plotting - Plot3D: ColorFunction depending on (x,y) or z

While trying to answer this question I found out "weird" behavior of ColorFunction

Plot3D[x y, {x, -10, 10}, {y, -10, 10}, PlotRange -> Full, 
 ColorFunction -> Function[{x, y, z}, Hue@(x y)]]

gives incorrect colors

the virtually identical code (from naive point of view)

Plot3D[x y, {x, -10, 10}, {y, -10, 10}, PlotRange -> Full, 
 ColorFunction -> Function[{x, y, z}, Hue@z]]

gives correct plot

Also if you disable ColorFunctionScaling as many posts (e.g. this) suggest

Plot3D[x y, {x, -10, 10}, {y, -10, 10}, PlotRange -> Full, 
 ColorFunction -> Function[{x, y, z}, Hue@(x y)], 
 ColorFunctionScaling -> False]

the colors are really messed up

So the question - why are the first two plots different? A bug?

I'm using Mathematica 11.1 on Linux.

Answer

As the documentation on ColorFunctionScaling says,

ColorFunctionScaling is an option for graphics functions that specifies whether arguments supplied to a color function should be scaled to lie between 0 and 1.

In the first plot, $x$ and $y$ are both scaled to the range $[0,1]$. This means that the ColorFunction does not receive $x$ and $y$, but rather $(x+10)/20$ and $(y+10)/20$, and the color is then the hue of $(x+10)(y+10)/400$.

In the second plot, $z$ is scaled to the range $[0,1]$. This means that the ColorFunction does not receive $z$, but rather $(z+100)/200$ as the third argument.

In the third plot, the number of PlotPoints is too small and thus you see Moiré effects.

Comments

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I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

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