Is Greenhill's calculation still used for approx twist rate or is there something new being used? With his formula for optimal twist rate calculation for a 308 bullet I get the following:
CALIBERS = Length of 168gr SMK = 1.1835 inches divided by .3065 inch diameter = 3.861
Take 150 (some constant) and divide by CALIBERS (3.861 above) = 38.85 calibers
Thus, ideal twist rate = one turn in 38.85 calibers.
My bullets = 3.861 calibers, therefore 38.85 divided by 3.861 = 1:10
My Savage FCP-K is a 1 in 10 inch twist which, by this calculation, appears to be optimal for this bullet. Is this coincidence or do these calculations make sense?
If I was to use a longer bullet I assume then that a 1:9 then be more optimal? While shorter bullets want slower twist?
Does mass play a role at all in optimal twist rate? This calculation doesn't seem to think so but a lot of websites say it does. Obviously you could keep the same dimensions but change the mass/density of the bullet by a considerable amount.
I still have some homework to do, obviously, but thought it might be a starting point to discuss.
CALIBERS = Length of 168gr SMK = 1.1835 inches divided by .3065 inch diameter = 3.861
Take 150 (some constant) and divide by CALIBERS (3.861 above) = 38.85 calibers
Thus, ideal twist rate = one turn in 38.85 calibers.
My bullets = 3.861 calibers, therefore 38.85 divided by 3.861 = 1:10
My Savage FCP-K is a 1 in 10 inch twist which, by this calculation, appears to be optimal for this bullet. Is this coincidence or do these calculations make sense?
If I was to use a longer bullet I assume then that a 1:9 then be more optimal? While shorter bullets want slower twist?
Does mass play a role at all in optimal twist rate? This calculation doesn't seem to think so but a lot of websites say it does. Obviously you could keep the same dimensions but change the mass/density of the bullet by a considerable amount.
I still have some homework to do, obviously, but thought it might be a starting point to discuss.