# Spinning bullet, why?

by Clausius2
Tags: bullet, spinning
 PF Patron Sci Advisor P: 1,481 I was wondering me why a bullet is shoot spinning around it axis of symmetry. The cannon of the gun has usually small channels in spiral shaped, to exert a torque on the bullet in order it can rotate about it axis (i suppose). I've been thinking about this, and i heard that it has to do with bullet trajectory stability, but i can't demostrate that. What is the true reason?. Missiles are shoot spinning too?.
 P: 1,560 A spinning bullet breaks the air better, and missles have guidance systems so they don't ususally need to spin....I suspect the spinning helps trajectory a little like a bicycle wheel spinning help to keep the bicycle upright, angular momentum..something like that (mentor, correct me please, if I am wrong)
 P: 231 So a spinning bullet is something of a mini gyroscope? I never thought about that before, but I guess it must have some gyroscopic effects.
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## Spinning bullet, why?

I think, and of course I may be wrong, that when the barrel of a gun is built in a way that forces bullets to spin as they are shot (using helical slices along the barrel) the bullet's vertical displacement is lower. So if you shoot something 100ft away, a spinning bullet will hit 2" below the target whereas a non-spinning bullet will hit 5" below (I made those numbers up, they are not necessarily true).
 P: 2,380 Spinning the bullet is called "rifling". Hence the name: Rifle. What happens to a football when you throw it if you don't add some spin? The same thing would happen to a bullet. How far the bullet drops has nothing to do with spinning other than that it keeps it in the most favorable aerodynamic position.
 P: n/a Just so you know, a 30-06 with a 150 grain pointed projectile fired (level) at a muzzle velocity of 2910 ft/sec will drop a total vertical distance of 470 inches at 1000 yds after a flight time of 1.824 seconds. If you use the equation $$\Delta x = 1/2at^2$$ for the given time you will find that delta x = 642 inches. This difference is not directly related to spin. The aerodynamic drag is a vector quantity. The dynamic pressure is exerted on the projectile from an angle below the horizontal as the bullet falls through the air during flight. This is also true of the wind drift correction, and must be taken into account to make any sort of accurate prediction of flight path. Wind drift is directly proportional to wind velocity. A wind velocity of 10 miles/hour perpendicular to the gun-target line will give a drift of 140 inches at 1000 yds. I realize that the units I have used here are a pain(not metric, not consistant), but they are the units that are in common use for ballistic calculations. Without spin, the bullet would tumble in flight. Modern rifles are only capable of such amazing accuracy because the bullet is stable in flight. Even spherical projectiles must have spin to acheive any sort of acceptable accuracy. The rate of spin can be very critical. If the bullet is "over stabilized" the bullet will retain a nose up attitude after reaching it's highest point in the trajectory giving an unpredictable point of impact and very, very poor accuracy. Also, bullets often leave the muzzle of the gun with quite a bit of yaw; "Overstabilization" can cause accuracy problems with that as well. There are formulas that are used with very good success to predict the proper spin rate for a bullet of a given shape fired at a given velocity. The old formula(1879) was called the Greenhill formula but has been changed to take into account the different shapes of modern bullets. If anyone is interested I will give you some links. -Mike
 P: 1,560 Not to be questioning the authority of the knowledge as you present it, but I had a 'hunter', and another, look at the figure, a 40 FOOT drop in 1000 yards seems rather lots....but I can accept it is the sense of the knowledge of what you do to the gunsighting, to adjust to that kind of range, essentially 'shooting it uphill' so to speak...still seems like alot to me too....your sure of it, aren't you?
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 Quote by Mr. Robin Parsons Not to be questioning the authority of the knowledge as you present it, but I had a 'hunter', and another, look at the figure, a 40 FOOT drop in 1000 yards seems rather lots....but I can accept it is the sense of the knowledge of what you do to the gunsighting, to adjust to that kind of range, essentially 'shooting it uphill' so to speak...still seems like alot to me too....your sure of it, aren't you?
Mr. Robin Parsons,
For someone who doubts my data, the tone of your post was very nice. Thank you.

I assure you that the information that I gave you is correct. I am not asking you to believe me alone. I will provide some links. I do want to say that I know that the information given in these links is correct, not just because it is correct mathematically, or agrees with Newton's laws, but also because I have used it myself on the range many times. The equations given on these sites are mathematically correct, and are derived from Newton's laws. I have gone over them very carefully to be sure that I understood exactly how they were derived. They are also in agreement with data from Remington Arms, Winchester, Sierra, and the Army Ballistics Research Laboratory. Also, I have a personal shooting journal that confirms the data on the range. I assure you, this information is rock solid(in one of the links there are a couple of typos in the charts, I have the corrected numbers if you need them).

M-1 rifles were sighted in for 200 yds, which is up 5 clicks of elevation from level. At the 1000 yd line you put on 41 clicks of elevation for a total of 46 clicks. From there you have two sighters to fine-tune your wind calculation and temperature correction, and center your zero on the target. If you are old enough to remember, on an M-1 rifle, a click is 1 inch at 100 yds, or about 1 minute of angle. 46 clicks is 460 inches at 1000 yds.

-Mike

www.aeroballisticsonline.com
www.ballistics-software.com
www.ballistics.org
www.eskimo.com/~jbm
www.remington.com
www.shootingsoftware.com
www.sierrabullets.com
www.snipercountry.com
www.winchester.com
www.arl.army.mil
US Army Research Lab, Aberdeen Proving Grounds Md.
www.usamu.com
US Army Marksmanship Unit, Ft Benning Ga.
 P: 1,560 (YIKES!) THANKS! (sorry I questioned it, then again, look at all the new information you gave, Thanks again!)
 P: 1,560 If you would forgive me Mr. Sewell, you would then know the answer to the 'philosophical' question, "Would A bullet, hitting a snowflake, it gain in accuracy of flight, from the wetting? (Clearly Sir, your are way past my 'league')
 Sci Advisor P: 789 Mr. Sewell, very informative posts! One question: What's a click? Isn't that what the SF guys used in Vietnam used as a distance measurement or something to that effect as in "...the target is 3 clicks past that hill..." or have I been led astray by pop culture? Cliff P.S. Mr. Parsons - have you ever seen the classroom experiment with a dart gun fired horizontally at a stuffed animal dropped at the same moment? Very effective teaching tool, trick is getting the timing correct but once sorted very simple to repeat and a regular video camera offers enough resolution to see the effect.
 P: 2,380 Guys, just remember this: A speeding bullet doesn't drop any slower than something that is dropped stationary.
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 Quote by Averagesupernova Guys, just remember this: A speeding bullet doesn't drop any slower than something that is dropped stationary.
This is not correct. Re-read the posts, and check out the links.
 PF Patron Sci Advisor P: 1,481 Well, I was not imagining such a conversation due to bullet spinning. If this spinning provides stablity to bullet trajectory, as I have undertood in this posts, it would not work for space rockets, spacecrafts, missiles and bombs?. I have seen a little spinning when the space shuttle takes off. Does it do with that?. I have not cleared myself one thing: the bullet runs more yards with an spin around its axis?. I mean, is it possible that for the same initial force, the spinning bullet goes faster than another that has no spinning?. To answer this question I will be grateful if you consider the aerodynamic drag. Does the spinning reduces this drag?.
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 Quote by Cliff_J Mr. Sewell, very informative posts! One question: What's a click? "...the target is 3 clicks past that hill..." P.S. Mr. Parsons - have you ever seen the classroom experiment with a dart gun fired horizontally at a stuffed animal dropped at the same moment?
Cliff_J,
A click is used in that way is military jargon for 1 kilometer. The Army seems to like clicks very much because everywhere we went on foot was always several clicks away.

The meaning of a click in shooting has to do with the sight, or scope on a rifle or pistol. As the sight is adjusted, it "clicks". Each click represents a given amount of elevation or windage at a given range. One click on a military rifles often corresponds to one inch at 100 meters(used to be Yards).

The stuffed animal demonstration is a great one, especially if it is set up ahead of time. It is a logical place to start when teaching parabolic motion.
With hi-speed projectiles it doesn't always hold, for the reason I mentioned in my first post.(it could work depending on the individual teddy bear's coefficient of aerodynamic drag in free-fall). Aerodynamic forces in the classroom experiment are not a significant factor because of the short range and low velocity. The aerodynamic forces involved with hi-speed projectiles are tremendous. If you use the charts and graphs in the links I gave, and do the math to check out the acceleration due to aerodynamic drag, the results will probably come as quite a surprise to you.

I Hope you have fun with this stuff,
-Mike
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 Quote by Clausius2 Well, I was not imagining such a conversation due to bullet spinning. If this spinning provides stablity to bullet trajectory, as I have undertood in this posts, it would not work for space rockets, spacecrafts, missiles and bombs?. I have seen a little spinning when the space shuttle takes off. Does it do with that?. I have not cleared myself one thing: the bullet runs more yards with an spin around its axis?. I mean, is it possible that for the same initial force, the spinning bullet goes faster than another that has no spinning?. To answer this question I will be grateful if you consider the aerodynamic drag. Does the spinning reduces this drag?.
Clausius2,
I was not imagining such a conversation either, but each answer seems to bring more questions. So I'll do my best to help.

I am not in any way, shape, or form an expert on rockets or spacecraft but I'll answer your questions the best I can.

Once a bullet leaves the muzzle of a gun, the shooter's influence on it is over. Nothing can be done to change the path of the projectile once it is in flight. This is not the case with many of the other types of projectiles you mentioned, most are guided, some have fins for stability(like the fletching on an arrow).

The space shuttle's roll maneuver is not like the spin of a rifle bullet at all. I'll leave any further explanation of this to someone who knows what they're talking about.

The spin of a rifle bullet is for stability in flight only, which is in turn for accuracy.

As for the questions on maximum range and velocity...well...If the bullet were to tumble in flight it would lose velocity quickly and would also have a much shorter maximum range. The spin on a bullet does not in any way violate the the law of conservation of momentum. It doesn't give you a free lunch. As a matter of fact, you have to "pay" for the spin with velocity, a study of rotational motion will show you why.

Spinning increases the total aerodynamic drag.

The art and science of precision shooting is very complex, and I have left out a lot of information, and simplified things as much as possible. If you want to go deeper into the subject start with the links. If I can be of any help, just let me know.

I hope this helps,
-Mike
PF Patron