Data storage requirements for an image.

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This is a question from one of my review problems. I thought I had it right until another student posted an answer completely different. Hoping someone can check my logic.

A drawing in an 8.5-by-11-inch sheet is digitized by means of a 300 dpi scanner.

a) What is the visual resolution of the resulting image (number of dots in each dimension)?

b) If 8 bits are used for the quantization of each pixel, how much data storage space is needed to store the image as raw data?

For (a) I simply did 8.5 *300 and 11 * 300 to get
2550 * 3300

For (b) 2550 * 3300 * 8
= 67,320,000 bits

67,320,000/8
= 8,415,000 bytes.

8,415,000/2^20
= 8.025Mbytes (or there abouts)

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If think you did one step too many.

2550 * 3300 = 8,415,000 dots

This is the number of pixels (there is no reason to multiply by 8 to get pixels).

The factor of 8 (if I understand the question correctly) is the number of bits needed to represetn 1 pixel.

If there is 8 bits per pixel, then there is 1 byte per pixel, meaning I think the answer is 8,415,000 bytes.

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ebrown_p wrote:

This is the number of pixels (there is no reason to multiply by 8 to get pixels).

No, I multiplied by 8 to get the number of bits. It's co-incidental to this question that it then requires a division by 8 to get the number of bytes. If the question asked for 16 bits for quantization (or 24 for that matter) then I would have been doing 2550 * 3300 * 16 to get the number of bits. Then a division by 8 to get the number of bytes, and a division by 2^20 to get Mbytes.

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Hmmm.. I just scanned a 8.5"x11" black and white document at 300 dpi w/8-bit color depth and it comes up as 25,178.77 Kb as an uncompressed file size.

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A rather silly but important question. Is dpi a linear measurement?

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dots per inch would be per square inch I would have thought.

Edit: from wikipedia
Dots per inch (DPI) is a measure of printing resolution, in particular the number of individual dots of ink a printer or toner can produce within a linear one-inch (2.54 cm) space.

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Wilso and I came up with the same (I believe) final answer. I went back (and I feel silly) and checked and we are basically saying the same thing in slightly different ways.

The unit dpi is certainly a linear measure. Consider that 300 dots per square inch would be 17 rows of 18 dots each in a square inch. This would not be readable at all.

Fishin is a factor of 3 higher than the calculation that Wilso and I did. I don't know why that is... but it is interesting.

Fishin... what format are you storing this image in?

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ebrown_p wrote:

Fishin... what format are you storing this image in?

25,178.77 Kb is the uncompressed image size that both Photoshop and HP's ImageScan utility reported since he mentioned "raw data" in the inital post. (I assumed he meant "raw data" vs. ".raw image format")

Of course, if I save the file in jpg, bmp, gif, etc... it would get significantly smaller since they'd all compress the image (in different ways). If I save it as a jpg it stores on the HD as 1,999 KB.

I don't know why it comes up as the file size I listed. I saw the question and decided to do the test. Maybe there is something else being saved with the scan itself? (Metadata??)

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The number you gave is almost perfect for 3 bytes per pixel... 3 bytes per pixel is the value for "true color".

I might suspect that your scanner is saving in true color... at least internally and then failing to take advantage of the fact the format you specified can be saved in 1 byte per pixel.

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ebrown_p wrote:

The number you gave is almost perfect for 3 bytes per pixel... 3 bytes per pixel is the value for "true color".

I might suspect that your scanner is saving in true color... at least internally and then failing to take advantage of the fact the format you specified can be saved in 1 byte per pixel.

I think you are right. I opened it with Micorsoft's Image Editor and played with it there and it looks like the numbers corrected.

I haven't got Image Editor down pat but I can get the image to 8.44" x 10.97" @ 300 dpi w/1-bit grayscale and it comes up as 8,332,812 bytes of data uncompressed so it looks like the mathmatical formula is correct.

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Re: Data storage requirements for an image.

Wilso wrote:

This is a question from one of my review problems. I thought I had it right until another student posted an answer completely different. Hoping someone can check my logic.

A drawing in an 8.5-by-11-inch sheet is digitized by means of a 300 dpi scanner.

a) What is the visual resolution of the resulting image (number of dots in each dimension)?

b) If 8 bits are used for the quantization of each pixel, how much data storage space is needed to store the image as raw data?

For (a) I simply did 8.5 *300 and 11 * 300 to get
2550 * 3300

For (b) 2550 * 3300 * 8
= 67,320,000 bits

67,320,000/8
= 8,415,000 bytes.

8,415,000/2^20
= 8.025Mbytes (or there abouts)

I get the same answer.

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Data storage requirements for an image.

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