Disk Storage

Hard Disk Structure In OS

Hard disk is secondary storage which stores the large amount of data. The Hard disk drive contains a dozens of disks.  These disks are also known as platters. These platters are mount over the spindle which rotates in any direction i.e. clockwise or anti-clockwise. Lets look at hard disk structure in OS.

Platter

• The Platter is made of aluminum or iron oxide.
• Platter diameter range is 1.8 inches to 5.25 inches
• Each platter contains 2 surfaces and 2 Read/Write Head. One Read/Write head requires for one surface of the platter. So, other R/W head use for other surface to store the information’s.
• Every platter holds the same no. of tracks.
• Multiple platters increase the storage capacity.

R/W Head

• R/W Heads moves forth and back over the platter surfaces to Read or Write the data on sectors.
• Read/Write heads does not touch to platters surface. The data written over the platter surface is done through magnetic field. If R/W Head touches over the surface of platter then bad sectors may creates. Had disk may damage due to these bad sectors.

Tracks

• Circular areas of disk are known as tracks.
• There may be more than a 1000 tracks on a 3.5 inch hard disk and sector size. Track Numbering start with zero at outermost track.

Sectors

Tracks are further divided into number of small units; these small units are known as sectors. Sectors are the smallest physical storage units on disk. Size of each sector is almost always 512 Bytes. Sector Numbering start with 1 at outermost tracks.

Cylinder

All Corresponding tracks with same radius of all platters in the Hard disk are known as cylinders.  In simple words we say

“Each tack of all platters with same radius is called a cylinder”

So, Number of tracks in platter is always equal to number of cylinders. For example, a hard disk, where each platter contains 600 tracks then the number of cylinders will also be 600 in the hard disk.

Cylinder Numbering start with zero at outermost cylinder.

Cluster

Cluster is also known as blocks. Group of sectors makes a cluster. There may be 64 or sectors in a cluster. These clusters are used by OS to Read/Write the data.

Disk Capacity

As we know there are number of platters in the hard disk. Each platter contains two R/W heads. There are number of cylinder/tracks in the hard disk. Each track is divided into number of sectors. And each sector has some size but mostly size of sector is 512 Bytes.

Structure of a magnetic disk

A magnetic disk contains several platters. Each platter is divided into circular shaped tracks. The length of the tracks near the centre is less than the length of the tracks farther from the centre. Each track is further divided into sectors, as shown in the figure.

Tracks of the same distance from centre form a cylinder. A read-write head is used to read data from a sector of the magnetic disk.

The disk is divided into tracks. Each track is further divided into sectors. The point to be noted here is that outer tracks are bigger in size than the inner tracks but they contain the same number of sectors and have equal storage capacity. This is because the storage density is high in sectors of the inner tracks where as the bits are sparsely arranged in sectors of the outer tracks. Some space of every sector is used for formatting. So, the actual capacity of a sector is less than the given capacity.

Read-Write(R-W) head moves over the rotating hard disk. It is this Read-Write head that performs all the read and write operations on the disk and hence, position of the R-W head is a major concern. To perform a read or write operation on a memory location, we need to place the R-W head over that position. Some important terms must be noted here:

1. Seek time – The time taken by the R-W head to reach the desired track from it’s current position.
2. Rotational latency – Time taken by the sector to come under the R-W head.
3. Data transfer time – Time taken to transfer the required amount of data. It depends upon the rotational speed.
4. Controller time – The processing time taken by the controller.
5. Average Access time – seek time + Average Rotational latency + data transfer time + controller time.

Note:Average Rotational latency is mostly 1/2*(Rotetional latency).

In questions, if the seek time and controller time is not mentioned, take them to be zero.

If the amount of data to be transferred is not given, assume that no data is being transferred. Otherwise, calculate the time taken to transfer the given amount of data.

The average of rotational latency is taken when the current position of R-W head is not given. Because, the R-W may be already present at the desired position or it might take a whole rotation to get the desired sector under the R-W head. But, if the current position of the R-W head is given then the rotational latency must be calculated.

Question 01: Let suppose there are 8 platters in hard disk drive. As each platters has two surfaces so 16 surfaces will be there in hard drive. Therefore, required R/W head will also be 16. Suppose, there are 1,024 cylinders and 128 sectors in each track. The sector size is 512 bytes.Then

A). Calculate Disk Size?
B). How many number of bits required to represent the disk size?

Part A. Solution

Size of Hard Disk = Cylinder x Heads X Sectors x Sector-Size 
= 1,024 x 16 x 128 x 512 Bytes
=  210 + 24 + 27 + 29 Bytes = 230Bytes = 2GB

Part B. Solution

As 2GB = 230 So, 30 bits are required to represent  2GB hard disk.

Question-02:  Find the size of Single platter if track size 1KB and there are 1024 cylinder? 

Solution: As there is a single platter So, Heads requirement will be 2.

Size of Platter = Cylinder x Heads X track size (because, Track size = Sectors x Sector-Size)
= 210 + 21 + 210 Bytes
=21+220 Bytes = 2MB

Question 03: Why does sector number addressing in CHS (Cylinder Head Sector) start at sector 1 and not 0?

Answer: For all translation modes, Cylinder (C) =0, Head (H) =0 and Sector (S) =1 is equivalent to Logical Block Address (LBA) =0. So the value of Sector is always 1.

Question-03: 

Consider a hard disk with:

4 surfaces

64 tracks/surface

128 sectors/track

256 bytes/sector

1. What is the capacity of the hard disk?
   Disk capacity = surfaces * tracks/surface * sectors/track * bytes/sector
   Disk capacity = 4 * 64 * 128 * 256
   Disk capacity = 8 MB
2. The disk is rotating at 3600 RPM, what is the data transfer rate?
   60 sec -> 3600 rotations
   1 sec -> 60 rotations
   Data transfer rate = number of rotations per second * track capacity * number of surfaces (since 1 R-W head is used for each surface)
   Data transfer rate = 60 * 128 * 256 * 4
   Data transfer rate = 7.5 MB/sec
3. The disk is rotating at 3600 RPM, what is the average access time?
   Since, seek time, controller time and the amount of data to be transferred is not given, we consider all the three terms as 0.
   Therefore, Average Access time = Average rotational delay
   Rotational latency => 60 sec -> 3600 rotations
   1 sec -> 60 rotations
   Rotational latency = (1/60) sec = 16.67 msec.
   Average Rotational latency = (16.67)/2
   = 8.33 msec.
   Average Access time = 8.33 msec.

Logical structure

Logical division of hard disk is generally divided into 5 basic terms which are

• MBR (Master Boot Record)
• DBR (DOS Boot Record)
• FAT (File Allocation Tables)
• Root Directory
• Data Area