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Thread: How to Scale Image + maintain its exact original Aspect Ratio?

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    How to Scale Image + maintain its exact original Aspect Ratio?

    Hi
    When resizing an image down, even if maintaining its proportions, the result can be a size that differs in Aspect Ratio from its original Aspect Ratio.
    Example:
    https://pacoup.com/2011/06/12/list-o...9-resolutions/

    How do i scale an image 'OriginalWith x OriginalHeight' down and still maintain its EXACT aspect ratio?

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    I'm confused. Could you give an example of an image and its proportions, what results you're getting now and what the desired proportions would be for it to be its original aspect ratio?

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    Confused as well, proportionally scaling an image should keep the aspect ratio.

    W/H = aspect ratio

    W/H = x

    (W/y) / (H/y) = x

    should remain the same

    The article could possibly deal with problems in encoding videos due to encoders working with multiples of 8?

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    Say we originally have an image of.
    1920(x) x 1080(y) ( Aspect ratio 16:9 ).

    I open GIMP2 and create a Frame of 1920(x) x 1080(y).
    I then scale the Frame ( and lock its proportions ) so that x = 750, yielding a Frame of 750(x) x 422(y) ( Aspect ratio 375 : 211 ).

    Im using this tool to calculate the Aspect Ratio for me:
    http://andrew.hedges.name/experiments/aspect_ratio/

    ...
    Like the example link i posted:
    https://pacoup.com/2011/06/12/list-o...9-resolutions/

    There appears to be intervals in the scaling range where the aspect ratio is 'slightly' different.
    Maybe im just missing something.

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    your 750(x) x 422(y) is still quite close to 16:9
    you would have had 421,875 to make for a perfect 16:9

    but this can't (generally) be done with pixels

    aspect ratio of 16:9 is 1,777777777777777777777777777777...

    don't know if you are talking about these sub-pixel differences?


    EDIT:
    just checked the online tool you were using,
    I think it's a little bugged / gives little useful results in some situations
    (375:211 is just your original numbers /2)

    you simply need to divide W/H to get the aspect ratio
    (can be expressed as fraction 16:9 or decimal 1,7777777777..)
    so when you scale W to any value
    to get new H simply divide scaled W by 1,777777777777...
    to preserve 16:9 aspect ratio

    and this is automatically done when you divide both W/H by the same value
    because their ratio will be preserved

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    Well, maybe im being difficoult here, but how do i find these 'perfect' ratio points?
    It should be logical enough maybe, but i dono

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    I would say perfect ratio points, within a monitor pixels,
    are those who does not result in decimals popping out.

    All those in that table of the link you referenced are perfect 16:9 ratios
    closest one to the image resizing example you did above would be: 752 x 423

    Can't think of a quick method to procedurally find them (without looping )

    I guess the guy there "hard-coded" those..?

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    Thanks, but 16:9 was only an example, and i was fishing for a method.

    Im actually looking for a true aspect of 226:155 as close to the size 'Something x 504'

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    734,86451612903225806451612903226
    x 504


    I would do this way:

    find the closest integer number to 504 that multiplied x 1,4580645161290322580645161290323
    gives smaller decimal amount...
    if you find one that gives no decimals at all - you got a perfect "pixel-renderable" ratio

    edit__

    729 * 500
    could be a decent choice?

    734 * 524
    gives also low decimals
    but is slightly less adherent to true 226:155


    closest perfect to 226:155 is probably 678:465

    (ok - we've been very blind - we got a method-
    simply multiply/divide one of your starting ratio numerators by an integer/perfect divisor for both
    and multiply/divide the other by the same...
    ...was pratically doing the inverse )

    full method: find smaller integer divisor common to both numbers,
    individually divide W and H for this number
    and then multiply result of this division by any given number

    to quickly find the smallest divisor, you should search between numbers < sqr(starting_value)

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