Computer hardware info please?

Question

A hard magnetic disk has two surfaces. The storage area on each surface has an inner radius of 1 cm and an outer radius of 5 cm. Each track holds the same number of bits, even though each differs in size from every other. The maximum storage density of the media is 10,000 bits/cm. The spacing between corresponding points on adjacent tracks is .1 mm, which includes the intertrack gap. Assume that the intersector gaps are negligible and that a track exists on each edge of the storage area. a.What is the maximum number of bits that can be stored on the disk? b.What is the data transfer rate from the disk to the head in bits per second at a rotational speed of 3600 RPM

Answer 1

You have to break a problem like this up into smaller sections and try to visualize each part.

For example, lets start by figuring out HOW MANY tracks there are: The outer radius is 5cm (50mm) and the inner one is 1cm (10mm), and the spacing between corresponding points on adjacent tracks is 0.1mm. The total distance from the inner track to the outer track then would be 4cm (40mm).

So, there is a track at 10mm, one at 10.1mm, one at 10.2mm, and so on -- the 11th track is at 11mm, the 21st track at 12mm, and so on -- this means that there are a total of 401 tracks. So there id calculated fact #1:

401 tracks per side, times 2 sides = 802 total tracks of data.

"Each track holds the same number of bits, even though each differs in size from every other." This means that they are all arranged in a big fan-like or starburst-like pattern. "Assume that the intersector gaps are negligible," means that there is no wasted space, one bit bumps up right against the bit next to it.

The most closely-spaced bits are going to be on the inner track, the one whose radius is 1cm. The circumference of that track would be:

C = 2*Pi*r = 2 * 3.14159.... * 1cm = 6.283185307.... cm

Call Pi 3.1416 cm, the circumference of this track is 2.2832 cm, and the maximum bit density is given as 10,000 bits per cm, so this track holds 62,832 bits. So does every other track. That means the total is:

62,832 bits per track times 802 tracks = a little over 50 million bits. I'll let you calcualte the exact answer. Remember to substitute whatever value of Pi your instructor wants you to use.

For the second part, you first need to convert rpm (revolutions per minute) to rps (revolutions per second) -- do this by dividing 3600 rpm by 60 seconds per minute, and you come out with 60 revolutions per second.

Because all parts of the disk spin at the same rpm (or rps), you just need to multiply the number of bits per track (revolution) times 60 revolutions per second, to get the number of bits per second that you can read PER SIDE OF THE DISK -- remember to double this, because you are reading two sides of the disk at the same time, presumably.

Answer 2

Sorry, my brain is not ready for this... PLEASE re-submit into the Computers category... you'll get the answer there... I hope.

Or since it's a math problem, stuff it into the Mathematics category too!