In order to participate in the GunBroker Member forums, you must be logged in with your GunBroker.com account. Click the sign-in button at the top right of the forums page to get connected.
Bullet Rotational Energy and Reaction Torque
spirits
Member Posts: 363 ✭✭✭
While reading the post concerning if "Mr. Lazzaroni" as in Lazzaroni Rifles was really that Lazzaroni, one post expressed the hope that maybe now some ballistic questions would be answered. One concerned rotational kinetic energy of a bullet. Here's a simple equation, which can be applied for any bullet weight, speed, twist, diameter, etc., for rotational kinetic energy if you assume the bullet's rotational inertia is treated like that of a simple cylinder
Bullet's Rotational Kinetic Energy (BRKE) in Foot Pounds (Ft-Lbs) equals
Bullet's Translational Kinetic Energy (BTKE) divided by [2 X [twist / (Pi X diameter)]^2 ] ; where [ ]^2 means square of [ ] and the twist and bullet diameter are in inches.
BRKE Example: 150 grain 0.308 inch diameter bullet at 3000 ft/s in a rifle with a 10 inch twist.
BTKE = m v^2 /2 = 2995 (Ft-Lbs)
[2 X [twist/(Pi X diameter)]^2] = 214
Thus, the BRKE = 14 (Ft-Lbs).
For the average reaction torque on the rifle which ultimately acts on you too. From the torque impulse angular momentum theorem and making the same assumption as in the BRKE concerning the bullet's rotational inertia
Average Reaction Torque (ART) = Bullet's Rotational Kinetic Energy multiplied by [twist/(Pi X barrel length)]
ART Example: Same as the BRKE example but now we require the barrel length in inches too - so lets equate it to 24 inches in this example.
[twist/(Pi X barrel length)] = 0.133
Thus, the ART = 1.86 (Ft-Lbs) or 22.3 (Inch-Lbs).
This torque tends to twist the rifle counter-clockwise out of your hands and off your shoulder; and if you are a right-handed shooter, the butt stock smacks you in the face too but in this example the ART is so small it probably isn't noticeable. Also, for a rifle with right-hand rifling twist and barrel threads with a right-hand twist too, the reaction torque tightens the receiver-barrel joint. Now, if you own a 577 T-Rex rifle which shoots a .577 inch diameter 750 grain bullet at 2460 ft/s with a 12 inch twist, the bullet's BTKE = 10,068 (Ft-Lbs) and its BRKE = 115 (Ft-Lbs) which totals 10,183 (Ft-Lbs) for the bullet's energy. For a 24 inch barrel length, the ART = 18.3 (Ft-Lbs) or 220 (Inch-Lbs) for a big smack in the face for you right-handed shooters and sore shoulders for everyone.
Bullet's Rotational Kinetic Energy (BRKE) in Foot Pounds (Ft-Lbs) equals
Bullet's Translational Kinetic Energy (BTKE) divided by [2 X [twist / (Pi X diameter)]^2 ] ; where [ ]^2 means square of [ ] and the twist and bullet diameter are in inches.
BRKE Example: 150 grain 0.308 inch diameter bullet at 3000 ft/s in a rifle with a 10 inch twist.
BTKE = m v^2 /2 = 2995 (Ft-Lbs)
[2 X [twist/(Pi X diameter)]^2] = 214
Thus, the BRKE = 14 (Ft-Lbs).
For the average reaction torque on the rifle which ultimately acts on you too. From the torque impulse angular momentum theorem and making the same assumption as in the BRKE concerning the bullet's rotational inertia
Average Reaction Torque (ART) = Bullet's Rotational Kinetic Energy multiplied by [twist/(Pi X barrel length)]
ART Example: Same as the BRKE example but now we require the barrel length in inches too - so lets equate it to 24 inches in this example.
[twist/(Pi X barrel length)] = 0.133
Thus, the ART = 1.86 (Ft-Lbs) or 22.3 (Inch-Lbs).
This torque tends to twist the rifle counter-clockwise out of your hands and off your shoulder; and if you are a right-handed shooter, the butt stock smacks you in the face too but in this example the ART is so small it probably isn't noticeable. Also, for a rifle with right-hand rifling twist and barrel threads with a right-hand twist too, the reaction torque tightens the receiver-barrel joint. Now, if you own a 577 T-Rex rifle which shoots a .577 inch diameter 750 grain bullet at 2460 ft/s with a 12 inch twist, the bullet's BTKE = 10,068 (Ft-Lbs) and its BRKE = 115 (Ft-Lbs) which totals 10,183 (Ft-Lbs) for the bullet's energy. For a 24 inch barrel length, the ART = 18.3 (Ft-Lbs) or 220 (Inch-Lbs) for a big smack in the face for you right-handed shooters and sore shoulders for everyone.
Comments
It would be interesting to take this further and be able to calculate where the velocity and rotational velocity degrades to the point the bullet can no longer remain stable. I'm not talking the point of transonic transition, where the bullet gets disrupted and may or may not return to stability. I'm talking about the point where it flat runs out of steam.
That's the point where it hits the dirt. Watch for the dust cloud and you'll have your answer.
quote:Originally posted by sandwarrior
It would be interesting to take this further and be able to calculate where the velocity and rotational velocity degrades to the point the bullet can no longer remain stable. I'm not talking the point of transonic transition, where the bullet gets disrupted and may or may not return to stability. I'm talking about the point where it flat runs out of steam.
That's the point where it hits the dirt. Watch for the dust cloud and you'll have your answer.
Ummm NO...I'm looking for the point in the air when a bullet becomes radically unstable. NOT the point at which it stops. Which would be the ground, as you suggested.
Charliemeyer007,
That would be pretty cool if you could find that. I have read or was told that the rotational inertia usually slows down about the rate of 10% of the forward velocity.