real number of sectors on HDD
This is part of output of fdisk -l
Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes
255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 4096 bytes
Disk identifier: 0x000140bd
Device Boot Start End Blocks Id System
/dev/sdb1 2048 167774207 83886080 7 HPFS/NTFS/exFAT
/dev/sdb2 167774208 1740638207 786432000 7 HPFS/NTFS/exFAT
/dev/sdb3 * 1740638208 1935673343 97517568 83 Linux
/dev/sdb4 1935673344 1952448511 8387584 82 Linux swap / Solaris
But when I multiply 255*63*121601 i get 1953520065 sectors which is less than 1953525168 sectirs reported by fdisk. The difference is 5103 sectors.
I know that the number of heads, sectors/track and cylinders reported does'n probably refer to real hardware properties and that is was somehow calculated for backward compatibility. But I still wonder what did hapen with those 5103 sectors? Can I use them?
Later I wanna move filesystems, do some changes, change partition table and use combination of GPT and MBR. That is why I want to know everything about sectors on my disk. My changes will include backing up with dd utility before start and after completing changes. I want to back up GPT sectors. Even those ones which should reside in last 34 sectors of HDD.
But now I wonder where last 34 sectors are. On the reported end or on the place I calculated?
linux hard-drive partitioning gpt fdisk
asked Mar 17 '13 at 10:22
RoboRobo
3125
add a comment |
This is part of output of fdisk -l
Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes
255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 4096 bytes
Disk identifier: 0x000140bd
Device Boot Start End Blocks Id System
/dev/sdb1 2048 167774207 83886080 7 HPFS/NTFS/exFAT
/dev/sdb2 167774208 1740638207 786432000 7 HPFS/NTFS/exFAT
/dev/sdb3 * 1740638208 1935673343 97517568 83 Linux
/dev/sdb4 1935673344 1952448511 8387584 82 Linux swap / Solaris
But when I multiply 255*63*121601 i get 1953520065 sectors which is less than 1953525168 sectirs reported by fdisk. The difference is 5103 sectors.
I know that the number of heads, sectors/track and cylinders reported does'n probably refer to real hardware properties and that is was somehow calculated for backward compatibility. But I still wonder what did hapen with those 5103 sectors? Can I use them?
Later I wanna move filesystems, do some changes, change partition table and use combination of GPT and MBR. That is why I want to know everything about sectors on my disk. My changes will include backing up with dd utility before start and after completing changes. I want to back up GPT sectors. Even those ones which should reside in last 34 sectors of HDD.
But now I wonder where last 34 sectors are. On the reported end or on the place I calculated?
linux hard-drive partitioning gpt fdisk
asked Mar 17 '13 at 10:22
RoboRobo
3125
add a comment |
This is part of output of fdisk -l
Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes
255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 4096 bytes
Disk identifier: 0x000140bd
Device Boot Start End Blocks Id System
/dev/sdb1 2048 167774207 83886080 7 HPFS/NTFS/exFAT
/dev/sdb2 167774208 1740638207 786432000 7 HPFS/NTFS/exFAT
/dev/sdb3 * 1740638208 1935673343 97517568 83 Linux
/dev/sdb4 1935673344 1952448511 8387584 82 Linux swap / Solaris
But when I multiply 255*63*121601 i get 1953520065 sectors which is less than 1953525168 sectirs reported by fdisk. The difference is 5103 sectors.
I know that the number of heads, sectors/track and cylinders reported does'n probably refer to real hardware properties and that is was somehow calculated for backward compatibility. But I still wonder what did hapen with those 5103 sectors? Can I use them?
Later I wanna move filesystems, do some changes, change partition table and use combination of GPT and MBR. That is why I want to know everything about sectors on my disk. My changes will include backing up with dd utility before start and after completing changes. I want to back up GPT sectors. Even those ones which should reside in last 34 sectors of HDD.
But now I wonder where last 34 sectors are. On the reported end or on the place I calculated?
linux hard-drive partitioning gpt fdisk
asked Mar 17 '13 at 10:22
RoboRobo
3125
This is part of output of fdisk -l
Disk /dev/sdb: 1000.2 GB, 1000204886016 bytes
255 heads, 63 sectors/track, 121601 cylinders, total 1953525168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 4096 bytes
Disk identifier: 0x000140bd
Device Boot Start End Blocks Id System
/dev/sdb1 2048 167774207 83886080 7 HPFS/NTFS/exFAT
/dev/sdb2 167774208 1740638207 786432000 7 HPFS/NTFS/exFAT
/dev/sdb3 * 1740638208 1935673343 97517568 83 Linux
/dev/sdb4 1935673344 1952448511 8387584 82 Linux swap / Solaris
But when I multiply 255*63*121601 i get 1953520065 sectors which is less than 1953525168 sectirs reported by fdisk. The difference is 5103 sectors.
I know that the number of heads, sectors/track and cylinders reported does'n probably refer to real hardware properties and that is was somehow calculated for backward compatibility. But I still wonder what did hapen with those 5103 sectors? Can I use them?
Later I wanna move filesystems, do some changes, change partition table and use combination of GPT and MBR. That is why I want to know everything about sectors on my disk. My changes will include backing up with dd utility before start and after completing changes. I want to back up GPT sectors. Even those ones which should reside in last 34 sectors of HDD.
But now I wonder where last 34 sectors are. On the reported end or on the place I calculated?
linux hard-drive partitioning gpt fdisk
linux hard-drive partitioning gpt fdisk
asked Mar 17 '13 at 10:22
RoboRobo
3125
asked Mar 17 '13 at 10:22
RoboRobo
3125
asked Mar 17 '13 at 10:22
RoboRobo
3125
asked Mar 17 '13 at 10:22
RoboRobo
3125
asked Mar 17 '13 at 10:22
RoboRobo
3125
3125
add a comment |
add a comment |
2 Answers
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CHS is a very old way to address sectors on a hard drive and usually limits the total number of sectors addressable to numbers divisible by the Cylinder and Head count. So in this case it will be a number divisible by 63 * 255 or 16065. The sectors on the drive in excess of the closest such number won't be used.
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
add a comment |
Using the default geometry of 255 head/ 63 sector will cause misaligned partitions. This will cause the next partitions to start on the next pysical sector. So you are losing those sector in between partitions. Reset your disk geometry to 256 heads/32 sector if you want to be always optimal.
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
CHS is a very old way to address sectors on a hard drive and usually limits the total number of sectors addressable to numbers divisible by the Cylinder and Head count. So in this case it will be a number divisible by 63 * 255 or 16065. The sectors on the drive in excess of the closest such number won't be used.
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
add a comment |
CHS is a very old way to address sectors on a hard drive and usually limits the total number of sectors addressable to numbers divisible by the Cylinder and Head count. So in this case it will be a number divisible by 63 * 255 or 16065. The sectors on the drive in excess of the closest such number won't be used.
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
add a comment |
CHS is a very old way to address sectors on a hard drive and usually limits the total number of sectors addressable to numbers divisible by the Cylinder and Head count. So in this case it will be a number divisible by 63 * 255 or 16065. The sectors on the drive in excess of the closest such number won't be used.
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
CHS is a very old way to address sectors on a hard drive and usually limits the total number of sectors addressable to numbers divisible by the Cylinder and Head count. So in this case it will be a number divisible by 63 * 255 or 16065. The sectors on the drive in excess of the closest such number won't be used.
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
answered Mar 17 '13 at 11:57
BrianBrian
8,2711833
8,2711833
add a comment |
add a comment |
Using the default geometry of 255 head/ 63 sector will cause misaligned partitions. This will cause the next partitions to start on the next pysical sector. So you are losing those sector in between partitions. Reset your disk geometry to 256 heads/32 sector if you want to be always optimal.
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
add a comment |
Using the default geometry of 255 head/ 63 sector will cause misaligned partitions. This will cause the next partitions to start on the next pysical sector. So you are losing those sector in between partitions. Reset your disk geometry to 256 heads/32 sector if you want to be always optimal.
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
add a comment |
Using the default geometry of 255 head/ 63 sector will cause misaligned partitions. This will cause the next partitions to start on the next pysical sector. So you are losing those sector in between partitions. Reset your disk geometry to 256 heads/32 sector if you want to be always optimal.
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
Using the default geometry of 255 head/ 63 sector will cause misaligned partitions. This will cause the next partitions to start on the next pysical sector. So you are losing those sector in between partitions. Reset your disk geometry to 256 heads/32 sector if you want to be always optimal.
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
answered Dec 9 '14 at 16:34
JonathanCJonathanC
32
32
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add a comment |
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