fuzzy math?
Ok assuming an areal density of "509Gb/in" as quoted in the article and an assumption of one side of each platter being utilized for data storage, how is it they reached this capacity per sq/in. By my math, assuming that the center inch roughly of the platter cannot be used as that's where they mount from I get roughly 12.4sq/in that can be used for data on this drive. Even assuming my math is off, on the high side, by 40% there is no way that they are getting that much per sq/in as they appear to only be using 20% ish of the drive for data. Hope this makes sense anyway, correct me if i am wrong though as I am not fully upto speed on the exact dimensions of the platters in these drives.