Here's a way to get an idea of the kinds of energy we are talking about. Pretend you are firing a 62-grain .224" double-ended wadcutter. The formula for the axial moment of inertia of a cylinder of homogenous density is very simple and can be accomplished without resorting to pendulums and the like. It is I=½mr². So, r=0.112". Since we are working in the lbf-ft-s system of measure, we divide r by 12 to convert it to feet, so 0.09333…ft, and we square that to get 0.00008711…ft². Mass in this system is in slugs, and grains are converted to slugs by dividing by 225218. So 62/225218=0.0002753 slugs.
So, where Ia is the axial moment of inertia
Ia = ½ × 0.0002753 slug × 0.00008711 ft² = 0.00000001199 slug-ft²
If the bullet is fired at 3000 fps, we all are familiar with the formula for RPM which is 720×MV/T where T is the rifling pitch in inches. Revolutions per second are just 60 times smaller than that, so 12×MV/T. In this case we want angular velocity (rotational speed), ω, in radians per second which we get by multiplying revolutions per second by 2× π that will be:
Where T is in inches:
ω = 12MV×2×π/T
ω7” = 32314 rad/s
ω9” = 25133 rad/s
For kinetic energy you have linear and angular (spin energy)
Muzzle Energy Linear:
KElin = ½ × m × v²
Angular looks the same except moment of inertia takes the place of mass and ω takes the place of v. So:
KEang = ½ × I × ω²
So let’s solve them for this bullet at 3000 fps for both T = 7” and T = 9”.
KElin = ME = ½ × 0.0002753 slugs × (3000 fps)² = 1239 ft-lb
7” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (32314 rad/s)² = 6.260 ft-lbs
9” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (25132.74 rad/s)² = 3.787 ft-lbs
The difference in those two rotational KE number is 2.473 ft lbs. If you subtract that to or from the ME, we get:
1239 ft-lb * 2.473 ft-lb = 1236.527 ft-lb. Solving the two MEs for velocity we get:
MV₉ = √(1239 ft-lb / (½ × 0.0002753 slug)) = 3000 fps
MV₇ = √(1236.527 ft-lb / (½ × 0.0002753 slug)) = 2997 fps
So going from a 9” to a 7” twist with this bullet will make a difference of 3 ft/s, assuming the extra load doesn’t make the powder burn faster and tend to compensate with extra pressure. In other words, it is less difference than firing in two different chambers is likely to make, and hence insignificant to my way of looking at it.
So, where Ia is the axial moment of inertia
Ia = ½ × 0.0002753 slug × 0.00008711 ft² = 0.00000001199 slug-ft²
If the bullet is fired at 3000 fps, we all are familiar with the formula for RPM which is 720×MV/T where T is the rifling pitch in inches. Revolutions per second are just 60 times smaller than that, so 12×MV/T. In this case we want angular velocity (rotational speed), ω, in radians per second which we get by multiplying revolutions per second by 2× π that will be:
Where T is in inches:
ω = 12MV×2×π/T
ω7” = 32314 rad/s
ω9” = 25133 rad/s
For kinetic energy you have linear and angular (spin energy)
Muzzle Energy Linear:
KElin = ½ × m × v²
Angular looks the same except moment of inertia takes the place of mass and ω takes the place of v. So:
KEang = ½ × I × ω²
So let’s solve them for this bullet at 3000 fps for both T = 7” and T = 9”.
KElin = ME = ½ × 0.0002753 slugs × (3000 fps)² = 1239 ft-lb
7” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (32314 rad/s)² = 6.260 ft-lbs
9” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (25132.74 rad/s)² = 3.787 ft-lbs
The difference in those two rotational KE number is 2.473 ft lbs. If you subtract that to or from the ME, we get:
1239 ft-lb * 2.473 ft-lb = 1236.527 ft-lb. Solving the two MEs for velocity we get:
MV₉ = √(1239 ft-lb / (½ × 0.0002753 slug)) = 3000 fps
MV₇ = √(1236.527 ft-lb / (½ × 0.0002753 slug)) = 2997 fps
So going from a 9” to a 7” twist with this bullet will make a difference of 3 ft/s, assuming the extra load doesn’t make the powder burn faster and tend to compensate with extra pressure. In other words, it is less difference than firing in two different chambers is likely to make, and hence insignificant to my way of looking at it.