How much power does the action consume

I may be wrong on this, but my understanding is the barrel remains locked to the SLIDE until the bullet has left the muzzle. The slide/barrel have already started to move backwards together however so there is some (albeit an insignificant amount) of energy being bled away.

The slide does not start to move backwards until it is unlocked form the barrel which does not happen until the bullet leaves the muzzle. The barrel does not move backwards at all except slightly to disengage from the slide.

The barrel can not disengage from the slide until the pressure drops enough for the barrel to unlock. The pressure is not low enough until the bullet exits.

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No, when the barrel and slide are locked the action does not cycle. Everything is locked up until the bullet has left the muzzle. No wasted energy on the action.

I'm curious as to what magical goblin is sitting there, inside the action, telling the slide to not move backwards until the bullet has left the muzzle, as opposed to the bullet being 1 inch out from the end of the case-head.
 
I'm curious as to what magical goblin is sitting there, inside the action, telling the slide to not move backwards until the bullet has left the muzzle, as opposed to the bullet being 1 inch out from the end of the case-head.

The pressure goblin. The extreme pressure locks them up and they stay locked until the pressure drops. The pressure drops enough when the bullet has exited because the inside of the barrel is open and not blocked by the bullet.

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I'm curious as to what magical goblin is sitting there, inside the action, telling the slide to not move backwards until the bullet has left the muzzle, as opposed to the bullet being 1 inch out from the end of the case-head.

It's not a magic goblin, but a well known principle of physics that things under pressure behave differently than when not under pressure. You can take advantage of that, as browning did.

But this is not free, just because the bullet left the chamber before the slide moves doesn't mean that at some point in the discharge, energy wasn't directed in some direction in order to take advantage from it later.
 
The pressure goblin. The extreme pressure locks them up and they stay locked until the pressure drops. The pressure drops enough when the bullet has exited because the inside of the barrel is open and not blocked by the bullet.

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Um I'm afraid that you're slightly mistaken. The lugs in a 1911 lock the barrel to the slide - the breach-block in a Hi-Power style. The purpose of locking them that way is to allow the bullet time to exit the barrel and the pressure to drop to a safe level before the breach opens. The slide/barrel have already started to move backwards before the bullet leaves the barrel.
 
It's not a magic goblin, but a well known principle of physics that things under pressure behave differently than when not under pressure. You can take advantage of that, as browning did.

But this is not free, just because the bullet left the chamber before the slide moves doesn't mean that at some point in the discharge, energy wasn't directed in some direction in order to take advantage from it later.

I thought the whole purpose of the lugs/edge of the block were to create a mechanical lock???
 
I'd say that 2 cases of Wolf and a Yugo 59/66 SKS could be a test. One case with the gas lock open, one case with it closed. Then you would only have to deal with the hysterically funny statistical spread resulting from the precise loading of Wolf ammo.

At least you'd get to blast off 2 cases in the name of science. Or submit it to Mythbusters.
 
I thought the whole purpose of the lugs/edge of the block were to create a mechanical lock???

It does, but it doesn't open under pressure. The slide is the counter weight which allows the whole action to operate, but the pressure controls the lock. The barrel does not move until the bullet leaves the chamber. I am trying to think of a good example to illustrate but am drawing blanks atm.
 
Well, assuming that the pressure on the slide/barrel is zero, and then the bullet starts travelling...

So the slide/barrel will not move until the pressure inside the barrel comes down to a certain level. Let's say the slide/barrel will start moving backwards at 1,000 ft/lbs of energy. But the pressure must start rising to that point before exceeding it. So the barrel/slide starts to move backwards just a hair, stops, and then continues again after the pressure drops back below that threshold?
 
No, when the barrel and slide are locked the action does not cycle. Everything is locked up until the bullet has left the muzzle. No wasted energy on the action.

You fail to understand the basic concept here, that energy is conserved. I don't care what kind of action the gun has, even if the bullet leaves the muzzle ten seconds before the barrel or slide moves.

The amount of energy that's available is determined by the amount and type of powder in the cartridge (assuming it all burns etc). You can either use all that energy to propel the bullet (as in a bolt action rifle), or you can use some of it to cycle a semi-auto action of some type.

Bullet energy = Powder energy - energy used by action

If you use a semi-auto, then there will be less energy available to propel the bullet and that bullet MUST be slower than an identical bullet (in an identical barrel.... thanks for correcting me earlier lupis42) from a fixed-action gun. It's basic physics.

ETA: All this is based on the assumption that the gun does not recoil at all, e.g. that it's clamped to a bench. If the gun moves then the acceleration of the gun also steals some energy away from the bullet.

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You fail to understand the basic concept here, that energy is conserved. I don't care what kind of action the gun has, even if the bullet leaves the muzzle ten seconds before the barrel or slide moves.

The amount of energy that's available is determined by the amount and type of powder in the cartridge (assuming it all burns etc). You can either use all that energy to propel the bullet (as in a bolt action rifle), or you can use some of it to cycle a semi-auto action of some type.

Bullet energy = Powder energy - energy used by action

If you use a semi-auto, then there will be less energy available to propel the bullet and that bullet MUST be slower than an identical bullet from in a fixed-action gun. It's basic physics.

I disagree. If the bullet is on its way to the target before any of that happens it does not matter. The semi is "fixed action" up until that point.

I apologize, but I have to leave now. Thanks for the discussion!!!
 
What that video shows me is that all this speculation is moot in the case of handguns, heh. The bullet is gone before any substantial back-pressure has a chance to exert itself on the slide/barrel.

Thx Terra.

No, wrong. There is energy (which will become movement) stored up in the mass of the slide. Energy takes time to move. There is a spring and the mass of the slide to overcome. By the time it does the bullet has left the chamber.
 
No, wrong. There is movement stored up in the mass of the slide. Energy takes time to move. There is a spring and the mass of the slide to overcome. By the time it does the bullet has left the chamber.

We'll agree to disagree on this point, Terra. This thread is not the place to get into it. We should get a beer and talk shop, I find your analysis of many topics to be thorough to a fault. I like that.
 
that energy is conserved.

This, more than anything, is what people are bickering over. The truth is, no matter what mechanical cleverness you come up with, if you're powering your action with energy from a cartridge, you're trading that energy from either:

A. Bullet velocity
B. Muzzle flash
C. Sound
D. Heat
E. Matter/Energy Conversion

Even with clever engineering, you're going to have traded some of A, B, C, and D's energy to do *something* before the bullet becomes irrelevant to the system. Otherwise your slide would never move.

Really, when things happen doesn't matter. If you had some magical 20 minute delay before the action completely unlocked, you'd still have to get the energy from some place.
 
I disagree. If the bullet is on its way to the target before any of that happens it does not matter. The semi is "fixed action" up until that point.

I apologize, but I have to leave now. Thanks for the discussion!!!

Sorry, but your premise is flawed.

As soon as the bullet starts moving, the barrel/slide MUST also start moving, albeit much more slowly. That is defined by conservation of momentum:

(mass of bullet) x (bullet velocity) = (mass of barrel/slide) x (barrel/slide velocity)

This assumes the gun itself does not move, only the barrel/slide, and we ignore the mass of all the gases involved.

Therefore in a semi-auto some of the energy has already been lost to the motion of the barrel/slide EVEN BEFORE THE BULLET LEAVES THE BARREL. Thus the bullet is moving more slowly than in a fixed action gun at every point in time.

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No, wrong. There is energy (which will become movement) stored up in the mass of the slide. Energy takes time to move. There is a spring and the mass of the slide to overcome. By the time it does the bullet has left the chamber.

Sorry, but there is no such thing as "energy stored in mass".

Energy = 1/2 (mass)(velocity)^2

If there is no velocity there is no energy.

As I stated in an earlier post, the INSTANT the bullet starts moving the slide does too (or the barrel/slide if you like in a Browning action where they're locked together). True, the slide will only move a very small distance before the bullet leaves the barrel (because its mass is much greater than the bullet), but it MUST move. And it's that small slide velocity that's "stealing" energy from the bullet, hence making it slower than a fixed action gun.

ETA: This is ignoring any friction effects. It takes a certain amount of gas force to break the static friction between slide and frame, so in reality the slide motion may be delayed by a very short time relative to the bullet, but I don't think that's significant here. Actually, now that I think about it, the bullet friction in the barrel rifling is probably much larger, so the slide friction is probably irrelevant.

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You guys should read Hatcher's Notebook. He has an excellent chapter on the theory of recoil. Yes the gun moves when the bullet moves.

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B
 
Sorry, but there is no such thing as "energy stored in mass".

Energy = 1/2 (mass)(velocity)^2

If there is no velocity there is no energy.

As I stated in an earlier post, the INSTANT the bullet starts moving the slide does too (or the barrel/slide if you like in a Browning action where they're locked together). True, the slide will only move a very small distance before the bullet leaves the barrel (because its mass is much greater than the bullet), but it MUST move. And it's that small slide velocity that's "stealing" energy from the bullet, hence making it slower than a fixed action gun.

ETA: This is ignoring any friction effects. It takes a certain amount of gas force to break the static friction between slide and frame, so in reality the slide motion may be delayed by a very short time relative to the bullet, but I don't think that's significant here. Actually, now that I think about it, the bullet friction in the barrel rifling is probably much larger, so the slide friction is probably irrelevant.

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I was trying to explain this without resorting to equations. If you think through what I said, I was trying to illustrate kinetic energy and inertia. That as energy is imparted on the slide, it retains that energy as velocity. Keeping in mind the need for the slide to absorb enough energy to break the friction of the frame, the inertia of the stationary mass of the slide and the main spring pushing the slide forward. It just seemed easier to say that the energy was stored in the mass of the slide and move on.
 
Keeping in mind the need for the slide to absorb enough energy to break the friction of the frame, the inertia of the stationary mass of the slide and the main spring pushing the slide forward.

I really should let this go, as we're on the same side of the argument, but since this is a physics thread, I hope you'll forgive the nitpick.

We're confusing force, friction and inertia a bit here. There is no physical inertia quantity that needs to be overcome before the slide will start moving. All it needs is sufficient force to overcome static friction. Once this is complete, the force will start pushing on the locking lugs, and once those are released, the slide will begin to move and recoil spring will begin to compress immediately.

TheRoland said:
{Calculations}
I also would like to retract my previous calculations, as I've failed to take into account the fact that we're bleeding energy into the locking lugs in the form of friction, and it's actually a pretty significant quantity. This must be accounted for, and I did not do so.

Lastly, the OP's thread title actually wanted power, not energy. Strap some magnets to a slide, put the gun in a copper coil, toss on a resistive load, and fire away.
Code:
P = I^2 * R


NOTE: That last is a physics joke.
 
Sorry, but your premise is flawed.

As soon as the bullet starts moving, the barrel/slide MUST also start moving, albeit much more slowly. That is defined by conservation of momentum:

(mass of bullet) x (bullet velocity) = (mass of barrel/slide) x (barrel/slide velocity)

This assumes the gun itself does not move, only the barrel/slide, and we ignore the mass of all the gases involved.

Therefore in a semi-auto some of the energy has already been lost to the motion of the barrel/slide EVEN BEFORE THE BULLET LEAVES THE BARREL. Thus the bullet is moving more slowly than in a fixed action gun at every point in time.

.

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.
 
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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.

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, 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.
 
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.

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.
 
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