This relates to going from dvd to dvd-r because I need to know what to set the compressions at to make my backups.
They say "4.7 GB" dvd-rs, but my computer actually reads it as having only 4.33 GBish of space. Something like the discrepency between hardware GB and softwear GB? Like my computer reads my "20 GB iPod" as only have 18.35 GB of space.
So how much exactly can "4.7 GB" Dvd-rs hold?
On DVD Shrink 3.2, it likes having the files be a total of 4.4-4.5GB ish Will this fit and play on my 4.7 GB?
Right now, Im trying to compress my movies so that the disc size will be under 4.3GB, because i want it to work.
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4.38gb is the max. cause an actual mb on a computer is 1024kb. anyway 4.38 is the max and usually wanna set whatever you using to compress just under that mark to compesate for some cheap media having rought outter edgesPhenII firstname.lastname@example.org - GA-790XTA-UD4 AM3 - 2x4 Corsair Vengeance@1600 - Radeon 5770 - Corsair 550VX - OCZ Agility 3 90GB WD BLACK 1TB - LiteOn 24x - Win 8 Preview - Logi G110+G500
The amount of space on a DVDR is very close to 4,700,000,000 bytes (something around 4,711,500,000 for DVD-R, 4,706,000,000 gor DVD+R). That's why they're labeled 4.7 GB, but windows reports them as 4.38 GB (4,700,000,000/1024^3), and DVDShrink wants less than 4,482 MB (4,700,000,000/1024^2) - same number of bytes represented different ways.
4.3 GB for movie data, I believe you only can use the whole 4.7GB if it's just normal data.
Um... no, Thunder, that's quite wrong.
There is no difference -- none -- between "movie data" and "normal data" on a DVD-R. Glockjr and Skbenin have it correct -- the discrepancy comes from the way a computer calculates kilo, mega, and gigabytes vs. the way hard-drive manufacturers advertise it.
The "proper" metric definitions of a kilo, mega, and giga are, of course:
K = kilo = 10^3 = 1,000 (one thousand)
M = mega = 10 ^6 = 1,000,000 (one million)
G = giga = 10^9 = 1,000,000,000 (one billion)
However, these are "proper" only because we humans calculate in base 10, so we prefer measurements which are powers of 10. Computers, on the other hand, calculate in base 2, and the metric definitions of K, M, and G do not fit neatly into powers of 2.
Therefore, way back in the "dark ages" of computing, it was decided that for computing purposes, a "Kilobyte" would be 1,024, or 2^10, since that fit neatly into the base-2 numbering system as a power of 2. When data capacity started getting large enough to require measuring in "Megabytes", we defined it as 2^20, or 1024^2, or 1,048,576 bytes; and when we went to the "Gigabyte", we followed the same pattern, making it 2^30, or 1024^3, or 1,073,741,824 bytes.
Now, for a while, hard drive manufacturers followed the accepted computing definitions, and all was good. Then, somewhere along the line, one of them realized that they could make their products seem to have a higher capacity than the competition simply by reverting to the standard base-10 metric definition and advertising their products' capacity that way. Since the majority of computer hardware was, by that point, being sold to home and office users who didn't know the difference, instead of to geeks and engineers who wouldn't be fooled, they got away with it, and their competitors quickly followed suit in order to keep up.
However, your PC and its operating system still calculate drive capacity based on the base-2 definitions. Therefore, what the manufacturers call a 4.7Gb DVD under the base-10 definition of a "giga", actually works out to about 4.377Gb under the base-2 definition of a "Gigabyte."
wow solar, you went off on this one
bottom line the 4.7 gig is just a marketing skeem... of course they are going to advertise the bigger number. think of your hard drive. do you ever get the full 160 gig you bought? kinda but not reallyPhenII email@example.com - GA-790XTA-UD4 AM3 - 2x4 Corsair Vengeance@1600 - Radeon 5770 - Corsair 550VX - OCZ Agility 3 90GB WD BLACK 1TB - LiteOn 24x - Win 8 Preview - Logi G110+G500
In a related question, why is it when you use UDF packet writing (i.e. Nero InCD) with DVD-/+RW that you get the full 4.37 GB with no "overhead" space needed, whereas when you use CD-RW for the same you only get 572 MB out of 700 MB useable space. (Not complaining, just wondering why.)
Well, IIRC there's packet-writing built into the DVD spec, whereas on a CD it was an after-the-fact sort of idea... but I could be entirely off-base with that, it's just something I remember hearing.
dvd-r = 4489 mb's
dvd+r = 4482 mb's
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