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one of us |
Could some one explain the relationship between barrel twist, bullet weight and velocity. Especially how it relates to choosing the right bullet size for the best accuracy? I guess caliber also comes into play in this question also. Thanks, Simdow | ||
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One of Us |
I'll give you a small explaination. Someone will explain it in more detail I'm sure. The bullet must be spun fast enough to keep it stabilized during it's flight. Just like how a quarterback puts a spiral on the ball when he throws it. There is no point in over spinning a bullet, but, most gun makers use a barrel that will stabilize just about every available bullet for that caliber/cartridge. An example, a 308 usually comes with a 10 twist barrel. It will stabilize all the factory ammo you find. But, with up to a 150 grain flat base bullet, a 15 twist will do fine out to 300yds. Don't try it with a heavier, or even a boattail bullet with that slow twist. Bullet jacket length plays a part when you get into fine accuracy shooting. Go to Shilen's website and look at their chart, it will give you some ideas. And, Berger's site has recommendations of twist for each bullet they offer. | |||
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one of us |
That's pretty much enough said, but one correction to the previous. The issue is bullet LENGTH, not weight. Sufficient gyroscopic stability must be imparted by twist rate to overcome the aerodynamic moments resulting from the displacement between center of gravity and center of pressure(aerodynamic). With all else equal, the longer the bullet, the longer the moment arm, thus faster twist is required. If yuro'e corseseyd and dsyelixc can you siltl raed oaky? | |||
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one of us |
So are you saying that in my 300WSM Browning Stalker, a 150gr TSX would not be inherantly more, or less accurate than a 180grTSX because they are both common factory weights. Simdow | |||
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one of us |
The following is the Greenhill formula for calculating apprx min rifle twist to stabilize bullets 150 divide by (length divide by caliber)caliber. example .308 cal 110 gr hornady V-max is .924" long divide by .308 = 3 150 divide by 3 = 50 multiply by .308= 15.4 so 15.4 would be min twist required to stabilize said bullet. HAVE FAITH IN GOD. | |||
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one of us |
If the bullets are of the same quality, that is of the same weight distribution around their geometric axis, they will both be equally accurate once they are spun fast enough to achieve stability. Since bullets are not perfect, the heavier bullet will be ever so slightly more accurate than the 150. Ever so slightly, because both bullets have extremely high uniformity (being all copper helps). For all effects and purposes, twist is irrelevant for all but bench rest type application, and some extreme range applications. Providing, of course, that you have enough of it to begin with. FWIW, Dutch. Life's too short to hunt with an ugly dog. | |||
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one of us |
AOWM, the greenhill formula is designed for copper jacketed lead bullets traveling somewhere around 2400 fps. And it over estimates twist a little. If you want to get really nit-picky, the JBM has a great site that gets into the meplat shape and the whole shebang. http://www.eskimo.com/~jbm/ FWIW, Dutch. Life's too short to hunt with an ugly dog. | |||
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