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Posted: 6/1/2001 5:51:24 AM EDT
Got bored last nite so I tried to calculate the rearward force acting on an AR15 bolt when a round is fired. Basic assumption was rearward force equals the surface area of the base of the cartridge times the pressure. Actual force should be less since I used the outside diameter of the cartridge vs inner diameter. Also during firing the brass casing 'sticks' to the chamber wall, reducing the forces some. Anyway here goes. Diameter of cartridge just above web = 0.376" Radius = 0.1880" Area = Pi x 0.1880 x 0.1880 = 0.1110 sq in. Average pressure = 50,000 psi Force on the Bolt = 50,000 x 0.1110 = 5,550 lbs. Ideally this force is evenly distributed across 8 locking lugs = about 700 lbs / lug I gotta get out more often [:D]
Link Posted: 6/1/2001 6:07:09 AM EDT
I don't have my bolt with me right now but I believe that most modern bolts have the lug accross from the extracter relieve to ballance the stresses evenly around the bolt. Is that still 8 lugs?
Link Posted: 6/1/2001 6:10:36 AM EDT
yup i just counted eight......HAHAHAHAHAHA
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