Will somebody figure this out: The muzzle velocity of that round is 1200 feet per second, that barrel spins the bullet once for every 9.84 inches the bullet travels. How many RPM is that bullet spinning at when it leaves the barrel?
I got 87,805 RPM, or 1,463 rotations per second.
If barrel is 9.84 inch long it spins 1RPM per second
Here is the formula for determining bullet RPM:MV x (12/twist rate in inches) x 60 = Bullet RPMQuick Version: MV X 720/Twist Rate = RPMMV = muzzle velocity