Raw capacity for the stated one-petabyte erasure-code placement case
about 1.8 PB rawderived
| Value kind | derived — Derived values are guide calculations; their result is only as fixed as the stated inputs, scope, and method. |
|---|---|
| Scope | Worked-example value, not a measurement: 1.0 PB of logical data under an (8,3) erasure code (11/8 = 1.375x) with 25% of raw capacity held free. Triple replication at the same reserve needs 4.0 PB raw. Rerun it with your own code, logical size and reserve. |
| As of | 2026-09 |
| Source | Worked example in Chapter 9.6 (erasure-code placement case; assumptions stated in the chapter) |
| Derivation | 1.0 PB × (8+3)/8 / (1−.25) = 1.8333333333333333 PB raw; select the next eligible hardware configuration. |
| Review | checking…review by 2027-03-05 · standard cadence |
| Recorded changes | last 2026-10-19 · 2 revisions tracked |
| Claim id | guide2-t23-object-raw-capacity |
Where the guide uses it
← Full numbers register — every date-stamped figure in the guide, with revision history.