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How to calculate partition Start End Sector?



The 2019 Stack Overflow Developer Survey Results Are InDisk problems prevent me from booting, or set the disk to read-only. How do I fix the disk?How to format a 1GB USB stick to FAT32 with 512 bytes sector?Expanding root partition CentOS 6 With using fdiskHow to extend logical & extended partition with fdiskExtend partition using LVMHost unreachable after resizing partitionCorrupted ntfs volume mounting problemI have a dedicated with 2 SSDs, how to I group them to behave as 1?Wrong fs type, bad option, bad superblock on /dev/sdXExternal HDD failure due to bad blocks



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








6















I am wondering what Start and End value to choose when partitioning my ext. SSD using fdisk.



fdisk suggests 2048-250069679, default 2048 but 250069679 cannot be divided by 512 nor by 2048. Wouldn't it be better to set the Start and End value to a number that can be divided by 512 or 2048 or 4096?



For example: Start 4096 and End 250068992




Command (m for help): p

Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300


Command (m for help): n
Partition type
p primary (0 primary, 0 extended, 4 free)
e extended (container for logical partitions)
Select (default p): p
Partition number (1-4, default 1):
First sector (2048-250069679, default 2048):
Last sector, +sectors or +sizeK,M,G,T,P (2048-250069679, default 250069679):

Created a new partition 1 of type 'Linux' and of size 119,2 GiB.


Command (m for help): p
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux

Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63


mkfs.ext4 /dev/sda1
mke2fs 1.43.4 (31-Jan-2017)
Ein Dateisystems mit 31258454 (4k) Blöcken und 7815168 Inodes wird erzeugt.
UUID des Dateisystems: fdce9286-4545-447c-9cca-7d67f5bb9f43
Superblock-Sicherungskopien gespeichert in den Blöcken:
32768, 98304, 163840, 229376, 294912, 819200, 884736, 1605632, 2654208,
4096000, 7962624, 11239424, 20480000, 23887872


fdisk -l
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux


And how can it be that the Sectors number is lower than the End value?




Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63










share|improve this question
























  • Sectors is End minus Start. Usually for alignment, only Start matters.

    – frostschutz
    Apr 6 at 8:59












  • have you considered using a higher level (therefore easier to use) tool, such as gparted?

    – ctrl-alt-delor
    Apr 6 at 10:34











  • @ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

    – AlexOnLinux
    Apr 6 at 12:01











  • Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

    – ctrl-alt-delor
    Apr 6 at 14:58











  • @ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

    – AlexOnLinux
    Apr 7 at 6:37

















6















I am wondering what Start and End value to choose when partitioning my ext. SSD using fdisk.



fdisk suggests 2048-250069679, default 2048 but 250069679 cannot be divided by 512 nor by 2048. Wouldn't it be better to set the Start and End value to a number that can be divided by 512 or 2048 or 4096?



For example: Start 4096 and End 250068992




Command (m for help): p

Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300


Command (m for help): n
Partition type
p primary (0 primary, 0 extended, 4 free)
e extended (container for logical partitions)
Select (default p): p
Partition number (1-4, default 1):
First sector (2048-250069679, default 2048):
Last sector, +sectors or +sizeK,M,G,T,P (2048-250069679, default 250069679):

Created a new partition 1 of type 'Linux' and of size 119,2 GiB.


Command (m for help): p
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux

Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63


mkfs.ext4 /dev/sda1
mke2fs 1.43.4 (31-Jan-2017)
Ein Dateisystems mit 31258454 (4k) Blöcken und 7815168 Inodes wird erzeugt.
UUID des Dateisystems: fdce9286-4545-447c-9cca-7d67f5bb9f43
Superblock-Sicherungskopien gespeichert in den Blöcken:
32768, 98304, 163840, 229376, 294912, 819200, 884736, 1605632, 2654208,
4096000, 7962624, 11239424, 20480000, 23887872


fdisk -l
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux


And how can it be that the Sectors number is lower than the End value?




Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63










share|improve this question
























  • Sectors is End minus Start. Usually for alignment, only Start matters.

    – frostschutz
    Apr 6 at 8:59












  • have you considered using a higher level (therefore easier to use) tool, such as gparted?

    – ctrl-alt-delor
    Apr 6 at 10:34











  • @ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

    – AlexOnLinux
    Apr 6 at 12:01











  • Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

    – ctrl-alt-delor
    Apr 6 at 14:58











  • @ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

    – AlexOnLinux
    Apr 7 at 6:37













6












6








6








I am wondering what Start and End value to choose when partitioning my ext. SSD using fdisk.



fdisk suggests 2048-250069679, default 2048 but 250069679 cannot be divided by 512 nor by 2048. Wouldn't it be better to set the Start and End value to a number that can be divided by 512 or 2048 or 4096?



For example: Start 4096 and End 250068992




Command (m for help): p

Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300


Command (m for help): n
Partition type
p primary (0 primary, 0 extended, 4 free)
e extended (container for logical partitions)
Select (default p): p
Partition number (1-4, default 1):
First sector (2048-250069679, default 2048):
Last sector, +sectors or +sizeK,M,G,T,P (2048-250069679, default 250069679):

Created a new partition 1 of type 'Linux' and of size 119,2 GiB.


Command (m for help): p
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux

Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63


mkfs.ext4 /dev/sda1
mke2fs 1.43.4 (31-Jan-2017)
Ein Dateisystems mit 31258454 (4k) Blöcken und 7815168 Inodes wird erzeugt.
UUID des Dateisystems: fdce9286-4545-447c-9cca-7d67f5bb9f43
Superblock-Sicherungskopien gespeichert in den Blöcken:
32768, 98304, 163840, 229376, 294912, 819200, 884736, 1605632, 2654208,
4096000, 7962624, 11239424, 20480000, 23887872


fdisk -l
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux


And how can it be that the Sectors number is lower than the End value?




Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63










share|improve this question
















I am wondering what Start and End value to choose when partitioning my ext. SSD using fdisk.



fdisk suggests 2048-250069679, default 2048 but 250069679 cannot be divided by 512 nor by 2048. Wouldn't it be better to set the Start and End value to a number that can be divided by 512 or 2048 or 4096?



For example: Start 4096 and End 250068992




Command (m for help): p

Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300


Command (m for help): n
Partition type
p primary (0 primary, 0 extended, 4 free)
e extended (container for logical partitions)
Select (default p): p
Partition number (1-4, default 1):
First sector (2048-250069679, default 2048):
Last sector, +sectors or +sizeK,M,G,T,P (2048-250069679, default 250069679):

Created a new partition 1 of type 'Linux' and of size 119,2 GiB.


Command (m for help): p
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux

Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63


mkfs.ext4 /dev/sda1
mke2fs 1.43.4 (31-Jan-2017)
Ein Dateisystems mit 31258454 (4k) Blöcken und 7815168 Inodes wird erzeugt.
UUID des Dateisystems: fdce9286-4545-447c-9cca-7d67f5bb9f43
Superblock-Sicherungskopien gespeichert in den Blöcken:
32768, 98304, 163840, 229376, 294912, 819200, 884736, 1605632, 2654208,
4096000, 7962624, 11239424, 20480000, 23887872


fdisk -l
Disk /dev/sda: 119,2 GiB, 128035676160 bytes, 250069680 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0xa4b57300

Device Boot Start End Sectors Size Id Type
/dev/sda1 2048 250069679 250067632 119,2G 83 Linux


And how can it be that the Sectors number is lower than the End value?




Command (m for help): i
Selected partition 1
Device: /dev/sda1
Start: 2048
End: 250069679
Sectors: 250067632
Cylinders: 15566
Size: 119,2G
Id: 83
Type: Linux
Start-C/H/S: 0/32/33
End-C/H/S: 206/29/63







hard-disk fdisk external-hdd mkfs






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 7 at 9:40









ctrl-alt-delor

12.4k52662




12.4k52662










asked Apr 6 at 8:53









AlexOnLinuxAlexOnLinux

1706




1706












  • Sectors is End minus Start. Usually for alignment, only Start matters.

    – frostschutz
    Apr 6 at 8:59












  • have you considered using a higher level (therefore easier to use) tool, such as gparted?

    – ctrl-alt-delor
    Apr 6 at 10:34











  • @ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

    – AlexOnLinux
    Apr 6 at 12:01











  • Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

    – ctrl-alt-delor
    Apr 6 at 14:58











  • @ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

    – AlexOnLinux
    Apr 7 at 6:37

















  • Sectors is End minus Start. Usually for alignment, only Start matters.

    – frostschutz
    Apr 6 at 8:59












  • have you considered using a higher level (therefore easier to use) tool, such as gparted?

    – ctrl-alt-delor
    Apr 6 at 10:34











  • @ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

    – AlexOnLinux
    Apr 6 at 12:01











  • Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

    – ctrl-alt-delor
    Apr 6 at 14:58











  • @ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

    – AlexOnLinux
    Apr 7 at 6:37
















Sectors is End minus Start. Usually for alignment, only Start matters.

– frostschutz
Apr 6 at 8:59






Sectors is End minus Start. Usually for alignment, only Start matters.

– frostschutz
Apr 6 at 8:59














have you considered using a higher level (therefore easier to use) tool, such as gparted?

– ctrl-alt-delor
Apr 6 at 10:34





have you considered using a higher level (therefore easier to use) tool, such as gparted?

– ctrl-alt-delor
Apr 6 at 10:34













@ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

– AlexOnLinux
Apr 6 at 12:01





@ctrl-alt-delor normaly i use gparted, but this time i have no gui installed.

– AlexOnLinux
Apr 6 at 12:01













Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

– ctrl-alt-delor
Apr 6 at 14:58





Live OS (boot off of USB, use ssh -X and run it remotely, or parted.

– ctrl-alt-delor
Apr 6 at 14:58













@ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

– AlexOnLinux
Apr 7 at 6:37





@ctrl-alt-delor first: it's a raspberry pi using raspbian lite without gui. second: whats the problem of understanding how to properly align third: did i use the tools fdisk and mkfs.ext4 incorrect?

– AlexOnLinux
Apr 7 at 6:37










1 Answer
1






active

oldest

votes


















5














Alignment doesn’t matter for the end sector. Sectors are numbered from 0; fdisk is suggesting the last sector on your disk (which has 250069680 sectors).



 Start: 2048
End: 250069679
Sectors: 250067632


is correct, 250069679 minus 2048 plus one is 250067632: the partition contains 250067632 sectors, starting at offset 2048.






share|improve this answer

























  • i am wondering why the end-value is not important. do you know why perhaps?

    – AlexOnLinux
    Apr 6 at 12:01











  • @AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

    – icarus
    Apr 6 at 17:35











  • @icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

    – AlexOnLinux
    Apr 7 at 6:41











  • @AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

    – Stephen Kitt
    Apr 7 at 6:56











  • Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

    – ctrl-alt-delor
    Apr 7 at 9:44











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














Alignment doesn’t matter for the end sector. Sectors are numbered from 0; fdisk is suggesting the last sector on your disk (which has 250069680 sectors).



 Start: 2048
End: 250069679
Sectors: 250067632


is correct, 250069679 minus 2048 plus one is 250067632: the partition contains 250067632 sectors, starting at offset 2048.






share|improve this answer

























  • i am wondering why the end-value is not important. do you know why perhaps?

    – AlexOnLinux
    Apr 6 at 12:01











  • @AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

    – icarus
    Apr 6 at 17:35











  • @icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

    – AlexOnLinux
    Apr 7 at 6:41











  • @AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

    – Stephen Kitt
    Apr 7 at 6:56











  • Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

    – ctrl-alt-delor
    Apr 7 at 9:44















5














Alignment doesn’t matter for the end sector. Sectors are numbered from 0; fdisk is suggesting the last sector on your disk (which has 250069680 sectors).



 Start: 2048
End: 250069679
Sectors: 250067632


is correct, 250069679 minus 2048 plus one is 250067632: the partition contains 250067632 sectors, starting at offset 2048.






share|improve this answer

























  • i am wondering why the end-value is not important. do you know why perhaps?

    – AlexOnLinux
    Apr 6 at 12:01











  • @AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

    – icarus
    Apr 6 at 17:35











  • @icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

    – AlexOnLinux
    Apr 7 at 6:41











  • @AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

    – Stephen Kitt
    Apr 7 at 6:56











  • Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

    – ctrl-alt-delor
    Apr 7 at 9:44













5












5








5







Alignment doesn’t matter for the end sector. Sectors are numbered from 0; fdisk is suggesting the last sector on your disk (which has 250069680 sectors).



 Start: 2048
End: 250069679
Sectors: 250067632


is correct, 250069679 minus 2048 plus one is 250067632: the partition contains 250067632 sectors, starting at offset 2048.






share|improve this answer















Alignment doesn’t matter for the end sector. Sectors are numbered from 0; fdisk is suggesting the last sector on your disk (which has 250069680 sectors).



 Start: 2048
End: 250069679
Sectors: 250067632


is correct, 250069679 minus 2048 plus one is 250067632: the partition contains 250067632 sectors, starting at offset 2048.







share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 6 at 10:05

























answered Apr 6 at 9:21









Stephen KittStephen Kitt

181k25414493




181k25414493












  • i am wondering why the end-value is not important. do you know why perhaps?

    – AlexOnLinux
    Apr 6 at 12:01











  • @AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

    – icarus
    Apr 6 at 17:35











  • @icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

    – AlexOnLinux
    Apr 7 at 6:41











  • @AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

    – Stephen Kitt
    Apr 7 at 6:56











  • Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

    – ctrl-alt-delor
    Apr 7 at 9:44

















  • i am wondering why the end-value is not important. do you know why perhaps?

    – AlexOnLinux
    Apr 6 at 12:01











  • @AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

    – icarus
    Apr 6 at 17:35











  • @icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

    – AlexOnLinux
    Apr 7 at 6:41











  • @AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

    – Stephen Kitt
    Apr 7 at 6:56











  • Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

    – ctrl-alt-delor
    Apr 7 at 9:44
















i am wondering why the end-value is not important. do you know why perhaps?

– AlexOnLinux
Apr 6 at 12:01





i am wondering why the end-value is not important. do you know why perhaps?

– AlexOnLinux
Apr 6 at 12:01













@AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

– icarus
Apr 6 at 17:35





@AlexOnLinux your choices are either to use all the sectors available or not. If you want them all and the disk has a size which is not a multiple of 512/2048/4096 then the end will not be aligned.

– icarus
Apr 6 at 17:35













@icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

– AlexOnLinux
Apr 7 at 6:41





@icarus Having the Start and End aligned gives any performance advantages? Is it usefull to properly align the End-value when using multiple partitions?

– AlexOnLinux
Apr 7 at 6:41













@AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

– Stephen Kitt
Apr 7 at 6:56





@AlexOnLinux all partitions should have their start aligned for performance (and wear-and-tear on SSDs, although that’s less of an issue than people make it out to be). If you create partitions with GPT and no space between them, the end of each partition will be aligned, apart from the last one in some cases (as in your situation).

– Stephen Kitt
Apr 7 at 6:56













Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

– ctrl-alt-delor
Apr 7 at 9:44





Miss alignment, of the start, will affect the whole partition, but of the end only affects the end and the next partition.

– ctrl-alt-delor
Apr 7 at 9:44

















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