Megabytes required for storage of an image given matrix size

[mitch_1211]

Homework Statement

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.

1
37
600
695
2500

Homework Equations

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 :)

Mitch

 

[SteamKing]

How do you store half a byte?

 

[mitch_1211]

How do you store half a byte?

Ya i was just thinking that actually. You got to round up to an integer value for a byte right?

(no rhyme intended)

 

[Borek]

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.