Just to stirr the muck a little . . . .
Can anyone tell me the required twist to stablize a 155grain 30 calibre boolit.cheers
Here are a couple of examples, from calculations, using the Tioga Eng. (the Late Bill Davis') bullet design program:
1) For a gyroscopic stability factor of 1.5, a ten caliber, tangent ogive, 0.060" diameter meplat, 150 Gr. FB, 1.170" long (specific gravity 9.70), air density of 1.02 atmospheres, and a MV of 3000 FPS, needs a 1-14.2" twist rate.
2) For the same Sg, a 150 BT, from the same point-die, but featuring a 0.090" x 9.0 Deg. BT, and running 1.190" long, has a specific gravity of 9.6, and needs a 1-13.5" twst-rate. The additional length reduces the specific-gravity, and the BT reduces DRAG (a stabilizing force), a combination necessitating a 0.7" faster twist rate to stabilize bullets of the same same weight, but differing geometry.
When using JBM (an excellent twist-rate/BC calculator) one must enter the proper specific-gravity, or, risk a faulty result.
I don't have time to do it now, but one could go to JBM, and plug in the specific-gravity, and calculate the minimum twist rate which would assure Sg 1.4 or, better . . . and do so for just about any earthly environment.
Here's an example of what the addition of a BT does to Sg: Adding the same BT to bullet #1, above, while artificially maintaining the SAME length as the FB version, creats a specific gravity of 9.8, & results in a twist requirement of 1-13.8". So, for any of the above, a 1-14" will work ok, keeping Sg above 1.4. So, just adding the BT necessitates a 0.6" FASTER twist - worthy of consideration.
This is probably why the BIG bullet/barrel makers prefer to talk about bullet weight and twist-rate . . . but that is just, "ball-park" information!
RG