Hi,
  I am totally dumb on this question. Please someone can help me to
know how to calculate this one.

Question:
-->>>A disk unit has 20 recording surfaces. It has a total of 14,000
cylinders. There is an average of 400 sectors per track and each
sector contains 512 bytes of data. Find the storage capacity of the
disk unit. If disk rotates at a speed of 7200 rpm then what is the
data transfer rate in bytes per second?


I am loking forward to you.

Thanks
 Chnag

I'll give it a try:

capacity: 20x14.000x400x512 bytes = 57.344.000.000 bytes.

speed: 20x400x512x7200/60 = 491.520.000 bytes/sec.

Bye,
  Skybuck.

"chang" <chinmoy.chittaranjan@gmail.com> wrote in message 
news:8f0e55f8-3d9b-40cb-a7d4-3594b66c5065@z6g2000pre.googlegroups.com...
>
> Hi,
>  I am totally dumb on this question. Please someone can help me to
> know how to calculate this one.
>
> Question:
> -->>>A disk unit has 20 recording surfaces. It has a total of 14,000
> cylinders. There is an average of 400 sectors per track and each
> sector contains 512 bytes of data. Find the storage capacity of the
> disk unit. If disk rotates at a speed of 7200 rpm then what is the
> data transfer rate in bytes per second?
>
>
> I am loking forward to you.
>
> Thanks
> Chnag 

On 5 Feb, 09:33, chang <chinmoy.chittaran...@gmail.com> wrote:
> Hi,
> =A0 I am totally dumb on this question. Please someone can help me to
> know how to calculate this one.
>
> Question:
> A disk unit has 20 recording surfaces. It has a total of 14,000
> cylinders. There is an average of 400 sectors per track and each
> sector contains 512 bytes of data. Find the storage capacity of the
> disk unit.

1 track =3D 400 sectors of 512 bytes =3D 204,800 bytes.

1 surface =3D 14,000 tracks =3D 2,867,200,000 bytes.

1 unit =3D 20 surfaces =3D 57,344,000,000 bytes.

> If disk rotates at a speed of 7200 rpm then what is the
> data transfer rate in bytes per second?

7200 revs per minute =3D 120 revs per second.

120 trackfuls per second =3D 24,576,000 bytes per second.

My answer assumes that only one of the 20 reading heads
is selected to be active during a transfer.  Skybuck's
is correct if all heads are active, reading interleaved
data in parallel from all 20 surfaces.
--

On Feb 5, 6:39=A0am, bert <bert.hutchi...@btinternet.com> wrote:
> On 5 Feb, 09:33, chang <chinmoy.chittaran...@gmail.com> wrote:
>
> > Hi,
> > =A0 I am totally dumb on this question. Please someone can help me to
> > know how to calculate this one.
>
> > Question:
> > A disk unit has 20 recording surfaces. It has a total of 14,000
> > cylinders. There is an average of 400 sectors per track and each
> > sector contains 512 bytes of data. Find the storage capacity of the
> > disk unit.
>
> 1 track =3D 400 sectors of 512 bytes =3D 204,800 bytes.
>
> 1 surface =3D 14,000 tracks =3D 2,867,200,000 bytes.
>
> 1 unit =3D 20 surfaces =3D 57,344,000,000 bytes.
>
> > If disk rotates at a speed of 7200 rpm then what is the
> > data transfer rate in bytes per second?
>
> 7200 revs per minute =3D 120 revs per second.
>
> 120 trackfuls per second =3D 24,576,000 bytes per second.
>
> My answer assumes that only one of the 20 reading heads
> is selected to be active during a transfer. =A0Skybuck's
> is correct if all heads are active, reading interleaved
> data in parallel from all 20 surfaces.
> --

Thanks to all .I got it .

Chnag