- #1

mitch_1211

- 99

- 1

## 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