MadRat
Lifer
A typical 3.5" hard drive (assuming 3" dia. platter) that rotates at 7200rpm has a tangential velocity of 68000 inches/minute. At the same rate, the tangential velocity of a 5.25" hard drive (assuming a 4.75" dia. platter) would require just over 4500rpm. If we assume the 3.5" platter's read/write strip to be 1" wide then it makes approx. 6.28" square available for data. If we assume the 5.25" platter's read/write strip to be 1" wide then it makes approx. 11.8535" square available for data. If we assume the 5.25" platter's read/write strip to be 2." wide then it makes approx. 17.113490625" square available for data.
Check my math:
3.5" platter area w/1" datastrip = (3.14(1.5"^2)) - (3.14(.5"^2)) = 7.065 - 0.785 = 6.28" square
5.25" platter area w/1" datastrip = (3.14(2.3875"^2)) - (3.14(1.3875"^2)) = 17.898490625 - 6.044990625 = 11.8535" square
5.25" platter area w/1.8875" datastrip = (3.14(2.3875"^2)) - (3.14(.5"^2)) = 17.898490625 - 0.785 = 17.113490625" square
I find it extremely difficult to believe that a 3.5" platter is significantly more expensive to manufacture. Single platter drives (7-12mm thick) built on 5.25" form factor should most certainly compete with much wider 3.5" dual-platter drives. The slower rotational speed of the platter should mean a quieter drive with a longer MTBF rate; I assume the wider platter would be inherently more stable due to its lower rotational speed. Perhaps a whole stack of single-platter 5.25" drives could be able to be stacked inside of a regular 5.25" slot. These benefits are to name just a few. The drawback would probably be seek and read/write times, but the increased time should be insignificant to most end-users.
Just seems that hard drive technology should be developing in several physical directions, but conventional wisdom is mostly limited to just shrinking them. The current drives are shrinking at a fantastic rate and its beginning to look like 2.5" will become the norm in laptops. But for desktops and server farms it would make sense to also increase the physical size of the drives. It should NOT require seven to twelve drives just to get the redundant capacity necessary for a terabyte of storage. Only few 9-12" platters (@20 GB per inch square) should more than suffice to run 1 terabytes in a safely striped set. Wouldn't the price per terabyte be less for large capacity platters, too?
10" platter area w/9" datastrip = (3.14(4.5"^2)) - (3.14(.5"^2)) = 63.585 - 0.785 = " square
Now take " sqaure times 20 GB per inch and we get a capacity of 1271.7 GB.
12" platter area w/11" datastrip = (3.14(5"^2)) - (3.14(.5"^2)) = 78.5 - 0.785 = 77.715" square
Now take " sqaure times 20 GB per inch and we get a capacity of 1554.3 GB.
14" platter area w/13" datastrip = (3.14(6.5"^2)) - (3.14(.5"^2)) = 132.665 - 0.785 = 131.88" square
Now take " sqaure times 20 GB per inch and we get a capacity of 2637.6 GB!
16" platter area w/15" datastrip = (3.14(7"^2)) - (3.14(.5"^2)) = 153.86 - 0.785 = 153.075" square
Now take " sqaure times 20 GB per inch and we get a capacity of 3061.5 GB!!
I couldn't even begin to imagine the tangential velocity on a 16" platter. 🙂
Check my math:
3.5" platter area w/1" datastrip = (3.14(1.5"^2)) - (3.14(.5"^2)) = 7.065 - 0.785 = 6.28" square
5.25" platter area w/1" datastrip = (3.14(2.3875"^2)) - (3.14(1.3875"^2)) = 17.898490625 - 6.044990625 = 11.8535" square
5.25" platter area w/1.8875" datastrip = (3.14(2.3875"^2)) - (3.14(.5"^2)) = 17.898490625 - 0.785 = 17.113490625" square
I find it extremely difficult to believe that a 3.5" platter is significantly more expensive to manufacture. Single platter drives (7-12mm thick) built on 5.25" form factor should most certainly compete with much wider 3.5" dual-platter drives. The slower rotational speed of the platter should mean a quieter drive with a longer MTBF rate; I assume the wider platter would be inherently more stable due to its lower rotational speed. Perhaps a whole stack of single-platter 5.25" drives could be able to be stacked inside of a regular 5.25" slot. These benefits are to name just a few. The drawback would probably be seek and read/write times, but the increased time should be insignificant to most end-users.
Just seems that hard drive technology should be developing in several physical directions, but conventional wisdom is mostly limited to just shrinking them. The current drives are shrinking at a fantastic rate and its beginning to look like 2.5" will become the norm in laptops. But for desktops and server farms it would make sense to also increase the physical size of the drives. It should NOT require seven to twelve drives just to get the redundant capacity necessary for a terabyte of storage. Only few 9-12" platters (@20 GB per inch square) should more than suffice to run 1 terabytes in a safely striped set. Wouldn't the price per terabyte be less for large capacity platters, too?
10" platter area w/9" datastrip = (3.14(4.5"^2)) - (3.14(.5"^2)) = 63.585 - 0.785 = " square
Now take " sqaure times 20 GB per inch and we get a capacity of 1271.7 GB.
12" platter area w/11" datastrip = (3.14(5"^2)) - (3.14(.5"^2)) = 78.5 - 0.785 = 77.715" square
Now take " sqaure times 20 GB per inch and we get a capacity of 1554.3 GB.
14" platter area w/13" datastrip = (3.14(6.5"^2)) - (3.14(.5"^2)) = 132.665 - 0.785 = 131.88" square
Now take " sqaure times 20 GB per inch and we get a capacity of 2637.6 GB!
16" platter area w/15" datastrip = (3.14(7"^2)) - (3.14(.5"^2)) = 153.86 - 0.785 = 153.075" square
Now take " sqaure times 20 GB per inch and we get a capacity of 3061.5 GB!!
I couldn't even begin to imagine the tangential velocity on a 16" platter. 🙂