RAID 0 performance increases on Canterwood

Tavoc

Member
Nov 30, 2002
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What exactly is the performance increase of 2 Drives in RAID 0 on a Canterwood in comparison to just 1 Drive?

The reason I ask is because I am trying to decide wether to get 2 WD Raptors for my IC7 or save the money and just get 1 instead.
 

bgeh

Platinum Member
Nov 16, 2001
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in generalization of RAID 0, we can conclude that
access times increase due to latency, but Raptors have low access times to beat that
transfer rates increase, by a factor of almost x2
 

OverVolt

Lifer
Aug 31, 2002
14,278
89
91
See the Storage Review Main for details on WD Raptor vs other ATA/SCSI drives.

I don't think it has any articles on RAID vs Single... But no matter what RAID WILL give you a performance boost on loading, or HD intensive tasks. Like when i load games like BF1942 + CNC Renegade, and multiplayer maps, it's sooo much faster. It depends if you think it warrants the cost.

If your mobo has RAID 0 in SATA already, and u have the money, go for it!!! It's the setup i'm using right now. But i've never used a single Raptor setup before so i can't compare.

And the latency thing... i think it doesn't add a noticable amount of latency, much faster than any 7200rpm drive i've owned. Depends on what ur using for, and only you can really decide that.

Good luck!
 

DynaOne

Senior member
Jan 30, 2001
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If you look at the overall benchmarks on PCMAG.com - they did a test of desktop 875 systems and also of "gaming" systems (separate articles). Though its clear Raid0 will give significantly higher disk-only benchmarks - the overall benchmarks are hardly any different at all. Same ole story - as system is only as fast as its weakest link.
 

OverVolt

Lifer
Aug 31, 2002
14,278
89
91
Yes, but RAID 0 makes the system feel a lot "zippier" if you know what i mean. Just everyday stuff is a little faster. Not something that will show in synthetic benchies.
 

DynaOne

Senior member
Jan 30, 2001
393
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Yes - I understand that software launch etc can feel faster - just have to weigh against the cost.