and velocity. And if you have the velocity and mass of the rifle, you can calculate it's recoil FPE, which is NOT going to be anywhere near the bullet's FPE.
As I said earlier, a AA battery contains 7,000 FPE, just as your recoiling rifle on your shoulder contains 3,000 FPE. Do you think that if you put an AA battery up to your shoulder that you should recoil with 7,000 FPE just because it contains that much energy?
I'll go over the recoil math, but you're going to have to be pretty bored to follow it! Most people won't like it!
First, check out this table of recoil energy. Note that a .30-06 firing a 180 grain bullet at 2700 FPS only has 20 FPE of recoil energy, although if you do the calculations for the bullet it has 2900 FPE. Why is that?
http://www.chuckhawks.com/recoil_table.htm
Hopefully you know that Muzzle Energy (Kinetic Energy) of a bullet is Mass times Velocity Squared divided by 2. Sometimes written like this:
KE(bullet) = M(bullet) * V^2 / 2
The little "^" sign means raised to that power, squared in this case.
Likewise, KE(rifle) = M(rifle) * V^2 / 2
Ok, it's easy to find the mass of the rifle, but we also need the recoil velocity of the rifle.
Using Conservation of Momentum would be much simpler, but we'll go all the way down to the basic F = MA (Force = Mass * Acceleration) level.
Weight of bullet = 180 grains.
Since Force = Mass * Acceleration, we need the mass of the bullet. We'll use English units, and the English unit of mass is the "slug". Weight(lbs) = Force = Mass * Acceleration. On Earth, acceleration due to gravity = 32.2 ft/s^2
Knowing that good stuff we can figure out that Mass(slugs) = Weight(lbs) / Acceleration, so Mass(bullet) = (180 grains / 7000 grains/lb ) / 32.2 ft/s^2 = .0008 slugs.
Now we know the Mass(bullet), we need the Acceleration of the bullet so we can use F = M * A to calculate the Force on the bullet.
We're going to say that we have an 18" barrel on our rifle. We know that the bullet starts at a Velocity of 0, and accelerates to a velocity of 2700 ft/s in 18". So we can calculate the acceleration required to make it go from 0 to 2700 ft/s in 18".
The basic position equation is:
X = X(initial) + ((Velocity(initial)) * time)) + ((acceleration * time^2) / 2)
Usually written: X = X(i) + V(i) * T + (1/2) * AT^2 (This will be Equation 1)
The first derivative of position with respect to time is velocity, and second derivative of position (therefore the first derivative of velocity) is acceleration. I'm not going to try to teach you calculus, either take my word for it or stop reading now!
What this means is:
V = AT (This will be Equation 2)
In other words, Velocity = Acceleration * Time.
We currently have two unknowns for our bullet, Acceleration and Time. But the good thing is, we have two independent equations we can apply to our bullet. If you remember your basic algebra, you can solve for any number of unknowns if you have the same number of independent equations.
So, Equation 1:
X = X(i) + V(i) * T + AT^2 / 2
We're going to say that the zero point (that's X(initial), or X(i)) for our bullet is it sitting in the chamber. The end point "X" will be the end of the barrel. So X(i) = 0, and X = 18". We need to use the same units, so since we're using ft/sec we need to use feet for the barrel. 18" = 1.5 feet.
We also know that the initial velocity of our bullet is zero, that means that V(i) = 0.
So we can write:
X = X(i) + V(i) * T + AT^2 / 2 (which is Equation 1)
or
1.5 = 0 + 0 * T + AT^2 / 2 (which is still Equation 1)
which, if you remember algebra, is:
1.5 = AT^2/2 (which is still Equation 1)
Now we've got one equation and two unknowns. We need another equation, we can use Equation 2:
V = AT (Equation 2)
We know the velocity of our bullet at the end of the 1.5' barrel, it's 2700 ft/sec. That means Equation 2 is:
2700 = AT (Equation 2)
or
A = 2700 / T (still Equation 2, both sides divided by T)
Now we have two equations and two unknowns. We solve for A and T.
We'll plug our value for A from Equation 2 into Equation 1
1.5 = AT^2/2 (Equation 1)
1.5 = (2700 / T) * T^2 / 2 which reduces to:
3 = 2700 * T
T = 3/2700 = .00111 seconds
So our bullet goes from 0 to 2700 ft/sec in 18 inches over a time of .00111 seconds.
We can now plug this value for T back into Equation 2 to find the Acceleration:
V = AT (Equation 2)
2700 = A * .00111
A = 2700 / .00111 = 2,432,432 ft/sec^2
As a side note, if we want to know how many G's it's pulling, that would be
=2,432,432 / 32.2 = 75,541 G's
If you're designing some fancy electronics (self guiding, maybe?) for a .30-06 bullet, you better make them capable of withstanding a force of over 75,000 times their weight!
Back to recoil:
Now we know the bullet's acceleration, so we can finally calculate the average force over the length of the barrel on it:
F = MA
F = .0008 slugs * 2,432,432 ft/s^2 = 1,946 pounds
Finally! The Force on the bullet is 1,946 pounds.
As a side note, this is the same way you calculate the rocket burn if you're trying to rendezvous with the space station. Slightly different numbers but it works the same way. Figure out the difference in velocity you're trying to compensate for, the thrust(force) of your motor, and weight of your rocketship. A motor exerting 1,946 pounds of force on the base of a 180 grain rocketship for .00111 seconds will accelerate it to 2700 FPS in 18 inches.
Anyway, since we know about equal and opposite reactions, we know that this means that the Force on the rifle from the bullet also equals 1,946 pounds.
But to balance the Forces, there's also an additional Force shoving the rifle back. There's also the mass of the powder times the acceleration of the burning powder that acts on the gun.
Velocity of the gas is usually taken to be 1.5 times the velocity of the bullet. If we say that we used 50 grains of powder in our .30-06, we can figure out the Force due to the acceleration of the burning powder just like we did for the bullet. You can do it yourself, or just accept my calculations that 50 grains accelerated to 4050 ft/s in 18 inches requires a Force of 800 pounds.
This means that the total Force on the rifle is 1946 + 800 = 2746 pounds
2,746 pounds shoving on my .30-06 you say! I bet you figure I'm pretty stupid to claim that. But if you've made it this far, keep going another few minutes and you'll see!
Now we can calculate the Acceleration of the rifle.
We know that Force = Mass * Acceleration, and we have Force acting on the rifle. It's easy to get Mass, same way we got the bullet mass. We'll say that we have an 8 pound rifle. We need rifle mass in slugs which is:
M = F(weight) / A(gravity) = 8 / 32.2 = .25 slugs
Now it's easy to find the rifle's Acceleration:
F = M * A
2746 = .25 * A
A = 2746 / .25 = 10,984 ft/sec^2
This means that our rifle accelerates into our shoulder at 10,984 ft/sec^2 for the .00111 seconds that the bullet is in the barrel.
Remember Equation 2, the Velocity Equation? Now we can figure out the recoil velocity of our rifle:
V = A * T
V = 10,984 * .00111 = 12 ft/sec
So our rifle is recoiling at 12 FPS when the bullet leaves the barrel at 2700 FPS.
Look at the recoil velocity of the 8 pound .30-06 in the Recoil Table. Notice how they show 12.8 FPS. The difference between the 12.0 FPS we calculated and their 12.8 FPS is due to round-off and/or assuming a different powder charge.
http://www.chuckhawks.com/recoil_table.htm
Now that we know the recoil velocity of the gun, it's easy to compute the recoil energy:
E(gun) = (Mass(gun) * Velocity(gun)^2) / 2
E(gun) = .25 * 12^2 / 2 = 18 ft-lb of energy.
The recoil table shows 20.3 FPE, due to the fact that our V(gun) was 12 FPS and their V(gun) was 12.8 FPS.
Bottom line is your 180 grain bullet at 2700 FPS has 2900 FPE, and your 8 pound gun at 12 FPS recoil velocity has 18 FPE.
Told you that you wouldn't like it!
All this stuff is over 400 years old. Gun materials, strengths, and thicknesses as well as rocket trajectories, structural loads, power and fuel requirements, burn times, etc, are all based on Newton's Laws from 1687.