How to calculate partition Start End Sector?





.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 devided 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 +size{K,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
    19 hours ago













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

    – ctrl-alt-delor
    17 hours ago











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

    – AlexOnLinux
    16 hours ago











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

    – ctrl-alt-delor
    13 hours ago


















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 devided 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 +size{K,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
    19 hours ago













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

    – ctrl-alt-delor
    17 hours ago











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

    – AlexOnLinux
    16 hours ago











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

    – ctrl-alt-delor
    13 hours ago














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 devided 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 +size{K,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 devided 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 +size{K,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










asked 19 hours ago









AlexOnLinuxAlexOnLinux

1686




1686













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

    – frostschutz
    19 hours ago













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

    – ctrl-alt-delor
    17 hours ago











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

    – AlexOnLinux
    16 hours ago











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

    – ctrl-alt-delor
    13 hours ago



















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

    – frostschutz
    19 hours ago













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

    – ctrl-alt-delor
    17 hours ago











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

    – AlexOnLinux
    16 hours ago











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

    – ctrl-alt-delor
    13 hours ago

















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

– frostschutz
19 hours ago







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

– frostschutz
19 hours ago















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

– ctrl-alt-delor
17 hours ago





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

– ctrl-alt-delor
17 hours ago













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

– AlexOnLinux
16 hours ago





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

– AlexOnLinux
16 hours ago













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

– ctrl-alt-delor
13 hours ago





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

– ctrl-alt-delor
13 hours ago










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
    16 hours ago











  • @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
    10 hours ago












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "106"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f510868%2fhow-to-calculate-partition-start-end-sector%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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
    16 hours ago











  • @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
    10 hours ago
















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
    16 hours ago











  • @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
    10 hours ago














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 18 hours ago

























answered 18 hours ago









Stephen KittStephen Kitt

180k25408487




180k25408487













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

    – AlexOnLinux
    16 hours ago











  • @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
    10 hours ago



















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

    – AlexOnLinux
    16 hours ago











  • @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
    10 hours ago

















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

– AlexOnLinux
16 hours ago





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

– AlexOnLinux
16 hours ago













@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
10 hours ago





@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
10 hours ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Unix & Linux Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f510868%2fhow-to-calculate-partition-start-end-sector%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Statuo de Libereco

Tanganjiko

Liste der Baudenkmäler in Enneberg