Calculate usable RAID capacity, parity or mirror overhead and how many drive failures RAID 0, 1, 5, 6 and 10 survive, in TB and the TiB Windows shows.
By AdminDrafted with AI writing toolsUpdated Checked examples: 6
RAID 5 on 4 drives of 8 TB gives 24 TB usable (21.83 TiB in the binary units Windows shows), 75% of the raw space, and always survives 1 failed drive.
Drive labels use 1 TB = 10¹² bytes. Windows divides by 2⁴⁰ but still prints “TB”, so a new drive looks about 9% smaller there; macOS reports decimal TB.
One drive's worth of parity is spread across all drives. Every member counts as the smallest drive in the set.
1 drive
In the formula, C is usable capacity, s is the smallest drive size and n is the number of drives. Usable capacity is the drive size times the number of drives' worth of space that holds data: n for RAID 0 striping, 1 for a RAID 1 mirror, n − 1 for RAID 5 single parity, n − 2 for RAID 6 double parity and n ÷ 2 for RAID 10 striped mirrors. The rest of the raw space holds copies or parity, which is what lets the array keep running after a drive fails.
People use it to size a NAS or server before buying drives. The default, four 8 TB drives in RAID 5, gives 24 TB usable out of 32 TB raw, 75% efficiency, and survives one failed drive. In Windows the same space appears as 21.83 TB, because Windows divides by 2^40 (the binary TiB) but prints "TB".
The results assume identical drives; with mixed sizes, each drive counts as the smallest one. File-system metadata, hot spares and reserved space are not subtracted, so the room left for files is slightly lower.
Multiply the smallest drive's size by the drives' worth of space that holds data: n for RAID 0, 1 for RAID 1, n − 1 for RAID 5, n − 2 for RAID 6 and n ÷ 2 for RAID 10. Four 8 TB drives give 32, 8, 24, 16 and 16 TB; the rest of their 32 TB holds parity or copies.
With n drives of size s:
Redundancy also costs writes. A small random write takes 1 disk operation on RAID 0, 2 on RAID 10, 4 on RAID 5 and 6 on RAID 6, while RAID 1 writes once to every member. SNIA's Data Protection Best Practices describes the levels.
Raw capacity is 4 × 8 = 32 TB. One drive's worth holds distributed parity, so usable space is 3 × 8 = 24 TB, an efficiency of 24 ÷ 32 = 75%. The array keeps running after any single drive failure.
Windows reports that 24 TB as about 21.8 TB, because it divides by 2⁴⁰ bytes (a tebibyte, TiB) but still prints "TB": 24 × 10¹² ÷ 2⁴⁰ = 21.83. No space is missing. NIST's binary-prefix page defines both units.
| Scenario | Usable TB | Usable TiB | Failures always survived |
|---|---|---|---|
| Default | 24 | 21.83 | 1 |
| RAID 1 on 3 | 8 | 7.28 | 2 |
| RAID 6 on 6 | 32 | 29.10 | 2 |
| RAID 10 on 8 | 32 | 29.10 | 1 |
| Drive unit TiB | 3.298534883328 | 3.00 | 1 |
The first four rows use 8 TB drives: four in RAID 5, three in one mirror, six in RAID 6 and eight in RAID 10. The last row is four 1 TiB drives in RAID 5, which gives exactly 3 TiB, or 3.298534883328 TB.
RAID 6 on six drives and RAID 10 on eight both give 32 TB, but RAID 6 survives any two failures, while RAID 10 is only guaranteed to survive one.
To estimate how long copying the data will take, use the bandwidth and transfer calculator.
Checked against: (4 − 1) × 8 = 24 TB
Checked against: (6 − 2) × 4 = 16; 16/24 = 66.67%
Checked against: 4 / 2 × 2 = 4 TB; with 2 mirror pairs one failure per pair is survivable, but two in the same pair are not
Checked against: Definition: every member holds a full copy
RAID 5 gives (n − 1) × drive size, because one drive's worth of space holds parity. Four 8 TB drives give 24 TB (75%) and eight give 56 TB (87.5%). It needs at least 3 drives and survives exactly one failure. Efficiency rises as you add drives, but so does the amount of data a rebuild must read without hitting a second failure.
RAID 6 stores two independent parities (P and Q) instead of one, so it survives any 2 failed drives at the cost of a second drive's capacity. With six 4 TB drives, RAID 5 gives 20 TB and RAID 6 gives 16 TB. With read-modify-write, a small random write costs 4 disk I/Os on RAID 5 and 6 on RAID 6. Actual throughput depends on the controller, cache and workload.
RAID 10 can benefit small random writes and rebuilds, while RAID 5 stores more; actual performance depends on the workload and controller. RAID 10 always keeps 50% of raw space, so 4 × 8 TB gives 16 TB against 24 TB for RAID 5. A small random write costs 2 disk I/Os instead of 4, and a rebuild copies one mirror partner instead of reading every drive. RAID 10 survives 1 failure for certain and up to half its drives if each is in a different pair.
Drive makers use decimal units, 1 TB = 10^12 bytes, while Windows divides by 2^40 = 1,099,511,627,776 bytes and still labels the result TB. 8 × 10^12 ÷ 2^40 = 7.28, so no space is missing; the gap is about 9% at terabyte scale. IEC 80000-13 names the binary unit the tebibyte (TiB). Macs have reported decimal units since OS X Snow Leopard.
No. RAID keeps an array running through drive failures, but deletions, ransomware encryption and file-system corruption reach every drive at once, and RAID 0 has no redundancy at all. Keep separate backups: CISA recommends the 3-2-1 rule, which means 3 copies of important files, on 2 different media types, with 1 copy stored off-site.
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