In Which way only 6 samples will be missing at the end of the cd when reading

vbimport
#1

Well,
This is the situation:

I’m read cds with EAC.
I’m burning with EAC
I use only one DVD RW (ASUS 24B1ST)
read offset correction: 6 +
write offset: -6
Lead-Out/in: none (because it the rip would not be “perfect rip”)
Read with a combined read/write sample offset correction 0 (no other option because the drive Lead-Out/in is none)

That way I’ll lose only 6 samples at the end of the cd when readnig?

If not, in which other way I’ll lose only 6 samples at the end of the cd when reading?

jonathan.

#2

There is a lot of information in the other topic you started.
The + or - samples are not just at the beginning or end . They are offset by that much throughthe entire CD.
In EAC you can set the read offset & the write offset & have those two selected.
That is the first option.
If you select the combined read/write sample offset it overrides the other two…

What you need to do is get a CD -RW & do some test writes & reads . This will let you test the results without creating coasters.
In both examples you listed above the offset should be 0(zero) [B]IF[/B] the actual offset of your drive is 0(zero).

#3

I did all the tests.
These are all the correct data.

With which method of extrication fewer samples will be missed as much as possible in relation to the original cd?

jonathan.

#4

No offense, just out of curiosity:
You know that 6 samples are 136 millionths of a second ?

#5

I was not offended at all.
I did not know what is the number but I knew I could not hear any difference at all.
I just want to know how many samples I’d lose in any way and where on the cd?
1.read and write with with combined read / write offset correction
2.read and write with separate read offset on extraction and write offset on writing.
3.without offsets at all.

The details are:
I’m read cds with EAC.
I’m burning with EAC
I use only one DVD RW (ASUS 24B1ST)
read offset correction: 6 +
write offset: -6
overread Lead-Out/in: none

that’s all.