Author Topic: Bullte spin rate (Read 711 times)

Donaldo · « **on:** January 04, 2005, 08:53:08 AM »

To those interested..... A while back I asked a question about stablility of bullets at different twist rates. I found on the 6mmbr site this formula for calculating the rpm of a bullet given two knowns. This in a round about way will answer these questions. Also by knowing what rpm a certain bullet is stable at, one can calculate the unknown or needed twist rate, or velocity needed. Its a three part formula, you only have to know two to solve the the third. Here it is: Velocity in fps times 720 divided by twist in inches, equals bullet rpm. Example: Say you have been shooting a .223 50 grain bullet at 3200 fps out of your 1/12 twist barrel and know it is stable to a certain range. To find the rpm.... 3200 x 720 = 2,304,000. 2,304,000/12 = 192,000 rpm. Past this certain range, X, the bullet is beginning to get unstable, i.e., larger groups, keyholes, whatever, you know the bullet is getting unstable from lower velocity. So you can manipulate the formula to find the twist or bullet weight or velocity to get you where you want to be. I think I got this all right, I know Fred or someone will check it for me. As I understand, this has some leeway in it, as the bullet will be stable over a range of velocity and rpm, but it will answer most of the questions I had about it.

Fred M · « **Reply #1 on:** January 04, 2005, 09:56:59 AM »

Donaldo.
Your analogy has nothing to do with stability it only tells you the rotational speed, which could be a lot more than needed.

http://www.nennstiel-ruprecht.de/bullfly/index.htm#Contents

http://www.nennstiel-ruprecht.de/bullfly/gyrocond.htm

Quote

. An practical example is shown in a figure .
Thus, if a bullet is gyroscopically stable at the muzzle, it will be gyroscopically stable for the rest of its flight. The quantity sg also depends on the air density r and this is the reason,

A VLD bullet can be lauched at relatively low velocity as long it is gyroscopically stable. Gyroscopic or rotational speed does not exponentionally follow the reduction of velocity. That is to say the bullet is still stable when it hits the ground at whatever distance that is.

Over-stabilization of a bullet is most probable, if a bullet has excessive static stability (a high value of sg and a low value for the tractability factor ) and is fired at a high angle of departure, especially when fired vertically. An over-stabilized bullet on a high-angle trajectory lands base first.

There are all sorts of equations you can set up on an MS Excel program on which you can do " if what". The computer will do the rest.

Donaldo · « **Reply #2 on:** January 04, 2005, 04:11:27 PM »

FredM.
Yes Fred, I read the articles you suggested several months ago. Highly technical and relate to special situations, IMHO. I for one do not plan on shooting vertically. Maybe you do. I am just trying to simplify things a bit. I asked a question the other day and you really avoided it by stating some gibberish. (Check that and see if that is a real word). I asked what velocity would be needed out of a 1/12 twist to stabilize a certain bullet. You never did answer the question, inferred it was out of your "abilities" or something like that. In fact there probably isn't a firm answer. The formula I presented is for calculating the rpm of a bullet. I think it is reasonably accurate to assume that if a bullet is stabilized at a certain rpm and velocity, you can extrapolate to another variable. Not with absolute certainty but pretty close. Well to get back to the original question I asked, the approximate answer is 3,900 fps or a little more. Now I can't get there with my cartridge and bullet, so there is my answer. As a matter of fact, Fred, rpm has a great deal to do with stability. Any fool can tell that by watching a football being thrown or passed. Those that wobble either don't have enough rpm or came out of a bad barrel. :-) So don't pull that engineering crap on me, I may not have a degree in engineering but I am not stupid either. There are a bunch of variables in "what is stability". If only to get close is to understand it a little bit more. On second thought I should qualify these statements, my football team is loosing right now and I am not in a good mood. No offense meant Fred. Love ya man. But I stand by what I said. Maybe I overreacted, if so I appolgise.

cheatermk3 · « **Reply #3 on:** January 04, 2005, 04:51:12 PM »

Donaldo, maybe you should stick with football.

I answered your question, partially, by giving you the RPS (revolutions per second) expressed in the simplest terms. 1-in-12 twist rate is one turn in 12 inches. 12 inches=1 foot, so 1000 fps out of a 1/12 twist gives 1000 rps. Times 60 (60 seconds being one minute) gives you 60,000 rpm. For 3200 fps you'd multiply that by 3.2=192,000 rpm.

This only tells you how fast your bullet is spinning when it leaves the muzzle. It don't tell you whether it will be stabilized. In fact it probably won't settle down and fly right, "go to sleep", until somewhere between 60-200 yds. Going back to your spiral pass analogy, often the ball precesses or wobbles about it's axis of spin for a few yards, then "settles down" and flies "straight" (hopefully to where the intended receiver will be).

What I was too polite to say was that the short answer to your question is, "you can't". You ain't gonna get there just by going faster.

Which, if Walt Berger says you need a 1/9 twist for that bullet, you can take to the bank. (as Fred politely suggested).

while99 · « **Reply #4 on:** January 04, 2005, 05:01:28 PM »

The Greenhill Formula predicts the required barrel twist rate (number of inches for one complete rotation) to stabilize a given bullet, based on the bullet caliber and length. This formula is somewhat dated (it's over 70 years old), but still provides a good 'rule of thumb' for determining the required twist rate for proper bullet stabilization. The important thing to remember is that this formula is a good simplification of the more accurate, complex equations.

The formula is :

minimum required twist rate = (150 * bullet_diameter2) / bullet_length

It will quickly become clear to all who use this formula that the general rule is : the longer the bullet (i.e. in general, the heavier the bullet, in any given caliber), the quicker the twist required to effectively stabilize the bullet.

It is also quite simple to determine the longest bullet of a given caliber that a given twist rate will suitably stabilize by manipulating the formula :

maximum bullet length = (150 * bullet_diameter2) / twist_rate

The 'rec.guns.FAQ' reference below states that the '150' constant works well at velocities in the vicinity of 1500 fps or greater, while a value of 180 can be used with good results at velocities at 2800 fps. Using 180 as the constant in the equation will give you a 20% greater maximum bullet length, or a 20% slower required twist.

Twist Rate

FASTER MEDIUM SLOWER

1-in-7 1-in-10 1-in-12
(1 turn in 12 inches)

--------------------------------------------------------------------------------

Examples
80 grain Sierra 0.224 = 1.064" in length
required twist rate = (150 * 0.2242) / 1.064
required twist rate = (150 * 0.050176) / 1.064
required twist rate = (1 turn in) 7.07 (inches)

150 grain Nosler 0.277 = 1.362" in length
required twist rate = (150 * 0.2772) / 1.362
required twist rate = (150 * 0.076729) / 1.362
required twist rate = 8.45

168 grain Sierra 0.308 = 1.210" in length
required twist rate = (150 * 0.3082) / 1.210
required twist rate = (150 * 0.094864) / 1.210
required twist rate = 11.76

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Fred M · « **Reply #5 on:** January 04, 2005, 07:54:31 PM »

Domaldo

You obviously want to read only what suits you and only halve at that.. The point I tried to make was that you can't establish a bullet's stability by multiplying or interpolating a bullets rotational velocity.

Yes the Greenhills formula is quite usable if you use an 180 constant instead of the 150. It serves most ordinary circumstances very well. As you can see velocity is not part of the Greehill equation but is rather build in in the equation's constant.

I could present you with a complete and very complex formula developed by Eric Willliams a ballistic scienetist, but I don't think that would do much good.

The 180 constant by the way stabizes a bullet about 125%or so while the 150 constant will stabilize a bullet 170%, 100% represents stability.
Another point is that a bullet is either stable, unstable or over stabilized.

Velocity is for sure a part of the stability equation, but not as a determening factor. it serves more to adjust for other influences. Really there is nothing simplistic about bullet stability.

If you consider my writings gibberish, then I assume you know more about the subject then I do. So please answer you own questions I just try to be polite.

Enough said.

Donaldo · « **Reply #6 on:** January 04, 2005, 07:55:14 PM »

OK....Before this thing gets out of hand, let me apologize up front if I stepped on toes, especially FredM. I know what Berger says on the twist for the 75 grn VLD, it is 1/9. My question, to put it another way, was can you overcome the lack of twist with velocity. In re-reading the posts, I guess the answer was not really but no one knows for sure. Or that is the way it seems. At least not practically, as Fred said. Yes the Greenhill formula has been around for many years and does not adequately work with the new low drag bullets. I don't think these things can be pinned down firmly, especially by calculations. But I am pretty darn sure my 223 with a 1/12 twist will not stabilize the long bullet. That was what I wanted to know, as I also have a 22-250 with 1/12 twist, hence the velocity question. That is also why I ordered the new 1/9 barrel. Thanks for all who contributed. I will now get off my soap box and beg forgiveness from all, :oops: it is past my bedtime anyway. And my team lost.

Well it was not really my team, was just an Oklahoma team. I went to college at the "other" university in Oklahoma and they lost their bowl game also.

Have a Happy New Year All.

Edit: You right Fred, enough said.

Fred M · « **Reply #7 on:** January 04, 2005, 08:48:10 PM »

Eh man you answered your own question, When a quarterback throws a ball he has learnt how to put a spin on the ball.Once he has established his twist he can throw a ball fast and far or slow and short, always point on. That is stability of the ball, the right spin is the thing.

You will notice when he gets sacked and throws the ball away in hurry he has no time to put a spin on or don't want to, because a wobbling ball is hard to catch.

And spin does wonderful things to a golf ball or a base ball.

Badnews Bob · « **Reply #8 on:** January 04, 2005, 09:20:45 PM »

You guys are amazing sometimes, all I know about it is that a 69gr bullet outta my 1-9 barrel key holes at almost anything under 80yrds or so but punchs pretty little holes at anything over or so 100 or so which in my pea brian tells me its spinning to fast when it leaves the muzzle and settles down after a little drag gets it. Og don't do numbers well Og hate to do numbers. :grin: