Consider a 15000 RPM disk This has a tracktotrack seek time

Consider a 15,000 RPM disk. This has a track-to-track seek time of
1 msec. The disk has 400 sectors of 512 bytes on each track.
How much cylinder skew is required?

Solution

Please follow the data and description :

In general the disk rotates at a speed of 120 rpm, so for 1 rotation it takes around 1000/120 msec of time. With 400 sectors per rotation, the sector time is given as 1/400 of this number or simplified as

1/400 * 1000/120 = 5/240 = 1/48 msec.

During the 1-msec seek, 48 sectors pass under the head. Thus the cylinder skew should be 48.

Hope this is helpful.