# Not Homework (but it sure looks like it though)

I am a machine-gun enthusiast and am trying to understand the function of the “Grease-gun”.

The M3 Greasegun is a 'blow-back' operated device. In this case, this means it's very simple. The portion of it which I am concerned about at the moment consists of a pistol or rifle barrel, the receiver (a piece of tube about 2 inches in diameter and roughly 16 inches long), the bolt (a large piece of solid cylindrical steel which slides inside the receiver), and a compression spring which pushes the bolt forward towards the barrel.

The spring forces the bolt forward until it comes into contact with the bullet, which has already been inserted in the back of the barrel. At the time of contact, the bullet fires and starts moving down the barrel. While the the expanding gases in the chamber are forcing the bullet through the barrel, the same gases are trying to push the bullet casing out the back of the barrel, and to a certain extent are succeeding. Energy from the casing is imparted to the bolt which now starts moving backward into the now collapsing spring. Eventually the rearward motion of the bolt is halted by the spring, and this is all the farther I need to go right now.

The point is to find how far the bolt has traveled backward at the time the bullet leaves the barrel.

Where: s = 1/2 at*t (read: t squared) then dv = 2s/t

When the bullet is fired, the casing, and thus the bolt, accelerate back into the receiver until the bullet leaves the barrel, relieving the pressure in the chamber. From here the bolt decelerates due to the action of the compression spring until the bolt stops.

This part of the cycle can be broken down into two parts, the acceleration of the bolt, followed by the deceleration of the same. At the beginning and end of this part of the cycle, the bolt has zero velocity. At the end of the acceleration, and the beginning of the deceleration, the velocities of the bolt are by definition the same.

Where s1 and t1 refer to the acceleration phase and s2 and t2 refer to the deceleration phase,
and where in real life t1 is known to be about 0.007s, and t2 is known in real life to be about 0.085s, and s2 is known in real life to be about 6.125 inches;

then: 2 s1 / t1 = dv = 2 s2 / t2 or 2(s1)/(0.007s) = 2(6.125”)/(.085s) or s1 = 0.504”.

That's ridiculous. S1 should be closer to 0.010” – 0.030”.

Where did I go wrong?

SpectraCat
I am a machine-gun enthusiast and am trying to understand the function of the “Grease-gun”.

The M3 Greasegun is a 'blow-back' operated device. In this case, this means it's very simple. The portion of it which I am concerned about at the moment consists of a pistol or rifle barrel, the receiver (a piece of tube about 2 inches in diameter and roughly 16 inches long), the bolt (a large piece of solid cylindrical steel which slides inside the receiver), and a compression spring which pushes the bolt forward towards the barrel.

The spring forces the bolt forward until it comes into contact with the bullet, which has already been inserted in the back of the barrel. At the time of contact, the bullet fires and starts moving down the barrel. While the the expanding gases in the chamber are forcing the bullet through the barrel, the same gases are trying to push the bullet casing out the back of the barrel, and to a certain extent are succeeding. Energy from the casing is imparted to the bolt which now starts moving backward into the now collapsing spring. Eventually the rearward motion of the bolt is halted by the spring, and this is all the farther I need to go right now.

The point is to find how far the bolt has traveled backward at the time the bullet leaves the barrel.

Where: s = 1/2 at*t (read: t squared) then dv = 2s/t

When the bullet is fired, the casing, and thus the bolt, accelerate back into the receiver until the bullet leaves the barrel, relieving the pressure in the chamber. From here the bolt decelerates due to the action of the compression spring until the bolt stops.

This part of the cycle can be broken down into two parts, the acceleration of the bolt, followed by the deceleration of the same. At the beginning and end of this part of the cycle, the bolt has zero velocity. At the end of the acceleration, and the beginning of the deceleration, the velocities of the bolt are by definition the same.

Where s1 and t1 refer to the acceleration phase and s2 and t2 refer to the deceleration phase,
and where in real life t1 is known to be about 0.007s, and t2 is known in real life to be about 0.085s, and s2 is known in real life to be about 6.125 inches;

then: 2 s1 / t1 = dv = 2 s2 / t2 or 2(s1)/(0.007s) = 2(6.125”)/(.085s) or s1 = 0.504”.

That's ridiculous. S1 should be closer to 0.010” – 0.030”.

Where did I go wrong?
The forces on the bolt are time-dependent, so you cannot use the assumption of constant acceleration, which your analysis seems to be based on (if I follow it). So, to find the answer, I am pretty sure you are going to need to make some assumptions about the time dependent forces from the spring (harmonic treatment is a good place to start), and from the expanding gas (seems much trickier). Once you know those, you can set up the appropriate integrals over the time intervals you mentioned.

Just out of curiosity .. does the bolt really travel 6+ inches during recoil?

"Just out of curiosity .. does the bolt really travel 6+ inches during recoil? "

If you're firing hot loads, yes.