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Question for Alamy users

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samueldilworth
10 Dec 2012 - 8:04 PM

To really understand this, it might help to go back to first principles.

As you said earlier, your D60 puts out an image with pixel dimensions 3872 × 2592. Multiply those sides together and you get 10.03 million pixels in total, or about 10 megapixels.

Each of those pixels has a red, green, and blue (RGB) channel, which together describe all the colours in the photo. When using 8-bit colour depth, as requested by Alamy, the R gets 8 bits, the G gets 8 bits, and the B gets 8 bits. So each pixel gets a total of 24 bits.

Since you have 10.03 million pixels, each with 24 bits, the file size is about 241 million bits. Dividing by 8 gives you the file size in bytes, since there are 8 bits in a byte. Dividing that by 1024 gives you kilobytes. And dividing again by 1024 gives you megabytes. The result is 28.7 megabytes.

(Since kilo- in everyday language means 1000, not 1024, there’s currently a move to redefine kilobyte and megabyte in those terms. To get this newfangled megabyte value you’d divide by 1000 twice, above, instead of 1024. That would give you 30.1 megabytes for the D60. That’s what recent versions of OS X would report, for example; but if you opened the same file in Photoshop you’d be told it’s 28.7 megabytes. Which kilo- is Alamy using? Almost certainly the traditional 1024.)

This size, 28.7 megabytes (or 30.1 megabytes), is the so-called uncompressed file size for 8-bit colour depth. Since Alamy asks for 24 megabytes, you’re comfortably over their threshold.

How much can you crop? That’s trickier, because the minimum pixel dimensions depend on the aspect ratio of your crop. It’s easier to calculate the minimum number of total pixels needed for a 24-megabyte 8-bit file.

Doing so, we find the minimum file size accepted by Alamy is 24 × 1024 × 1024 × 8 = 201 million bits. Dividing that by 24 (the number of bits per RGB pixel) gives us exactly 8388608 pixels.

If you get a number equal to or bigger than 8388608 when you multiply the pixel dimensions of your photo (width × height), then your photo passes Alamy’s file-size requirement. The smallest crop with a 3:2 aspect ratio that would pass is 3547 × 2365, but your crop might be squarer or more elongated than that. And just to be clear, this number 8388608 works for any camera.

On second thoughts, Alamy SizeCheck sounds alright! I started this post thinking the above could be explained in three sentences.

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10 Dec 2012 - 8:04 PM

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User_Removed
10 Dec 2012 - 8:14 PM


Quote: "uncrompessed size should be greater than 24mb (8bit)"

Can't believe you've greened LeftForum for telling you about hovering your mouse over a JPEG. JPEG is definitely not uncompressed. Also none of what he says about tonal range etc applies to uncompressed.

cats_123
cats_123 e2 Member 104009 forum postscats_123 vcard Northern Ireland25 Constructive Critique Points
11 Dec 2012 - 9:37 AM


Quote: In LR I think you just need to select EXIF & IPTC frm the Metadata dropdown menu in the Library mode. File size is near the top of the list of data that is then displayed.

thanks again, John...just got back from holidays....I hadn't got `Exif 7 IPTC' selected...I have now and can see the file size SmileSmile

cats_123
cats_123 e2 Member 104009 forum postscats_123 vcard Northern Ireland25 Constructive Critique Points
11 Dec 2012 - 9:41 AM


Quote: If you get a number equal to or bigger than 8388608 when you multiply the pixel dimensions of your photo (width × height), then your photo passes Alamy’s file-size requirement. The smallest crop with a 3:2 aspect ratio that would pass is 3547 × 2365, but your crop might be squarer or more elongated than that. And just to be clear, this number 8388608 works for any camera.

Hi Samuel...thanks for the detailed explanation...whilst (as I said) I'm not too good at this sort of maths, I can now see how it's constructed and will be able to calculate the outcome after cropping.....or, use the easy program GrinGrin

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