Originally Posted By: cpayne5
Originally Posted By: eljefino
Originally Posted By: Traction
Just to add, to clean up the beads on 4 16 inch wheels, it is over 33 feet of surface area to deal with.
16 inches x 2 inches effective "depth" x 2 beads per rim x pi x 4 wheels * (1ft^2/144in^2) = 5.58 sq feet. How do you figure?
He meant 33 feet of linear bead, not literal surface area.
Aha! I thought too deeply into it.
Originally Posted By: eljefino
Originally Posted By: Traction
Just to add, to clean up the beads on 4 16 inch wheels, it is over 33 feet of surface area to deal with.
16 inches x 2 inches effective "depth" x 2 beads per rim x pi x 4 wheels * (1ft^2/144in^2) = 5.58 sq feet. How do you figure?
He meant 33 feet of linear bead, not literal surface area.
Aha! I thought too deeply into it.