Disk scan for deleted entries

Cluster chain recovery for a deleted entry

After a cluster chain is defined, either automatically or manually, the only task left is to read and save the contents of the defined clusters to another place for verifying their contents.

Having a chain of clusters; each cluster offset from the beginning of the drive can be calculate using standard formulas.

After that, the next step is to copy the amount of data equal to the cluster size, starting from the calculated offset into a newly created file.

Not all clusters are copied but only from the file size minus the number of copied clusters multiplied by cluster size.

Formulas for calculating a cluster offset could vary depending on file system.

Here is the information required to calculate the cluster offset for a FAT volume:

• Boot sector size
• Number of FAT supported copies
• Size of one copy of FAT
• Size of main root folder
• Number of sectors per cluster
• Number of bytes per sector

NTFS volumes use space linearly so to calculate a cluster's offset simply as the multiply the cluster number by the cluster size.

Example of recovering cluster chains on FAT16

In continuing with a previous example, here is the deleted file MyFile.txt again.

By now, its been determined that a chain of clusters 3, 4, 5, 6 are ready for recovery. The cluster consists of 64 sectors, sector size of 512 bytes, so the cluster size is: 64*512 = 32,768 bytes = 32 Kb. The first data sector is 535 (there is 1 boot sector, plus 2 copies of FAT by 251 sectors each, plus a root folder of 32 sectors, the total 534 occupied by system data sectors).

Clusters 0 and 1 do not exist, so the first data cluster is 2. Cluster number 3 is next to cluster 2, i.e. is located 64 sectors behind the first data sector (535). i.e. 535 + 64 = 599 sector, equal offset of 306,668 bytes from the beginning of the drive (0x4AE00).

With the help of a low-level disk editor to view the disk, here is the data starting with offset 0x4AE00, or 3 cluster, or 599 sector:

   Offset      0  1  2  3  4  5  6  7   8  9  A  B  C  D  E  F
   0004AE00   47 55 49 20 6D 6F 64 65  20 53 65 74 75 70 20 68   GUI mode Setup h
   0004AE10   61 73 20 73 74 61 72 74  65 64 2E 0D 0A 43 3A 5C   as started...C:\
   0004AE20   57 49 4E 4E 54 5C 44 72  69 76 65 72 20 43 61 63   WINNT\Driver Cac

Because the cluster chain is consecutive, all that needs to be done is to copy the 112,435 bytes of data starting from this point forward. If the cluster chain were not consecutive, the offset of each cluster found would be re-calculate. 3 times by 64*512 = 32768 bytes starting from each cluster offset, and then from the last cluster copy reminder: 14,131 bytes that is calculated as 112,435 bytes - (3 * 32768 bytes).

Example of recovery cluster chains on NTFS

In this example 110 clusters need to be copied starting from the cluster 312555.

The Cluster size is 512 byte, so the offset of the first cluster would be 512 * 312555 = 160028160 = 0x0989D600

Offset      0  1  2  3  4  5  6  7   8  9  A  B  C  D  E  F
0989D600   D0 CF 11 E0 A1 B1 1A E1  00 00 00 00 00 00 00 00   ÐÏ.ࡱ.á........
0989D610   00 00 00 00 00 00 00 00  3E 00 03 00 FE FF 09 00   ........>...þÿ..
0989D620   06 00 00 00 00 00 00 00  00 00 00 00 01 00 00 00   ................
0989D630   69 00 00 00 00 00 00 00  00 10 00 00 6B 00 00 00   i...........k...
0989D640   01 00 00 00 FE FF FF FF  00 00 00 00 6A 00 00 00   ....þÿÿÿ....j...
0989D650   FF FF FF FF FF FF FF FF  FF FF FF FF FF FF FF FF   ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ

Here is the data. What's left is just reading 110 clusters (56320 bytes) from this point and then copying them to another location. Data recovery is now complete.