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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. GOOGLE HOTLINK FIX FOR BLOCKED PHOTOBUCKET IMAGES https://chrome.google.com/webs...inkfix=1516144253810 | ||
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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." | |||
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