Why the different arc length sector have the same size data capacity?
The sector-A and sector-B, the arc length are different, but they representing data capacity are same, such as 512KB.
you see, whether the different cylinder sector's density is different?
and if we make the disk more large, we waste more materials?
hard-drive sectors
asked Jan 14 at 8:16
244boy244boy
1485
add a comment |
The sector-A and sector-B, the arc length are different, but they representing data capacity are same, such as 512KB.
you see, whether the different cylinder sector's density is different?
and if we make the disk more large, we waste more materials?
hard-drive sectors
asked Jan 14 at 8:16
244boy244boy
1485
1
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
1
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
In short, the image is full of lies.
– grawity
Jan 14 at 11:04
add a comment |
The sector-A and sector-B, the arc length are different, but they representing data capacity are same, such as 512KB.
you see, whether the different cylinder sector's density is different?
and if we make the disk more large, we waste more materials?
hard-drive sectors
asked Jan 14 at 8:16
244boy244boy
1485
The sector-A and sector-B, the arc length are different, but they representing data capacity are same, such as 512KB.
you see, whether the different cylinder sector's density is different?
and if we make the disk more large, we waste more materials?
hard-drive sectors
hard-drive sectors
asked Jan 14 at 8:16
244boy244boy
1485
asked Jan 14 at 8:16
244boy244boy
1485
asked Jan 14 at 8:16
244boy244boy
1485
asked Jan 14 at 8:16
244boy244boy
1485
asked Jan 14 at 8:16
244boy244boy
1485
1485
1
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
1
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
In short, the image is full of lies.
– grawity
Jan 14 at 11:04
add a comment |
1
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
1
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
In short, the image is full of lies.
– grawity
Jan 14 at 11:04
1
1
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
1
1
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
In short, the image is full of lies.
– grawity
Jan 14 at 11:04
In short, the image is full of lies.
– grawity
Jan 14 at 11:04
add a comment |
2 Answers
2
active
oldest
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The sector-A and sector-B, the arc length are different, but they representing data capacity are same,
True, because the angular speed is constant, whereas the linear speed it not constant as the cylinder changes.
This is old technology that provided a consistent disk layout to the host device (aka CHS (cylinder, head, sector) for disk geometry), and was implementable with available technology and economic parameters.
such as 512KB.
Umm no, disk sectors are never that large.
Such a sector size exceeds the raw number of bytes per track of older HDDs.
The ECC (error correction code) would not be reliable with such a large sector size.
The IBM PC/XT established the 512 byte sector as a de-facto standard.
you see, whether the different cylinder sector's density is different?
No, there is a fixed number of sectors per track, so the "sector density" is constant.
and if we make the disk more large, we waste more materials?
No, there is no "wasted material" just because the bit density is variable.
You seem to be unaware that modern disk drives employ zone bit recording, which has a variable number of sectors per track and aims for a consistent bit density for all cylinders.
Zone bit recording only became feasible after the disk controller was integrated with the disk drive, i.e. the IDE drive, and a LBA (logical block address) could
be used as an abstraction to conceal the variable disk geometry.
With a modern disk drive employing zone bit recording, the disk geometry is essentially unknown (other than the number of R/W heads). Old techniques such as cylinder alignment of partitions and seek optimizations have dubious value with modern HDDs.
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
add a comment |
To the extent it may represent a 512 block as traversing different lengths dont think that this image is correct (or at least not correct for non-ancient drives) - indeed im fairly certain that the outer tracks of drives hold more data Then the inner ones. +I wonder if it was true of the very early drives though)
Its well known that the first (outermost) part of the disk is a lot faster then the innermost part. You can verify this by doing a copy/read of a disk - ie block device - and watching the speeds (ive done this often using ddrescue). The speed typically slows down by half. As disks typically have a fixed rotation speed (7200 rpm for example) it stands to reason the outer tracks have more data.
This same line of reaoning is used in the well-established enterprised practice of "short stroking" disks to increase performance at the cost if wasting disk space (google it!)
Of-course all this is becoming less relevant woth the move to SSDs.
answered Jan 14 at 9:01
davidgodavidgo
44k75292
add a comment |
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2 Answers
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2 Answers
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The sector-A and sector-B, the arc length are different, but they representing data capacity are same,
True, because the angular speed is constant, whereas the linear speed it not constant as the cylinder changes.
This is old technology that provided a consistent disk layout to the host device (aka CHS (cylinder, head, sector) for disk geometry), and was implementable with available technology and economic parameters.
such as 512KB.
Umm no, disk sectors are never that large.
Such a sector size exceeds the raw number of bytes per track of older HDDs.
The ECC (error correction code) would not be reliable with such a large sector size.
The IBM PC/XT established the 512 byte sector as a de-facto standard.
you see, whether the different cylinder sector's density is different?
No, there is a fixed number of sectors per track, so the "sector density" is constant.
and if we make the disk more large, we waste more materials?
No, there is no "wasted material" just because the bit density is variable.
You seem to be unaware that modern disk drives employ zone bit recording, which has a variable number of sectors per track and aims for a consistent bit density for all cylinders.
Zone bit recording only became feasible after the disk controller was integrated with the disk drive, i.e. the IDE drive, and a LBA (logical block address) could
be used as an abstraction to conceal the variable disk geometry.
With a modern disk drive employing zone bit recording, the disk geometry is essentially unknown (other than the number of R/W heads). Old techniques such as cylinder alignment of partitions and seek optimizations have dubious value with modern HDDs.
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
add a comment |
The sector-A and sector-B, the arc length are different, but they representing data capacity are same,
True, because the angular speed is constant, whereas the linear speed it not constant as the cylinder changes.
This is old technology that provided a consistent disk layout to the host device (aka CHS (cylinder, head, sector) for disk geometry), and was implementable with available technology and economic parameters.
such as 512KB.
Umm no, disk sectors are never that large.
Such a sector size exceeds the raw number of bytes per track of older HDDs.
The ECC (error correction code) would not be reliable with such a large sector size.
The IBM PC/XT established the 512 byte sector as a de-facto standard.
you see, whether the different cylinder sector's density is different?
No, there is a fixed number of sectors per track, so the "sector density" is constant.
and if we make the disk more large, we waste more materials?
No, there is no "wasted material" just because the bit density is variable.
You seem to be unaware that modern disk drives employ zone bit recording, which has a variable number of sectors per track and aims for a consistent bit density for all cylinders.
Zone bit recording only became feasible after the disk controller was integrated with the disk drive, i.e. the IDE drive, and a LBA (logical block address) could
be used as an abstraction to conceal the variable disk geometry.
With a modern disk drive employing zone bit recording, the disk geometry is essentially unknown (other than the number of R/W heads). Old techniques such as cylinder alignment of partitions and seek optimizations have dubious value with modern HDDs.
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
add a comment |
The sector-A and sector-B, the arc length are different, but they representing data capacity are same,
True, because the angular speed is constant, whereas the linear speed it not constant as the cylinder changes.
This is old technology that provided a consistent disk layout to the host device (aka CHS (cylinder, head, sector) for disk geometry), and was implementable with available technology and economic parameters.
such as 512KB.
Umm no, disk sectors are never that large.
Such a sector size exceeds the raw number of bytes per track of older HDDs.
The ECC (error correction code) would not be reliable with such a large sector size.
The IBM PC/XT established the 512 byte sector as a de-facto standard.
you see, whether the different cylinder sector's density is different?
No, there is a fixed number of sectors per track, so the "sector density" is constant.
and if we make the disk more large, we waste more materials?
No, there is no "wasted material" just because the bit density is variable.
You seem to be unaware that modern disk drives employ zone bit recording, which has a variable number of sectors per track and aims for a consistent bit density for all cylinders.
Zone bit recording only became feasible after the disk controller was integrated with the disk drive, i.e. the IDE drive, and a LBA (logical block address) could
be used as an abstraction to conceal the variable disk geometry.
With a modern disk drive employing zone bit recording, the disk geometry is essentially unknown (other than the number of R/W heads). Old techniques such as cylinder alignment of partitions and seek optimizations have dubious value with modern HDDs.
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
The sector-A and sector-B, the arc length are different, but they representing data capacity are same,
True, because the angular speed is constant, whereas the linear speed it not constant as the cylinder changes.
This is old technology that provided a consistent disk layout to the host device (aka CHS (cylinder, head, sector) for disk geometry), and was implementable with available technology and economic parameters.
such as 512KB.
Umm no, disk sectors are never that large.
Such a sector size exceeds the raw number of bytes per track of older HDDs.
The ECC (error correction code) would not be reliable with such a large sector size.
The IBM PC/XT established the 512 byte sector as a de-facto standard.
you see, whether the different cylinder sector's density is different?
No, there is a fixed number of sectors per track, so the "sector density" is constant.
and if we make the disk more large, we waste more materials?
No, there is no "wasted material" just because the bit density is variable.
You seem to be unaware that modern disk drives employ zone bit recording, which has a variable number of sectors per track and aims for a consistent bit density for all cylinders.
Zone bit recording only became feasible after the disk controller was integrated with the disk drive, i.e. the IDE drive, and a LBA (logical block address) could
be used as an abstraction to conceal the variable disk geometry.
With a modern disk drive employing zone bit recording, the disk geometry is essentially unknown (other than the number of R/W heads). Old techniques such as cylinder alignment of partitions and seek optimizations have dubious value with modern HDDs.
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
edited Jan 14 at 9:29
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
answered Jan 14 at 9:01
sawdustsawdust
13.9k12438
13.9k12438
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
add a comment |
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
You have "there is a fixed number of sectors per track" in one paragraph, and "modern disk drives [have] a variable number of sectors per track" in another... I think OP meant "density of data in a sector" when they wrote "sector's density" – not "density of sectors in a track".
– grawity
Jan 14 at 11:04
add a comment |
To the extent it may represent a 512 block as traversing different lengths dont think that this image is correct (or at least not correct for non-ancient drives) - indeed im fairly certain that the outer tracks of drives hold more data Then the inner ones. +I wonder if it was true of the very early drives though)
Its well known that the first (outermost) part of the disk is a lot faster then the innermost part. You can verify this by doing a copy/read of a disk - ie block device - and watching the speeds (ive done this often using ddrescue). The speed typically slows down by half. As disks typically have a fixed rotation speed (7200 rpm for example) it stands to reason the outer tracks have more data.
This same line of reaoning is used in the well-established enterprised practice of "short stroking" disks to increase performance at the cost if wasting disk space (google it!)
Of-course all this is becoming less relevant woth the move to SSDs.
answered Jan 14 at 9:01
davidgodavidgo
44k75292
add a comment |
To the extent it may represent a 512 block as traversing different lengths dont think that this image is correct (or at least not correct for non-ancient drives) - indeed im fairly certain that the outer tracks of drives hold more data Then the inner ones. +I wonder if it was true of the very early drives though)
Its well known that the first (outermost) part of the disk is a lot faster then the innermost part. You can verify this by doing a copy/read of a disk - ie block device - and watching the speeds (ive done this often using ddrescue). The speed typically slows down by half. As disks typically have a fixed rotation speed (7200 rpm for example) it stands to reason the outer tracks have more data.
This same line of reaoning is used in the well-established enterprised practice of "short stroking" disks to increase performance at the cost if wasting disk space (google it!)
Of-course all this is becoming less relevant woth the move to SSDs.
answered Jan 14 at 9:01
davidgodavidgo
44k75292
add a comment |
To the extent it may represent a 512 block as traversing different lengths dont think that this image is correct (or at least not correct for non-ancient drives) - indeed im fairly certain that the outer tracks of drives hold more data Then the inner ones. +I wonder if it was true of the very early drives though)
Its well known that the first (outermost) part of the disk is a lot faster then the innermost part. You can verify this by doing a copy/read of a disk - ie block device - and watching the speeds (ive done this often using ddrescue). The speed typically slows down by half. As disks typically have a fixed rotation speed (7200 rpm for example) it stands to reason the outer tracks have more data.
This same line of reaoning is used in the well-established enterprised practice of "short stroking" disks to increase performance at the cost if wasting disk space (google it!)
Of-course all this is becoming less relevant woth the move to SSDs.
answered Jan 14 at 9:01
davidgodavidgo
44k75292
To the extent it may represent a 512 block as traversing different lengths dont think that this image is correct (or at least not correct for non-ancient drives) - indeed im fairly certain that the outer tracks of drives hold more data Then the inner ones. +I wonder if it was true of the very early drives though)
Its well known that the first (outermost) part of the disk is a lot faster then the innermost part. You can verify this by doing a copy/read of a disk - ie block device - and watching the speeds (ive done this often using ddrescue). The speed typically slows down by half. As disks typically have a fixed rotation speed (7200 rpm for example) it stands to reason the outer tracks have more data.
This same line of reaoning is used in the well-established enterprised practice of "short stroking" disks to increase performance at the cost if wasting disk space (google it!)
Of-course all this is becoming less relevant woth the move to SSDs.
answered Jan 14 at 9:01
davidgodavidgo
44k75292
answered Jan 14 at 9:01
davidgodavidgo
44k75292
answered Jan 14 at 9:01
davidgodavidgo
44k75292
answered Jan 14 at 9:01
davidgodavidgo
44k75292
44k75292
add a comment |
add a comment |
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1
Where did you get the assumption that each arc gives the same data capacity?
– Mokubai♦
Jan 14 at 8:55
before, every sector is 512KB. do you mean in the snapshot, every arc has many sectors? not only one?
– 244boy
Jan 14 at 8:57
1
No, I'm asking where you got the information that this is how it works. Up until a few years ago actors were 512B, not KB. I would expect a hard disk to use a fixed (linear) sector size, or variable sector size depending on the area of disk in order to achieve maximum capacity. Look up CAV (constant angular velocity), CLV (constant linear velocity) and ZBR (zoned bit recording) to see various solutions and benefits of various disk formats.
– Mokubai♦
Jan 14 at 9:05
In short, the image is full of lies.
– grawity
Jan 14 at 11:04