Round balls and obturation

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Out of all the tens of thousands of round balls, loosely patched and horribly tight patched, not a single one ever had rifling marks on it 😆
Patching material is spongy, it conforms nicely to the shape of its container, under pressure. Good patch material is able to grip both the rifling and the round ball, thus imparting the rifling’s twist, creating a stabilized projectile. As the patch exits the muzzle, the hot expanding gases blow out the patch. Though a lot of evidence left by the rifling is then lost, not all of it is….

Burs and sharp-edged rifling will leave cuts in the patching, those macro cuts are from rifling. Under magnification the tiny retained impressions of the rifling can be seen in the patch material. Micro damages from imperfections in the rifling also become visible.

Old timers that have been shooting blackpowder for a spell, can tell a lot about what’s happening inside a bore by looking at the patching after each shot.

Ask yourself this… why is it that pure lead round balls are inherently more accurate than round balls cast from hardened lead?
 
Round balls have been around for hundreds of years, yet here we are today, trying to reinvent the wheel and think we know more about it now than they ever did 🤣🤣

I find round balls with the patch material imprint, but NEVER any rifling.

If you find rifling on a recovered round ball, you either bounced it down the bore. Both ways, or used an aggressively oversized ball that you literally pushed through the rifling.

That ball does not have enough bearing surface period.
 
Round balls have been around for hundreds of years, yet here we are today, trying to reinvent the wheel and think we know more about it now than they ever did 🤣🤣

I find round balls with the patch material imprint, but NEVER any rifling.

If you find rifling on a recovered round ball, you either bounced it down the bore. Both ways, or used an aggressively oversized ball that you literally pushed through the rifling.

That ball does not have enough bearing surface period.
Exactly…
 
Look at the bearing surface on this 300gr conical.

Ain't no way a round ball, wrapped in a thick cloth patch, will ever encounter rifling under proper conditions.

Even with a blown patch, id say itll be out that bore before the ball touches the rifling.
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The energy vectors in the diagram are not correct. .....
Yes, and also if length represents force vector magnitude, they're not all equal, as diagrammed by OP. The magnitude of the force exerted perpendicular to the side of the ball where it contacts the bore is only a fraction of the force exerted on the back of the ball.
 
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Friction creates drag in the barrel..

Bullets being longer as mentioned create more drag/ friction than a patched roundball.

Thus,
Apples vs Oranges 🤷‍♂️
 
I agree on the slow mo vids of the gas exiting the muzzle ahead of the ball but wouldn't that argue in favor of my concept of the ball not obturating the bore??
The gas escaping ahead of the projectile is actually compressed atmospheric air. The "smoke" from this that is seen in the videos is due to condensing moisture, vaporized residual components in the bore, and aerial contaminates that normally cannot be seen with the naked eye.
 
I'm in the mood to agitate so here goes.

We probably all have read about and understand the concept of bullets "obturating" the bore. It's a pretty simple concept. The charge ignites and the weight of the bullet column resists movement and consequently the pressure from the charge causes the bullet to collapse on itself and press against the sides of the barrel. So the expansion of the projectile obturates (plugs) the bore and holds (hopefully) the expanding gasses behind the bullet.

But then we have the round ball. We often speak of the ball obturating the bore just as a conical bullet does. I've always wondered about this and I've made a diagram of why I think the round ball may not expand and obturate the bore.

What I'm trying to illustrate here is the distribution of the pressure on the base (bottom half) of the ball. Since the bottom half is a half round shape, the pressures are not applied in direct opposition to the weight or inertia of of the whole ball. The pressure applied on the semi circle is actually opposing the collapse of the ball and keeping it round.

So, that's just what I'm thinking on this.

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This is correct. The round ball does not obturate from the exploding gasses.

well the rifling marks I find on recovered roundballs tells the tale that the ball obturated

Well I've never recovered a patched round ball that obturated from the gasses, but I have seen round ball swaged onto the rifling when fired from a revolver. I've also seen folks whang their ramrod down onto their round ball, which might deform them.

I've also seen perfectly round smooth bore balls fired with a tight patch. Now IF the balls were obturating then although no grooves, there would be a visible band on the smooth bore fired ball as it reacted to the ignition of the powder, and tried to compress, forcing a small amount of the led to flatten against the side of the smoothbore barrel. But his does not happen.

LD
 
Yes, and also if length represents force vector magnitude, they're not all equal, as diagrammed by OP. The magnitude of the force exerted perpendicular to the side of the ball where it contacts the bore is only a fraction of the force exerted on the back of the ball.
The force on the ball is pressure, which is exerted in all directions equally in a closed chamber, and normal (perpendicular) to the interior surfaces of that chamber. In this case, the chamber is the barrel and the rear surface of the projectile. The force diagram, if I'm remembering my engineering correctly (it's been 50 years), is exactly as shown as regards the ball.
 
The force on the ball is pressure, which is exerted in all directions equally in a closed chamber,.....
I don't think you're taking into account that it's not a closed chamber. The force from the expanding gas is parallel to the bore axis. Force = mass * acceleration. The acceleration is along the bore axis. Therefore since the force exerted normal to the circumference of the ball is not parallel to the bore axis. The normal direction force against the ball from the expanding gas at that point is much less. (Been more than 50 years for me since Statics 101 in EE major, and I never got a degree, hah!!)
 

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I believe I own an old Lyman Blackpowder Manual that has high speed photos from the side view, of round balls exiting the muzzles by only inches.
I recall these photos clearly showing round balls that were flattened on powder facing arc of the ball (had obturated)…inertia held the front face of the ball motionless for a microsecond as the pressure wave smacked the rear of the ball with enough force to deform it before overcoming the inertia of the whole mass of the lead ball starting its way out the barrel.

I don’t pretend to know with certainty that is in fact what’s happening but Lyman’s point was that heavy charges were unnecessary and unpleasant to use.
 
Yes, and also if length represents force vector magnitude, they're not all equal, as diagrammed by OP. The magnitude of the force exerted perpendicular to the side of the ball where it contacts the bore is only a fraction of the force exerted on the back of the ball.
I believe this is correct. It certainly works this way in hydraulics and mechanics.
 
I don't think you're taking into account that it's not a closed chamber. The force from the expanding gas is parallel to the bore axis. Force = mass * acceleration. The acceleration is along the bore axis. Therefore since the force exerted normal to the circumference of the ball is not parallel to the bore axis. The normal direction force against the ball from the expanding gas at that point is much less. (Been more than 50 years for me since Statics 101 in EE major, and I never got a degree, hah!!)
Think of the instant the powder ignites, before the ball starts moving. The barrel and ball form a static system for some very small time, the duration of which is immaterial. There are no preferred directions for the force to be exerted, because it's exerted equally in all directions. When the ball starts to accelerate down the barrel, there are still no preferred directions for force to be exerted since at any instant, the ball may be considered to be motionless for analysis purposes. (Oh, Lord, my calculus isn't up to this!! Or much of anything else to be honest.) You remember how every force vector is actually the summation of all of the forces being exerted on a point of the containment vessel, and it all sums up to a force perpendicular to the containment wall at that point. Again, since this is an instantaneous calculation, the pressure will be normal to the walls of the containment, including the base of the ball.

And please, keep in mind that I'm NOT a mechanical engineer! I just couldn't deal with alternating current. I'm just messing around here, trying to remember stuff from over half a century ago, early 1970's. I switched to Mechanical Technology, as it was called then, and finally graduated. I've entirely shot my wad on this thing. Any more detailed analysis of the force vectors on that ball are well beyond me!

Happy shooting!
 
....

And please, keep in mind that I'm NOT a mechanical engineer! ....
Seeing as how my local draft board decided to interrupt my college education, I never became any kind of an engineer. Better they did so probably, as I'm better at doing stuff with my hands than with my brain anyways!! 😄
 
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