[Solved] WD 750 re3 raid edition missing 50 gb

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