Famous Monitor's XI-inch Dahlgren Shell Guns

trice · May 25, 2022

Saphroneth said:
Forgive me, but you don't seem to have provided the specifics of the effect which produces the reduced accuracy you describe at higher velocity.

I am not trying to be insulting when I say this, but I do want to point out an obstruction I believe you have in your approach. What I usually find in our discussions is that your concern on any subject is asserting that your position is correct and that anything anyone else brings up must be attacked and cast in a bad light if it leads to a result that is not what you desire. You routinely pursue only one factor to the exclusion of anything else. I find that this is the same on any topic at all, so I am not just talking about ballistics. I believe I am simply describing your routine point of view on any topic. I find that most situations we discuss are multi-faceted and believe that your single-issue focus serves you very badly.

Here you seem intent on asserting that a cannonball in flight will not be affected in any way during flight -- that a spinning sphere will always fly straight and true. Apparently nothing will change that. I do not believe that is true.

In the case of Dahlgren, his belief (starting back around 1850 or so) was that there was some point at which velocity became problematic. At that time he was working with US 32-lbers (one of which nearly killed him when it blew up under testing in 1849). He believed there was a difference in action between the light and heavy 32-lber caused by excess velocity. I have never seen any particular reference to the speed of sound in his work, but the figures he does mention as ranges would fall on opposite sides of what was later called the "sound barrier" (which would roughly be, in terms of naval artillery effects in that day, from about 183 FPS to 1130 FPS, depending on local conditions).

Saphroneth said:
Specifically, a ballistic coefficient (which you do mention) influences drag, but a poor ballistic coefficient simply means that a projectile sheds velocity more quickly - that is, the drag is higher - and so all that means is that a round 1,200 fps projectile decelerates into being a 1,000 fps projectile more quickly than an elongated one would.

Which, as I pointed out, changes the trajectory and thus the time of flight. Does a change in trajectory and time of flight not also affect accuracy?

Round spheres that are supersonic also experience destabilization effects when their speed drops below the speed of sound -- which are not experienced by round spheres which are only subsonic. By itself, this would seem to indicate that something what Dahlgren was groping towards in 1850 contains some relationship to muzzle velocity that he simply had no way to detect.

You can work through the drag coefficient for round spheres of different sizes at the reference i have posted a few times now if you want. The article Cannonball Aerodynamic Drag at A R Collins site may help.

Saphroneth said:
It's still the case that a projectile with a 1,200 fps starting velocity has no major difference in transverse forces (forces that would knock it off target) and reaches the target sooner so those transverse forces have less time to act on it; thus, it is more accurate.

No, that statement is based on blotting out anything that disagrees with your desired conclusion. You seem to have agreed that there will be a disruptive event when the cannonball drops below the speed of sound, but claim that the disruptive event will have no effect.

Saphroneth · May 25, 2022

trice said:
Which, as I pointed out, changes the trajectory and thus the time of flight. Does a change in trajectory and time of flight not also affect accuracy?

But how does undergoing the transsonic transition actually change the trajectory? What is the mechanism?

For a non-round projectile then the reason is obvious - by destabilizing and thus ending up pointing in a different direction, the projectile's cross-section has been altered. But a sphere is rotationally symmetrical, so even if it does rotate to point in a different direction there is no change.

trice said:
No, that statement is based on blotting out anything that disagrees with your desired conclusion. You seem to have agreed that there will be a disruptive event when the cannonball drops below the speed of sound, but claim that the disruptive event will have no effect.

Not quite; what I've said is that there may be a disruptive effect, because I am waiting for you to demonstrate what it actually is.

trice said:
You can work through the drag coefficient for round spheres of different sizes at the reference i have posted a few times now if you want. The article Cannonball Aerodynamic Drag at A R Collins site may help.

The changing drag does not change the trajectory of the sphere, except trivially in that it reduces velocity - but if the only way in which a 1200 fps sphere dropping to 1000 fps changes the trajectory is that it reduces the velocity, then that does not produce more inaccuracy. Inaccuracy is caused over time (that is, the ball has more time to drift out of the ideal trajectory that it "should" be following).

This is because if projectile A starts out at higher velocity than projectile B, and they are fired over the same distance, projectile A will finish the distance still at a higher velocity than projectile B. It may have lost a higher % of its starting velocity than projectile B, or it may not, but there's no way for the two projectiles to cross over in speed.

This means that projectile A takes less time to travel over the distance, and there is less time for the magnus effect (which is not amplified at higher projectile speeds) to act upon it.

The only way I can see in which a higher velocity projectile might be said to be less accurate is if by losing a higher % of its velocity over the course of the journey it varies more from the "ideal ballistic arc". But that's a pretty simple matter of allowing for elevation in the same way you do with any projectile, and if that's what you actually meant in the first place then I'm surprised you were talking about "number of rotations" and said that the magnus effect wasn't controlled by time (post 10). Is it in fact the case that that's what you mean?

If so then I can try actually simulating it for a 200mm diameter projectile (since the plot of drag coefficient by mach number is given in the link you provided) but it's far from what I understood you to be talking about before.

trice · May 25, 2022

Saphroneth said:
The changing drag does not change the trajectory of the sphere, except trivially in that it reduces velocity - but if the only way in which a 1200 fps sphere dropping to 1000 fps changes the trajectory is that it reduces the velocity, then that does not produce more inaccuracy. Inaccuracy is caused over time (that is, the ball has more time to drift out of the ideal trajectory that it "should" be following).

So you say it is trivial that the projectile travels less distance and hits short of the target? That accuracy is only about horizontal motion and vertical motion can be ignored?

trice · May 27, 2022

Saphroneth said:
But how does undergoing the transsonic transition actually change the trajectory? What is the mechanism?

For a non-round projectile then the reason is obvious - by destabilizing and thus ending up pointing in a different direction, the projectile's cross-section has been altered. But a sphere is rotationally symmetrical, so even if it does rotate to point in a different direction there is no change.

This has already been demonstrated to you in at least one respect, the increase in drag coefficient that will apply to the supersonic projectile when its' speed starts to fall below the "sound barrier". As you already know, this will affect the trajectory. There may be other effects as well, because what this really means is that force will be applied to the supersonic round that will not be applied to the subsonic round.

As a result of that, modern ballistics formulas include conditional values for the speed of sound.

From Dahlgren's perspective in 1850, he had no way of knowing about the implications for ballistics of the speed of sound (the "sound barrier"). He had no way of observing/detecting/measuring what was going on during the flight. He was, however, one of the most knowledgeable and skilled gun designers of his day. He could observe that shots fired with high muzzle velocities acted differently than expected/predicted.

As a result, back in 1850, Dahlgren theorized there might be some "golden rule" type relationship between the powder charge and the shot weight. By emphasizing muzzle velocity, he thought you might be throwing the balance between the two out of whack. He was wrong about the charge/weight relationship being the cause, but he was right about the supersonic shells not acting in the same way as the subsonic shells.

Saphroneth said:
Not quite; what I've said is that there may be a disruptive effect, because I am waiting for you to demonstrate what it actually is.

No, it has been demonstrated to you that there is at least one definite disruptive effect. There may be others.

Saphroneth said:
The changing drag does not change the trajectory of the sphere, except trivially in that it reduces velocity - but if the only way in which a 1200 fps sphere dropping to 1000 fps changes the trajectory is that it reduces the velocity, then that does not produce more inaccuracy. Inaccuracy is caused over time (that is, the ball has more time to drift out of the ideal trajectory that it "should" be following).

But it does.

That is why modern ballistics includes adjustments for the speed of sound in their formulas.

Saphroneth said:
This is because if projectile A starts out at higher velocity than projectile B, and they are fired over the same distance, projectile A will finish the distance still at a higher velocity than projectile B. It may have lost a higher % of its starting velocity than projectile B, or it may not, but there's no way for the two projectiles to cross over in speed.

This means that projectile A takes less time to travel over the distance, and there is less time for the magnus effect (which is not amplified at higher projectile speeds) to act upon it.

That is correct about velocity, but not correct about trajectory. If you use the same calculation for trajectory on a subsonic and a supersonic round, one of the results will be wrong.

Saphroneth said:
The only way I can see in which a higher velocity projectile might be said to be less accurate is if by losing a higher % of its velocity over the course of the journey it varies more from the "ideal ballistic arc". But that's a pretty simple matter of allowing for elevation in the same way you do with any projectile, and if that's what you actually meant in the first place then I'm surprised you were talking about "number of rotations" and said that the magnus effect wasn't controlled by time (post 10). Is it in fact the case that that's what you mean?

If so then I can try actually simulating it for a 200mm diameter projectile (since the plot of drag coefficient by mach number is given in the link you provided) but it's far from what I understood you to be talking about before.

Then you need to reread the article to correct your impression.

Start out by assuming that you are wrong and devote your energy to finding what you are missing. Abandon the attempt to make everything prove you are 100% right.

ADDED Later

On the magnus effect:

I am sure that I probably gave a poor description. I am neither a physicist nor a ballistician. Since college, my mathematics has only been used in application work on things like computer science, arbitrage work, metallurgy, etc. I have been wrong much more than once in my life and have no unusual problem in admitting it when I am. If I am here, I apologize for the confusion.

However, the magnus effect (known, but probably not clearly understood in Dahlgren's day) is not about time. It is about the pressure built up on the opposite side of the sphere from the direction of the spin rotation which is what causes the movement of the sphere. If that spin speeds up or slows down or changes direction in any way during flight, there will be a change in the magnus effect. More time simply means the magnus effect has longer to act.

The same is true of other effects that act on a projectile in flight. I have no idea if Dahlgren would have been aware of or included the coriolis effect and the resulting adjustment that should be made based on the direction you are firing and the rotation of the Earth. I doubt he would have known how to make adjustments to calculations for differing meteorological effects at different heights (or had the data to work with). If he knew about the magnus spin effect, he may have known about the gyroscopic spin effects on deflection, but I can't see how he would have included those in his calculations/projections. There are adjustments that should be made for not only wind but also air temperature and barometric pressure.

Many of the things Dahlgren would need to know were either concepts not developed yet in the 1850s or data that could not be measured/observed at the time. As a result, he is looking at a situation where the calculations do not really work, trying to figure out the contradictory results. As a result, he knows something is wrong and is reduced to experimental testing to narrow down the cause.

Saphroneth · Oct 25, 2022

trice said:
However, the magnus effect (known, but probably not clearly understood in Dahlgren's day) is not about time. It is about the pressure built up on the opposite side of the sphere from the direction of the spin rotation which is what causes the movement of the sphere. If that spin speeds up or slows down or changes direction in any way during flight, there will be a change in the magnus effect. More time simply means the magnus effect has longer to act.

I've come back to this after giving some significant thought to try and work out if there's something I'm misunderstanding, and I'm afraid that I'm going to have to lay out my reasoning to see if there's a point which is at issue. This is because this paragraph of yours - in particular - actually seems to confirm part of my reasoning.

So we start with two large, round cannonballs of identical weight and dimensions being fired. One of them (A) is fired (with the appropriate elevation) at 1,000 fps, and the other (B) is fired (with the appropriate elevation) at 1,400 fps. The target is a fixed distance away, say 600 yards.

Based purely on ballistic trajectory, without for the moment considering the effects of the air in general or inaccuracy in other ways, both will strike the target.

Causes of inaccuracy

One cause of inaccuracy for a smoothbore is that the trajectory of the cannonball as it leaves the muzzle is the trajectory of the "last bounce" out of the barrel.
This factor may or may not be invariant with the velocity of the cannonball as it leaves the barrel. If it is variable depending on the velocity of the cannonball, I would expect (A) to be more inaccurate as the forwards component of its velocity is smaller.

Another cause of inaccuracy is wind. For this one, we would expect (A) to be more inaccurate because the wind blows on the ball to push it off course. With a slower ball (A) the flight time is longer and so the force has longer to take effect.

Not directly causing inaccuracy but still relevant is air resistance. The air resistance for ball (B) is a larger force, and so (B) will slow down more quickly, but it will not "cross over" in speed with ball (A) - ball (B) will always be faster at a given point on the trajectory. This also means that it will always take less time.

For non-round projectiles, the destabilization that takes place owing to the transsonic regime also has to be considered, but I do not think it is relevant for round projectiles. This is because for a non-round projectile it can destabilize by physically ending up pointing in a different direction as it travels through the air, drastically changing the cross-section, but for a round projectile the most that can happen is that the spin direction changes - and a smoothbore's spin direction is random (unknown) when it comes out of the barrel anyway.

Then the final factor to consider (unless I've forgotten one) is the Magnus effect. This is the product of asymmetric vortices around a rotating projectile, and where it is generated it produces a force which is

Which is to say, the force is proportional to the cross product of the velocity vectors for the fastest side (relative to the fluid) and the slowest side.
(All the other values are the same for two identical cannonballs in the same air at different velocities.)

This is the formula for the magnus effect on a round ball. For a projectile moving fast enough that both sides are moving in the same direction relative to the air (that is, if it is moving north at 1400 fps but the spin is such that there is no point on the ball that is instantaneously moving south) then the cross product is the cross product of two vectors that are mostly aligned (i.e. they are mostly moving north). This puts the magnitude of the cross product as being low.

There are two factors which mean that this appears to be less at high (supersonic or near-supersonic) velocities, for the same projectile.
The first is that the addition of 100 metres per second in the direction of travel to all the vectors involved makes the two velocity vectors in the calculation more parallel. This means that the magnitude of the force is lower, and may in fact be zero if the two vectors in the calculation are fully parallel; the airflow charts 67th provided which show symmetrical vortices at high mach numbers would tend to support this.

The second is that the higher velocity of a faster projectile means that the force (which, remember, is not directly proportional to velocity) has less time to act on the projectile. This means the total deflection (which is the result of acceleration times time squared) is less, as the time is less.

If there's a part of this I've misunderstood, I would be grateful to hear about it. But none of the effects that could throw off the trajectory of a spherical cannonball seem to be greater in impact at higher velocities, except for the amount of velocity lost by the projectiles relative to the same projectiles moving in a vacuum.

Things which would not act as counterexamples

Anything that relies on the behaviour of a non spherical projectile would not act as a counterexample here, because we are discussing spherical cannonballs. Any non-spherical projectile presents a different drag profile in different orientations, but a sphere is always oriented such that it has the same drag profile.

Anything which confirms the existence of the magnus effect would not act as a counterexample here. I am concerned with why it would be stronger at 1400-1100 fps (the regime only experienced by the faster cannonball) than at 1000-800 fps.

trice · Oct 25, 2022

Saphroneth said:
I've come back to this after giving some significant thought to try and work out if there's something I'm misunderstanding, and I'm afraid that I'm going to have to lay out my reasoning to see if there's a point which is at issue. This is because this paragraph of yours - in particular - actually seems to confirm part of my reasoning.

So we start with two large, round cannonballs of identical weight and dimensions being fired. One of them (A) is fired (with the appropriate elevation) at 1,000 fps, and the other (B) is fired (with the appropriate elevation) at 1,400 fps. The target is a fixed distance away, say 600 yards.

Based purely on ballistic trajectory, without for the moment considering the effects of the air in general or inaccuracy in other ways, both will strike the target.

Causes of inaccuracy

One cause of inaccuracy for a smoothbore is that the trajectory of the cannonball as it leaves the muzzle is the trajectory of the "last bounce" out of the barrel.
This factor may or may not be invariant with the velocity of the cannonball as it leaves the barrel. If it is variable depending on the velocity of the cannonball, I would expect (A) to be more inaccurate as the forwards component of its velocity is smaller.

Another cause of inaccuracy is wind. For this one, we would expect (A) to be more inaccurate because the wind blows on the ball to push it off course. With a slower ball (A) the flight time is longer and so the force has longer to take effect.

Not directly causing inaccuracy but still relevant is air resistance. The air resistance for ball (B) is a larger force, and so (B) will slow down more quickly, but it will not "cross over" in speed with ball (A) - ball (B) will always be faster at a given point on the trajectory. This also means that it will always take less time.

For non-round projectiles, the destabilization that takes place owing to the transsonic regime also has to be considered, but I do not think it is relevant for round projectiles. This is because for a non-round projectile it can destabilize by physically ending up pointing in a different direction as it travels through the air, drastically changing the cross-section, but for a round projectile the most that can happen is that the spin direction changes - and a smoothbore's spin direction is random (unknown) when it comes out of the barrel anyway.

Then the final factor to consider (unless I've forgotten one) is the Magnus effect. This is the product of asymmetric vortices around a rotating projectile, and where it is generated it produces a force which is
View attachment 455861
Which is to say, the force is proportional to the cross product of the velocity vectors for the fastest side (relative to the fluid) and the slowest side.
(All the other values are the same for two identical cannonballs in the same air at different velocities.)

This is the formula for the magnus effect on a round ball. For a projectile moving fast enough that both sides are moving in the same direction relative to the air (that is, if it is moving north at 1400 fps but the spin is such that there is no point on the ball that is instantaneously moving south) then the cross product is the cross product of two vectors that are mostly aligned (i.e. they are mostly moving north). This puts the magnitude of the cross product as being low.

There are two factors which mean that this appears to be less at high (supersonic or near-supersonic) velocities, for the same projectile.
The first is that the addition of 100 metres per second in the direction of travel to all the vectors involved makes the two velocity vectors in the calculation more parallel. This means that the magnitude of the force is lower, and may in fact be zero if the two vectors in the calculation are fully parallel; the airflow charts 67th provided which show symmetrical vortices at high mach numbers would tend to support this.

The second is that the higher velocity of a faster projectile means that the force (which, remember, is not directly proportional to velocity) has less time to act on the projectile. This means the total deflection (which is the result of acceleration times time squared) is less, as the time is less.

If there's a part of this I've misunderstood, I would be grateful to hear about it. But none of the effects that could throw off the trajectory of a spherical cannonball seem to be greater in impact at higher velocities, except for the amount of velocity lost by the projectiles relative to the same projectiles moving in a vacuum.

Things which would not act as counterexamples

Anything that relies on the behaviour of a non spherical projectile would not act as a counterexample here, because we are discussing spherical cannonballs. Any non-spherical projectile presents a different drag profile in different orientations, but a sphere is always oriented such that it has the same drag profile.

Anything which confirms the existence of the magnus effect would not act as a counterexample here. I am concerned with why it would be stronger at 1400-1100 fps (the regime only experienced by the faster cannonball) than at 1000-800 fps.

It is a long time since we were discussing this. I am sorry you are wasting so much time and energy on it. The matter has been discussed. If you cannot accept the results, I do not know what I can do further to enlighten you. Please review what has already been posted without focusing on disproving what you do not like.

Saphroneth · Oct 25, 2022

trice said:
It is a long time since we were discussing this. I am sorry you are wasting so much time and energy on it. The matter has been discussed. If you cannot accept the results, I do not know what I can do further to enlighten you. Please review what has already been posted without focusing on disproving what you do not like.

But what you said here was, and I will quote:

trice said:
On the magnus effect:

I am sure that I probably gave a poor description. I am neither a physicist nor a ballistician. Since college, my mathematics has only been used in application work on things like computer science, arbitrage work, metallurgy, etc. I have been wrong much more than once in my life and have no unusual problem in admitting it when I am. If I am here, I apologize for the confusion.

However, the magnus effect (known, but probably not clearly understood in Dahlgren's day) is not about time. It is about the pressure built up on the opposite side of the sphere from the direction of the spin rotation which is what causes the movement of the sphere. If that spin speeds up or slows down or changes direction in any way during flight, there will be a change in the magnus effect. More time simply means the magnus effect has longer to act.

What you said here was that the Magnus effect was not about time, but you then said that more time simply means the magnus effect has longer to act.

This is exactly in line with what I said here:

Saphroneth said:
This means that for a given rotational rate the velocity of the whole projectile does not affect the force (since the issue is the relative vectors between the fastest and slowest side, and consequently they don't change if you accelerate the whole system) and obviously if the travel time is smaller then the force has less time to produce acceleration (i.e. vector change) and for that vector to have an effect. In fact since the suvat equations (under constant force) state that s = ut + 1/2 at^2 (i.e. displacement is the square of time) and a projectile with double the velocity travels in half the time, the displacement from the magnus effect (at constant rotation rate) should go as the inverse square of the velocity; in other words, increasing velocity drastically reduces the amount of deflection from the magnus effect.

And which you disagreed with and told me:

trice said:
You are trying to come at this from the wrong direction. Assume you might be wrong, step back, research and think through the problem.

I did step back, research, and think through the problem, and it seems now that your post (quoted) is to the effect that I was right in the first place - the magnus effect has longer to act if there is more time.

Now, a faster projectile reaches the target in less time, so the magnus effect has less time to act on the projectile. This should mean that a faster projectile is less affected by the magnus effect.

Remember, what started all this was the claim that a faster round ball was less accurate than a slower one. What mechanism causes this?

trice · Oct 25, 2022

Please review the thread. Your answer lies within. Stop trying to prove your own conclusion to yourself.

Saphroneth · Oct 25, 2022

trice said:
Please review the thread. Your answer lies within. Stop trying to prove your own conclusion to yourself.

I'm not "trying to prove my own conclusion to myself". I'm asking you to highlight where my reasoning was incorrect, which is why I laid it out in such a detailed way - in the hopes that you could point to the error (as you saw it) in my reasoning. That way we can actually look into the point of disagreement.

Over the course of the thread, you pointed to an article on the aerodynamic drag on a cannonball, but this article talks about aerodynamic drag at various velocities. While the aerodynamic drag is certainly higher at higher mach numbers, this does not provide a mechanism for higher velocity projectiles to be more inaccurate because all it does is cause a higher velocity projectile to become lower velocity.
It doesn't cause the higher velocity projectile to deviate from its expected trajectory more than a lower velocity projectile, unless the expected trajectory doesn't take drag into account at all.

To be clear - the fact that a faster projectile has more aerodynamic drag is not relevant here. All it means is that the faster projectile slows down more quickly, but it still covers the distance to the target faster than a projectile which started off slower.

You also mentioned the Magnus effect and the rotations that cause it (which the article, incidentally, does not mention). But I tried to outline how my understanding of the mathematics behind the Magnus effect was such that it could not cause a greater deviation from the target for a projectile with a higher velocity - either this is not the cause, or my understanding of the Magnus effect is flawed, and if the latter is the case then the place where I have made the mistake should be obvious to someone who knows how it does work.
When I cited the equation behind the Magnus effect, you said that I "seem(ed) to be trying to assert you were right instead of trying to discover what the real actions of physics on the projectile are."
Does this mean that the equation I cited behind the Magnus effect does not actually govern the real actions of physics on the projectile? Because if I've used the wrong equation that's an obvious thing to correct; if I've applied it wrongly, then it should be possible to explain how.

You also mentioned the destabilization of a projectile, but a spherical projectile can't destabilize except insofar as its direction of spin might change. But this wouldn't actually increase the deviation of a faster projectile, because a slower projectile starts off spinning anyway, and a single randomly assigned deviation is actually greater than two randomly assigned deviations each with a total magnitude equal to the single one (as the vector sum is no greater than the scalar sum and is usually less).

trice · Oct 25, 2022

We have discussed all this. Please review what has already been said.

While you are at it, note that the leading gunnery experts of the mid-nineteenth century knew that there was an accuracy difference. They knew because they were testing the guns and looking at the results. They had no way to determine what caused it, because many of the factors were unknown and undiscoverable in that time. Still, they could see that there was a difference in the results.

Saphroneth · Oct 25, 2022

trice said:
While you are at it, note that the leading gunnery experts of the mid-nineteenth century knew that there was an accuracy difference. They knew because they were testing the guns and looking at the results. They had no way to determine what caused it, because many of the factors were unknown and undiscoverable in that time. Still, they could see that there was a difference in the results.

Can you provide a citation to that effect? (i.e. that a higher velocity cannonball of the same diameter has a lower accuracy, after trajectories have been corrected for.) That might allow me to work out what on earth is going on...

trice said:
We have discussed all this. Please review what has already been said.

But I have. I apologize if this is something blindingly obvious and I'm not seeing it, but it really does seem like you've mentioned several different effects at different times as the root cause and none of them are related to velocity in the right way - is it all of them? Is it only one of them?

It surely can't be destabilization, can it? Round projectiles can't destabilize because they're round...

trice · Oct 25, 2022

Saphroneth said:
Can you provide a citation to that effect? (i.e. that a higher velocity cannonball of the same diameter has a lower accuracy, after trajectories have been corrected for.) That might allow me to work out what on earth is going on...

But I have. I apologize if this is something blindingly obvious and I'm not seeing it, but it really does seem like you've mentioned several different effects at different times as the root cause and none of them are related to velocity in the right way - is it all of them? Is it only one of them?

It surely can't be destabilization, can it? Round projectiles can't destabilize because they're round...

Please review all the discussion. You have already been told the things you are ignoring. I will not waste the time.

Saphroneth · Oct 25, 2022

trice said:
Please review all the discussion. You have already been told the things you are ignoring. I will not waste the time.

Then what is the citation? I have checked the links provided, and while Dahlgren's shell gun treatise compares the Lancaster projectile with the smoothbore he says specifically that the round ball has greater accuracy under certain conditions because of the directness of its trajectory (on account of having a higher velocity).

I hardly claim I've caught every possible page of Dahlgren's shell-gun treatise, but I'm asking in so many words for a citation to the effect that a higher velocity cannonball of the same diameter has a lower accuracy. In fact in multiple places Dahlgren specifies that a lower trajectory (through higher velocity or greater specific density of the round) increases accuracy.

So, to review the discussion:

trice said:
Beyond that, Dahlgren's mission for the IX-inch and XI-inch was to produce a shell gun to use against wooden warships. This made accuracy important, not muzzle velocity. High muzzle velocity guns were less accurate at longer ranges.

trice said:
To start out, there are two classes: subsonic (under 1100 FPS) and supersonic (above 1100 FPS).

Supersonic projectiles suffer from an inherent instability when the speed drops back below the sound barrier. This would apply to both small arms and cannons.

IIRR, an XI-inch Dahlgren firing solid shot with a 15-pound charge would have a muzzle-velocity of about 900 FPS or so, well under 1100 FPS. With a larger charge, the velocity goes up. I think that at about 25-30 pounds it would be above the 1100 FPS velocity.

The closer to the sound barrier the velocity is, the sooner the drop below the speed of sound, the earlier the instability occurs. The earlier it occurs, the more it is likely to affect accuracy.

trice said:
Yes. You are simply thinking wrong about the problem: all projectiles will be destabilized as they drop through the sound barrier.

trice said:
All ballistic projectile with a muzzle velocity above the speed of sound will be subject to a disruptive effect when it loses velocity and falls through the barrier.

Only ballistic projectiles with a muzzle velocity above the speed of sound will be subject to that effect.

No ballistic projectile that starts out below the speed of sound will be subject to that disruptive effect

Your 1200 FPS example will experience the impact as the velocity drops.
Your 1000 FPS example will not experience the impact as the velocity drops.

It seems quite clear here that you mean that the faster projectile will be destabilized as it drops below the sound barrier. This is a consistent claim and so it appears that that is what you mean.

Here is what I am not understanding: how?
Any non-round projectile has an orientation in which it is initially fired, and then it can destabilize when it drops below the sound barrier. This means that it is now facing in a random orientation, which is different to the orientation it started with (front first) and changes its ballistic characteristics.

However, a round projectile is the same in any orientation, so it has the same ballistic characteristics no matter how it is rotated.

Alternatively, it could be meant that the projectile's rotation after destabilization now applies a random magnus effect which alters its course.
However, a round projectile fired from a smoothbore already has a random rotation, when it's fired in the first place.

I have checked all the links you have provided, and none of them answer the questions which I've raised; nor do any of them demonstrate what I asked for a source to demonstrate. I'm hardly claiming that I must be right here - I'm begging you to explain how it is that I'm making a mistake, and laying out my complete working for you to point out the flaw, or alternatively I am asking you to provide a citation that the "greater velocity equals lower accuracy" effect for an identical cannonball actually existed at the time.

67th Tigers · Oct 25, 2022

trice said:
Please review the thread. Your answer lies within. Stop trying to prove your own conclusion to yourself.

Yes it does. It is exactly what Saphroneth says. If you have no counterpoint, we should consider that you can't defend a position you've asserted.

Saphroneth is right in that the Magnus effect is generating an impulse, and an impulse is the integral of force over time. Ergo it is obvious that the impulse is proportional to the time, and that at lower velocities, for a given range the time is greater.

It's why slower baseballs curve more than fast ones.

trice · Oct 25, 2022

Obviously, you understand that you have long-since been answered. Drop your POV and approach the problem differently.

You understand that there is an effect related to the muzzle velocity which must be accounted for, but you want it to somehow be quantified into the magnus effect. There are several different effects beyond the magnus effect. Even today, ballistics calculations take into account the difference for projectiles fired above and below a certain velocity. There are actually three:

projectiles with a muzzle velocity below this point
projectiles with a muzzle velocity above this point
an attempt to combine #1 and #2 within the same formula (essentially an if-test branching on velocity)

67th Tigers · Oct 25, 2022

trice said:
Obviously, you understand that you have long-since been answered.

but not satisfactorily. You answers don't actually address the major problems with your assertions.

How can a round ball be destabilised? It's a literal physical impossibility, because a sphere is dimensionally constant in all directions.

Saphroneth · Oct 25, 2022

trice said:
Obviously, you understand that you have long-since been answered. Drop your POV and approach the problem differently.

You understand that there is an effect related to the muzzle velocity which must be accounted for, but you want it to somehow be quantified into the magnus effect. There are several different effects beyond the magnus effect. Even today, ballistics calculations take into account the difference for projectiles fired above and below a certain velocity. There are actually three:

projectiles with a muzzle velocity below this point

projectiles with a muzzle velocity above this point

an attempt to combine #1 and #2 within the same formula (essentially an if-test branching on velocity)

Well, firstly I'm asking for a citation that it happens to round balls at all. I have quite a solid understanding of how going from the supersonic regime to the subsonic regime destabilizes a non-round projectile, but that's the point - it destabilizes because it is no longer guaranteed to be in a front-first configuration, and in other orientations it looks different.

A round ball has no front or back.

trice · Oct 25, 2022

So what is it you two guys think will happen when force is applied to the round ball in flight?

67th Tigers · Oct 26, 2022

trice said:
So what is it you two guys think will happen when force is applied to the round ball in flight?

What force?

Saphroneth · Oct 26, 2022

trice said:
So what is it you two guys think will happen when force is applied to the round ball in flight?

If there are two identical round balls at different velocities, upon which the same force acts (e.g. gravity), then the faster round ball will feel the same effect in a given time but will strike the target sooner and thus have less total effect.

If the force differs according to velocity, then of course the balls will be affected in a different way. Which force are you thinking of, and how does it scale with velocity? (That second bit is important.)

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