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...or would the computer see it as a damaged disc?

2006-12-20 17:10:32 · 8 answers · asked by cooltomboy 2 in Computers & Internet Hardware Other - Hardware

I know I can just buy MiniCDs. :)

2006-12-21 07:01:21 · update #1

8 answers

I Wouldnt Reccomend that. Because the Edgesof the disk that were cut, May be frayed, To The Microscopic hair..

that you cant see..

which will tear your laser lens coating to shreds.

Your welcome,
Hope This Helps.
Rpdemers
speaking for
Rpdemerscomputertechnology

www.myspace.com/Rpdemerscomputertech

2006-12-28 15:54:44 · answer #1 · answered by Robert D 1 · 0 0

It may. Looking at the cd you can actually differentiate the portion where data was written and portions which was left untouched. If the cutting damaged the part where data was written, the cd will not work. But if the cutting done within the untouched portion, it will work.

Look cd's use to clean the lens of the cd drive. When they put the brush, they damage the surface of the cd, but the cd still work.

But the best thing to do is to test my advice. Get a working cd and scratch the surface (the back portion, let the light shine through) within the unwritten portion and see if it works.

2006-12-21 02:40:49 · answer #2 · answered by dax 3 · 0 0

Damaged disc, no disc header area.

2006-12-21 01:13:21 · answer #3 · answered by Kasey C 7 · 0 0

Nope, it would work. :) But I would burn the disc, finalize it, and then cut off the part that is not being used.

2006-12-28 16:22:39 · answer #4 · answered by ninjamasterdave 1 · 0 0

Just buy the Mini CD's.

2006-12-21 12:19:21 · answer #5 · answered by ? 7 · 0 0

damaged disc of course..

2006-12-21 01:12:09 · answer #6 · answered by Anonymous · 0 0

before you write the cd and cut it and if you write it .. i ll work
but cutting after writting wont work...
and cuts made should be proper...

2006-12-21 01:45:54 · answer #7 · answered by sweetraskels 4 · 0 0

Im going to take a wild guess and say that it will not work...

2006-12-21 01:12:47 · answer #8 · answered by gamereaper3 3 · 0 0

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