I know there's a formula used to calculate the proper width of the magazine to keep the cartridge cases stacked evenly @ an equalateral triagle (60 degrees)

The factory magazines are made to this specification.

Does anyone have this formula handy?

I would appreciate if it it could posted here.

I am working on lengthening/widening 1904 Portuguese bottom metal for an 8X68S project.

Take a look at this thread.

http://forums.accuratereloadin...=901100711#901100711

Just as easy to copy the formula and print it here.

"The formula for optimum magazine width is:

(Cosine(30degrees) X base diameter) + base diameter



For a .30-06

(.866 x .473) + .473 = .882 inches



For a .300 Win Mag

(.866 X .532) + .532 = .993 inches."