So there are 41943040 bytes in each cylinder.

The original Model 9 had a capacity of 9 times the Model 1, or 10017 cylinders. Additional models raised that to 32760 cyls (model 27) and then to 65520 cyls (model 54).

A 3390-n device has a capacity of 56,664 bytes per track,of which 55,996 bytes are accessible by applications programmers.

* Storage Administrators refer to 3390 mod 9, 27, and 54 as large-volume support 10017 cylinders, 32760 cylinders, and 65520 cylinders respectively. These large volumes are all considered mod 9s with varying number of cylinders.

A track is a section of a cylinder. A drive with one platter and one head on a single surface of that track would have one track per cylinder. A drive with 7 platters would probably have a head on each surface of each , so that would make 14 tracks per cylinder.

Dividing the number of records expected by the number that will fit on one track gives the number of tracks required. Divide by 15 to find the number of cylinders required.

Code: 1 Cylinder = 55,996 * 15 = 839,940 bytes. 1 Megabyte = 1,048,576 (2 to the 20th power) bytes.

A 3390-n device has a capacity of 56,664 bytes per track,of which 55,996 bytes are accessible by applications programmers. And 1 cylinder is 15 tracks.

One track of 3390 DASD is divided into two half track blocks, each having a blocksize of 27798. If you take the integer valus of divide 27998 by the LRECL – you get the number of records that can be stored on one half of one track. So doubling that gives you the number of records that will fit onto one track.

Modern disks use FBA storage, but they have to emulate 3390 CKD format, so the terms ‘track’ and ‘cylinder’ are still used. IBM raised the bar on volume sizes with z/OS V1R10.

The most efficient block size for the 3390 would be 27998 bytes; two of these blocks would fit on a 3390 track. The maximum data length for a track multiplied by the number of tracks per cylinder produces the number of bytes available per cylinder for a device.