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.

 

[rezeew]

ell

I can confirm that the explanation provided is correct. The key difference between the two approaches is the unit conversion. While your calculation is correct in terms of the number of bytes, the question specifically asks for the storage capacity in megabytes. Therefore, the final answer needs to be converted from bytes to megabytes, which is done by dividing by 1024 twice (since 1 megabyte is equal to 1024 kilobytes, and 1 kilobyte is equal to 1024 bytes).

Additionally, the question states that each pixel requires 12 bits, which is equivalent to 1.5 bytes. Your calculation uses 1.5 bytes per pixel, whereas the given explanation uses 2 bytes per pixel. This may be a slight difference, but it can add up when dealing with a large number of images.

In summary, both approaches are correct, but the given explanation is more accurate in terms of unit conversion and the number of bytes per pixel. As a scientist, it is important to pay attention to the units and make sure they are consistent throughout the calculation.