Where does the energy go?

[hanshi]

Another thing to consider is that attempting to include variables renders the question irrelevant. The balls MUST have the same velocity for any logical assessment of results. A ball cannot go any faster than the velocity it has acquired at the point of breaking contact with the muzzle. If a ball stays in a rifled barrel longer, then the speed in the air will be less. In air it will start losing speed the micro instant contact is broken.

 

[Larry Pletcher]

The energy required to spin a bullet via rifling is very, very small, less than .5% of total energy in a modern fast twist CF cartridge gun.

Can this be quantified for a patched RB?? Could we calculate this from velocity differences between the 2 balls?

..gravity does not care whether the ball spins or not, the rate of downward acceleration is the same. Should the smoothbore ball not deviate vertically in any way due to the magnus force, it will strike the ground at the same time and distance as the rifled ball.

This comment begs an extension. If, for discussions sake, we fire the guns from a bench with the level bores 1 meter above level ground. The balls should both strike the ground .45 seconds after firing.

If they strike the ground at the same time, then no deviation can take place. If they do strike the ground at the same time then their velocities must be different.

It seems that the difference in range is caused by either the velocity variation or deviation -- or both. My gut leans toward velocity because I can't see a deviation so great that we would see such a big range difference. BTW if a downward deviation could shorten the range by more than 3 times, can an upward deviation extend the range by 3 times as well?

Regards,
Pletch

 

[H. E. Schwalm]

The balls strike the ground at the same time, out of horizontal barrels fired at the same time. Gravity works the same on both. If the rilfe's ball hits the ground further out it has to be due to its higher velocity. Black powder reacts to pressure. That makes me think that higher pressure will produce higher velocities....?

 

[hoochiepapa]

Mebbe that's my problem. You have to learn to read to understand this stuff? :hmm:

 

[Bon Sauvage]

When this topic finally dies I hope that someone will drive a stake through it so that it can never rise up again.

 

[Danbo]

The energy required to spin a bullet via rifling is very, very small, less than .5% of total energy in a modern fast twist CF cartridge gun.

Can this be quantified for a patched RB?? Could we calculate this from velocity differences between the 2 balls?

..gravity does not care whether the ball spins or not, the rate of downward acceleration is the same. Should the smoothbore ball not deviate vertically in any way due to the magnus force, it will strike the ground at the same time and distance as the rifled ball.

This comment begs an extension. If, for discussions sake, we fire the guns from a bench with the level bores 1 meter above level ground. The balls should both strike the ground .45 seconds after firing.

If they strike the ground at the same time, then no deviation can take place. If they do strike the ground at the same time then their velocities must be different.

It seems that the difference in range is caused by either the velocity variation or deviation -- or both. My gut leans toward velocity because I can't see a deviation so great that we would see such a big range difference. BTW if a downward deviation could shorten the range by more than 3 times, can an upward deviation extend the range by 3 times as well?

Regards,
Pletch

Pletch, on the first part I'd rather imagine it can be deduced in definitive manor after the variables are ironed out. The 1/2% figure I quoted came from one of the ballistic gurus at Lost River Ballistics several years back and though not supported by a formula at that time my other reading indicates the number is valid. A short form comparison would be found in establishing the difference in angular momentum of the projectile in comparison to linear momentum. Might not be totally correct but I'm not inclined to do a swan dive into the deep end of the thermodynamics pool at the moment. My gut tells me the energy used to spin a round ball is so small as to be insignificant.

On the second part, if all else is equal the two balls will strike the ground at the same time regardless of velocity. Should the smooth bore ball deviates vertically (either way) due to aerodynamic forces that alter the outcome. The acceleration of gravity is a constant regardless of velocity or rotation on the axis of the flight path.

Another way to view this is Ballistic Coefficient. For round balls of the same caliber and alloy, the BC is the same at a given velocity, therefore trajectory and/or drop is equal. Equal velocity means the same distance of travel if all else is equal. BC=M/I x D^2...no provision for spin in any direction within that formula.

I do not believe a deviation of a ball due to the magnus force will shorten the travel distance that much either. It would take a substantial difference in velocity to do that as well, not that represented by the difference in smooth and rifled barrels.

Now if the smooth bore ball hit a turkey flying by at an inopportune moment, that might do the trick.

 

[George]

Now if the smooth bore ball hit a turkey flying by at an inopportune moment....

Don't you just hate it when that happens?

Spence

 

[longbow1]

I find it very interesting that you you would see it that way :shake:

 

[Joel/Calgary]

I didn't know at the time that round balls don't obturate. Well not to any appreciable degree anyhow.

Kodiak, this is not directed at you personally, but if I may raise a matter of terminology, obturation is a term used in a wide variety of areas of science and technology. It refers to the sealing or stopping up of something, regardless of the mechanism, and the usage is transitive - some mechanism obturates some object. A sphincter obturates a capillary; a packing gland obturates the front (or rear) of the barrel in the water jacket of a Maxim/Vickers/Browning MMG. The obturator in a 155mm howitzer is a mechanism to seal the breech. The upset of a bullet, if it happens, may obturate the bore. OTOH, if the projectile upsets but not enough to seal the bore, there has been no obturation. It could be the tight fit of a patched ball or of a wad that obturates the bore. A breech-loading bullet may be squeezed down to obturate the bore.

Some consider this pedantry, and that this (mis)use of the term in internal ballistics (and it is misused thus ONLY in internal ballistics) is now acceptable, but parochial misuse of a surprisingly widely used technical term can only cause confusion.

[/rant] I now return you to your regularly scheduled postings.

Regards,
Joel

 

[TKelly]

When conicals (miniballs) are loaded into a muzzleloader and then shot. The internal breech explosion then causes the base of the conical to expand thereby creating a tighter seal within the bore. The act of the internal explosion creating enough pressure to expand the base of the projectile is obturation.

A round ball does not appreciably deform to quite this degree by comparison. I therefore, withdrew this particular terminology from that application.

 

[Zonie]

IMO, the localized swelling of a projectile is caused by more than just the high gas pressure acting on the base of the bullet.

It has a lot to do with old Newton's laws of physics.

The stationary things want to stay stationary and moving things want to keep on moving thing.

When the powder fires it pushes on the rear of the bullet but everything towards the front wants to stay stationary. That causes the rear to move forward while the front doesn't move at all so something has to give.
Because the bullet suddenly finds itself shorter it gets fatter and this fattening or swelling obturates (or plugs) the bore. ( :grin: How'd I do Joel? :grin: ).

The amount of swelling a bullet or ball does depends a lot on how hard the gas is pushing at the rear, how soft the lead is and how much lead is located forward of the rear surface of the bullet/ball.

In the case of a roundball, there isn't much material forward of the rear surface that is exposed to the gas pressure so not much swelling happens, even with a pure lead ball.
With a long slug there's a lot of material ahead of the rear surface so they can grow enough to seal off the rifling grooves provided the grooves aren't too deep.

The .005 deep grooves found on a lot of muzzleloaders are pretty easy for a slug to seal but the .013 deep grooves found on barrels designed to shoot roundballs are just too deep for even the softest lead in a long bullet to seal.

 

[Joel/Calgary]

The internal breech explosion then causes the base of the conical to expand thereby creating a tighter seal within the bore. The act of the internal explosion creating enough pressure to expand the base of the projectile is obturation.

That is exactly the problem in this informal misuse of the term in internal ballistics. Obturation is not the upsetting of the bullet, it is the sealing produced by the upsetting of the bullet.

Because the bullet suddenly finds itself shorter it gets fatter and this fattening or swelling obturates (or plugs) the bore. ( How'd I do Joel? ).

Prezackly! :thumbsup:

Regards,
Joel

 

[TKelly]

The internal breech explosion then causes the base of the conical to expand thereby creating a tighter seal within the bore. The act of the internal explosion creating enough pressure to expand the base of the projectile is obturation.

That is exactly the problem in this informal misuse of the term in internal ballistics. Obturation is not the upsetting of the bullet, it is the sealing produced by the upsetting of the bullet.

Because the bullet suddenly finds itself shorter it gets fatter and this fattening or swelling obturates (or plugs) the bore. ( How'd I do Joel? ).

Prezackly! :thumbsup:

Regards,
Joel

So you therefore, agree that some minimal obturation occurs with a round ball, but to a lesser degree than a conical. That was precisely my point. However, we could split hairs as to what effective degree obturation takes place. In general, round balls themselves make for poor objects for effective obturation to effectively engage muzzleloader rifling as Zonie has already stated.

The physics that takes place during obturation actually involves all three of Newtons laws of motion. At the intial point of explosion, potential energy in the form of the black powder (or chemical propellent) gets converted immediately to kinetic energy. Some of this kinetic energy is converted to thermal energy, some into rotational energy, and the bulk of it gets converted into translational energy.

Once the breech explosion takes place, the limited volume of the breech, causes very rapidly moving gaseous molecules to expand. The pressure at the rear of the breech causes the projectile to move forward towards the muzzle. This Newton's 3rd Law.

The more efficiently this is done, the greater the force applied to the projectile, and the greater the muzzle velocity realized. Therefore, the mass of the projectile will be under a greater accelerative force, resulting a greater change in velocity within the barrel. This is Newton's 2nd law. F = (m)(a)

As Zonie has already stated concerning inertia, the pressure and heat build against the base of the projectile, the head or nose of the projectile tends to remain stationary until, its inertia is overcome by the force of the pressure build-up. This is Newton's 1st law.

The muzzle velocity will primarily depend upon the efficiency with which the BP burns upon breech explosion, the effective degree of ball & patch obturation (tightness of seal), and the mass of the projectile.

Since both the rifled and smoothbore muzzleloaders have the same amount of propellant and we assume the same amount of efficiency of BP burn, or energy conversion, the relative muzzle velocities must be comparable.

Therefore, the two round balls should theoretically have very similar ranges. However, since we already know that the round ball from the smoothbore falls drastically short, we can only attribute this anomally to barrel deflection. Therefore its kinetic energy is primarily converted to translational energy, where upon barrel collision, this translational energy becomes obliquely detoured along some acutely angled linear path.

The round ball from the rifled muzzleloader was impacted by the breech explosion which propelled it forward, imparting a finite muzzle velocity which propelled it along a gyroscopically spin stabilized trajectory to its final destination. The potential energy is converted into kinetic energy, which then gets converted into rotational energy (as governed by the rate of twist) and translational energy. The translational energy is observed via its horizontal velocity which is verified by its range.

 

[TKelly]

What makes you think that rifling increases speed you can only increase speed by increasing the powder charge or decreasing the weight of the ball , KE is only used to calculate the impact of the ball on the target and nothing else . Unless the rifling was extremly fast it is not going to impart a noticable difference and a slow twist will not create that much drag in the barrel.Take it to a more realistic range say 50-60 yards and then do it

The rifling doesn't increase the speed directly. What it does do as the projectile obturates along the length of the barrel, is develop greater friction, which in turn allows for a greater amount of time for BP to burn more efficiently, thus converting more potential energy ( propellant), into kinetic energy, via translational and rotational motion.

So it's really more of a matter of increasing energy conversion efficiency, than actually increasing the speed or velocity in a more direct manner.

 

[TKelly]

As it turns out, you are quite correct here, Stumpkiller! :v

 

[TKelly]

Hey there Stumpkiller!

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.

Truer words were never spoken!

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.

This statement is only minimally accurate, if the powder has not been efficiently ignited.

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 is only minimally true as well. There should be greater resistance within the rifled bore due to greater obturation and therefore, more efficient ignition of powder.

This greater horizontal velocity therefore, accounts for a greater horizontal range.

If this is true, it doesn't occur to any significant degree.

 

[TKelly]

Like I've already stated, the one spinning, goes further, says me!LOL :rotf:

Here, I was thinking of the gyroscopic spin from the rifling...

 

[TKelly]

Danbo, theres very little if any irregular imbalance in the laminar flow, concentricity or any other big words on my 7.62 Sierra Matchking 168gr boatail round, but as good as that bullet is, if it aint spinnin, I aint hittin an Abrams M1 even at 500 metres

as to a round ball yep theyre irregular, especailly cast balls tend to have more mass on one side than a swager, still gyroscopic stabilization does occur (even if it dont "yaw" like a pointy bullet)

Ok so if I have miss understood you KH its still a valid question

which one goes farther?

.58 ball spinnin
.58 ball not spinnin

both equally irregular both shot from theoretical devices so they start the contest at exactly the same velocity, I thinks it be a good question

Discussion about conicals is irrelevant to this debate.

Given equal velocity betwixt smoothbore and rifled bores at the muzzle (all else being equal), the answer to the hypothetical question can be found via any online ballistic calculator. Methinks the question is moot. There is no magic imparted on the spinning ball that reduces drag and promotes higher velocity. Simple truth of the matter is that the spin of a round ball is nested in the boundary layer and thus does not interact with overall drag components.

The general consensus here is that smoothbore balls rotate randomly and are thus influenced by imperfections of form and construction as you point out. Quantify the spin they generate please, and explain why it does not contribute to higher retained velocity as well. Sorry, but KH's assumption/conclusion/question is simply unsupported by physics, new or old.

Tell me please, what is the gyroscopic stability factor(Sg)of a roundball at 1800 fps, 1300 fps and 900 fps? Pick a caliber of your choice, use pure lead density in applicable calculations.

Because I am fairly benevolent I'd like to tell you not to waste your time trying to figure that out. Round balls have center of pressure(cp) and center of gravity(cg) co-located, therefore there is no moment arm to consider in calculation. Does the spin of a RB impart gyroscopic stability? Yes. Does in mean anything insofar as drag is concerned? No. A sphere presents the same drag form regardless of orientation to path. When the axis of rotation is other than parallel to path the magnus effect influences that path via aerodynamic force. The amount of that force is very small and will not significantly affect the distance the ball travels unless the deviation is downward.

How I missed this response, I don't know, but later is sometimes better than sooner.

Danbo, I simply presented a problem. The problem was a practical problem. It would appear that the only incongruent variable overlooked was the greater vent clearance given to the flintlock as opposed to the smoothbore. All other varibles were just fine and to marginalize the set-up in the manner in which you did, I think was uncalled for.

Furthermore, round balls are subject to deformation upon the sudden explosion of heated gases pressing upon them from ignition. They then take more elongated shapes. The more elongated they become, the more probability there is for the shape of a nose to appear. Once the nose of a projectile is present, it has a distinct chance of pointing away from its trajectory which still has its center of mass on track. One breech ignition taks place, the perfect little round balls cease to exist. The center of pressure is no longer "co-located" with the center of mass. This will then, require that a certain amount gyroscopic spin be implemented for aerodynamic stability. Of course, the magnus effect will be a factor. We just don't know how much of a factor due to undisclose data concerning round ball deformation.

*Other than the fact that the percussion ML had a smaller vent than the flintlock, nothing else was assumed here.

The potential energy from the propellant still had to be converted to kinetic energy upon ignition. The kinetic energy still had to be transformed into thermal, rotational, and translational energy. All of these things had to be accomplished by both muzzleloaders.

The difference in efficiency with which they were done was negligible.

 

[TKelly]

Danbo, theres very little if any irregular imbalance in the laminar flow, concentricity or any other big words on my 7.62 Sierra Matchking 168gr boatail round, but as good as that bullet is, if it aint spinnin, I aint hittin an Abrams M1 even at 500 metres

as to a round ball yep theyre irregular, especailly cast balls tend to have more mass on one side than a swager, still gyroscopic stabilization does occur (even if it dont "yaw" like a pointy bullet)

Ok so if I have miss understood you KH its still a valid question

which one goes farther?

.58 ball spinnin
.58 ball not spinnin

both equally irregular both shot from theoretical devices so they start the contest at exactly the same velocity, I thinks it be a good question

Discussion about conicals is irrelevant to this debate.

Given equal velocity betwixt smoothbore and rifled bores at the muzzle (all else being equal), the answer to the hypothetical question can be found via any online ballistic calculator. Methinks the question is moot. There is no magic imparted on the spinning ball that reduces drag and promotes higher velocity. Simple truth of the matter is that the spin of a round ball is nested in the boundary layer and thus does not interact with overall drag components.

The general consensus here is that smoothbore balls rotate randomly and are thus influenced by imperfections of form and construction as you point out. Quantify the spin they generate please, and explain why it does not contribute to higher retained velocity as well. Sorry, but KH's assumption/conclusion/question is simply unsupported by physics, new or old.

Tell me please, what is the gyroscopic stability factor(Sg)of a roundball at 1800 fps, 1300 fps and 900 fps? Pick a caliber of your choice, use pure lead density in applicable calculations.

Because I am fairly benevolent I'd like to tell you not to waste your time trying to figure that out. Round balls have center of pressure(cp) and center of gravity(cg) co-located, therefore there is no moment arm to consider in calculation. Does the spin of a RB impart gyroscopic stability? Yes. Does in mean anything insofar as drag is concerned? No. A sphere presents the same drag form regardless of orientation to path. When the axis of rotation is other than parallel to path the magnus effect influences that path via aerodynamic force. The amount of that force is very small and will not significantly affect the distance the ball travels unless the deviation is downward.

Again, they do not remain as round balls after ignition.

 

[TKelly]

I was initially thinking along those lines, as well. However, round balls don't obturate and create that much more pressure buil-up, to manifest that much velocity.

However, they do obturate the degree that they are no longer perfectly round balls.