Does a RB travel with constant rotation, or accelerate?

[Larry (Omaha)]

Two other BP shooters I know have different opinions about RB rotation after it leaves the barrel. So, I am going to ask the experts here. A 50 cal, 48" barrel, with 1 revolution in 48" twist is fired using 60 grains of 2F.
When the RB leaves the muzzle it has already rotated 1 complete turn. Here is the question: Does the RB continue rotation of 1 turn every 4', or does it accelerate to a faster rotation?
Flintlocklar

 

[rdlowe]

Good question, Larry. The bullet won’t spin faster or slower until/ unless an external force acts upon it. This usually happens when it impacts something. If you could fire a round ball in outer space, it would keep spinning and spinning and spinning at the same rate as when it left the muzzle.

 

[Carbon 6]

The moment it leaves the barrel it begins to slow down.

The confusion comes from how you measure rotation. Often it is converted to RPMs
For example a round ball exiting the muzzle doing 1500 fps out of a 1/66 twist gun is rotating just over 16000 rpms.

 

[pmccoywss]

What would make it accelerate? It will slowly lose both velocity and rotational speed.

 

[tenngun]

A ball drops from the mold looking smooth and shiny, to use. Up close it looks like the moon. And air resistance grabs it.
There would have to be an energy source to speed it up, or even maintain it.

 

[Larry (Omaha)]

What would make it accelerate? It will slowly lose both velocity and rotational speed.

BP shooter #2 says it accelerates, not me, I heard both BP shooters thoughts and that is why I posted.
Flintlocklar

 

[tnlonghunter]

I believe it's Newton's 1st law - an object in motion stays in motion until acted upon by an outside and equal (or greater) force. In this case, the force of the expanded gas from your main charge produces the forward motion, and the rifling induces spin. But once the ball leaves the barrel, the explosion of the charge no longer exerts any force. Rather, gravity becomes the more powerful force, along with wind resistance against the surface of the ball. In combination, they will slow the ball's rate of rotation to some degree until it hits the target or the ground.

 

[Col. Batguano]

Pletch! Pletch!

 

[Larry (Omaha)]

The moment it leaves the barrel it begins to slow down.

The confusion comes from how you measure rotation. Often it is converted to RPMs
For example a round ball exiting the muzzle doing 1500 fps out of a 1/66 twist gun is rotating just over 16000 rpms.

What is the formula to convert rate of twist to RPMs?
Flintlocklar

 

[Boomerang]

What is the formula to convert rate of twist to RPMs?
Flintlocklar

FPS X 12 divided by rate of twist X 60

 

[Zonie]

To convert a rate of twist to RPM's you have to know the velocity of the ball and the rate of twist.
Since velocity in firearms is usually measured (in the USA) in feet per second and barrel twist is usually measured in inches we have to convert the twist rate to feet by dividing by 12. For instance a 1:48 twist equals 48 inches per revolution. 48/12 = 4 feet per revolution.

If the velocity of the ball was 1400 feet per second and the rate of twist was 4 feet, dividing 1400 fps by 4 will give the rotation of the ball per second. In this case, 1400/4 = 350 rotations per second.
Because RPM stands for Revolutions Per minute and there are 60 seconds in a minute we need to multiply that 350 RPS times 60 to get, 350 X 60 = 21,000 RPM.

Putting both of these calculations together we end up with (V/T) X 60 = RPM (Where V is the velocity in feet per second and T is the rate of twist in feet).

Let's do it again with a 1600 fps velocity and a rate of twist of 1:60. 60 inches/12 = 5 feet.

(V/T) X 60 = (1600/5) 60 = (320) 60 = 19,200 RPM.

 

[tenngun]

Throw in how GUI key the ball looses velocity and that would slow the rpm

 

[tenngun]

Quickly.... confused spell check with my custom letter order

 

[hawkeye2]

Throw in how GUI key the ball looses velocity and that would slow the rpm

Sorry but I respectfully disagree. The velocity of the ball will decrease mostly due to air resistance but the rate of rotation will remain constant as there are no forces acting on it to slow it's rotation. That holds true in Physics 101 as we were told to ignore friction in pulleys and things like that. Actually there is air resistance acting on the surface of the ball in opposition to its rotation which will slow it somewhat however if it left the muzzle at 1500 fps and 20,000 rpm it will still be spinning at (almost) 20,000 rpm when its velocity has dropped to 900 fps. The only place forward velocity influences rotational velocity is while the projectile is still in the barrel.

 

[tenngun]

Um, you think?
If it turns one turn in four feet and it is going 1600 fps it would take .0025 seconds to make one turn, cover that four foot distance. 24000 revolutions in one minute.
But it will slow in a short time to 1200 fps .0033 of a second to cover four feet and make one turn. 18000 revolutions per min.
So I’m thinking even with a ball keeping a spin rate of 1 in 48 the rpm would slow as the time it takes to go four feet increases
Of course a ball will rarely be one second in flight.

 

[hawkeye2]

The RPM is entirely independent of the forward velocity once the ball leaves the muzzle. The RPM or angular velocity is established by the rate of twist and the velocity of the ball at the time of firing but has nothing to do with the distance it travels any time after leaving the barrel. If we have a barrel with a twist of 1 turn in 48" (constant twist and not gain twist rifling) and the barrel is 48" long It uses that length to reach the maximum velocity and after the ball exits the muzzle it begins to loose velocity and to drop due to air resistance and gravity. Those two forces however do not oppose the rotation of the ball and in turn slow it. The momentum of its spin (see: spin angular momentum) will continue to keep its angular velocity constant and that momentum must be overcome by something to change the rate of spin. If the RPM were to drop the projectile would loose stability. That would have a considerable effect on an elongated projection and I suspect a lesser effect on a round ball.

 

[TNGhost]

Good question, Larry. The bullet won’t spin faster or slower until/ unless an external force acts upon it. This usually happens when it impacts something. If you could fire a round ball in outer space, it would keep spinning and spinning and spinning at the same rate as when it left the muzzle.

The "something" it impacts is the air the split second it leaves the barrel,, and it is also affected immediately by gravity.

The rate at which it accelerates in the barrel is also ******** by a degree by the air column in the barrel which has to be moved for the ball to escape.

Insofar as the rotation, it will be slowing as well, from friction, and the rate of slowing will be determined by the density of the air. While not as great as the resistance on forward velocity, it is still there.

 

[hawkeye2]

"Insofar as the rotation, it will be slowing as well, from friction, and the rate of slowing will be determined by the density of the air. While not as great as the resistance on forward velocity, it is still there."

That is very true but it is my opinion that the amount of slowing before the ball struck the ground would be almost negligible considering it was spinning at something like 20,000 RPM. Air resistance increases as the square of the velocity and the velocity we would be looking at here is the surface speed (rpm times circumference) of the projectile in the direction of rotation.

In my previous post (#16) please substitute projectile for projection in this sentence:
"a considerable effect on an elongated projection and I suspect a lesser effect on a round ball."

 

[TNGhost]

I agree, negligible in relation to forward velocity, but measurable, and significant in comparison to traveling in a vacuum.

 

[hawkeye2]

Yes measurable. Ballistics in space free of an atmosphere and gravity can be quite an interesting subject.