This question comes up because people realize it takes some of the powder's energy to spin the bullet, and therefore, logically, it should take away from what is available to drive the bullet forward. Real guns don't work quite that simply because more resistance to the bullet going down the tube causes the powder to burn faster and reach higher peak pressure, thus actually speeding the bullet up. For example, when moly-coating bullets was a craze, newbies to the game thought their bullets would go faster without friction holding them back, but they typically lost 20 to 50 feet per second off their pet load's velocity because the powder had less resistance to build pressure against, and therefore burned and converted its chemical energy to bullet kinetic energy less efficiently.
That said, even that effect is not apparent with changes in bullet spin. I have made this calculation several times for different bullets in different calibers and the amount of energy in their spin (rotational kinetic energy) at the muzzle. If you subtract it from the bullet's muzzle energy and calculate how much that subtraction would lower velocity, the difference is typically on the order of just a tenth of a percent of the muzzle velocity. A difference that small is lost in shot-to-shot velocity variation. The normal chamber-to-chamber velocity difference you find in two guns with same-length, same-twist barrels on them is also greater than that, so comparing two guns with different rifling twists on them will just show the gun-to-gun difference and not tell you anything about the energy that went into spinning the bullet up.
But, calculate for yourself to satisfy your curiosity. The forward-moving kinetic energy (translational kinetic energy) of the bullet has the same units as rotational kinetic energy (foot-pounds, in the avoirdupois system), so finding the difference is just performing the subtraction of one from the other. However, they are calculated somewhat differently because the translational motion has the whole mass moving forward at the same velocity, while rotation has the speed at the surface greater than what is underneath. The solution requires finding a mass-equivalent characteristic for ratation called the axial mass moment of inertia, which you will find challenging to work out for some bullet shapes. My CAD software does it for me automatically when I draw a 3D shape, but I have to get internal feature dimensions like hollow points and jacket thicknesses right for that to be accurate. The better approach is to measure the axial moment of inertia with a hanging torsion pendulum. I've done that, too. It took me a while to get a serviceable instrument for this, though, as the suspension wire has to be good. I wound up buying some 0.0029" tungsten wire that works. The traditional fused quartz filament is something I didn't have as readily available.
If you try this, keep in mind that mass in the Avoirdupois measuring system has units of
slugs, not pounds or grains, so you have to divide bullet weight in grains by 225,218 (225,000 is close enough for what we are doing here) to get slugs before plugging it into the
Ekt=½mV² equation if you want to get foot-lbs for translational KE. The rotational KE equation is analogous at
Ekr=½Iω². Follow the links for the equation details. The first link explains the concept of KE as well.
An example is made easier by using a cylindrical bullet, the double-ended wadcutter bullet (the axial moment of inertia of a cylinder is just
½mr²):
For a 38 DEWC weighing 148 grains and traveling at 750 fps from a gun with an 18.75-inch twist. I'll assume the radius, after allowing for rifling engraving, is the equivalent of about 0.177 inches, or 0.01475 feet.
Bullet mass = 148 grains / 225000 grains/slug = 0.000657 slug =
m
Bullet axial moment of inertia = I = ½mr² = 0.5 × 0.000657 × (0.01475 ft)² = 0.0000000715 slug-ft²
With forward velocity,
V, in ft/s and rifling pitch,
T, in inches:
Bullet rotation speed,
ω, in radians/second =
24πV/T = 24π × 750 / 18.75 = 3016 rad/s
Bullet translational kinetic energy =
½mV² = 0.5 × 0.00657 × 750² = 184.781 ft-lbs
Bullet rotational kinetic energy =
½Iω² =
0.5 × 0.0000000715 × 3016² = 0.325 ft-lbs
The difference in the two kinetic energies is:
184.781 - 0.325 = 184.456
The difference in velocity is proportional to the square root of the difference in energies:
184.156 / 184.781 = 0.998241
√0.998241 = 0.9912
0.99912 × 750 ft/s = 749.3 ft/s
So the rotational kinetic energy in the bullet is equal to a translational velocity of just 0.7 ft/s.
You'll have noticed some unjustifiable exctra decimal places above, but I felt they were useful to illustrate how small this is. In this case, 0.088% of velocity impact.
