0

Specifically 90 minute, 60fps, 1920x1080 movies using 128kbps audio. I did the math and got (60)x(5400^2)x(8^2073600)x(2^128000). The base 8 number refers to the number of possible pixel configurations on a 1920x1080 frame. The base 2 number refers to the number of possible audio tones per second at the 128kbps bit rate. the 5400^2 refers to the number of seconds in 90 minutes, and is to the power of 2 to reflect the existence of 2 data formats (video and audio). The 60 reflects the 60fps frame rate. Is the math legit? Thanks.

Daniel Pintor
  • 1

1 Answers1

2

No, the number you want is this: $C^{1920\cdot1080\cdot60\cdot5400}\cdot2^{128000\cdot5400}$

Where C is the number of colours in your display, which you don't specify but which is more than 8 surely. Typically it would be C=256x256x256=16777216

Of course, very large subsets of these movies would be indistinguishable from each other.

Steven Irrgang
  • 716
  • +1 but the $12800$ should either be $128000$ (if one kb is 1000b) or $131072$ (if one kb is 1024b). – 2'5 9'2 Feb 05 '18 at 06:52
  • So I shouldn't use however many configurations a pixel can be and instead use a number like 16.7 million? – Daniel Pintor Feb 05 '18 at 06:54
  • I don't understand your comment. 16.7 million is the number of different values a pixel can have, normally I'd call that the same thing as the "number of configurations" it can have so I'm not sure what you're referring to that's different. – Steven Irrgang Feb 06 '18 at 00:01