number of tracks

If you know the disk capacity, bytes per track, and number of surfaces, then how do you calculate the number of tracks?

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– Jonathan Leffler Mar 16 '10 at 11:13
Makework question that lost its relevance quite some time ago. – Fiasco Labs Nov 16 '14 at 1:11

Number of tracks is totally dependant on how the manufacturer laid out the disk internally, which you will not know. All modern disks use LBA (logical block addressing), in which the OS addresses the drive on a sector-by-sector basis, not knowing or caring how or where the sectors are physically located on the platters (nor how many platters there actually are).

Not only that, but the number of sectors per track depends on how far out from the spindle motor you are at the time; it's not a spiral like on a CD. The further from the spindle you are, the more sectors per track (and thus the higher the transfer rate).

Since you do not know the number of sectors per track (which, again, varies depending where on disk you are), you cannot determine how many tracks exist, given capacity, number of sectors, and number of surfaces.

In other words, you don't. Period.

UPDATE:

If your teacher wants a mathematically correct answer, it would be (CAPACITY_IN_BYTES / BYTES_PER_TRACK) to get the number of tracks on disk, and if you want tracks per surface, divide the number of tracks on disk by the number of surfaces.

However, note that this is totally inapplicable to modern drives. The real answer is that it cannot be calculated since you do not know how many sectors per track there are (and there are a variable number in different zones of the disk) -- hence, there are too many unknowns to be able to solve the formula.

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@Phenom: If this answers your question, please mark it as accepted! :) – Alexander Burke Mar 13 '10 at 6:19
Your answer is that it can't be calculated, so it doesn't answer my question. There should be a way to calculate it, since this is a question from my midterm study guide. – Phenom Mar 13 '10 at 8:00
@Phenom: Updated above. Upvote and accept. :) – Alexander Burke Mar 13 '10 at 8:26
The equation you gave doesn't have the number of surfaces in it. – Phenom Mar 13 '10 at 13:15