The basic reason he's off is he is assuming bullet rotation decays at the same rate as velocity. He says:
Iowegan said:
Yes! After the bullet exits the muzzle, the bullet's spin rate will decay at the same percent rate as velocity.
That's patently untrue. If the barrel length went all the way to the target then it would be true (if friction didn't stop the bullet in the bore on the way) because the rifling would then keep rotation rate tied to forward velocity. But, obviously, that's not what we shoot.
If you were to fix a microscopically small anemometer to the surface of the bullet, it would measure wind speed that was a vector product of the forward velocity and the rotational surface speed of the bullet. Only the vector component due to the rotational surface speed of the bullet has friction slowing the bullet's rotation. Moreover, because the boundary layer of air prevents the rifling from mattering to this, there is no frontal surface area crashing into the wind as the cross-sectional area of the bullet does going forward through the air, so the bullet rotation is slowed by what is called laminar fluid surface friction, alone. Where frontal area drag increases as the square of velocity times the drag coefficient of the shape at each velocity, the laminar drag is merely proportional to surface velocity. So we need to see how these velocities compare.
Taking the poster's example,
Iowegan said:
Example: a 223 Rem barrel has a 1:8 twist rate and drives a bullet to 2880 fps. 12/8=1.5; 2880 x 1.5=4320; 60 x 4320 =259,200 rpm.
But now let's do what he left out: calculate the surface speed of rotation whose friction acts to slow the spin of the bullet. Forward velocity is 2880 fps, so let's see how many fps the rotational component at the bullet surface has:
Bullet diameter = .224 in., so the circumference of the bullet is pi ×.224=0.704 in.
The bullet surface rotates through 0.704 in./turn
0.704 in./turn / 12 in./ft = 0.0587 ft./turn
The rifling pitch in 8 in/turn. or 0.667 ft/turn.
If we call feet of circumference ftc and feet of bullet forward travel ftd, then:
0.0587 ftc/turn / 0.667 ftd/turn = 0.0880 ftc/ftd, meaning 0.08796 feet of bullet surface rotation for each foot of bullet travel in the bore. Now we just multiply velocity at the muzzle by that ratio to get bullet surface speed of rotation:
2880 fps × 0.0880 ft/ft = 253 fps of surface rotational speed.
But there's more: remember the bullet presents cross-sectional area to the headwind its velocity creates, but in the rotation the rifling marks are not tall enough to present any sort of significant paddle area to the rotation, so bullet slowing forces are even greater, proportionally than just 11.4:1 (2880/253). For example:
An SS109 bullet fired at 2880 fps loses just about exactly 1 fps for each foot of travel right after leaving the muzzle.
Rotation loses about 0.042 ft/s of rotation for each foot of forward travel near the muzzle, or about 1/24 the rate of surface speed loss that the bullet has in forward speed loss.
Ballistician Geoffrey Kolbe has a formula for rotation loss that I've used to estimate the above numbers. You take the starting rotation rate in any unit you like, calling it
N. You then need the time of flight,
tof, from the ballistics program of your choice. To find rate of rotation,
Ntof, at the end of that time of flight in the same units:
Where
d = bullet diameter in inches
Ntof =
N × exp(-0.035 ×
tof /
d)
If you are not familiar with the exp() notation, it just means the natural logarithm base,
e, raised to the power inside the parenthesis. I used it here because we can't make superscripts or subscripts in text on the board. Use the
e^x function on the Microsoft calculator, where x is what's inside the parentheses. Excel accepts the exp() notation.
The other thing that post says that I find baseless is the assertion that a "correct" rotation would necessarily introduce too much torque to handle. He must think an ideal twist is very fast, but I don't know any reason to assume that automatically. If, like the .32 wadcutter example I gave, the best accuracy was produced by a very fast twist, it might be so, but as I said, I think that situation is peculiar to that bullet and the velocity range it is normally loaded to. I've had the standard 20" twist in a .44 Magnum group 240 grain soft points into under an inch at 50 yards from my Redhawk. I think that's down around the limit of my ability to hold a handgun shooting off bags, so I don't see how much more optimal it could get, and torque wasn't bothering me there.
Overall, Iowegun just seems to have a fictional sense of how exterior ballistics work.