Thumb rule for disks IOps

Since there were no takers for this question, I'm rephrasing my question.

I would like to calculate the disk IOps and plan for number of disks required based inputs from application IOps requirement who is looking for disk space from VNX. While calculating is there any requirement to take into account IO size (blocks)?

I understand that block size would have no impact on throughput (IOps) but would certainly need if we are concerned about bandwidth (MBps).

Re-posting the question in a hope to get answer or discussion on this topic

I was going through a excellent discussion "Ask the Expert: Performance Calculations on Clariion/VNX" and the first post by Rob saying:

.......... The magic word in performance discussions is "IOps". 1 IOps is 1 "input or output operation per second". A rule of thumb EMC uses is that a single disk rotating at 15k RPM can handle 180 IOps, a 10k disk can handle about 140 IOps, a 7200 RPM disk can do 80 and a power efficient 5400 RPM drive can only do 40 IOps. An EFD (flash drive or SSD) can handle 2500 IOps. As said, this is a rule of thumb based on small random (4kB) blocks. When the blocksize increases the number of IOps will go down, but the amount of MBps goes up.............

The discussion is closed long so i cannot ask a question directly in that discussion. I was wondering about that IOps thumbrule. According to this post the IOps thumbrule it is based on small random (4kb) blocks... so if i'm calculating IOps for following applications should i consider the disk IOps as follows:

15k RPM disk

4kb block (operating system) - 180 IOps

8kb block (database / mail) - 90 IOps

64kb block (file server / webserver / DB log) - 11.25 IOps

Or i should use the flat rate of 180 IOps for all cases?