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From: lindstrom@sgi.siemens.com
Newsgroups: comp.publish.cdrom.hardware
Subject: Transfer rate clarification
Date: 9 Mar 1994 17:11:12 -0600
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I need clarification on transfer rates.  Ignoring seek times, if I have a 
4x reader and want to know how fast I can read a 1024x1024x8 bit image, 
can I just use the following formula?

   1 Mbytes / 600 Kbytes/sec = 1.75 seconds
   
How about the overhead within each sector of the CD?  For example, if the
image is stored using the mode 1 format, each sector consists of 2048 bytes
of my image and 304 bytes of header and ECC/EDC data.  That's 13% overhead.
Does that mean that it will actually take 1.75 / .87 = 2.01 seconds?

What about switching to mode 2?  My overhead drops to less then 1% but I
don't have any error correcting codes.  How often will I see errors using
this mode?

Thanks,

Tom
Lindstrom@sgi.siemens.com