in CO and Architecture
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A hard disk has 63 sectors/track ,10 platters each with 2 recording surfaces and 1000 cylinders ,the address
of each sector is given as
where C is cylinder no , H is surface no and S is sector no , thus the 0 th sector is addressed
as <0,0,0>  , 1st sector is address as <0,0,1> and so on , so then the address <400 , 16, 29 > corresponds to
which sector no ?
in CO and Architecture
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2 Answers

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Number of sectors in a track = 63

Number of track surfaces in a cylinder = 10*2

Number of sectors in a cylinder = 63*10*2

<400,16,29> , sector number  = 400th cylinder  + 16th surface of 400th cylinder + 29th sector of 16th surface of 400th cylinder

Sector number of address <400,16,29> = 400*63*10*2 + 16*63 + 29 = 505037 sector .

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<400,16,29> address is given

step1. no. of sectors in 1 cylinder=no.of surfaces * no. of sectors in one track

                                                = 20 mul 63=1260 sectors

$\because$ we are marking no. from 0 so,(0-399) has 400 cylinder

so, 400 cylinder have no of sectors=400 mul 1260=504,000

 

step2: now to reach on 400th cylinder's 16th surface we have gopne through no. of sectors=

      no. of sectors on each surface of each cylinder

=16 mul 63=1008

step3: no. to reach on 29th sectors of 16th surface of 400th cylinder

(0-29)=30

now adding all =504,000 + 1008 + 30 =505038

so, the given address is 505038th sector's address

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