# Bullets in flight

1. Dec 18, 2009

### usdmgtr

Is this statement correct? As a bullet flies, it loses forward velocity MUCH faster than it loses rotational velocity.

If a barrel rotates a bullet 1 turn in 10 inches. Would it not exit rotating at the same rate regardless of velocity? Understanding as velocity is increased the RPM will increase. Why would not rotational velocity be reduced equally to forward velocity?

Useful info.
http://en.wikipedia.org/wiki/Rifling
http://kwk.us/twist.html

2. Dec 18, 2009

### Neo_Anderson

That's because most of the gunpowder energy is used up to push the bullet in its barrel, while only some of the gunpowder energy is used to spin the bullet as it travels down the barrel. The opposite of this would be a gun barrel that has a grooved spiral that would rotate the bullet 10 times for every one inch of travel. The bullet would leave the barrel spinning very quickly, but wouldn't travel very far...

3. Dec 18, 2009

### Staff: Mentor

Welcome to the PF. What are your thoughts on your question? What impedes the linear motion of a bullet? What impedes the rotational motion of a bullet?

4. Dec 18, 2009

### usdmgtr

My apologies, I was assuming a rifled barrel with 1 rotation per 10".

Quote from http://en.wikipedia.org/wiki/Rifling#cite_note-12
"Rifling is the process of making spiral grooves in the barrel of a gun or firearm, which imparts a spin to a projectile around its long axis. This spin serves to gyroscopically stabilize the projectile, improving its aerodynamic stability and accuracy.
Rifling is described by its twist rate, which indicates the distance the bullet must travel to complete one full revolution, such as "1 turn in 10 inches" (1:10 inches), or "1 turn in 30 cm" (1:30 cm). A shorter distance indicates a "faster" twist, meaning that for a given velocity the projectile will be rotating at a higher spin rate.
A combination of the weight, length and shape of a projectile determines the twist rate needed to stabilize it – barrels intended for short, large-diameter projectiles like spherical lead balls require a very low twist rate, such as 1 turn in 48 inches (122 cm).[1] Barrels intended for long, small-diameter bullets, such as the ultra-low-drag, 80-grain 0.223 inch bullets (5.2 g, 5.56 mm), use twist rates of 1 turn in 8 inches (20 cm) or faster.[2]"

5. Dec 18, 2009

### usdmgtr

Thank you for the warm welcome. This is the first time I have had to do some real thinking in a long time. It is fun.

I have not defined my thoughts yet. I am reading a lot currently trying to gather facts and ideas.

The main impedance of linear motion is air resistance. Rotational motion and its relation to linear motion is the bit I don’t understand.

If the bullet is always rotating at 1 per 10 inches and RPM is increased with velocity. Would it not decrease at and equal rate in deceleration.

6. Dec 18, 2009

### Neo_Anderson

I think what I was trying to relay to you is that the rate at which the bullet spins is dependent upon the twist rate. If you have a 1:10 twist rate, the bullet will travel quite far, while at the same time will not spin very fast. A barrel with a twist rate of 10:1 on the other hand will spin very quickly but not go very far, per your Wikipedia quote.
What does any of this mean? It means that with a 1:10 twist rate, the bullet leaves the barrel going forward much faster than it twists, and with the 10:1 twist rate, the bullet leaves the barrel going forward much slower than it twists.
With the 10:1 twist rate, the bullet stops spinning before it drops to the ground, and with the 1:10 twist rate, the bullet will drop to the ground while it's still spinning.

I hope this explains why the bullet can go either faster than its twist rate, or slower than its twist rate. Remember, you asked why both the twist and the forward velocity are not equal; my explination illustrates why.

7. Dec 18, 2009

### Staff: Mentor

The rotational rate (RPM, not rotations per distance) that the bullet leaves the barrel with will remain fairly constant until it impacts the target. There is not much resistance to the rotation. The spinning motion stays fairly constant, even as the velocity slows due to air resistance.

8. Dec 18, 2009

### Neo_Anderson

The bullet would not decrease its spinning at an equal rate it decreases its velocity, because if this were the case, all bullets--regardless of their twist rate--would drop to the ground at the same time they cease spinning. The rate at which the bullet decreases its spinning however is a function of the rate it decreases its velocity.

9. Dec 18, 2009

### Staff: Mentor

Sorry, what does that mean? I would think that the linear velocity and the rotational rate (in RPM) would be totally decoupled after the bullet leaves the muzzle.

10. Dec 18, 2009

### Neo_Anderson

There is much more resistance to rotation for the bullet in-flight than you realize. Air molecules--stationary with respect to the bullet--are vectors that act on the radial--or "front"--of the bullet. The angular momentum vector of the bullet must be subtracted by the vector which describes the air molecule, and this subtraction between the two vectors are perpendicular to each other.

Do a thought experiment in your mind, and you'll clearly see that the vector which keeps the bullet spinning gets reduced in magnitude by the vector which describes the stationary air molecule.

11. Dec 18, 2009

### Neo_Anderson

Absolutely not. To prove this using a sort of gendankenexperiment, consider a bullet with a flat face. Would the stationary air molecules not interact with the molecules that make up the bullet? And would those air molecules--traveling directly at the molecules of the bullet--strike the face of the bullet at a 90-degree angle relative to the bullet?
And since the air molecules are striking the bullet molecules at a 90-degree angle, any motion, whether angular or linear, would be affected.

Extend the gendankenexperiment ("thought experiment") to the bullet that's very aerodynamic. The effect the air molecule has on the linear flight of the bullet would be lessened, but so would the effect the air molecule would have on the angular spin of the bullet.

12. Dec 18, 2009

### usdmgtr

I get it now :)

A bullet will maintain angular momentum as it decelerates and can increase past its original rate of twist (in distance). Twist rate is used with velocity to get a bullet to a specific rotational velocity required to stabilize in flight.

Thank you, I plan to hang out on the forum to learn... Don't suspect I will contribute much. I hope that is ok.