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Old February 13, 2000, 10:39 PM   #1
Art Eatman
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Sometime last year, a very kind soul provided the formula for rifling twist, to determine the best twist for stability for various bullets of a given caliber.

I had a terminal attack of the stupids, and didn't write it down.

Come back, Shane! Pretty please? With sugar on it? Give us the formula, once again?

Thanks, Art
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Old February 14, 2000, 01:53 PM   #2
Coinneach
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Art,
Is this maybe what you're looking for?

D=Diameter
L=Bullet Length
T=Twist

L = 180 * D^2 / T

Solving for T:

T = L / 180 * D^2

DISCLAIMER: I've done no algebra for 13 years, so I cannot guarantee the accuracy of this formula.

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"If your determination is fixed, I do not counsel you to despair. Few things are impossible to diligence and skill. Great works are performed not by strength, but perseverance."
-- Samuel Johnson
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Old February 14, 2000, 06:23 PM   #3
Art Eatman
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Hey, thanks; it's at least something to check against "the numbers".

I assume that D^2 means D-squared? (I wish posting to sites allowed subscripts and superscripts without having to do umpteen other things, first.)

Later, Art
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Old February 14, 2000, 08:06 PM   #4
Coinneach
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Art, yep, D^2 == "D-squared."

Now if one of the administrivia types would turn HTML back on, we wouldn't have to deal with this. (hint)

------------------
"If your determination is fixed, I do not counsel you to despair. Few things are impossible to diligence and skill. Great works are performed not by strength, but perseverance."
-- Samuel Johnson
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Old February 14, 2000, 08:20 PM   #5
Bill Daniel
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Art:

If you can find the most recent issue of Rifle Shooter magazine, they addressed the question of optimal twist for bullet length. Sorry I don't have it with me now, but I will try to post it tomorrow.

Bill Daniel
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Old February 15, 2000, 09:41 AM   #6
Bill Daniel
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Art:
The article was by John Haviland. The Greenhill formula, which he states is an empirically developed formula for artillery adapted to small-bore bullets, is the number 150 divided by the length of the bullet in calibers. His example is to find the optimal rate of twist for a Nosler .257 115 grain Ballistic tip you measure the length of the bullet (1.2") and divide it by .257 to convert the length to calibers. In this example the length caliber for that bullet is 4.67 which is then divided into 150 yielding 32.18 which is reconverted to inches by multiplying it again by .257 resulting in 8.25 or 8.25 inches of barrel length per complete turn of rifle twist.
The validity of this I do not know. Perhaps others can comment. Bill Daniel
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Old February 15, 2000, 10:52 AM   #7
Art Eatman
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Thanks, Bill. Using your numbers and definitions with Coinneach' formula, the twist would be 9.9 compared to your 8.25--due to the 180 vs. 150 as the Constant.

So the formula works; the argument now is about what Constant to use...Off the cuff, since the 117-grain bullet in .257 is rather long, it would appear that using 150 as a constant might work better. I'm guessing that a 1-in-8 twist would be more stable than a 1-in-10 for that bullet.

Ain't it fun?

, Art
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Old February 16, 2000, 12:02 AM   #8
Tim Brooks
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Gentlemen I've a zipped file that contains The Greenhill formula as a program works very well. E-mail me off the forum if interested
Cheers, Tim
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Old February 16, 2000, 01:46 AM   #9
Mal H
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Here ya go Art. This should keep you busy for a few hours:
http://www.fulton-armory.com/fly/index.htm
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Old February 16, 2000, 09:48 AM   #10
Tim Brooks
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http://www.norma.cc/htm_files/framp1e.htm
At this address if you click on ballistics
you can while away some time with a lot of what-ifs
enjoy, Tim
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Old February 16, 2000, 10:08 AM   #11
Art Eatman
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MalH: Thanks.

If you read through the article, you now know why the leaf sight on an early 1903 Springfield has an angle-cut: When you raise the aiming-bar for long range, it is moved to the side--to compensate for the drift induced by the direction of rotation.

Back in the 1890s, before radar, doppler radar and shadow photography, a fellow named Mann did extensive research on exterior ballistics. "The Bullet's Flight From Powder To Target" was one result.

He built a 200-yard tunnel; after aiming at the target, he hung sheets of paper at intervals between the muzzle and the target. The bullet holes inh the sheets of paper showed the precession around the center-line between the muzzle and target, and the drift. (I vaguely recall that he used a .32 Ballard.)

The farther out from the muzzle, the farther off centerline that twist-induced drift will move the bullet. Note the experiment with the .308 with 100 yards of drift at maximum range.

Even on an absolutely windless day for a shot at 1,000 yards, any variable in the precession can open a group far beyond what is expected. Sorta leads me to believe that inside of 300-400 yards, it's not worth worrying about; out past 600 yards or so, it's a whole 'nother ball game.

Ah, science and technology!

, Art

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Old February 16, 2000, 12:12 PM   #12
STLRN
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If any of you want to look at the "big picture" dealing with ballistics go to the following link
http://www.fas.org/man/dod-101/sys/l...fm6-40-ch3.htm

It contains the FM 6-40 Manual Field Artillery Gunnery. Although it discusses items on a Macro scale, it is a good primer for the discussion of ballistics.

Oh- If the greenhill formaula is a extraction from research done by artillerymen, be aware of the fact that standard condition, as computed by BRL at Aberdeen, for artillery systems are different than their computed standard conditions for small arms.

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Old February 16, 2000, 12:15 PM   #13
STLRN
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Drift is a function of time of flight and projectile wheight, most small arms engagement occur under the range that drift comes into play. Even for indirect fire with small arms, the low projectile weight and relatively short TOF make drift a almost non-issue, the first weapon, in bore size that, mentions drift in TFT (tabular firing table) is the Mk-19 40mm grenade launcher.

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