Actually I wasn't thinking when I posted that. You will be doing a 1/4" radius. Still a lot of grinding. I was checking out an online surface area calculator. If I punched the numbers in right it would be the same as taking a 6 foot long 1/4" rod and grinding it to dust. That is if you ground a perfect radius. I could be wrong though, I'd like to see if anybody else comes up with the same thing.
Okay, you're on! A 1/4" rod has a cross sectional area of pi r^2, or 3.1416 * .125 * .125 = .049 sq". Multiply that by the length, 72" and the result is 3.53 cu".
Figuring how much material is removed doing a 1/4" roundover is a little trickier. Start with a rectangular solid the the thickness of the plate by the thickness by the length, or .5" x .5" x 144". Subtract the volume of the inscribed cylinder, .5" dia x 144". That gives you the volume of material to be removed to turn the rectangular solid into a cylinder. With me so far? Now, since we're only rounding over one edge, the material removed is on fourth of that.
So: Rect.Sol.Vol= .5" x .5" x 144" = 36 cu". Volume of the inscribed cylinder = pi r ^2 = 3.1416 * .25 * .25 = 28.27 cu". Subtract, and you get 7.73 cu". One fourth of that is 1.93 cu". Equal to the volume of a 39" long 1/4" rod. Still a fair amount of material to remove, but not so tough with a belt sander, which would be my choice. I'd make the final run using the free-running section between the front roller and the platen, with a fine grit (120 or so). There, the belt will follow the contour of the roundover and sand off the lateral ridges left by going back and forth. It would take several belts of at least 3 grits, but probably not more than a couple of hours.