ESX disk geometry math

[JamesConaway]

I am trying to understand the math in how the VMFS LUN's are structured.

If I do a fdisk -lu and get the following output:

Disk /dev/sdg: 536.9 GB, 536952700928 bytes

255 heads, 63 sectors/track, 65280 cylinders, total 1048735744 sectors

Units = sectors of 1 * 512 = 512 bytes

Device Boot Start End Blocks Id System

/dev/sdg1 128 1048723199 524361536 fb VMware VMFS

The math adds up when it comes to calculating the total disk size based on 1048735744 sectors output.

1048735744 * 512 = 536952700928 bytes. 536952700928 / 1024 /1024/1024 = 500gb LUN.

It states the total sectors as 1048735744.

I calculate the total sectors as heads * sectors * cylinders = 1048723200

Which is one more than the ending sector size.

Where is the difference between the 1048735744 and the 1048723200 be used? and how is the 1048735744 being calculated?

[VTsukanov · Accepted solution]

Since INT 0x13 interface supports drives of sizes up to about 528 MB (Physical CHS addressing), then to support larger disks was introduced in INT13h EXT, it uses 64-bit logical block addressing (LBA). The CHS Value in LBA mode are calculated. Take a look at LBA to CHS Translation

In your case

CYL = Trunc (1048735744 / (255 * 63)) = 65280