Energy loss from slide action/recoil?
- Thread starter Carbonator
- Start date
- Status
Carbonator
Member
Too bad I can't perform all of Archie's tests at this point and time. Anyone want to lend me some handguns and ammo? Can't be responsible for condition of returned guns though. 
This is one of those great threads! 
One factor that's scarcely been addressed is system compliance. This is impossible, but it will illustrate the point. If the gun/case combination weighed the same as the bullet AND was totally unconstrained by anything or anybody to prevent its movement, it would be analogous to the frictionless swivel chair demonstration where you can't rotate the chair by moving your arms or legs. The gun would go zinging in the opposite direction as the bullet at the same instantaneous velocity. By the same token, you can't jump from a compliant surface -- it's like jumping while you're in the air.
Just as guns cannot be made compliant, they're not noncompliant either. To be noncompliant, they would have to be immovable when fired.
On the presumption that what Jim Keenan said is true (and I believe it is) that the recoil impulse begins as soon as the bullet starts moving, then the more compliant the gun is vis-a-vis resisting the recoil impulse, the greater the energy loss imparted to the bullet.
Another factor to contemplate is elasticity. Conservation of energy dictates that when a perfectly elastic object in motion strikes a noncompliant surface, it will depart with the same energy it arrived with. A conceptual example is dropping a hardened steel ball on a very large hardened steel surface.
What's all this mean? Theoretically, the more compliant the gun system is, the greater the energy loss imparted to the bullet.
So is a semi a more compliant gun system than a revolver before the bullet leaves the muzzle? 'Fraid not....
One factor that's scarcely been addressed is system compliance. This is impossible, but it will illustrate the point. If the gun/case combination weighed the same as the bullet AND was totally unconstrained by anything or anybody to prevent its movement, it would be analogous to the frictionless swivel chair demonstration where you can't rotate the chair by moving your arms or legs. The gun would go zinging in the opposite direction as the bullet at the same instantaneous velocity. By the same token, you can't jump from a compliant surface -- it's like jumping while you're in the air.
Just as guns cannot be made compliant, they're not noncompliant either. To be noncompliant, they would have to be immovable when fired.
On the presumption that what Jim Keenan said is true (and I believe it is) that the recoil impulse begins as soon as the bullet starts moving, then the more compliant the gun is vis-a-vis resisting the recoil impulse, the greater the energy loss imparted to the bullet.
Another factor to contemplate is elasticity. Conservation of energy dictates that when a perfectly elastic object in motion strikes a noncompliant surface, it will depart with the same energy it arrived with. A conceptual example is dropping a hardened steel ball on a very large hardened steel surface.
What's all this mean? Theoretically, the more compliant the gun system is, the greater the energy loss imparted to the bullet.
So is a semi a more compliant gun system than a revolver before the bullet leaves the muzzle? 'Fraid not....
Archie
Member
I hate to disagree with Mr. Keenan, but....
To our already expensive experiment, we all more ammo and three Thompson-Center Contenders... or at least three differenct barrels per caliber. Sawed off to match the previously altered pistols and revolvers.
RedLeg 155 mentioned temperature control. Good point.
I should also add we have to determine and correct for the frictional co-efficient of each. Lord help us, the same lot of bullets should be the same all around.....
Hey, Carbonator! We could do this together.... I really think we could use a couple assistants, too. (Where did I put that grant request form......?)
More or less correct. In a 1911, the locking lugs on the barrel lock to the recesses on the slide... the cam is what locks it when closing and the link is what unlocks it at the appropriate place in the recoil cycle.
Here is the error. As the pressure builds in the case, the expanding gasses push in all directions. (Hopefully) the only vent is the bullet moving... but the recoil is generated when the bullet movement starts. That impulse to the slide opposite the movement of the bullet is what operates the pistol, as in "recoil operated". That energy used up in that movement is taken from the energy pushing the bullet out the barrel. Now that I think of it, the energy used to heat up the barrel and receiver are also deleted from the bullet pushing energy. Okay, let's forget about that part......
To our already expensive experiment, we all more ammo and three Thompson-Center Contenders... or at least three differenct barrels per caliber. Sawed off to match the previously altered pistols and revolvers.
RedLeg 155 mentioned temperature control. Good point.
I should also add we have to determine and correct for the frictional co-efficient of each. Lord help us, the same lot of bullets should be the same all around.....
Hey, Carbonator! We could do this together.... I really think we could use a couple assistants, too. (Where did I put that grant request form......?)
JiminCA
Member
Here is one point that may help clarify the concepts being discussed here. Stick with me - it's not as hard as it looks.
The equations of motion that govern the movement in this sort of circumstance are called "Impulse-Momentum" equations.
Energy is a different, but often confused concept. Energy is conserved in the total discharge, but does not properly account for the motion of the objects. This is because energy is dissipated in things like bullet deformation, heat loss, light generation, friction loss along barrel, linear and rotational energy imparted to the bullet, slide, and frame, etc. The total energy consumed equals the total potential chemical energy in the powder, or simply, the energy released when the powder burns.
You have to keep these concepts separate in your mind. Momentum governs motion, energy is conserved and influences motion, but does not govern it, because energy is dissipated in the process.
Going back to the equations of motion. Momentum is mass times velocity. Momentum is conserved in a collision, or separation of objects. As such the vector equation**
(where Mb and Vb is the mass and vector velocity of the bullet and Mg and Vg are the mass and vector velocity of the gun, and 1 and 2 denote before and after firing, respectively)
(Mb1+Mg1)V1=Mb2Vb2 + Mg2Vg2
(in english - mass of the gun and bullet times its intital velocity of zero equals the mass times the velocity of the bullet after firing plus the mass times the velocity of the gun after firing).
Note that the initial velocity of the gun plus bullet assembly (the lefthand side of the equation) is ZERO.
This means that the vector sum of the right hand side of the eqution is also ZERO. This means that Mb2Vb2=-Mg2Vg2, or in english "the gun and the bullet are going in opposite directions in proportion to their respective masses" The little bullet is going fast in one direction and the big gun is going slow in the opposite direction.
Or, the mass of the bullet times the velocity of the bullet is equally offset (in opposite directions) by the mass of the gun times the velocity of the gun.
BTW this basic 1st semester physics proves that the gun moves before the bullet exits the gun. It has to because the mass distribution of the gun has changed (or the center of gravity as previously noted) and it will move at every point in time according to the above equation, ie the equation is true at every point in the bullet's path down the barrel.
If you don't believe all that, note the post above about tall front sights on long barrelled revos. Its true!
Just remember, energy and momentum are different concepts and must be properly analyzed. They get tossed around interchangably, but they aren't the same thing by a long shot.
**If you don't know what vector equations are, just think of them as mathematical representations of the direction of the motion. In real life they can get a bit complicated but in concept they are very simple - they are math that accounts for the direction (in this case) of the momentum.
The equations of motion that govern the movement in this sort of circumstance are called "Impulse-Momentum" equations.
Energy is a different, but often confused concept. Energy is conserved in the total discharge, but does not properly account for the motion of the objects. This is because energy is dissipated in things like bullet deformation, heat loss, light generation, friction loss along barrel, linear and rotational energy imparted to the bullet, slide, and frame, etc. The total energy consumed equals the total potential chemical energy in the powder, or simply, the energy released when the powder burns.
You have to keep these concepts separate in your mind. Momentum governs motion, energy is conserved and influences motion, but does not govern it, because energy is dissipated in the process.
Going back to the equations of motion. Momentum is mass times velocity. Momentum is conserved in a collision, or separation of objects. As such the vector equation**
(where Mb and Vb is the mass and vector velocity of the bullet and Mg and Vg are the mass and vector velocity of the gun, and 1 and 2 denote before and after firing, respectively)
(Mb1+Mg1)V1=Mb2Vb2 + Mg2Vg2
(in english - mass of the gun and bullet times its intital velocity of zero equals the mass times the velocity of the bullet after firing plus the mass times the velocity of the gun after firing).
Note that the initial velocity of the gun plus bullet assembly (the lefthand side of the equation) is ZERO.
This means that the vector sum of the right hand side of the eqution is also ZERO. This means that Mb2Vb2=-Mg2Vg2, or in english "the gun and the bullet are going in opposite directions in proportion to their respective masses" The little bullet is going fast in one direction and the big gun is going slow in the opposite direction.
Or, the mass of the bullet times the velocity of the bullet is equally offset (in opposite directions) by the mass of the gun times the velocity of the gun.
BTW this basic 1st semester physics proves that the gun moves before the bullet exits the gun. It has to because the mass distribution of the gun has changed (or the center of gravity as previously noted) and it will move at every point in time according to the above equation, ie the equation is true at every point in the bullet's path down the barrel.
If you don't believe all that, note the post above about tall front sights on long barrelled revos. Its true!
Just remember, energy and momentum are different concepts and must be properly analyzed. They get tossed around interchangably, but they aren't the same thing by a long shot.
**If you don't know what vector equations are, just think of them as mathematical representations of the direction of the motion. In real life they can get a bit complicated but in concept they are very simple - they are math that accounts for the direction (in this case) of the momentum.
According to my faulty memory of viewing such a slo-mo of a .45 Auto firing, the first thing that happens is the gun jerks forward as the bullet digs into the rifling; then, the slide starts back in proportion to the bullet moving forward. True, they do not separate until the bullet leaves, but the slide has moved.
BluesBear
member
The slight initial forward motion could more properly be attributed to the force of the hammer falling.
- Status
