Re: color surface according to absolut value of the gradient
- To: mathgroup at smc.vnet.net
- Subject: [mg131460] Re: color surface according to absolut value of the gradient
- From: conrad.clausz at gmail.com
- Date: Wed, 10 Jul 2013 03:36:15 -0400 (EDT)
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- References: <20130708082302.A7B1B69DC@smc.vnet.net> <krgbh0$nvb$1@smc.vnet.net>
Thanks Bob, this did the trick. In my original data I had to rescale the range but then it worked perfectly.
Conrad
Am Dienstag, 9. Juli 2013 08:44:48 UTC+2 schrieb Bob Hanlon:
> Use ColorFunctionScaling -> False:
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> ListPlot3D[data,
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> ColorFunction -> (ColorData["Rainbow"][
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> Sqrt[D[f[x, #2], x]^2 + D[f[#1, y], y]^2] /.
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> {x -> #1, y -> #2}] &),
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> ColorFunctionScaling -> False]
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> Bob Hanlon
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> On Mon, Jul 8, 2013 at 4:23 AM, <conrad.clausz at gmail.com> wrote:
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> > I have a set of data, let's say
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> > data = Flatten[
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> > Table[{i, j,
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> > Sin[i] Cos[
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> > j]}, {i, -\[Pi], \[Pi], .25}, {j, -\[Pi], \[Pi], .25}], 1];
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> > If I plot this, everything is fine. To get the derivatives I interpolated
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> > the set of data
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> > f = Interpolation[data];
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> > If I then plot the absolute value of the gradient in the same range it
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> > also looks correct.
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> > ListPlot3D[
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> > Flatten[Table[{i, j,
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> > Sqrt[D[f[x, y], x]^2 + D[f[x, y], y]^2] /. {x -> i,
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> > y -> j}}, {i, -\[Pi], \[Pi], .25}, {j, -\[Pi], \[Pi], .25}], 1]]
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> > Now I want to plot the original data with the surface colored according to
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> > the values of the last plot. I tried
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> > ListPlot3D[data,
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> > ColorFunction -> (ColorData["Rainbow"][
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> > Sqrt[D[f[x, #2], x]^2 + D[f[#1, y], y]^2] /. {x -> #1,
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> > y -> #2}] &)]
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> > but the result is definitely off. What am I making wrong?
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> > Thanks in advance,
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> > Conrad
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> >
- References:
- color surface according to absolut value of the gradient
- From: conrad.clausz@gmail.com
- color surface according to absolut value of the gradient