Storage capacity of a disk system = Number of recording surfaces.
Number of tracks per surface.
Number of sectors per track.
Number of bytes per sector.
Let us assume that a disk pack has 10 disk plates, 2655 tracks per plate, 125 sector per track, and 512 bytes per sector. Since the disk pack has 10 disk plates, there are 18 recording surfaces (excluding the upper of the topmost disk and the lower surface of the bottommost disk). Hence, its capacity = 18×2655×125×512=3,05,85,60,000 bytes=3×10(9) bytes (approximately)= 3 GB (3 Giga Bytes). As one character is stored per byte (using 8-bit EBCDIC encoding), the disk pack can store over 3 billion characters.
For larger storage capacity, designers prefer to increase storage capacity by increasing the number of tracks per inch of surface and bits per inch of track, rather than increasing disk size. Hence, a constant goal of desigerns of disk surface and bits per inch of track, rather than increasing disk size. Hence, a constant goal of designers of disk systems is to increase the data density of each square inch of disk surface.
Cost-per-bit storage is very low for optical disks because of their low cost and high storage density. They come in various sizes ranging from 12.0-inch to 3.0-inch diameter. The most popular one is of 5.25-inch diameter with capacity of about 650 Megabytes, which is equivalent to about 2,50,000 pages of printed text. Optical disks are the most suitable storage media for multimedia applications because of their large storage capacity.
As optical disks have a single track, Storage Capacity of an optical disk=Number of sectors×Number of bytes per sector The 5.25-inch optical disk typically has 3,30,000 sectors each of 2352 bytes, making its storage capacity equal to 3,30,000×2352=776×bytes=776Megabytes. This is unformatted capacity. Formatted capacity is about 650 Megabytes.