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Old January 7, 2010, 09:37 AM
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Default Rumors of WD Raptor 600GB HDD

In the era of SSD's it becomes few and far between that new platter HDD's launch. However Western Digital looks to be revising and prepping a new version of their VelociRaptor Lineup. The premise behind this decision is simple and understandable. Fast large capacity SSD's do not come cheap and tend to be reserved for the top tier builders and e... [ Read full article ]
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Old January 7, 2010, 09:44 AM
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600GB raptor around 250 US...wow that would so nice! I want one!
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Old January 7, 2010, 09:47 AM
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300gb per platter isnt actually that great at all for aerial density. Both Seagate and WD have single-platter 500Gb drives nowadays. I'd like to see a 1TB Velociraptor @ 10RPM, two 500Gb platters, dual procs... then short stroke 'em and see how fast they can go at SATA3 speeds
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Old January 7, 2010, 09:53 AM
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Seagate and WD have single platter 500gb 3.5" HDDs....not single platter 2.5" HDDs. The Vraptor is a 2.5" hdd not a 3.5".
BIG difference.
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Last edited by AkG; January 7, 2010 at 10:00 AM.
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Old January 7, 2010, 09:55 AM
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If they keep their 2,5" size I guess it's more hard to make a 500 gb single plater , just a guess of course
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Old January 7, 2010, 10:01 AM
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hey hey , AkG have a faster connection than me.....
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Old January 7, 2010, 10:03 AM
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Quote:
Originally Posted by AkG View Post
Seagate and WD have single platter 3.5" HDDs....not single platter 2.5" HDDs. The Vraptor is a 2.5" hdd not a 3.5".
BIG difference.
Good point. But then again... unless im just too tired to understand phyics today. Wouldnt a single platter 3.5'' hard drive's outer edge spin faster at its outer edge running at 7200rpm than a 2.5''s outer edge running at 10k rpm? Perhaps my "lil fast and easy brain math" is off today.

the 300gb aerial density doesn't win it for the vraptor actually... dammit, math time:
Area of a 2.5'' disk would be Pi x 2.5''/2 = 3.92 inches cubed (not including the unuseable space in the very center)
Area of a 3.5'' disk would be = 5.497 inches cubed.

If the aerial density of the 2.5 is 300gb, then its managing approximately 76.53 gigs per cubic inch.
For the 3.5 @ 500gb aerial denisty, it would be 90.95 gigs per cubic inch

therefore, the 3.5'' disk with 500gb aerial density still has a marked advantage. I wonder if the faster RPM of the vraptor will make up for it?

PS: anyone can correct me on my math if im wrong - i dont like math. lol
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Old January 7, 2010, 10:09 AM
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The reason WD changed from 3.5 to 2.5 with the intro of the VRaptor was indeed the 10k rpm speed. And that "unusable space in the centre" is what is throwing your numbers off (not going to even touch on the math skillz you "used" ;) ). Thats why the itty bitty 1.8" hdds lag even further behind in size per platter than the 3.5". IMHO I wish 5.25" HDDs had stayed popular before these new fangled 3.5"s came onto the scene. I want my 4TB hdd NOW dang nab it :)~
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Old January 7, 2010, 10:09 AM
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Quote:
Originally Posted by CMetaphor View Post
Good point. But then again... unless im just too tired to understand phyics today. Wouldnt a single platter 3.5'' hard drive's outer edge spin faster at its outer edge running at 7200rpm than a 2.5''s outer edge running at 10k rpm? Perhaps my "lil fast and easy brain math" is off today.

the 300gb aerial density doesn't win it for the vraptor actually... dammit, math time:
Area of a 2.5'' disk would be Pi x 2.5''/2 = 3.92 inches cubed (not including the unuseable space in the very center)
Area of a 3.5'' disk would be = 5.497 inches cubed.

If the aerial density of the 2.5 is 300gb, then its managing approximately 76.53 gigs per cubic inch.
For the 3.5 @ 500gb aerial denisty, it would be 90.95 gigs per cubic inch

therefore, the 3.5'' disk with 500gb aerial density still has a marked advantage. I wonder if the faster RPM of the vraptor will make up for it?

PS: anyone can correct me on my math if im wrong - i dont like math. lol
You got it all wrong lol.

Area of 2.5" disk would be Pi*(2.5/2)^2 = 4.91 inch sq.
Area of 3.5" disk would be Pi*(3.5/2)^2 = 9.62 inch sq.

Areal density of 2.5" disk = 300/4.91 = 61GB/sq in.
Areal density of 3.5" disk = 500/9.62 = 52GB/sq in.

So the 2.5" platters are more dense.
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Old January 7, 2010, 10:11 AM
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Quote:
Originally Posted by bojangles View Post
You got it all wrong lol.

Area of 2.5" disk would be Pi*(2.5/2)^2 = 4.91 inch sq.
Area of 3.5" disk would be Pi*(3.5/2)^2 = 9.62 inch sq.

Areal density of 2.5" disk = 300/4.91 = 61GB/sq in.
Areal density of 3.5" disk = 500/9.62 = 52GB/sq in.

So the 2.5" platters are more dense.

Thanks. I need coffee... lol
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