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From my knowledge, ImageSize is an option that adjust the width of an image.

Plot[x, {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}, Axes -> False, 
  AspectRatio -> Automatic, ImageSize -> 100] // ImageDimensions

{167,167}


Plot[x, {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}, Axes -> False, 
  AspectRatio -> Automatic, ImageSize -> 1000] // ImageDimensions


{1667,1667}

The numbe 167, 1667 may be related to 5/3 = 1.66666......

I think it would be better if the result was {100,100} and {1000,1000}.

Where does the ratio 1.66666...... come from ?

J. M.'s missing motivation
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imida k
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1 Answers1

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UPDATE

Ian Hojnicki clarifies in the comments:

It’s not really magnification. It’s just how you convert points into pixels. They are different units after all. – ihojnicki

Therefore, what I call "magnification" in the original answer should be called "conversion factor" from points to pixels.

Original answer

By default under Windows Mathematica 13 uses hidden global magnification equal to 96/72:

SystemInformation["Devices", "ConnectedDisplays"]

screeenshot

On my system the display scale is set to 100% under the "Display" in the Windows "Settings" application, and your code returns Ceiling[size*96/72.]:

Plot[x, {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}, Axes -> False, AspectRatio -> Automatic,
   ImageSize -> 1000] // ImageDimensions

{1334, 1334}

Table[Ceiling[size*96/72.], {size, {100, 1000, 10000}}]

{134, 1334, 13334}

Probably your display scale is set to 125%, and hence you get the value Ceiling[size*1.25*96/72.]:

Table[Ceiling[size*1.25*96/72.], {size, {100, 1000, 10000}}]

{167, 1667, 16667}

See also this answer of mine for additional information:

Alexey Popkov
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