Disk Drive Basics

Article by Alan Johnson (220 pts ) , published Dec 11, 2008

This article discusses the fundamentals of disk drive technology.

Disk Geometry

Today's disk drives are based on magnetic recording principles. Data is stored on disks by magnetizing particles, which will ultimately correspond to the computer's data of binary ones and zeros. The disk drive itself contains a series of disks, which are termed platters. Each platter has two recording surfaces - top and bottom. These platters are stacked on top of one another in the same vertical plane with sufficient space between them to accommodate recording heads. The heads are responsible for reading and writing of data. Reading is the process of retrieving data from a disk drive, and writing is the process of storing data on a disk drive. The platters are usually made up of a lightweight material such as smooth glass or a light metal and coated with a substance that can be magnetized.

Data is stored on each surface in a series of invisible concentric rings, which are termed tracks. If each track is viewed within the same vertical plane, we get a structure that resembles a cylinder. The number of cylinders on a disk drive is the same as the number of tracks per recording surface. Each track is further divided up into sectors.

Capacity and recording

disk drive

A sector is further subdivided into a collection of bytes. Each byte is made up of 8 bits of information. Most Operating Systems have abstracted the three dimensional disk geometry into a series of logical blocks, where the total blocks on a disk is equal to the number of sectors multiplied by the number of data heads multiplied by the number of cylinders. This technique is known as Logical Block Addressing (LBA). Most blocks are 512 bytes in size. The capacity of a disk drive can be calculated by multiplying the number of sectors by the number of data heads by the number of cylinders by the number of bytes per sector, or by multiplying the number of blocks by 512. In the disk drive environment a Megabyte is equal to one million bytes rather than the binary Megabyte of 2**20 (1048576) bytes

The platters spin at a constant speed and the recording heads move laterally across the tracks. The heads do not come in contact with the surface of the disk but instead they ride on a cushion of air.

The closer the heads come to the surface of the disk the denser the recording becomes. The distance between the head and the surface is extremely close and is achieved by extremely precise mechanical tolerances. An instance of a head coming into physical contact with the recording portion of a disk is termed a head crash and usually has fatal consequences. When the disk is spun down the heads will retract to a specially lubricated portion of the surface which is termed the landing zone. Some disks when manufactured are not 100% error free but have a few minor blemishes. These blemishes are known as bad areas of the disk and are recorded on the disk in an area known as the Manufacturer's Defect Area.

More recent developments have introduced perpendicular recording where the magnetization takes place vertically rather than horizontally across the disk surface. This has led to greater bit densities.

Access times

Initially drives must be formatted before use. This process is analogous to preparing a blank set of pages for writing by ruling out lines and margins as well as the insertion of headers and footers. This extra information, which is held within each block, will actually take up some storage space, so the formatted capacity of a disk drive is less than the unformatted capacity. This distinction is very important to the user who is only concerned with usable capacity.

Because accurate retrieval of data is crucial for the end user, a method of incorporating error detection and correction is incorporated within each sector. A common method for detecting errors is a scheme known as Cyclic Redundancy Checking or CRC. A more advanced technique is to employ an Error Correcting Code, which contains redundant information to correct data errors. Some advanced forms of disk software will monitor this correction threshold and, when it exceeds a pre-determined limit, will automatically retire the defective block and transparently redirect the data to a good area of the disk.

The access time of a disk is the time taken to retrieve the data. This access time is made up primarily of two factors, one factor is the rotational latency and the other is the seek time. Rotational latency is the time taken for the desired sector to pass under the recording head. SATA disks typically spin at a rotational speed of 7200 R.P.M. so this means that it will take 1/60th of a second to make one revolution (1/60th of a second is equal to 8.3 milliseconds). When calculating the rotational delay we calculate the average delay, which is the time taken to perform one half of a disk revolution = 4.17 milliseconds.

The seek time is the time taken to move the heads to the desired track and this tends to vary from drive to drive but an average seek time of 10 milliseconds is not uncommon. When we add the rotational latency and the seek time together we get the access time. In this example, the access time would be quoted as 14 milliseconds. The time taken to actually transfer the data is known as the data transfer rate, but as this is normally very fast compared to the latency and seek times, we can usually ignore it. Modern disk drives have rotational speeds of up to 15,000 R.P.M., which result in very fast access times.

Comments

Oct 9, 2009 10:19 AM
alan johnson
RE: Disk Drive Basics
the rotational latency is the time taken to take 1/2 a revolution. Since the heads could we over any given sector at a point in time, the average is taken statistically to be 1/2 of the time to take a full revolution.
So in your case the disk is 1/2 rev per second so avg rot latency is 1/4 of a second or 250 ms so adding this to the seek time is 280ms.
Disks today normally spin at:
4200 RPM
5400 RPM
7200 RPM
10,000 RPM
15,000 RPM
Oct 9, 2009 1:39 AM
Biswajit
Disk Access Time
Sir,
I am having some problem understanding the calculation for this,
Assume that the average seek time for a disk is 50ms; the disk rotates at 12000rpm; and each track of the disk have 300 sectors. Can the access time be calculated for the disk? If so, how?

I will be expecting your help...
Thanking u
Biswajit
Jun 30, 2009 12:00 AM
Arun Ravindran
Query
Sir,

I have a doubt in the calculation of the disk access time.
If the seek time is given as 30 ms and the disk rotates 30 rotations per minute, then what is the access time?
Kindly help me solve this question with the appropriate formulae so that i can use them for future references.

Thanking you,

Arun
 
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