OT- Balistics.

[Gun Smith]

A bit OT, but it applies to all firearms.

In 1830 an English mathematician (Greenhill) created a table to calculate what the minimum twist (to spin the bullet)) is needed to stabilize a bullet. The table below is only part of his work. The table requires that the bullet length be converted to calibers.

Change the bullet length to calibers by dividing it's length by it's caliber. ie: a .50 caliber bullet 1 1/2" long = 3 calibers.

From the Table: 3.0 calibers requires 50.74 calibers of length for one turn of rifling. So, 50.74cal. x .50 cal = 25.37" or 1 in 25" twist.

.45 cal x 1.35" L. 1.35 divided by .45 = 3 cal. Table 50.74cal. x .45 = 22.83" twist.

.27cal x 15/16"L. (100 gr.) .9375 Div. by .27 = 3.47 cal. Table 42.4cal. x .27 = 11.448" twist.

.27 cal. x 1 3/8" L. (130 gr.) 1.375" Div. by .27 = 5.09cal. Table 28.44cal. x .27 = 7.6" twist

.45 cal. x 1" L. 1 Div.by .45 = 2.2 Cal. Table 74.32cal. x .45 = 33.44" twist.

.30 cal. x 15/16" L. (150gr.) .875 Div. by .30 = 2.91 cal. Table 52.72 x .30 = 15.8" twist.

.30 cal. x 1 3/8"L. (190gr.) 1.375 Div. by .30 = 4.58cal. Table 31.21cal. x .30 = 9.363" twist

.38 cal. x 5/8". 1.64 Div. by .38 = 4.32cal. Table 33.59cal. x .38 = 12.76" twist.

Round Ball: .45 cal. x .45". 1 div. by .45 = 2.2 Table 74.32cal. x .45 = 33.44" twist.

As we see from the table, longer bullets normally need tighter twists.

From the list above we see how manufacturerer's could determine barrel rifling twist. But because bullets vary by length in every caliber you would determine one length is more accurate in a 10 twist barrel than another, which is usually the case.
Interestingly, the table does not account for bullet weight or shape, ie. RN, SP, FP., etc. None of those shapes (except round nose) were even in existence in 1830. This must create a variable of some sort.

The Greenhill Formula is available on line.

[Nath]

Does vel play a part too?

Interesting, thanks.

Nath.

[flatnose]

I wonder how the greehill formula can be modified for future bullets made of less dense materials. eg. copper solids, or sintered materials?

[J Miller]

If I'm reading right this formula uses the entire length of the projectile rather than just the bearing surface, am I seeing correctly?

Joe

[flatnose]

Joe,
I think the bearing suface is only important in respect that there must be sufficient area to impart a rotation to the projectile. The amount of twist required is related to the mass, length and velocity of the projectile.

[Gun Smith]

Yes Joe, The length of the bullet is over all for his formula including the shape or bearing surface. The calculations are based on the length versus the diameter regardless of bullet shape.

I do feel that the shape has got to be a factor because of the center of mass varies with shape.

Maybe someone here can explain to us why this wouldn't be a factor.

Another question was asked about non-lead or part lead bullets. That too has to make a difference in my mind.

Perhaps the web sites available with the formula will explain this problem.

[KWK]

Some of the questions here are covered in a short article on my site. The velocity compensation I used (1960's reference) I later read is known to be inaccurate.

A fellow named Miller has a done work in this area, and he has a formula which compensates better for velocity and gives an approximate correction for bullet shape. I'll incorporate Miller's formula into my twist calculator some day (probably this spring).

[Gun Smith]

KWK, That's what I love about this forum, there is always someone around to update us. Great!

[AJMD429]

Some of the questions here are covered in a short article on my site.

The link embedded above, "http://kwk.us/twist.html", doesn't work for me - is it correct?

[KWK]

I had billing problems with my old hosting service, and I moved to a, ah, low end operation. I've noticed that especially in the evenings, the new hosting service isn't serving up pages promptly. Sorry. (You get what you pay for?) It worked for me just now.

[Gun Smith]

The link "in a short article" worked for me. It did answer some of the questions I had about Greenhill's original formula too.