5.7x28 case + 80gr projectiles = ???

[Artful]

Twist rate can be calculated via greenhills formula - but it wasn't designed for subsonic ranges really and you have to take into account bullet construction/shape velocity/air density.

http://www.frfrogspad.com/miscellb.htm

Q. How do you determine the rifling twist needed to stabilize a given bullet?

A. The needed rate of twist is effected by the diameter of the bullet, the bullets weight, and the bullet's overall length. Longer bullets need a faster twist to stabilize. As an example, a 1:12 twist in .30 caliber will adequately stabilized most commercial bullets of up to about 200 grains. To use a heavier (longer) bullet or to obtain optimum stability and accuracy with long pointed or boat tailed bullets of that weight requires a 1:10 twist. For best accuracy the slowest twist that will stabilize the bullet should be used.

Modern bullet stability calculations are based upon the work of the late Robert L. McCoy who was a ballistician with the Ballistic Research Laboratory at Aberdeen Proving Ground. His work, now used in advanced ballistics programs, accurately takes into account all of the factors involved in bullet stability and accurately describe the bullet's behavior.

There is an old formula called the Greenhill Formula that, while it was designed for estimating twists for boat tailed lead core bullets of moderate velocity, does a pretty good job of estimating twist required for flat based bullets under "normal" conditions.

T = Twist in inches
K = Greenhill's constant = 150 (This has to do with the specific gravity of a jacketed lead bullet)
D = Bullet diameter in inches
L = Bullet length in inches

T = (K * D2) / L

Using a 1.35 inch long .308 bullet (200 gr) and crunching the numbers we get about 10.5 (One turn in 10.5", which is pretty close to the 1:10 twist normally used in .30-06 rifles. The twist for the .308 is nominally 1:12 because it was based on the shorter bullet of the150 gr military ammunition from which the .308 commercial round was developed. Most match rifles in .308 have a 1:10 twist to stabilize the 180 and 190 boat tailed match bullets better. The results from the Greenhill formula are on the conservative side--indicating a faster twist than probably needed. That doesn't cause any problems because a little too much stabilization is better than too little.

However, the one catch with the Greenhill formula is that it does not account for the effects of temperature or muzzle velocity. As temperature or velocity decreases a faster twist is needed to maintain the same level of stability. Colder and thus denser air has a more destabilizing affect than warmer air. A lower muzzle velocity results in a slower rotational speed of the bullet and thus less stability.

As an example of the effect of temperature, the original M16 rifle for the M193 55 gr ball came out with a 1:14 twist which was barely stable at 68 degrees and which was totally unstable below about 40 degrees. They changed to a 1:12 twist to get stability (barely) at colder temperatures. The new M855 62 gr round is unstable below about 65 degrees with the 1:12 twist and requires a 1:9 twist to be stable. They went to a 1:7 twist because the M856 tracer round has a very long bullet, but the ball round does just fine in 1:9.

For velocity issues long bullets are more susceptible than short ones. For example a particular 210 gr .30 cal low drag bullet from a 1:11 twist barrel is stable from a muzzle velocity of about 2800 f/s and higher. Below that muzzle velocity it becomes unstable. With a 1:10 twist barrel the bullet will be stable at muzzle velocities of 1400 f/s and higher.

Another catch is that Greenhill assumes that the bullet's specific gravity is 10.9 (a lead cored jacketed bullet). For other bullet construction such as a steel core you need to apply a fudge factor by determining the bullets specific gravity. The formula would be:

Twist = [Square Root (10.9 / specific gravity of the new bullet)] * twist derived for a lead core bullet

You can determine the specific gravity of a bullet thusly:

1) Suspend the bullet at its balance point from the pan of a scale.
2) Weight the bullet.
3) Place a container of water under the scale so the bullet hangs fully in the water and weigh the bullet.
4) Subtract the weight obtained in step 3 from the weight obtained in step 2
5) Divide the weight obtained in step 2 by the difference obtained in step 4.

As an interesting exercise you can also determine the spin needed in mediums other than air under "standard conditions" by multiplying the spin for air by the square root of the number obtained by dividing the density of the medium in question by the density of air. As an example water is about 900 times as dense as air: 900 / 1 = 900 and the square root of 900 is 30. Thus you need a twist 30 times as fast to stabilize a bullet in water.

http://www.z-hat.com/twistrate.htm

calculator on the net
http://kwk.us/twist.html

downloadable
http://www.nfa.ca/content/view/134/197/

purchase
http://www.corbins.com/dc-twist.htm