I've a feeling there are many members out there having difficulties to understand all this talk about 8ECC, 1ECC, PIE, PIF, POS, samples, aso.
[Note aimed at all "expertise" on this subject; I'll try as long as possible to simplify ECMA terms, wording, aso in my expressions and explanations to keep this post handable and maybe also better understandable.
If and when you find simplier explanation/s, please post them.]
Cont. There isn't any easy answer to all that I posted at start (and I have no intention to do this for you here), most of the time senior members just refer to ECMA without any further (simple) explanation. When visiting that site, I and I suppose also many other members just drawn in information, an information very complex, highly techical and hard to understand.
To somehow ease this feeling and make this more understandable I can recommend you this document. Not that easy either but much more simple/understandable then the very same information at ECMA.
We have learned that at 1ECC drive collects much more samples then at 8ECC. You may ask; why is it like this, what are samples, how is disc structured to provide samples, how many samples can be taken on disc?
For in-depth studies, definitions like (DVD) bytes, sectors, pockets aso can be found with google and at ECMA, so I'm not going into details here.
I just like to explain and show how samples on a DVD are calculated. You can open the document I was linking to right above now.
We know that a fully burned DVD disc contains approx. 4.38GB (gigabytes) of data. These bytes are aranged in sectors (or pockets). Every data sector contains 2048 bytes of (user) data and is arranged in 16 rows making in total 32768 bytes of raw data for every ECC block (see example in document).
When we say 1ECC this means drive takes a sample of every ECC sector on drive. 8ECC means equally only one sample is taken in/of any of these 8 ECC sectors in a row, thus also less samples taken by drive (compared to 1ECC).
How many samples can there then be sampled on a DVD disc? To get this value we have to go back and check discs total amount of data and devide this value with the value of a ECC block.
Example with 1ECC on a DVD-R, total data devided with ECC block makes; ~4707100000/32768 = 143649 samples. A drive only able to sample at 8ECC can only collect or sample 1/8 of this value, or 17956 (randomly) samples.
Using calculator you can also calculate the values for DVD+R media.
I know all this might be difficult and hard to understand but I hope some of you out there know a little more now what we are talking about here. Thanks.
@rolling56, wanna trade?