ATA/RAID:
benchmark results
Figure 1.
Figure 1. Maxtor drive / internal Apple ATA66 bus / slave / Apple ATA driver 3.2.6
Figure 2.
Figure 2. Maxtor drive / internal Apple ATA66 bus / slave / HDST ATA driver 3.2.1
Figure 3.
Figure 3. Maxtor drive / Sonnet ATA bus / master / Apple SCSI driver 8.1.4
Figure 4.
Figure 4. Maxtor drive / Sonnet ATA bus / master / HDST SCSI driver 3.2.0
Figure 5.
Figure 5. Maxtor drive / Sonnet ATA bus / master / SoftRaid SCSI driver 2.2.2
Figure 6.
Figure 6. Maxtor drive / Sonnet ATA bus / master / ATTO SCSI driver Rev. 2.5.1
(driver NOT optimized)
Figure 7.
Figure 7. Maxtor volume / Sonnet ATA bus / / Apple SCSI driver 8.1.4
Figure 8.
Figure 8. Maxtor volume / Sonnet ATA bus / / HDST SCSI driver 3.2.0
Figure 9.
Figure 9. Maxtor volume / Sonnet ATA bus / / SortRaid SCSI driver 2.2.2
Figure 10.
Figure 10. Maxtor volume / Sonnet ATA bus / / ATTO SCSI driver Rev. 2.5.1
Do try this driver / ACard combination!!
See Note on Page 1.
Figure 11.
Figure 11. Just for comparison, here is my 10Krpm Cheetah (ST39204 - 9GB) Ultra2 SCSI / Apple SCSI driver 8.1.4 / Initio Miles2 SCSI Host
Egan's Formula for Theoretical Track Count
Egan has been kind enough to send me his formula, which can be used to determine the theoretical track count for any given drive configuration. You first need to use a benchmarking program (I use ATTO ExpressPro-Tools 2.5) to determine the data thruput of the drive in question.
I must remind you that I am not technically minded enough to know if this is completely scientific and accurate or not, but I must also state that, based on the fact that I have found all his other advice to translate to my own real world experience, I have no reason to doubt him.
Here is that formula, along with Egan's comments.
Bit depth/8 <for bytes> x sample rate x # of tracks = required throughput
( I think you said you were using 24bit/48KHz Sample Rate...)
24 bits (divided by 8 = 3 bytes) so
3<bytes> x 48Khz <sample rate> x 200 Tracks = 28,800KB/sec
<1024KB = 1MB, 28800 / 1024 = 28.125 MB/sec>
So, I think your System drive could have done 200 tracks, leaving an extra 7MB/sec for seek times, etc.
Theoretically, ATA-66 36MB/sec could do 250 24/48KHz tracks 36000KB/sec or
35.15625MB/sec
At the 70MB/sec + your ATA RAID does? Let's do 24/96 and 24/48, okay?
(24 bits = 3 Bytes) 3 x 96Khz x 250 tracks = 72,000KB or 70.3125MB/sec @ 96KHz
(24 bits = 3 Bytes) 3 x 48Khz x 500 tracks = 72,000KB or 70.3125MB/sec @ 48KHz
OK, so if this is "Theoretical" what is real world? I think using approx 1/2 this # should be 100% safe.......and if you DO *NEED* more than 125 24/96 tracks (or 250 24/48)....let me know what you are doing!!!!
However; You can probably add up to 70%+ in "Real world" so ;
175+ 24/96Khz tracks or 350 24/48Khz tracks.
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