bullet velocity increasing beyond the muzzle??????

[KFhunter]

25 feet was the standard because muzzle blast is hard on old school light chronographs

[Curly]

Epstein didn't kill himself.   

[Stein]

F=ma aka Newton's Second Law.

If we rearrange, we get a =F/m.  If we then ignore any losses in mass of the bullet due to deposition of the jacket in the rifling, we see that acceleration will be positive when the force is positive.  If acceleration is positive, velocity will increase (vector quantities, but assuming positive acceleration with respect to where the barrel is pointed.

So, the question is when does the force from the burning propellant go to zero?  If you have ever watched the people ride horses around and shoot blanks at balloons, or watch the muzzle blast affect things around the muzzle like grass and dirt, we know that there is some force beyond the muzzle, but that it falls off very quickly.

Finally, there is the force of wind resistance and gravity impacting the bullet when it leaves.  Assuming a level shot, gravity only pulls down so that leaves air resistance as a force acting against the force of the propellant.

Wrap it all together and there is a point where the force of the propellant equals the force of air resistance and at that point, the bullet is going as fast (downrange) as it will be going.  It would be fairly difficult to measure and nearly impossible to calculate, but I would say that it is within a few inches to a few feet and not 25 yards.

[Alchase]

F=ma aka Newton's Second Law.

If we rearrange, we get a =F/m.  If we then ignore any losses in mass of the bullet due to deposition of the jacket in the rifling, we see that acceleration will be positive when the force is positive.  If acceleration is positive, velocity will increase (vector quantities, but assuming positive acceleration with respect to where the barrel is pointed.

He's trying to speak to me, I know it? 

This graph shows shows the curve of pressure increasing in the barrel (pressure = velocity of the bullet) then dropping off to zero as the bullet leaves the barrel. Once the bullet leaves the barrel nothing can increase it's velocity short of hitting it from behind.

I know my goes intas and my what fors.

[Bofire]

Well now guys science and all: I met a guy 30 years ago, he was hunting bear out Bald Hills road out of Yelm. He had two 30-30s. one a pump and one a lever action, he informed me that the pump and lever left the barrel about the same speed but the pump speeded up at about 200 yards and hit harder. so there you go, facts!!
Carl

[Alchase]

Well now guys science and all: I met a guy 30 years ago, he was hunting bear out Bald Hills road out of Yelm. He had two 30-30s. one a pump and one a lever action, he informed me that the pump and lever left the barrel about the same speed but the pump speeded up at about 200 yards and hit harder. so there you go, facts!!
Carl

 

 

[Stein]

F=ma aka Newton's Second Law.

If we rearrange, we get a =F/m.  If we then ignore any losses in mass of the bullet due to deposition of the jacket in the rifling, we see that acceleration will be positive when the force is positive.  If acceleration is positive, velocity will increase (vector quantities, but assuming positive acceleration with respect to where the barrel is pointed.

He's trying to speak to me, I know it? 

This graph shows shows the curve of pressure increasing in the barrel (pressure = velocity of the bullet) then dropping off to zero as the bullet leaves the barrel. Once the bullet leaves the barrel nothing can increase it's velocity short of hitting it from behind.

I know my goes intas and my what fors.

Once it leaves the chamber, the chamber pressure is irrelevant.  There is still some force on the bullet as the pressure from the propellant spreads out beyond the barrel.  If you don't believe me, set an open can of a couple inches beyond the muzzle and to the side and I bet you a sixer it doesn't stay standing when you pull the trigger.   

Cowboy mounted shooters break balloons with the force of powder expansion at up to 20' or so.

Here is another example:

[JimmyHoffa]

A slow bullet with a large cross section would probably be most noticeable for this.  Drag force would primarily be from the square of the velocity, so slower would require less force for fighting drag.  Larger area would require less pressure to drive it, but also mean a higher drag.

[jasnt]

The force behind the bullet after it leaves the bore is not nearly enough to speed up the bullet. Not even a fps. It’s like a strong wind for a few feet at most.  Even with a magnum cartridge.  Negligible at best.  A bullet will never be faster than it is at the muzzle.

[Alchase]

F=ma aka Newton's Second Law.

If we rearrange, we get a =F/m.  If we then ignore any losses in mass of the bullet due to deposition of the jacket in the rifling, we see that acceleration will be positive when the force is positive.  If acceleration is positive, velocity will increase (vector quantities, but assuming positive acceleration with respect to where the barrel is pointed.

He's trying to speak to me, I know it? 

This graph shows shows the curve of pressure increasing in the barrel (pressure = velocity of the bullet) then dropping off to zero as the bullet leaves the barrel. Once the bullet leaves the barrel nothing can increase it's velocity short of hitting it from behind.

I know my goes intas and my what fors.

Once it leaves the chamber, the chamber pressure is irrelevant.  There is still some force on the bullet as the pressure from the propellant spreads out beyond the barrel.  If you don't believe me, set an open can of a couple inches beyond the muzzle and to the side and I bet you a sixer it doesn't stay standing when you pull the trigger.   

Cowboy mounted shooters break balloons with the force of powder expansion at up to 20' or so.

Here is another example:

I was actually agreeing with your previous post.

The only way a bullet could increase in velocity after it leaves the barrel, is if that pressure actually increased after the bullet left the barrel. That is impossible.
While in the barrel the bullet creates a seal with the barrell, optimizing the pressure behind the bullet. Once that seal is broken by leaving the barrel, the pressure decreases.
As soon as the pressure decreases, the bullet velocity will also decrease.
As you can see in my graph above, the pressure on the bullet after leaving the barrel drops to zero pretty quick.

[KFhunter]

The muzzleblast creates a cavitation hole and shock pulse, nullifying the resistance of the atmosphere and continues to push on the bullet as it displaces around the bullet and sheds off the the sides much like a supersonic torpedo

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[Magnum_Willys]

The muzzleblast creates a cavitation hole and shock pulse, nullifying the resistance of the atmosphere and continues to push on the bullet as it displaces around the bullet and sheds off the the sides much like a supersonic torpedo

Sent from my SM-G965U using Tapatalk

  Exactly.  Just like a cork popgun - the cork is at the end of the barrel.   

[Pegasus]

If you have one of these you can have a round accelerate after leaving the barrel:

[Stein]

F=ma aka Newton's Second Law.

If we rearrange, we get a =F/m.  If we then ignore any losses in mass of the bullet due to deposition of the jacket in the rifling, we see that acceleration will be positive when the force is positive.  If acceleration is positive, velocity will increase (vector quantities, but assuming positive acceleration with respect to where the barrel is pointed.

He's trying to speak to me, I know it? 

This graph shows shows the curve of pressure increasing in the barrel (pressure = velocity of the bullet) then dropping off to zero as the bullet leaves the barrel. Once the bullet leaves the barrel nothing can increase it's velocity short of hitting it from behind.

I know my goes intas and my what fors.

Once it leaves the chamber, the chamber pressure is irrelevant.  There is still some force on the bullet as the pressure from the propellant spreads out beyond the barrel.  If you don't believe me, set an open can of a couple inches beyond the muzzle and to the side and I bet you a sixer it doesn't stay standing when you pull the trigger.   

Cowboy mounted shooters break balloons with the force of powder expansion at up to 20' or so.

Here is another example:

I was actually agreeing with your previous post.

The only way a bullet could increase in velocity after it leaves the barrel, is if that pressure actually increased after the bullet left the barrel. That is impossible.
While in the barrel the bullet creates a seal with the barrell, optimizing the pressure behind the bullet. Once that seal is broken by leaving the barrel, the pressure decreases.
As soon as the pressure decreases, the bullet velocity will also decrease.
As you can see in my graph above, the pressure on the bullet after leaving the barrel drops to zero pretty quick.

You are close.  As soon as the bullet leaves the barrel, the pressure behind the bullet drops off quickly, but the bullet does not begin to decelerate until the force of the pressure behind the bullet is lower than the force of air resistance.  The pressure doesn't have to increase to accelerate the bullet, it just has to be more than anything slowing the bullet down (friction in the barrel and air resistance).

Here is another example.  Load a blank into the rifle, point it up and put an empty soup can over the barrel.  Pull the trigger and the soup can will go upwards because the force does not drop to zero at the end of the barrel.

Take a .308, it has about 60,000 psi SAAMI.  The area of a .308 bullet is about 0.3 inches square.  Thus, the force on the bullet from the powder is in the neighborhood of 17,881 pounds at the peak.  When it leaves the barrel, even as that pressure drops off quickly, even at only 5% of energy left you are still looking at 900 pounds of force on the bullet.  Going further, at 99% of the pressure gone, the bullet still has 178 pounds of pressure which is more than the air resistance and the bullet is still accelerating.

I think we all agree that the bullet doesn't accelerate very far past the barrel, but it actually can't stop accelerating at the end of the barrel as F=mA has to apply just before and just after the end of the barrel.  Since I showed above that F doesn't go to zero exactly at the end of the barrel, A can't be zero as well or we just crossed the beams.

[theleo]

If you guys had been around for Epstein we'd know who was ordered to commit his suicide. This thread should have been dead after 2-3 posts.