graphics - AspectRatio option works unpredictably?

Let us examine the agreement between specified value for the AspectRatio option and the actual aspect ratio of generated plot. For this purpose I define a function which uses the Rasterize trick for determination of the actual aspect ratio:

realAspectRatio = #2/#1 & @@ (#2 - #1) & @@ 
    Rasterize[
      Show[#, Epilog -> {Annotation[
          Rectangle[ImageScaled[{0, 0}], ImageScaled[{1, 1}]], "Two", 
          "Region"]}], "Regions"][[-1, 2]] &;

(Instead of realAspectRatio@ one can use much more basic but lesser precise function #2/#1 & @@ ImageDimensions@ what gives the same results.)

At first, we check the agreement for simple graphics Graphics[{Point[{{0, 0}, {1, 1}}]}]:

table = Table[{10^p, realAspectRatio@
     Graphics[{Point[{{0, 0}, {1, 1}}]}, AspectRatio -> 10^p]}, {p, -2, 2, .1}];

lm = LinearModelFit[table, x, x];
lm["BestFit"]
ListPlot[lm["FitResiduals"], Frame -> True, 
         FrameLabel -> {"AspectRatio", "fit residual"}]

1.26421 + 0.603143 x

One can see that actual aspect ratio substantially differs from the requested via the AspectRatio option and the dependence between them is nonlinear!

In this case the situation becomes MUCH better if we provide at least horizontal PlotRange specification PlotRange -> {{0, 1}, Automatic}:

table = Table[{10^p, 
    realAspectRatio@
     Graphics[{Point[{{0, 0}, {1, 10}}]}, AspectRatio -> 10^p, 
      PlotRange -> {{0, 1}, Automatic}]}, {p, -2, 2, .1}];
lm = LinearModelFit[table, x, x];
lm["BestFit"]
ListPlot[lm["FitResiduals"], Frame -> True, 
 FrameLabel -> {"AspectRatio", "fit residual"}]

0.00391481 + 1.00225 x

But for a little more complex graphics ListPlot[Prime[Range[25]]] this remedy does not help even if we vary AspectRatio in much lesser diapason from 1/5 to 5:

table = Table[{ar, realAspectRatio@
      ListPlot[Prime[Range[25]], PlotRange -> {{0, 25}, {0, 100}}, 
       AspectRatio -> ar]}, {ar, 1/5, 5, .1}];
lm = LinearModelFit[table, x, x];
lm["BestFit"]

ListPlot[lm["FitResiduals"], Frame -> True, 
 FrameLabel -> {"AspectRatio", "fit residual"}]

0.248369 + 0.768459 x

How this behavior can be explained? Is it intended behavior? And more importantly: is there a way to make the actual aspect ratio predictable?

---

Summary of the discussion and the conclusion

The source of confusion was the first sentence under the "Details" section on the Documentation page for AspectRatio: "AspectRatio determines the scaling for the final image shape." This statement is clearly wrong and more correct statement can be found as the first point under the "Properties and Relations" subsection in the "Examples" section on the same page: "AspectRatio determines the ratio of PlotRange, not ImageSize." But as it is clearly demonstrated in dedicated answer, even this statement is not precise because actually AspectRatio determines the aspect ratio of the plotting area which is specified by both PlotRange and PlotRangePadding.

Answer

It seems to me that there is little point in making the image size so small that the image cannot possibly have a close to accurate aspect ratio. If I also get rid of the superfluous parts outside of the plot range by setting PlotRangePadding, ImagePadding, and ImageMargins all to zero, I can use simple ImageDimensions to check aspect ratio. Doing all this I find that it is quite precise in 10.1 under Windows:

table = Table[{10^p, 
    N[#2/#] & @@ 
     ImageDimensions@
      Image@Graphics[{Disk[]}, AspectRatio -> 10^p, PlotRangePadding -> 0, 
        ImagePadding -> 0, ImageMargins -> 0, ImageSize -> {2000}]}, {p, -2, 2, .1}];

lm = LinearModelFit[table, x, x];

lm["BestFit"]
ListPlot[lm["FitResiduals"], Frame -> True, FrameLabel -> {"AspectRatio", "fit residual"},
  DataRange -> {10^-2, 10^2}, PlotRange -> All]

-0.00279057 + 1.00046 x

Likewise your ListPlot example:

table = Table[{ar, 

    N[#2/#] & @@ 
     ImageDimensions@
      Image@ListPlot[Prime[Range[25]], PlotRange -> {{0, 25}, {0, 100}}, 
        AspectRatio -> ar, PlotRangePadding -> 0, ImagePadding -> 0, ImageMargins -> 0, 
        ImageSize -> {2000}]}, {ar, 1/5, 5, .1}];

lm = LinearModelFit[table, x, x];
lm["BestFit"]
ListPlot[lm["FitResiduals"], Frame -> True, FrameLabel -> {"AspectRatio", "fit residual"},
 DataRange -> {1/5, 5}]

0.000344083 + 0.999798 x