Robert Egler
50 Cal.
- Joined
- Jul 17, 2007
- Messages
- 1,319
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I've been thinking about this from a physics point of view, what is the physics that would cause the ball to increase it's rate of spin?
I think that were actually talking about the wrong rate of spin and that’s causing the confusion.
As the ball leaves the muzzle it will have a certain spin in revolutions per second. Ignore for a moment the revolutions per inches traveled.
By conservation of angular momentum the ball will keep this rate a spin (in revolutions per sec) with a slow decrease due to friction. But that’s the friction of the air perpendicular to the direction of travel. In other words, the friction of the spin, not the drag as the ball travels forward. This friction of spin is small in comparison to the drag on the ball, and the drag on the ball as it travels downrange will have only a very small effect on the rate of spin. BUT the drag will slow the speed of the ball considerably.
At the muzzle a ball shot through a barrel with a 1/66 twist with a velocity of 2200 feet per second will have a rate of SPIN of 400 revolutions per second. It will keep this rate of SPIN pretty much (with a very slight slowdown as mentioned above) until it hits the target.
But the distance traveled in 1 second will decrease significantly as the velocity decreases due to drag.
So if we measure the rate that ball is turning in revolutions per inch (let’s call that the rate of twist) we find that the rate of twist increases after the ball leaves the barrel BECAUSE the velocity of the ball decreases faster than the rate of spin decreases. It’s still turning at 400 turns per second, but it’s traveling less distance in that second, so the ball is spinning faster in relation to the distance traveled, ie, the rate of twist.
For example, consider the same ball we discussed about, 1/66 rate of twist at 2200 f/s at the muzzle. The ball moves 66 inches in the time it takes to revolve once,: 1/66. When the velocity drops to 1900 f/s the ball is STILL turning at very close to 400 rev/s, but now because its velocity is lower it’s only moving forward 57 inches while it makes that revolution, so the rate of TWIST is now 1/57, a faster rate of TWIST than the ball had at the muzzle.
SO, to sum it up, the rate of TWIST in revolutions per inch traveled DOES increase after the ball leaves the muzzle. That's because the velocity decreases faster that the rate of spin in rev/sec. The rate of SPIN in revolutions per second does not increase.
That help any?
I think that were actually talking about the wrong rate of spin and that’s causing the confusion.
As the ball leaves the muzzle it will have a certain spin in revolutions per second. Ignore for a moment the revolutions per inches traveled.
By conservation of angular momentum the ball will keep this rate a spin (in revolutions per sec) with a slow decrease due to friction. But that’s the friction of the air perpendicular to the direction of travel. In other words, the friction of the spin, not the drag as the ball travels forward. This friction of spin is small in comparison to the drag on the ball, and the drag on the ball as it travels downrange will have only a very small effect on the rate of spin. BUT the drag will slow the speed of the ball considerably.
At the muzzle a ball shot through a barrel with a 1/66 twist with a velocity of 2200 feet per second will have a rate of SPIN of 400 revolutions per second. It will keep this rate of SPIN pretty much (with a very slight slowdown as mentioned above) until it hits the target.
But the distance traveled in 1 second will decrease significantly as the velocity decreases due to drag.
So if we measure the rate that ball is turning in revolutions per inch (let’s call that the rate of twist) we find that the rate of twist increases after the ball leaves the barrel BECAUSE the velocity of the ball decreases faster than the rate of spin decreases. It’s still turning at 400 turns per second, but it’s traveling less distance in that second, so the ball is spinning faster in relation to the distance traveled, ie, the rate of twist.
For example, consider the same ball we discussed about, 1/66 rate of twist at 2200 f/s at the muzzle. The ball moves 66 inches in the time it takes to revolve once,: 1/66. When the velocity drops to 1900 f/s the ball is STILL turning at very close to 400 rev/s, but now because its velocity is lower it’s only moving forward 57 inches while it makes that revolution, so the rate of TWIST is now 1/57, a faster rate of TWIST than the ball had at the muzzle.
SO, to sum it up, the rate of TWIST in revolutions per inch traveled DOES increase after the ball leaves the muzzle. That's because the velocity decreases faster that the rate of spin in rev/sec. The rate of SPIN in revolutions per second does not increase.
That help any?