Ok. I have been doing a little bit of research and studying. This may already be posted somewhere if so I apologize but I find this helpful.
Professor Sir Alfred George Greenhill of the British Royal Military Academy, who devised a formula for determining twist rate which, simplified here, multiplies the square of the bullet diameter by 150 and then divides the result by the length of the bullet, and looks like this: (C x D2) ÷ L = T.
C is a constant, 150; D2 is bullet diameter multiplied by itself, L is bullet length and T is the result, twist rate. Using it is simple; let’s stick with our 5.56 mm/.223 Rem. example. The bullet diameter is .244" and length of a 55 grainer is .740". Following Greenhill:
Step 1—Finding D2: .244 x .244 = .05953
Step 2—Finding C x D2: 150 x .05953 = 8.9295
Step 3—Divide result by bullet length: 8.9295 ÷ .740 = 12.06
So, 1:12.06-inch, rounded to 1:12-inch, is our standard twist rate for the .223 Rem. 55-grain bullet.
May be programs out there for this but I see it helpful to me as a newbie.
Bwhntr53
Professor Sir Alfred George Greenhill of the British Royal Military Academy, who devised a formula for determining twist rate which, simplified here, multiplies the square of the bullet diameter by 150 and then divides the result by the length of the bullet, and looks like this: (C x D2) ÷ L = T.
C is a constant, 150; D2 is bullet diameter multiplied by itself, L is bullet length and T is the result, twist rate. Using it is simple; let’s stick with our 5.56 mm/.223 Rem. example. The bullet diameter is .244" and length of a 55 grainer is .740". Following Greenhill:
Step 1—Finding D2: .244 x .244 = .05953
Step 2—Finding C x D2: 150 x .05953 = 8.9295
Step 3—Divide result by bullet length: 8.9295 ÷ .740 = 12.06
So, 1:12.06-inch, rounded to 1:12-inch, is our standard twist rate for the .223 Rem. 55-grain bullet.
May be programs out there for this but I see it helpful to me as a newbie.
Bwhntr53
