Mike;
Somewhere I am not following your math. Is that 4 holes 2" deep by .875" diameter. If so I get about 6 cu/in of mat'l removed for a weight reduction of about 2.3oz. It appears you have just multiplied the dia of the hole by total depth. Area of a 7/8" hole (.875) is .6013 sq/in times 10 inches of hole . You can't just multiply the dia of the hole by its depth to get cu/in's it takes the area (πR or πD/4) of it times the hole depth.


Miller/TN
I Didn't Say Everything I Said, Yogi Berra