Re: .308 twist rate?
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Turk</div><div class="ubbcode-body">http://www.6mmbr.com/barrelfaq.html#24640
I assume if one sees it as a function of time you are correct; but the bullet spins an x amount of revolutions for a given distance if I am not wrong? </div></div>
You can pick time or distance, it doesn't matter.
For the example below, let's just assume that the velocity doesn't change for the duration of it's flight for simplicity sakes.
A bullet at 2640 FPS (31,680 RPM) with a flight time of 2 seconds will have completed 1,056 revolutions in that 2 second span.
RPM (31,680)/second (60 in a minute) X (flight time(2) = 1,056
Same bullet at 2640 FPS spinning at 31,680 RPM will rotate 1,056 revolutions at the mile marker (5,280 feet)
31,680 / 60 = 528 revolutions per second
Since there is a 2 second flight @ 2640 to reach 5280, just multiply by 2 and you get the same answer, 1,056
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Turk</div><div class="ubbcode-body">http://www.6mmbr.com/barrelfaq.html#24640
I assume if one sees it as a function of time you are correct; but the bullet spins an x amount of revolutions for a given distance if I am not wrong? </div></div>
You can pick time or distance, it doesn't matter.
For the example below, let's just assume that the velocity doesn't change for the duration of it's flight for simplicity sakes.
A bullet at 2640 FPS (31,680 RPM) with a flight time of 2 seconds will have completed 1,056 revolutions in that 2 second span.
RPM (31,680)/second (60 in a minute) X (flight time(2) = 1,056
Same bullet at 2640 FPS spinning at 31,680 RPM will rotate 1,056 revolutions at the mile marker (5,280 feet)
31,680 / 60 = 528 revolutions per second
Since there is a 2 second flight @ 2640 to reach 5280, just multiply by 2 and you get the same answer, 1,056