Is There Anything To This???

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Agent_C

Someone posted this in another group and on the surface it seems
simply ridiculous. Essentially what she's saying, is that you can
approximate the data transfer rate of a 15,000 RPM SCSI drive if you
use an IDE drive of double the storage capacity and half the rotation
speed. Is there anything to this at all?


---- Quoted Text ----
All other things being equal, a 74 GB 15,000 rpm drive would
have about the same data transfer speed as a 350 GB 7,200
rpm drive.

IOW you should be able to approximate the kick of a 74 GB
15,000 rpm drive with a common 250 GB 7200 rpm drive, and
have the benefit of an extra 176 GB for storing data.
--- End
 
Agent_C said:
Someone posted this in another group and on the surface it seems
simply ridiculous. Essentially what she's saying, is that you can
approximate the data transfer rate of a 15,000 RPM SCSI drive if you
use an IDE drive of double the storage capacity and half the rotation
speed. Is there anything to this at all?
Click to expand...

Larger drives usually have more platters, not just higher data density.
 
Agent_C said:
Someone posted this in another group and on the surface it seems
simply ridiculous. Essentially what she's saying, is that you can
approximate the data transfer rate of a 15,000 RPM SCSI drive if you
use an IDE drive of double the storage capacity and half the rotation
speed. Is there anything to this at all?


---- Quoted Text ----
All other things being equal, a 74 GB 15,000 rpm drive would
have about the same data transfer speed as a 350 GB 7,200
rpm drive.

IOW you should be able to approximate the kick of a 74 GB
15,000 rpm drive with a common 250 GB 7200 rpm drive, and
have the benefit of an extra 176 GB for storing data.
Click to expand...

Depends on your usage patterns. All else being equal the sequential
transfer rate under ideal conditions should be about the same. Latency on
the 7200 will be twice as high though, and seek times are generally slower
as well, so access time will be much slower on the 7200, which means that
for random access performance will not be nearly as high.
 
Transfer rate is a meaningless indicator of performance.
Everyone would use RAID 0 if it were so important.

To double the bit rate, you need 4X the bit density.
Since 15K platters are smaller, you need a 300GB 3" platter
to match the 36GB 2.5" platter of 15K drives (bit rate wise).
 
In comp.sys.ibm.pc.hardware.storage Agent_C said:
Someone posted this in another group and on the surface it seems
simply ridiculous. Essentially what she's saying, is that you can
approximate the data transfer rate of a 15,000 RPM SCSI drive if you
use an IDE drive of double the storage capacity and half the rotation
speed. Is there anything to this at all?
Click to expand...

Not necessarily. It assumes twice the data per track on the IDE
drive. Since 15000RMP drives are usually 2.5" drives (track length at
the outer edge about 200 mm) and 7200 RPM drives are 3.5" (track
length at the outer edge about 280mm) that would mean an about 1.4
times higher data density on the IDE disk. I am doubtful about that. I
think that the IDE disk will instead use more platters, resulting in
it being slower by a factor of 1.4 in linear reads (!!) on the outer
edge (!). (Border condition: Reading is from one head at a time.
That is true for allmost all HDDs.)

On the other side, for random access the 15000 RPM disk will have an
average latency that is half that of the IDE disk and it will likely
have faster seek time as well, both because it has better machanics
and less distance to travel for the heads because the disks are
smaller. In addition it will have the native command queuing that
works (SCSI has had that for ages) which allows the disk to optimize
access patterns. Also on the inner edge of the disk, the 15000 RPM
disk will likely be twice as fast, under the assumption of the same
data density.

Frankly I don't see why the data density should be much lower on the
15000 RPM disk.

Also in many applications random read performance is much, much more
important than linear read performance. If you really have an unusual
access pattern that is mostly long linear reads, go for a n-way RAID0,
as that will give you almost n times the reading speed.

Arno
 
Someone posted this in another group and on the surface it seems
simply ridiculous. Essentially what she's saying, is that you can
approximate the data transfer rate of a 15,000 RPM SCSI drive if you
use an IDE drive of double the storage capacity and half the rotation
speed. Is there anything to this at all?


---- Quoted Text ----
All other things being equal, a 74 GB 15,000 rpm drive would
have about the same data transfer speed as a 350 GB 7,200
rpm drive.

IOW you should be able to approximate the kick of a 74 GB
15,000 rpm drive with a common 250 GB 7200 rpm drive, and
have the benefit of an extra 176 GB for storing data.
--- End
Click to expand...

Sequential (Data Read) Transfer Rate does not say everything about disk
performance.
For example:
Maxtor Atlas 15k (73 GB Ultra320 SCSI) - 73.7 MB/s
Seagate Barracuda 7200.8 (400 GB SATA) - 69.8 MB/s
but SR File Server DriveMark 2002 numbers are 351 and 108 IO/s respectively.

It all depens on how you "kick" particular hard drive.
 
Arno Wagner said:
Not necessarily. It assumes twice the data per track on the IDE
drive. Since 15000RMP drives are usually 2.5" drives (track length at
the outer edge about 200 mm) and 7200 RPM drives are 3.5" (track
length at the outer edge about 280mm) that would mean an about 1.4
times higher data density on the IDE disk. I am doubtful about that. I
think that the IDE disk will instead use more platters, resulting in
it being slower by a factor of 1.4 in linear reads (!!) on the outer
edge (!). (Border condition: Reading is from one head at a time.
That is true for allmost all HDDs.)

On the other side, for random access the 15000 RPM disk will have an
average latency that is half that of the IDE disk and it will likely
have faster seek time as well, both because it has better machanics
and less distance to travel for the heads because the disks are
smaller. In addition it will have the native command queuing that
works (SCSI has had that for ages) which allows the disk to optimize
access patterns. Also on the inner edge of the disk, the 15000 RPM
disk will likely be twice as fast, under the assumption of the same
data density.

Frankly I don't see why the data density should be much lower on the
15000 RPM disk.
Click to expand...

'Much' is such a big word but in order to have very good seek times
it may be prudent to have larger track spacing AND less bits/inch.
Combined that can make quite a difference to the 'data density'.
 
Eric Gisin said:
Transfer rate is a meaningless indicator of performance.
Everyone would use RAID 0 if it were so important.

To double the bit rate, you need 4X the bit density.
Huh?

Since 15K platters are smaller, you need a 300GB 3" platter
to match the 36GB 2.5" platter of 15K drives (bit rate wise).
Click to expand...
 
chrisv said:
He must mean 4X the areal density, which is about twice the linear density.
Click to expand...

Presumably you mean 2 squared?
Yeah, except that it doesn't work that way:

BPI and TPI aren't linearly/equally expanded with the areal density expansion.
Here's an example from Seagate:

7200.7 200GB 160GB and lower
Recording density in BPI (bits/inch max) 671,500 595,000
Track density TPI (tracks/inch max) 98,000 94,600
Areal density (Mbits/inch 2 max) 68.5 56.3

7200.8
Recording density, BPI(bits/in max) 763,000
Track density, TPI(tracks/in avg.) 120,000
Areal density,(Mbits/in 2 avg) 91,560
 
Folkert said:
Presumably you mean 2 squared?
Yeah, except that it doesn't work that way:

BPI and TPI aren't linearly/equally expanded with the areal density expansion.
Here's an example from Seagate:

7200.7 200GB 160GB and lower
Recording density in BPI (bits/inch max) 671,500 595,000
Track density TPI (tracks/inch max) 98,000 94,600
Areal density (Mbits/inch 2 max) 68.5 56.3

7200.8
Recording density, BPI(bits/in max) 763,000
Track density, TPI(tracks/in avg.) 120,000
Areal density,(Mbits/in 2 avg) 91,560
Click to expand...

Well, this example, if I'm reading it correctly, is actually pretty
close to "4X the areal density gives twice the linear density", i.e. a
squaring relationship. Dividing the linear density (BPI) numbers, we
get 763/672=1.14. Dividing the areal density numbers and taking the
square-root, we get sqrt(91.5/68.5)=1.16. Pretty close.
 
chrisv said:
Well, this example, if I'm reading it correctly, is actually pretty
close to "4X the areal density gives twice the linear density", i.e.
a squaring relationship. Dividing the linear density (BPI) numbers,
we get 763/672=1.14. Dividing the areal density numbers and taking
the square-root, we get sqrt(91.5/68.5)=1.16. Pretty close.
Click to expand...

Except that you obviously missed that double BPI doesn't come with double TPI,
so 4=x*y but x NE y NE 2.
 
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