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Some Algebra

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    This question isn't that hard just confusing:

    Write a program to compute the per-minute storage requirements for “full
    HD” TV at 60 HZ (60 frames per second) where each frame is captured at
    a resolution of 1920×1080 pixels using 24-bit color.

    2. Relevant equations

    1 pixel = 3 integers
    1 integer = 8 bits

    3. The attempt at a solution

    I said:

    Number of pixels one frame = (1920)(1080) = 2073600

    Number of integers one frame = (2073600)(3) = 6220800

    Number of bits one frame = (6220800)(8) = 49766400

    Number of bytes one frame = 49766400/8 = 6220800

    Number of kilobytes one frame = (6220800)/1024 = 6075

    Number of kilobytes one second = (6075)(60) = 364500

    Number of kilobytes one minute = (364500)(60) = 21870000

    Number of megabytes one minute = (21870000)/(1024) = 21357.42 approx. = 20.86 gigabytes approx. This looks like a very large number for one minute.
     
  2. jcsd
  3. Oct 5, 2009 #2

    jbunniii

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    I get the same answer. It IS a very large number for one minute. That's why DVDs and cable TV use such heavy compression. e.g., a dual-layer Blu-Ray DVD has 50GB storage capacity TOTAL, enough to hold 2.5 minutes of uncompressed video, not even including audio. The Blu-Ray FAQ claims that you can store over 9 hours of HD video on a 50GB disc, so that shows how extreme the compression is. (Source: http://www.blu-ray.com/faq/ )
     
  4. Oct 5, 2009 #3
    Ok that makes sense as the question also says "assume no other encoding is done" so it would be uncompressed video.
     
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