how can i be sure my ssd are perfectly alignment ?

hi
today i did a great mistake , i used Paragon Alignment Tool 4.0 Professional (linux based ) , to fix an old hard disk (no ssd)

i boot it via Paragon Alignment Tool 4.0 Professional (linux based) i wait for a while ,without have the possibility to select any option and after 6 or 8 munutes i had only the option to reboot shut down or save the log , i saved the log is incomprensible , paragon support is sleeping by the way

well i used BootIt BM , and i checked if my ssd are perfectly alignment

with bimb i selected the first partition of my ssd , properties and starting lba is 2048 and the cluster size is 4096 bytes

if i'm right i should divide them to obtain 0 , 0 should be perfectly alignment , Am I right?

are my ssd perfectly alignment ?

thanks

 

I use this online calculator to make sure my partitions are aligned: https://www.techpowerup.com/articles/other/157

 

hi thanks Minimalist
my operation system is not in english
look they have the same names
according to the first should be ok


Great link!
http://i.imgur.com/VOCqMia.png

 

Crystal diskmark ? benchmarking program has the alignment analysed for HDs and SSDs

 

They have little to do with each other. The 4096 is the NTFS DEFAULT Cluster size and the 2048 is the starting disk sector address for the 1st partition on a SSD.

The important thing for a SSD is that the initial partition start on a boundary divisible by 1mB. Since a disk sector for most disks is 512-bytes, a starting sector of 2048 (2048 x 512 bytes = 1mB) is perfect for a SSD.

 

If you desire cylinder alignment you want the start LBA divided by 16065 to be a integer. An integer is a whole number such as 20001. 20001.1 is not an integer. 16065= 255*63 (2048 255 63 CHS geometry)

If you desire 1 MiB alignment on a drive with 512 byte sectors you want the start LBA divided by 2048 to be a integer. There are 2048 sectors in a MiB.

If you desire 1 MiB alignment on a drive with 4096 byte sectors (4K native) you want the start LBA divided by 512 to be a integer. There are 512 sectors in a MiB.

As TRF has pointed out, this has nothing to do with 4096 byte clusters which is the default NTFS cluster size for any sized drive we will be using.

 

hi TheRollbackFrog
so my ssd are perfect seeng my result , aren't they?

Brian
i did not understand you
according to bibm the cluster size is 4096byte , my drives are 512-byte as the logical and physical sector size.
so ? what's should i do?
thanks Brian

 

hi
where is it ?
thanks trott3r

 

You should forget the 4096 cluster size number as it has nothing to do with alignment.
Which partition alignment do you desire? Cylinder or 1 MiB? Either are OK for you. Is the OS WinXP?

 

hi
i have 3 ssd , on each ssd i have a operation system
on the first ssd i have windows 10 64bit pro , on the second windows 7 64bit and on the third windows 8.1 pro 64bit
i want them alignment
thanks

 

For SSDs you need 1 MiB alignment. What does BIBM say about the Start LBA for each OS?

 

hi
i selected on each drives ,propriertie and is 2048

 

Then 2048/2048=1
An integer, so the partitions are 1 MiB aligned.

Alignment refers to the partitions, not the SSD or HD.

 

hi
so they are perfectly are not they?
bibm has a feature align ,what is it?
thanks Brian

 

That Align feature has nothing to do with partition alignment. See page 44 in the User Guide.

Let's say a partition had a start LBA of 3000...

3000/2048= 1.464 which is not an integer so the partition is not 1 MiB aligned. To correct this with BIBM...

Make sure you have these selections in BIBM Settings
tick in Align on 1 MiB Boundaries
no tick in Align on Cylinder

In Partition Work, select the partition
click Resize and make it 3 MiB smaller (this is fast)
you should see 3 MiB of Free Space below the partition
select the partition and click Slide. Accept Free Space Before= 0 (the slide could take much longer than a few minutes as all the "sectors in use" in the partition have to be moved)
when the slide has completed select the partition and click Resize. Make it Max Size (this is fast)

Click View MBR. Or Partition Properties. The start LBA divided by 2048 should be an integer.

 

thanks Brian but about my drives are aligned ?

 

mantra,

Rather than saying Yes or No, let me restate the method. If the start LBA/2048 is an integer then the partition is 1 MiB aligned. Are you having trouble with the definition of integer?

 

Brian
barely i understand english ,and google translator doesn't help me , i don't know what is it an integer

according to terabyte

but i did not understand it

 

OK. An integer is a whole number. Such as 1, 2, 23, 45624587, 122554788899655551

1 MiB contains 2048 sectors so to convert sectors to MiB you divide by 2048. Say your start LBA was 129,552,384 then
129,552,384/2048 = 63258

63258 is an integer so this partition is 1 MiB aligned. There is exactly 63258 MiB from the start of the drive to the start of the partition. The partition is aligned on 1 MiB blocks.

What TeraByte Support mean is if the start LBA/2048 was to equal 256.233 then you would call the remainder 0.233 which is not zero. In other words 256.233 is not an integer.

 

hi
almost clear , Brian , i should be ok , i guess

 

mantra, I'm sure you will be OK. Just remember start LBA divided by 2048 will be an integer if the partition is 1 MiB aligned.

 

hi
but bibm start lba is 2048 , divided by 2048 is 0 , where can i get Partition Offset:1048576 in bibim ?
i guess that integer are 1,2,3 ,100 ,23333
not integer are 0.4 234.3 right ?

thanks

 

Correct.

2048 divided by 2048 is 1, not 0

That was just an example used by TeraByte Support. You won't find that number in BIBM.

 

In your case the partition offset is 1 MiB which is equal to 2048 sectors which is equal to 1048576 bytes.

 

thanks Brian
1048576 i can find only in msinfo32