Weight reduction for back boring: assume steel has 4.16 oz/cubic inch, ID = 0.655 so circumference (pi X D) = 2.06", for 30" bbls assume you can bore 26" (chamber @ 2 3/4" + cone @ 1 1/4"), bore area = 26" X 2.06" = 53.5 sq inches. Removed volume per thousandth = 53.5 in sq X 0.001" = 0.05 in cubed, removed weight per thousandth per barrel = 0.05 in cubed X 4.16 oz/in cubed = 0.22 oz --- say 1/4 oz per thousandth back bore per barrel. Going from 0.655 to 0.662 = 7 thousandths X 1/4 oz per thousandth X 2 barrels = 3.5 oz.
Questions?
I think there's a math error in there. To enlarge the bore 7 thou, the wall is reduced by 3.5 thou. So the volume of the rectangular sheet in the above example comes from 2.06 x 26 x .0035, cutting the answer in half.
I skin the cat another way...
I figure the volume of a solid cylindrical rod with the larger diameter and subtract the volume of the solid rod with the smaller diameter, giving the volume of the 26" tube with .0035" walls that is the difference.
In any event, I believe the correct answer is 1.6 ounces.
