DrB

08-16-2011, 02:35 AM

Don't know that many folks will care, but a member asked what the ratio between rotational and translational kinetic energies are for a bullet.

Here it is, assuming a cylindrical bullet. Ogive and other bullet shapes will decrease it, maybe by as much as half or a bit more for very long VLD type spitzers whereas a HBWC type design would increase it by a bit. This ratio will be in the close ball park, though, order-of-magnitude-wise.

Note that the ratio is constant irrespective of bullet weight or velocity, depending only on caliber and twist length.

Rotational Kinetic Energy/Translational Kinetic Energy = (1/2)*(Pi*C/TL)^2

Where

C = Caliber (in inches, for example)

TL = Twist Length (in inches, for example)

Pi = A circles circumference divided by it's diameter

For example, for a .224 with a 10 inch twist, it turns out that the rotational kinetic energy is about 0.25%, or 0.0025 of the translational kinetic energy. FYI, at 3000 fps with a 60 grain bullet, translational kinetic energy is about 1.625 kJoules (which is about the energy equivalent of a 26 pound lead brick hanging 50 foot up over your head :)). At the same velocity the equivalent height for rotational energy would be about an inch and a half worth of drop.

For a .501 with a 10 inch twist, the answer would be about 1.24%.

Best regards,

DrB

Here it is, assuming a cylindrical bullet. Ogive and other bullet shapes will decrease it, maybe by as much as half or a bit more for very long VLD type spitzers whereas a HBWC type design would increase it by a bit. This ratio will be in the close ball park, though, order-of-magnitude-wise.

Note that the ratio is constant irrespective of bullet weight or velocity, depending only on caliber and twist length.

Rotational Kinetic Energy/Translational Kinetic Energy = (1/2)*(Pi*C/TL)^2

Where

C = Caliber (in inches, for example)

TL = Twist Length (in inches, for example)

Pi = A circles circumference divided by it's diameter

For example, for a .224 with a 10 inch twist, it turns out that the rotational kinetic energy is about 0.25%, or 0.0025 of the translational kinetic energy. FYI, at 3000 fps with a 60 grain bullet, translational kinetic energy is about 1.625 kJoules (which is about the energy equivalent of a 26 pound lead brick hanging 50 foot up over your head :)). At the same velocity the equivalent height for rotational energy would be about an inch and a half worth of drop.

For a .501 with a 10 inch twist, the answer would be about 1.24%.

Best regards,

DrB