Solved Installation for Advanced Format Hard Drives?

[COMPUTIAC]

Mine shows 102MB is this incorrect ?

 

[whs]

Mine shows 102MB is this incorrect ?

No problem with this disk. It is well aligned.

 

[COMPUTIAC]

Thank's for the quick reply.

Why is it not 101MB ?

 

[whs]

Thank's for the quick reply.

Why is it not 101MB ?

That is a good question. But it does not matter. The disk is fine.

 

[AddRAM]

It has to be an even number, right Wolfgang ?

 

[whs]

It has to be an even number, right Wolfgang ?

It has to be a number divisible by 4 at the KB level. So 1024KB is good but 102MB is also good because that is 102.000KB which is divisible by 4.

 

[King Arthur]

Assuming that 1MB == 1024KB, 102MB == 104,448KB which is still divisible by 4.

 

[whs]

Assuming that 1MB == 1024KB, 102MB == 104,448KB which is still divisible by 4.

Thanks, that looks like the more exact conversion in the binary system.

 

[lehnerus2000]

I'm still confused

Thanks whs, but I'm still confused.

Verification

If you want to verify the alignment (e.g. for a SSD where you are not certain whether the proper alignment was done), you use the following commands.

Diskpart
List disk
Select disk n
List partition

Now you should see a result like this.

Partition ### Type Size Offset
------------- ---------------- ------- -------
Partition 1 Primary 59 GB 1024 KB - but 64KB or any number divisible by 4 is also good

The offset (in KBs) has to be divisible by 4.


Note: Some readers and users of this tutorial got confused because the alignment numbers in a typical Windows7 installation are shown as:

1024KB for the 100MB partition
101MB for the next partition - which is most likely the C partition

They think that 101MB is not divisible by 4 and that there must be a problem. But that is not so. If you convert 101MBs into KBs (multiply by 1024), then the number is divisible by 4 and the partition is aligned.

By definition any integer multiplied by 1024 will be divisible by 4 (e.g. n x 1024 = n x 256 x 4).

Are you saying that the minimum offset must be 64 KB or more?
As a result, does this mean that I can't clone my OS partitions directly to a SSD (as the offset on Disk 0 is 31 KB)?

​

I always pre-partition my HDDs (using GParted) before installing any operating systems.
Is the end result, modern formatting tools (e.g. GParted) automatically create the correct format (which they should) without any input required from the user?

 

[whs]

1. On your cmd picture, disk0 and disk1 are not aligned. Disk2 is aligned.

2. I am not sure whether all imaging programs will maintain the original alignment. Macrium does if you recover the image a certain way. See here under Advanced Topics > Partition Alignment

 

[lehnerus2000]

Thanks whs

1. On your cmd picture, disk0 and disk1 are not aligned. Disk2 is aligned.

2. I am not sure whether all imaging programs will maintain the original alignment. Macrium does if you recover the image a certain way. See here under Advanced Topics > Partition Alignment

Oops, I edited my post before I saw your reply (the page hadn't been refreshed).

Thanks for that Macrium link.
It seems that as long as you use "Drag and Drop", the target disc alignment will be maintained.
It's a pity there is no indication of that fact in the GUI.

It makes sense that disk0 and disk1 are not aligned, as they pre-date "Advanced Format" (i.e. the 2 TB HDD was the first one that mentioned it).

So the minimum offset must be 64 KB?

 

[King Arthur]

Assuming that 1MB == 1024KB, 102MB == 104,448KB which is still divisible by 4.

Thanks, that looks like the more exact conversion in the binary system.

It's all a confusing mess, really, since both numbers are in fact correct. I've always gone with the 1MB == 1024KB rule which Windows has always used, but some other people and software sometimes insist on calculating 1MB as 1000KB. It's not a big deal usually, but it becomes a headache when precise calculations need to come into play. :banghead:

 

[whs]

It seems that as long as you use "Drag and Drop", the target disc alignment will be maintained

So the minimum offset must be 64 KB?

Both statements are correct. Except it works only on disks, not on discs, LOL

 

[lehnerus2000]

Thanks again

It seems that as long as you use "Drag and Drop", the target disc alignment will be maintained

So the minimum offset must be 64 KB?

Both statements are correct. Except it works only on disks, not on discs, LOL

Thanks for the clarifications whs.

Unfortunately, I'm not currently allowed to Rep you.

BTW, in Australia, disc = disk.

 

[whs]

BTW, in Australia, disc = disk

Hmm, the way I learned it is a disc is e.g. a CD and a disk is the storage device in the computer. In French it is even more confusing because a disque can be both but also a music record (e.g. vinyl)

 

[Lady Fitzgerald]

BTW, in Australia, disc = disk

Hmm, the way I learned it is a disc is e.g. a CD and a disk is the storage device in the computer. In French it is even more confusing because a disque can be both but also a music record (e.g. vinyl)

In the U.S., the rules get tossed because no one seems to care. It also depends where in the U.S. you are.