Where does the energy go?

[TKelly]

Hey there guys!

So here's the deal. You have a true Flintlock smoothbore and a percussion muzzleloader. Both are .58 caliber.

Both have the same amount of BP or charge. They are both fired from the same position, at the same time, and from the same angle. Therefore, the same amount of Kinetic Energy is imparted to each breech to launch each RB, respectively.

Due to the fact that the percussion ML has a rifled barrel, its round ball is located 250 yds. away. It is able to remain airborn for this distance due to its higher velocity and flatter trajectory, because of its rate of twist and rotational motion which assures greater accuracy. The smoothbores round ball is located only 70 yds away.

Questions:

How does the Kinectic Energy not get imparted to the smoothbore ML as it did the rifled ML?

Where did this Energy go?

If the KE doesn't go towards the linear propulsion of the RB, to which other forms of energy does it get converted?

 

[Josh Smith]

Hello,

A spinning projectile is harder to move from its path due to gyroscopic action.

A smoothbore imparts no such action on the ball. It therefore drops.

I'm seeing the physics but am having a hard time describing them.

OK, throw a baseball. As long as it has a spin on it, it stays up, right? Right.

Now throw a baseball using a knuckleball style. It goes straight then drops fast.

I guess maybe you could say that there is more energy imparted to the rifled ball as it travels on two axis in a rifled barrel - not only on the X axis but on the Z axis as well. The Y axis is a constant, gravity, or G sub n.

Guess the simplest answer to your question is that greater energy is imparted due to a further distance traveled due to rifling, but that energy is only seen on the Z axis and is not necessarily imparted in quantifiable manner upon a target on the X axis, though the energy from the spin on the Z axis can be observed when a bullet strikes a Kevlar vest as the fibers are caught up in the rotational force and help to stop the projectile.

Make any sense whatsoever?

Josh

 

[TKelly]

Hey there Josh!

Ok, what I'm thinking, is that both round balls leave their respective muzzles at relatively the same time.
Gravity begins to immediately work on them at the same time. They should both hit the ground simultaneously. However, the percussion round ball has greater linear velocity (for some reason) and therefore, is more able to traverse more ground along the horizontal within the same time period.
Therefore, either the rifling is increasing the efficiency of energy conversion from
KE(explosive) to KE(translational) + KE(rotational) or the smoothbore KE is just getting wasted in some manner or another.

 

[biliff]

The smoothbores round ball is located only 70 yds away.

Huh?? Are you saying a smooth bore will only reach to 70 yards? Mine will shoot a lot farther than that. Where are you getting your numbers from?

 

[TKelly]

Yeah, It was severely yawed!

 

[Josh Smith]

However, the percussion round ball has greater linear velocity (for some reason) and therefore, is more able to traverse more ground along the horizontal within the same time period.

I thought we were assuming the same KE for both, and both are .58 caliber.

By definition of a lead round ball with the same kinetic energy, both would have to weigh the same and have the same velocity.

So given the same velocity, same caliber, same weight, the only variable remaining is the gyroscopic action.

I'm assuming this is a purely hypothetical question as I would not expect the difference in range to be as great as you state.

Josh

 

[TKelly]

See there Josh,

That right there alone! I'm of the opinion that things must be happening inside of the smoothbore barrel to impart either a lower muzzle velocity or it's yawing downward like crazy. Otherwise the gyroscopic thing or the rotational energy is not going to be enough to defy gravity. Gravity is the only force really operating on the round other than air resistance. Therefore, either its direction is downward due to yaw, while exiting the muzzle, or it's moving slower from the muzzle.

 

[robtattoo]

One important variable in the equasion; How tight were the balls in their respective barrels?

The picture I've got in my head is a loose ball in a blowgun. Give it a blow & it won't get very far. Drop a bigger ball in the same blowgun, it'll go a lot further.

I'm sure words like 'inertia' 'obturation' & 'pressure build' are pertinent, but I'm dumb as a hammer & don't really know where they fit!

 

[TKelly]

Hi there Robtattoo!

Yes for certain, obturation comes into play with the percussion ML due to the round ball obturating along the length of the rifled barrel. Inertia comes into play along the horizontal after the round ball exits the muzzle of each ML. However, what imparts greater inertia to the percussion propelled round ball and why is that variable not present in the smoothbore ML.
Perhaps, you're right!
Somehow, the pressure build up is greater in the rifled barrel than the smoothbore, therefore KE becomes more efficiently translated into translational and rotational energy. That's what accounts for the increased muzzle velocity, and that allows for the greater range in trajectory.

That's it! That's got to be it!

Thanks Robtattoo!

 

[Stumpkiller]

Hey there guys!

So here's the deal. You have a true Flintlock smoothbore and a percussion muzzleloader. Both are .58 caliber.

Both have the same amount of BP or charge. They are both fired from the same position, at the same time, and from the same angle. Therefore, the same amount of Kinetic Energy is imparted to each breech to launch each RB, respectively.

OK so far . . .

Due to the fact that the percussion ML has a rifled barrel, its round ball is located 250 yds. away.

? Located AFTER firing?

It is able to remain airborn for this distance due to its higher velocity and flatter trajectory, because of its rate of twist and rotational motion which assures greater accuracy.

Now you lost me. Why would the rifled barrel produce more velocity? There is a slight loss from vent pressure of the flintlock, but the rifling does not increase speed. The twist gives better accuracy, but whether it adds speed (because it retards the ball and allows pressure to build???) is arguable.

The smoothbores round ball is located only 70 yds away.

Huh? Why would it be that much closer? Something more than ignition and rifling at work here. The range should be about identical for identical launch angles and loadings. I know guys who do quite well with smooth rifles using tightly packed balls. The accurate (and therefore effective) range may be less, but the ball gets there just as fast if not in as small a group.

Questions:

How does the Kinectic Energy not get imparted to the smoothbore ML as it did the rifled ML?

Where did this Energy go?

If the KE doesn't go towards the linear propulsion of the RB, to which other forms of energy does it get converted?

The M1A1 Abrams tank uses a smooth barrel and can accurately pick off another tank a mile away; but it ain't throwing a round ball. ;-)

 

[token tory]

I'm not sure if this is what you're looking for as I find your OP a bit confusing, but here goes anyway.

Energy is mass X velocity, not powder load X velocity, so your OP is confusing because the faster ball would have more energy than the slower one (assuming identical weight), no matter how much powder was used to push it.

To get equal energy you'd need to adjust the powder charge to give the same velocity to make the comparison as you've posted it.

So let's assume you've imparted the same energy by getting the same velocity, however you get there. Now the math starts to make more sense.

Once a ball, bullet or anything else has left the barrel you have started dealing with "exterior ballistics" as opposed to "internal ballistics", which is what was happening inside the bore.

Now the moment you go to "exterior ballistics" the only forces acting on the projectile are gravity & air resistance. NOTHING ELSE MATTERS AT ALL which is why external ballistics tables make no reference to the firearm used to fire the bullet, it is no longer relevant.

So your ball/bullet is using it's mass to push forward against the resistance of the tube of air it has to push out of it's way. This resistance saps energy and some is transferred to heat from friction too. The more mass you have pushing the least air the more efficient the projectile is & the longer it will retain it's initial energy.

Gravity has nothing to do with energy transfer, at least not directly. It simply pulls vertically down on the projectile steadily untill the projectile hits ground, or something & tops.

Again I'm not sure if that is what you're looking for but I hope you found it helpful in some way.

 

[TKelly]

Hey there Stumpkiller!

The 70 yds. thing was just a typo. Forget it. The point is that the smoothbore round fell dreadfully short.

Now, they both possessed equal amounts of charge, therefore we should expect an equal amount of KE within each barrel at the point of explosion. Now here's where facts begin to differentiate!

In the smoothbore ML, gases from the explosion are not able to impart as much pressure to the round ball due to microscopic clearances around the round ball itself. Therefore the muzzle velocity is less in the smoothbore than the percussion ML. That causes the round ball to have a shorter flight.

In the percussion ML, the gases are better able to apply greater pressure to the round ball due to obturation, creating a much better seal around the round ball, after the initial explosion. This in turn, imparts a greater muzzle velocity to the round ball because of greater pressure build-up within the percussion ML. This greater horizontal velocity therefore, accounts for a greater horizontal range.

At least, I think that's right!

 

[Larry Pletcher]

One possibility could be that the patched ball in the smoothbore has less resistance in its travel down the bore compared to the rifled barrel. The ball in the smoothbore leaves the barrel before all the powder is consumed. If this is true, some of its powder is burned beyond the muzzle.

In the rifled bore, with more resistance, more of the charge burns inside the barrel. More complete combustion means a higher velocity.

Obviously this is all theory. It would be interesting to add a chronograph to the hypothetical problem. As long as we're guessing, I'd suggest that if the ignition methods were exchanged the result wouldn't change.

Regards,
Pletch

 

[wy0mn]

Pletch is sort of on the right track I think, kinda, maybe...

Look at the ballistic chart on the Goex website, up toward the upper end of the caliber comparisons. Compare diameter/weights vs velocities.
I think the smoothbore roundball would travel faster, farther, and hit harder. Its just that at extended ranges no one knows where the heck it DID hit.
The friction from the rifling actually slowed down the projectiles, allowing gravity to take effect sooner.

I think the basic premise is flawed, especially given the fact that both the smoothie and the rifle could share the same type of ignition system. It was superfluous to even bring that into the equation.

 

[Zonie]

Rather than playing with percussion vs flintlock, lets assume that one .58 caliber ball is fired from a rifled barrel and the other .58 caliber ball is fired from a smoothbore and both balls have exactly the same muzzle velocity as they leave the barrel. They also are fired from the same height above the ground at the same target.

With the same mass and same velocity they will fly the same distance before hitting the ground assuming the ball fired from the smoothbore travels in a straight direction like the ball from the rifled gun will do.

As Colonialist said, the only thing that effects the distance the ball will fly is wind resistance (which is equal, regardless of whether one is spinning and the other is not) and gravity.

The only possible reason the ball fired from a smoothbore might hit the ground sooner than the one fired from a rifled barrel is due to deflection from the straight path that the spinning ball fired from a rifle would have.

As a ball fired from a smoothbore is not spinning it acts like a "knuckle ball" in baseball and the slightest variation of the surface of the ball will cause it to depart from a straight line of flight due to the various air pressures acting on the forward side.

Because the deflection of the smoothbores ball is unpredictable, it may move upwards or sideways or downwards. If it moves upwards, it will travel further than the ball shot from the rifle.

 

[Josh Smith]

OK, so Kodiak, Zonie, All,

Are we playing hypotheticals here, or are we talking about an actual event?

I guess I'm confused here.

I'm about ready to shout CHAOS THEORY and let it be! :idunno:

Josh

 

[John Wasmuth]

As you said Zonie, they will go the same distance except that the smoothie may be deflected off course by wind faster than the rifled ball. Drop two identical bowling balls off the same roof at the same time leave one going straight down and the other add a twist or spiral to it and they will both hit the ground at the same time.

 

[kelleyjk]

The M1A1 Abrams tank uses a smooth barrel and can accurately pick off another tank a mile away; but it ain't throwing a round ball. ;-)

I think this was a result after testing the saboted penetrator round in both rifled and smooth bores they found it functioned better ie penetrated better when not spinning, I reckon I read that somewhere

 

[paulvallandigham]

The fly in the ointment here is the Assumption that two equal powder charge, driving a PRB of the same weight, out of two barrels the same length, will produce the same MV, even when one is rifled, and the other is a smoothbore. That is simply NOT TRUE>. YOU CAN increase the chamber pressure, and MV if you use a OP wad between the powder charge and the PRB in a smoothbore. But, its highly doubtful that you can get the exact same MV out of both barrels.

With that in mind, what has been said about projectiles that spin vs. those that don't spin becomes a major factor affecting DRAG, that will cause the ball fired out of a smoothbore to drop sooner, and the spinning ball from the rifle to travel further.

Where does the Kenetic energy go? It flies away from the ball in the air, due to friction, and " DRAG". While drag may appear to be the same for both balls, the spinning ball creates a layer of vacuum around its surface, as a result of the spinning, allowing it to slide through the air easier, or be less subject to cross winds, or any other disturbance of the air around it. :thumbsup:

If you convert the MV to MPH, and then do the same with some chosen down range velocities, you can see that the "slip-stream" that we talk about with fast cars, is also present around those balls, but the spinning ball has more of the air NOT rubbing against its actual surfaces. This is then reflected in the differences in speed as the two balls travel down range. The sound barrier is also a major factor on drag coefficients, and we know that balls that do not exceed the speed of sound lose less velocity over distance, than do those balls that are sent out of the muzzle in excess of the speed of sound. The Faster a ball hits the air, the faster it slows down.

So, what holds true, or even seems to hold true for how balls act when spun or not, when the MV is over 1135 fps. is different than if the same balls are sent out at 1100 fps. Very few Smoothbores shoot loads with PRBs that exceed the speed of sound. That is why they can shoot fairly accurately out to about 80 yards. After that, the drag factors gain more influence on the ball, and groups open up. Some of the smaller gauge guns, like the 28 gauge( .550 caliber) smoothbores have been known to shoot some pretty tight groups out to 100 yds, simply because you can load up the gun with lots more powder, exceed the speed of sound, and still have your shoulder left intact. :hmm:

 

[Larry Pletcher]

I suspect that I was misunderstood. I would not expect that the velocities to be the same because of various internal variables. The variable I mentioned was friction. The ball with the most resistance (friction) is likely to burn the charge more completely. There are various ways to try to control this and exceptions to this, but that wasn't the question.

Regards,
Pletch