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Drive Format and Setup

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Cole
Drive Format and Setup
by
on Jan 26, 2003 at 5:42:19 pm

Hi,
I have a new 80GB ATA 100 IDE 7200 Seagate on a new Dell Precision (win 2000) that will be used as storage for video. I have another drive for op system and software.

I formatted the 80GB dive. It asked if I wanted to use
Default Allocation size
512 Bytes
1024 Bytes
2048 Bytes
4096 Bytes

and

Format options
Quick
Enable Compression

I chose 512 Bytes
Is this the correct way to format this drive? Should I have used the "quick" or "enable compression" options?

I did a properties on this drive and it has a couple of options.
1. Compress drive to save space
2. Allow indexing
Should I check either of these?

Also it says free space 74.4GB and used space 83mbs. What are the 83 MBs on a newly formatted drive?

Thanks in advance
Cole


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John Q
Re: Drive Format and Setup
on Jan 27, 2003 at 5:57:14 am

The larger allocation size would be slightly faster, requiring less directory access. The smaller size is better for a drive containing lots of small files. Video files are not small. Use the largest allocation size you can select.

The Quick option creates the directory structure without physically testing each sector. You're probably safe in doing a Quick format, as modern drives only have bad sectors when the drive is failing. They even have smart features built in to automatically test the drive and substitute bad sectors transparently.

You do not want to enable compression or the indexing service. Trying to compress the video would actually result in larger files and slow everything down.

The indexing service allows slightly faster access during read operations of large directory structures, but slows things down during the write. For video work, you want it turned off, as you usually have small directories of very large files.

The drive says it's 80 GB, but drive manufacturers assume that 1000 = 1 kb, where Microsoft assumes 1024 = 1 kb. If you divide 80 GB by 1.024**3, you get 74.5 The 83 MB used is the directory structure required to address the drive with the small cluster size. If you reformatted with the 4096 allocation size, this would be reduced.


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