WD 750 re3 raid edition missing 50 gb

dadiobob

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I purchased 2 wd 750gb raid edition hard drives to run in a raid 0 configuration. In the BIOS they show up as having 698.6gb of unassigned storage capacity per drive. The drives are brand new with nothing on them. I realize I'm not going to get exactly 750gb of storage but a 50gb difference is extreme. Is there something wrong? am I missing something or am I just being a total noob.
This is my first build so I don't have any experience dealing with new hdd's. the motherboard is a EVGA 141-BL-E757-TR X58 northbridge and ICH10R southbridge. It has a Pheonix Award bios I'm not sure what revision level it's at (could not find where that is identified). I've spent several hours seaching thru the forums trying to find a post with a similar issue but could not find anything.

If there is anyone out there who has experience with this issue and is willing to take the time out of their busy lives to post a reply it would be greatly appreciated. THANK YOU in advance for your reply.

General note: To all of the people who post replies to toms hardware. The knowledge base on this forum is truly amazing. Thank you all for your past and future posts.
 
Solution
If you right-click on a drive letter in that RAID 0,
then Properties, you'll see two numbers.

For example, we also recently installed a pair of WD 750GB RE3 HDDs,
and formatted the C: partition at 50GB -- to host Windows XP/Pro SP3.

Here's what we get from "Properties", using the
sequence above:

Capacity: 52,427,898,880 bytes 48.8 GB <--- NOTE HERE


Here's another exercise that will illustrate these
different ways of displaying bytes in decimal digits:

1K = 1024 (a little more than one thousand)
2K = 2048 (more than two thousand)
4K = 4096 (almost four thousand one hundred)
and so on.

Now, just use CALCULATOR to keep multiplying 2 by itself e.g.:

2 x 2 x 2 = 8
2 x 2 x 2 x 2 = 16
2 x 2 x 2 x 2 x 2 = 32

If you keep...
You're seeing the difference between the capacity as measured in decimal GB (1024^3) and binary GB (1000^3). Drive manufacturers quote the decimal figure because it results in a bigger number. Most computer utilities use binary GB for reasons that are beyond me...

It's the same capacity, just shown in different ways.
 

dadiobob

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Thank you sminlal for such a quick reply. I thought I was not understanding what I was reading. You have cleared up my confusion. And thanks for all of your previous replies to countless numbers of posts I have learned more from you than you will ever know. Thanks
 

MRFS

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If you right-click on a drive letter in that RAID 0,
then Properties, you'll see two numbers.

For example, we also recently installed a pair of WD 750GB RE3 HDDs,
and formatted the C: partition at 50GB -- to host Windows XP/Pro SP3.

Here's what we get from "Properties", using the
sequence above:

Capacity: 52,427,898,880 bytes 48.8 GB <--- NOTE HERE


Here's another exercise that will illustrate these
different ways of displaying bytes in decimal digits:

1K = 1024 (a little more than one thousand)
2K = 2048 (more than two thousand)
4K = 4096 (almost four thousand one hundred)
and so on.

Now, just use CALCULATOR to keep multiplying 2 by itself e.g.:

2 x 2 x 2 = 8
2 x 2 x 2 x 2 = 16
2 x 2 x 2 x 2 x 2 = 32

If you keep doing this, when you get into the really large
numbers, you'll see that "64 Megabytes" is not 64,000,000,
it's 67,108,864.

Thus, terms like "32 MB" or "8 GB" are industry conventions
that don't accurately express the real decimal number in question.

If you want a fast way to get the exact number,
change CALCULATOR View to Scientific, then
use x^y e.g. 2^3 is 2 cubed; 2^8 is 2 to the power of 8;
and so on.

Thus, if you want to know the real maximum number of bytes
that Windows XP 32-bit can address, calculate 2^32 = "4GB"
(actually 4,294,967,296 :)

And, the way XP sees this is by using binary arithmetic
also known as "base 2 representation":

00000000000000000000000000000000 = first byte address
00000000000000000000000000000001 = second byte address
00000000000000000000000000000010 = third byte address
00000000000000000000000000000011 = fourth byte address
...
11111111111111111111111111111111 = "4G" address (2^32-1)

This sequence produces 2^32 unique binary digits, or permutations
just as 2^2 produces four unique binary digits:

00 = zero in decimal or "base 10"
01 = one in decimal i.e. 0 x 2^1 + 1 x 2^0 (base 2)
10 = two in decimal i.e. 1 x 2^1 + 0 x 2^0 (base 2)
11 = three in decimal i.e. 1 x 2^1 + 1 x 2^0 (base 2)


MRFS
 
Solution

baddad

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You have it backwards.