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Megabytes required for storage of an image given matrix size

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data

    The minimum storage capacity required to store the information in a digital radiographic exam that consists of a 1024 – 1024 matrix, 12 bit deep, acquired at 30 frames per second for 10 seconds is _____ megabytes.


    2. Relevant equations

    3. The attempt at a solution

    I assumed that seeing each pixel is 12bits or 1.5 bytes that you would simply find it by doing

    1024 x 1024 x 1.5 x 30 x 10 = 471859200 bytes

    divide by 1024 = 460800 kbytes

    divide by 1024 = 450 mb

    However the answer is apparently 600mb with this explanation:

    Each pixel requires 12 bits. Each images consists of 1024 x 1024 pixels. Since a Kbyte is 1024 bytes and a Mbyte is a Kbyte times a Kbyte, each images requires 2 Mbytes of storage. There are 30 images times 10 seconds or 300 images. The total storage required is: 300 x 2 Mbytes = 600 Mbytes.

    Could anyone explain why this is the case?

    thanks :)

  2. jcsd
  3. Sep 15, 2013 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    How do you store half a byte?
  4. Sep 15, 2013 #3
    Ya i was just thinking that actually. You gotta round up to an integer value for a byte right?

    (no rhyme intended)
  5. Sep 15, 2013 #4


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    Staff: Mentor

    There is no problem with storing the data in 450 MB, you just need to use some bit shifting and combine two 12-bit values to store them in 3-bytes. Pretty trivial exercise in most programming languages. Whether it makes sense, whether it is worth the effort is another question, but I don't see anything wrong in your answer.
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