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Fitting mutiple hours of content on CD-ROM

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Damian McDonald
Fitting mutiple hours of content on CD-ROM
on Dec 7, 2010 at 7:03:00 pm

I'm not sure where this post should go so I apologize if I'm in the wrong spot.
My question is, How does manage to fit over 7 hours of video content on a CD-ROM. Are they using Flash or are they using some other combination of products to accomplish that?
If there is a better place to post this question, please let me know.

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Brodd Nesset
Re: Fitting mutiple hours of content on CD-ROM
on Dec 8, 2010 at 11:29:02 am

Actually I think there is a dedicated video compression forum here, somewhere, but:

7 hours = 420 minutes = 25.200 seconds

On a CD-ROM disk there is about 720 mB available storage. Let's take away 20 mB for the application, which should be ample if it's made in Flash for instance. That leaves us with 700 mB for video.

700 mB = 700.000 kB (million to thousand)

700.000 kB / 25.200 seconds = 27.777 kB per second, on average.

Video compression isn't measured in kilo Bytes, but in kilo Bits, per second (kbps). There are 8 bits in one byte, and 27.777 * 8 = 222.22 kbps.

This compression rate isn't 'great', but totally feasable. Especially sufficient for video with little movement in it. For instance software tutorials.

But: are you sure this is delivered on CD-ROM - not DVD-ROM? It has nearly six times the capacity of it's predecessor, and can be read by 'any' computer.

Not everything that can be counted counts, and not everything that counts can be counted.

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