An optical disk consists of two thousand concentric tracks. The diskis 5.2 inche

Question

An optical disk consists of two thousand concentric tracks. The diskis 5.2 inches in diameter. The innermost track is located at a radius of 1/2 inch from the center. The outermost track is located 2 1/2 inches from the center. The density of the disk is specied as 1630 bytes per inch along the track. The transfer rate is specied as 256,000 bytes per second. The disk is CLV. All blocks are of equal size. (a). The innermost track consists of ten blocks. How many bytes are contained in a block? (b). How many blocks would the outermost track contain? (c). The capacity of the disk is approximately equal to the capacity in bytes of the middle track times the number of tracks. What is the approximate capacity of the disk? (d). What is the motor rotation speed when reading the innermost track? the outermost track?

Explanation / Answer

a) Number of bytes in the innermost track = perimeter of the innermost track * 1630
                                          = 2 * 3.14 * 0.5 * 1630
                                          = 5118.2 bytes
   Number of bytes in a block = 5118.2/10 = 511.82 bytes

b) Number of blocks in the outermost track = (2 * 3.14 * 2.5 * 1630)/511.82
                                           = 49.99 blocks

c) capacity of the disc = 2000 * 2 * 3.14 * 1630   (Middle track will be at 1 inches from the center as the innermost is 0.5 and outermost is 2.5
                        = 20472800 bytes

d) As per the access speed of 256,000 bytes per seconn 157 inches of the track should move below the head.

    For innermost track 157 inches correspond to 157/0.5 radians = 314 radians
    speed = 314 radians/second

    For outermost track 157 inches correspond to 157/2.5 radians = 62.8 radians
    speed = 62.8 radians/second

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