Megabytes required for storage of an image given matrix size

• mitch_1211
In summary, the minimum storage capacity required for a digital radiographic exam with a 1024x1024 matrix, 12-bit depth, and acquired at 30 frames per second for 10 seconds is 600 megabytes. This is because each pixel requires 12 bits, resulting in 2 megabytes per image. With 300 images in 10 seconds, the total storage required is 600 megabytes. While it is possible to store the data in 450 megabytes by converting two 12-bit values into 3 bytes, it may not be worth the extra effort.
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

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

How do you store half a byte?

SteamKing said:
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)

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.

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.

What is the formula for calculating the megabytes required for storage of an image?

The formula for calculating the megabytes required for storage of an image is (matrix size in pixels * bit depth) / 8 / 1024 / 1024.

How do I determine the matrix size of an image?

The matrix size of an image can be determined by multiplying the width in pixels by the height in pixels.

What is bit depth and how does it affect the megabytes required for storage?

Bit depth refers to the number of bits used to represent each pixel in an image. The higher the bit depth, the more colors and shades can be represented, which results in a larger file size and more megabytes required for storage.

Can I estimate the megabytes required for storage of an image based on its dimensions?

Yes, you can estimate the megabytes required for storage of an image by using the formula mentioned above. However, the actual file size may vary depending on the image's content and compression.

Does the type of file format affect the megabytes required for storage?

Yes, different file formats use different compression methods, which can affect the file size and the amount of megabytes required for storage.

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