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Why you may not want to run out and buy a Pioneer 103 DVD writer!
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The key issue I was unaware of is the two "standards" for laser wavelength. If the Pioneer 103 to be available for <$1000 "soon" is based on the current 635nm "standard" then its a buy. If it uses the modified standard's "general use" DVD 650 nm then playback in existing players will be very iffy!
At least this is my understanding! I've zero intrest in the RW variations of DVD based on my poor experiences with CD-RW. For data backup its hard to compete with 80GB Maxtor's on a $/GB basis when you price the removable drive and media cost vs a firewire Maxtor.
--wally.
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From the links and discussions going on at rec.video.desktop , it appears that it will work on most current DVD settop players.
At least I hope so.
Here's one describing why the price is so much lower than the DVR-S201:
http://www.abcdv.com/pctech/pioneer_dvdr_whycheap.html
Also be sure to check out the other links on that page for additional info/press releases.
Here's one that shows the compatibility with different players. This site seems to be making DVDs for you:
http://www.homemovie.com/compatibility.htm
(Yippee. My player is listed!)
From what I gathered, it appears that older DVD players do not support the 4.7 GB. Only the 3.95 GB.
Please correct me if I'm wrong.
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Not to get off the wavelength issue but here's a link for some DVD information:
http://www3.toshiba.co.jp/dvd/e/whats/
And this one answers my question as to how much video a 4.7 GB DVD-R will hold at a certain Mb/s:
http://www3.toshiba.co.jp/dvd/e/whats/wh01/w0103idx.htm
<font face="Verdana, Arial, Helvetica" size="2">
3.5 megabits/second is dedicated to images, 1,152 kilobits/second to sound, and 40 kilobits/second to subtitles, giving a total of 4.692 megabits/second. 135 minutes playing time requires 4.692 * 135 * 60 / 8, which comes to about 4,750 megabytes. Since the capacity of one side of a DVD is 4.7 gigabytes, the requirement is satisfied for all practical purposes.
135 minutes playing time requires
4.692 * 135 * 60 / 8 = 4,750 megabytes.
A disc with a capacity of 4.7 gigabytes on one side is required.
</font>
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