I am operating on a Windows 7 OS. I performed a chkdsk and i know that the cluster space/bytes per allocation unit is 4096 bytes. I scanned my hard disk using a JDiskReport and the report is in the image
Distribution of sizes in C:\
Size Interval SumofFileSizes(KB) %ofTotal Files % of Files
Over 16 GB 0 0.0% 0 0.0%
4 GB – 16 GB 4,256,564 4.2% 1 0.0%
1 GB – 4 GB 16,592,054 16.4% 8 0.0%
256 MB – 1 GB 17,179,989 17.0% 23 0.0%
64 MB – 256 MB 18,418,314 18.2% 188 0.0%
16 MB – 64 MB 7,141,803 7.1% 231 0.1%
4 MB – 16 MB 11,427,285 11.3% 1,514 0.4%
1 MB – 4 MB 13,756,667 13.6% 6,482 1.6%
256 KB – 1 MB 5,891,778 5.8% 11,619 2.8%
64 KB – 256 KB 3,558,129 3.5% 29,668 7.1%
16 KB – 64 KB 1,764,479 1.7% 51,534 12.4%
4 KB – 16 KB 670,344 0.7% 80,269 19.3%
1 KB – 4 KB 220,179 0.2% 104,563 25.2%
0 KB – 1 KB 60,361 0.1% 129,148 31.1%
if you look at the last row for the file sizes of 1 kb and for 34,255 files the amount of space occupied in the hard disk will be 129148*4096=504 MB because my cluster size is 4096 bytes.
So I was wondering how my interpretation should change for file sizes beyond 4 KB ? and is there a way for me to find out the amount of space occupied for each of the different divisions mentioned below.
Edit 1:
Please point me to the correct forum that i should post this message on within the stack exchange group, if this is not the right place