> This will take a long time, since you will be writing to basically this whole drive as part of distributing its spare space. (Plus probably shuffling chunks a

> This will take a long time, since you will be writing to basically this whole drive as part of distributing its spare space. (Plus probably shuffling chunks around between many other drives) Yes. With the caveat that you'll be *reading* from all of the disks to satisfy those writes. In a traditional setup, you'd have multiple striped raidz{,2,3} vdevs. When you swap in a spare, you're only pulling reads from the disks in the same vdev. This will drive the performance from this vdev to the ground. In this draidz setup, you'll be pulling from all of the disks, so the entire pool is stressed at the same rate. I think the way I'm starting to think about it is that each drive has X amount of spare capacity or buckets. But they are otherwise used. When a drive goes out, the primary data from that drive is distributed across the array to the already-in-place spare capacity. So, yes, after the primary data is mirrored out, the pool is no longer degraded. But, I think the point being made here is that you'll still want to replace the drive, which will still cause *that* drive to be fully written.