How much power does the action consume

[Cato]

Cato, the internal mechanism doesn't really matter due to conservation of energy. That energy must come from somewhere. Were does it come from? Energy lost to friction, sound, or light? I think it improbable that some of that energy doesn't come from the bullet, or blanks would cycle just fine.

No one is saying that the energy does not come from somewhere. Obviously energy is used to cycle the action. There is a lot of energy from the powder that is not used to propel the bullet. There is wasted energy that you can feel as heat on the barrel. There is energy is used to cycle the slide. More wasted energy transferred to you hand and arm etc. The energy used to cycle the slide is why perceived recoil is less on a semi auto.

My comments and explanation are strictly related to the question asked by the original poster.

Again, this is unnecessary detail due to simple energy conservation, but I don't see how this could be correct. My understanding is that movement of the slide-barrel-cartridge system relative to the barrel link unlocks the barrel once the bullet is away. That is, the barrel unlocks once pressure is low, not because pressure is low. That is, while the bullet is still in the barrel, the locked slide-barrel-cartridge has already accelerated, even if it has only moved 1 millimeter

You seem to also under the assumption that the barrel cycles. It does not. If you are saying that the slide and/or barrel has moved one millimeter because of the loose parts on a semi auto as compared to, lets say, a Thompson Contender then I think you are being picky. For all practical purposes, the slide does not move until the bullet has exited.

 

[Cato]

Cato, look at your own diagram. The barrel moves back a very short distance before it unlocks & drops. That's why they call it Short Recoil.

Yes, I pointed that out in two earlier posts. See my last post also.

 

[TheRoland]

You seem to also under the assumption that the barrel cycles. It does not.

Sure it does. It cycles slightly back and slightly down, following the movement of the barrel link. At the end of its cycle, it unlocks from the slide.

Previous to this unlocking, it still moves some distance, not related to any loose parts.

 

[lupis42]

More wasted energy transferred to you hand and arm etc. The energy used to cycle the slide is why perceived recoil is less on a semi auto.

I knew I forgot another source of energy loss!. Thank you sir!

You seem to also under the assumption that the barrel cycles. It does not. If you are saying that the slide and/or barrel has moved one millimeter because of the loose parts on a semi auto as compared to, lets say, a Thompson Contender then I think you are being picky. For all practical purposes, the slide does not move until the bullet has exited.

The slide *must* start accelerating as soon as the bullet leaves the case, well before the bullet has left the barrel. The slide's speed is a tiny fraction of the bullets, but it is accelerating over the same time that the bullet is. If the slide did not accelerate with the bullet, the energy from the burning powder would have to be stored in some intermediate mechanism for the time it takes the bullet to leave the barrel.

 

[TheRoland]

I knew I forgot another source of energy loss!. Thank you sir!

Assume clamped to a block of lead.

The slide *must* start accelerating as soon as the bullet leaves the case, well before the bullet has left the barrel. ... If the slide did not accelerate with the bullet, the energy from the burning powder would have to be stored in some intermediate mechanism for the time it takes the bullet to leave the barrel.

This is more or less what we've both been saying .

 

[TheRoland]

Yes, I pointed that out in two earlier posts. See my last post also.

I'm confused. This is what I'm saying. The barrel, while still locked to the slide and cartridge, moves back some distance, say 5 millimeters, while the bullet is still in the barrel. The energy used to move these 5mm must be robbed directly from the projectile.

It's only 5mm, but if these 5mm are covered at a high speed, then a real amount of energy must be used to accelerate the barrel-slide-cartridge so fast.

 

[lupis42]

Assume clamped to a block of lead.

Even clamped directly to a block of steel the size of the moon, energy is still transferred through the clamps into the object. So even before the bullet leaves the gun, the gun + bullet system is not closed.

 

[Cato]

Sure it does. It cycles slightly back and slightly down, following the movement of the barrel link. At the end of its cycle, it unlocks from the slide.

Previous to this unlocking, it still moves some distance, not related to any loose parts.

Of course it does, as I pointed out several times. But you said "slide-barrel-cartridge has already accelerated".

As i stated, my comments are strictly related to the OP's question. He thought that when the powder ignites, the energy immediately starts to send the bullet one way and the slide the other. This does not happen on a semi auto handgun. The slide remains tight against the brass until the bullet exits.

 

[wheelgun]

No, you're mistaken and I know what the problem is. Where you are missing it is in the mechanical operation of the gun. You would be correct if the gun operated in the way that you think it does. You keep saying that the barrel and slide moves rearward. The barrel never moves rearward. The barrel is fixed. It only swivels slightly to disengage from the slide. When the slide cycles it leaves the barrel behind. Barrels do not cycle in a semi auto handgun.

So when the slide is locked to the barrel neither one starts to move until the bullet has exited and the pressure has dropped enough for them to disengage. Then the slide moves rearward by it self without the barrel.

Again, it doesn't matter what kind of semi-auto action the gun has. I don't care if the barrel is locked and the slide moves, or the barrel moves and slide doesn't, or whatever you like. If ANY part of the action moves, then some energy has been diverted that would otherwise have gone to the bullet, so that bullet is therefore slower than a fixed action gun.

All your detailed discussion of how the barrel is locked to the slide, when it disengages, how soon the bullet leaves the barrel, etc, etc, is irrelevant. If some part of the action moves when the round is fired then the bullet is slower, all other things being equal.

It's a mechanical system - we put X amount of energy into it (from the powder), of which Y gets applied to the bullet and Z to the action (we also have heat and light but I'm ignoring those). Overall, X=Y+Z where X is fixed. If Z increases, Y must decrease. Conservation of energy must be obeyed. No detailed knowledge of the action is required.

.

 

[TheRoland]

Even clamped directly to a block of steel the size of the moon, energy is still transferred through the clamps into the object. So even before the bullet leaves the gun, the gun + bullet system is not closed.

If there's no acceleration because of an equal, opposite force through the clamp, it doesn't really matter in this system, does it? It's just more energy lost to heat.

Of course it does, as I pointed out several times. But you said "slide-barrel-cartridge has already accelerated".

Hasn't it? If the barrel is moving, and if the barrel is locked to the slide, and if the cartridge is still locked to the slide-barrel, then the slide-barrel-cartridge has already accelerated.

He thought that when the powder ignites, the energy immediately starts to send the bullet one way and the slide the other. This does not happen on a semi auto handgun.

Doesn't it? That the length of travel is very small is not the point.

 

[Cato]

Again, it doesn't matter what kind of semi-auto action the gun has. I don't care if the barrel is locked and the slide moves, or the barrel moves and slide doesn't, or whatever you like. If ANY part of the action moves, then some energy has been diverted that would otherwise have gone to the bullet, so that bullet is therefore slower than a fixed action gun.

C'mon now. You are adjusting your position. I pointed out many times now that there is slight movement in the barrel in order to disengage. That is being way to picky in relation to the OP's original question.

 

[Cato]

Doesn't it? That the length of travel is very small is not the point.

Again, My comments were in response to the original question. The original poster could not measure any difference in the depth that his bullet went in to the wood because of that.

I think that you have lost sight of the original question.

 

[HooVooLoo]

Here you go...

Ejecta mass×ejecta velocity=recoiling mass×recoil velocity.

The answer is 42. Thread over, lol.

 

[wheelgun]

C'mon now. You are adjusting your position. I pointed out many times now that there is slight movement in the barrel in order to disengage. That is being way to picky in relation to the OP's original question.

Adjusting my position? This is physics. There is no "position". It's not an opinion poll. It's the basic laws of conservation of energy and momentum in a closed system. I've been trying to explain how those laws apply and their consequences in each of my replies here. My "position" has never deviated.

 

[Bob P]

Here you go...

Ejecta mass×ejecta velocity=recoiling mass×recoil velocity.

The answer is 42. Thread over, lol.

So long and thanks for the fish!

 

[HooVooLoo]

So long and thanks for all the fish!

Fixed it for you. ;-)

 

[TheRoland]

Again, My comments were in response to the original question. The original poster could not measure any difference in the depth that his bullet went in to the wood because of that.

I think that you have lost sight of the original question.

In this, we're in total agreement. It's a physics thread! How can you not lose sight of the original question?

I don't think anyone here would assert you're going to see any easily repeatable difference in velocity, even in a lab. But there absolutely is an energy difference, and it's probably higher than energy lost to heat or sound.

EDIT: And I like 1911s, so bickering over locking lugs has a certain appeal....

 

[Cato]

In this, we're in total agreement. It's a physics thread! How can you not lose sight of the original question?

I don't think anyone here would assert you're going to see any easily repeatable difference in velocity, even in a lab. But there absolutely is an energy difference, and it's probably higher than energy lost to heat or sound.

EDIT: And I like 1911s, so bickering over locking lugs has a certain appeal....

Agreed, and you made a lot of good points. See ya.

 

[wheelgun]

How about this example. I have a piece of pipe with a plug at each end. One plug is light and the other one is heavy. I fill the center of the pipe with gunpowder and ignite it by heating the center. The pipe itself is clamped to a bench. What happens? Assuming the pipe is strong enough and doesn't burst, and I've used sufficient powder, one plug or both plugs will shoot out of the pipe.

What can we say about the kinetic energy of those plugs? That's easy:

1/2 M1*V1^2 + 1/2 M2*V2^2 = powder energy

I can either have V1 very high, with V2 low, or vice versa, or something in between. But two must "add up" to the original powder energy (heat, sound, and light ignored here).

This is the most simple analogy I can think of, where you can think of M1/V1 as the bullet and M2/V2 the action.

ETA: Sorry I forgot to mention.... if V2 is zero (i.e. a fixed-action gun) then V1 is maximized. But as soon as V2 is becomes non-zero (as in a semi-auto), V1 becomes smaller.

 

[HooVooLoo]

How about this example. I have a piece of pipe with a plug at each end. One plug is light and the other one is heavy. I fill the center of the pipe with gunpowder and ignite it by heating the center. The pipe itself is clamped to a bench. What happens? Assuming the pipe is strong enough and doesn't burst, and I've used sufficient powder, one plug or both plugs will shoot out of the pipe.

What can we say about the kinetic energy of those plugs? That's easy:

1/2 M1*V1^2 + 1/2 M2*V2^2 = powder energy

I can either have V1 very high, with V2 low, or vice versa, or something in between. But two must "add up" to the original powder energy (heat, sound, and light ignored here).

This is the most simple analogy I can think of, where you can think of M1/V1 as the bullet and M2/V2 the action.

ETA: Sorry I forgot to mention.... if V2 is zero (i.e. a fixed-action gun) then V1 is maximized. But as soon as V2 is becomes non-zero (as in a semi-auto), V1 becomes smaller.

The mass of the plugs has no bearing on it, since it is the strength of the threads that keeps the plugs in place.

 

[wheelgun]

The mass of the plugs has no bearing on it, since it is the strength of the threads that keeps the plugs in place.

Assume no threads, just press fit.

HOWEVER, even if the plugs were threaded in, and the force of the gases was still enough to eject them, then the velocity of the two plugs relative to each other would still be as I decribed. I'm running out of ideas as how to better explain this conservation concept.

- We start with the energy in the powder (potential energy).
- There is no kinetic energy in the system to begin with (i.e. nothing is moving)
- We burn the powder (assume it all burns and its energy is 100% turned into gas pressure)
- Assume we're doing this in space (i.e. no air friction) for simplicity
- One or both plugs eject at some velocity
- The end result is a system with the plugs (one or both) moving
- The end energy is now the sum of the kinetic energy of both plugs.... i.e. all the powder potential energy has been transformed into kinetic energy.

That's it. You start with a certain amount of potential energy and you end with an equivalent amount of kinetic energy. This of course ignores heat and light etc.

It wouldn't matter if plugs were threaded or not. You start with nothing moving and you end up with something moving. All the energy is in that "something".

.

 

[HooVooLoo]

I'm just messing with ya', wheelgun. I understand what you are trying to demonstrate, but I think most folks in the thread already understand that principle.

 

[EddieCoyle]

This is an interesting discussion.

I bet there's more energy lost/wasted when someone limp wrists a gun than is used by the action. Think about it.

 

[wheelgun]

I'm just messing with ya', wheelgun. I understand what you are trying to demonstrate, but I think most folks in the thread already understand that principle.

Hey, if everyone's just messing with me, then I admit I've been PUNKED!

But from some of the replies I really don't think everyone understands the basics here.

 

[TheRoland]

I bet there's more energy lost/wasted when someone limp wrists a gun than is used by the action. Think about it.

It's possible. When held static, some amount of energy is transmitted to the frame through the rails, friction on the barrel, and as the recoil spring is compressed, etc, etc. This energy, you could say, would be lost to accelerating the frame backwards when limp-wristing.

I suspect if it's more or less than the action would depend on the particular gun and design, and, actually, I bet the two quantities are actually pretty close on average. Interesting thought.

 

[strangenh]

So let me get this straight. You physicists characterize the momentum of the "equal and opposite reaction" as "loss?" Do I have that right? How, pray tell, do you propose to have a kinetic system in which a projectile is accelerated without an equal and opposite reaction? Paging Sir Newton!

Ejecta mass × ejecta velocity = recoiling mass × recoil velocity​

Either the whole gun recoils as one, or some part on a spring or other system recoils first. Are you all confusing this question with the energy lost from a gas port system (like an AR-15) vs. a system without a gas port?

By some folks thinking on here, if we make the recoiling mass super large (clamp the gun in a lead sled or something else that will hold it still), we magically increase the total momentum on left side of the equation? By increasing the stationary mass of the recoiling mass, the system total increases? Really? Is that what you think? It can't shift it to the other side of the equation - the two sides are and will remain equal in total.

Here you go...

Ejecta mass×ejecta velocity=recoiling mass×recoil velocity.

The answer is 42. Thread over, lol.

Yup. Spot on.

 

[TheRoland]

So let me get this straight. You physicists characterize the momentum of the "equal and opposite reaction" as "loss?" Do I have that right? How, pray tell, do you propose to have a kinetic system in which a projectile is accelerated without an equal and opposite reaction? Paging Sir Newton!

Ejecta mass × ejecta velocity = recoiling mass × recoil velocity​

Either the whole gun recoils as one, or some part on a spring or other system recoils first. Are you all confusing this question with the energy lost from a gas port system (like an AR-15) vs. a system without a gas port?

By some folks thinking on here, if we make the recoiling mass super large (clamp the gun in a lead sled or something else that will hold it still), we magically increase the total momentum on left side of the equation? By increasing the stationary mass of the recoiling mass, the system total increases? Really? Is that what you think? It can't shift it to the other side of the equation - the two sides are and will remain equal in total.

"Loss" has been used in two ways in this discussion:
1. The standard usage: Conversion of useful energy to entropy.
or
2. The usage from the OP: A decrease in energy available to the bullet, because it's being converted to entropy elsewhere in the system.

I guess it must be me that you're responding to in the second part of your post, but I think you've misread something. Of course a static gun will still recoil, but for purposes of this discussion, we can neglect its affects on the rotation of the earth or whatever it's clamped to. We've been primarily interested in the behavior of the system before the energy is fully transferred to the frame.

 

[strangenh]

"Loss" has been used in two ways in this discussion:
1. The standard usage: Conversion of useful energy to entropy.
or
2. The usage from the OP: A decrease in energy available to the bullet, because it's being converted to entropy elsewhere in the system.

I guess it must be me that you're responding to in the second part of your post, but I think you've misread something. Of course a static gun will still recoil, but for purposes of this discussion, we can neglect its affects on the rotation of the earth or whatever it's clamped to. We've been primarily interested in the behavior of the system before the energy is fully transferred to the frame.

And there is your fallacy. Momentum is transferred equally to both sides of the system.

What do you think would happen if you fired a rifle suspended by a spring? Would that cut the velocity in half? Hint: No, only by up to the tiny fps the rifle is moving backwards before the round exits the barrel. And then only because of friction in the barrel.

What does it matter whether a "part" of the rifle is set up to recoil first, so long as the gas pressure is not changed? How is the slide recoiling first different from using a smaller non-SA gun of the same mass as the bolt?

You can't gain or lose energy in the recoiling of the gun on a closed bolt. You can't make one side's MV magically higher than the other side's MV before bullet/barrel separation.

Or do you think long recoil artillery systems are just plain bad designs?

ETA: For the picky: Even after separation [TOTAL ejecta mass] x [ejecta V] = recoil mass x recoil V - and note that the gun is -as a whole- already recoiling before the unlock and slide's own movement. The unlock just permits the slide to move semi-independently of the frame.

 

[TheRoland]

You can't gain or lose energy in the recoiling of the gun on a closed bolt.

Sure you can. You're using useful energy, thus converting it to heat, in more places than you are otherwise. You're doing mechanical work not just on the projectile, but on internals as well. Your gas pressure will be slightly lower. It is similar to mounting your rifle on a spring: it's that small amount of FPS we're talking about.

 

[strangenh]

Sure you can. You're using useful energy, thus converting it to heat, in more places than you are otherwise. You're doing mechanical work not just on the projectile, but on internals as well. Your gas pressure will be slightly lower. It is similar to mounting your rifle on a spring: it's that small amount of FPS we're talking about.

Show me the math.