Re: color surface according to absolut value of the gradient

*To*: undisclosed-recipients:;*Subject*: [mg131456] Re: color surface according to absolut value of the gradient*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Tue, 9 Jul 2013 06:50:29 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130708082302.A7B1B69DC@smc.vnet.net>

Use ColorFunctionScaling -> False: ListPlot3D[data, ColorFunction -> (ColorData["Rainbow"][ Sqrt[D[f[x, #2], x]^2 + D[f[#1, y], y]^2] /. {x -> #1, y -> #2}] &), ColorFunctionScaling -> False] Bob Hanlon On Mon, Jul 8, 2013 at 4:23 AM, <conrad.clausz at gmail.com> wrote: > I have a set of data, let's say > > data = Flatten[ > Table[{i, j, > Sin[i] Cos[ > j]}, {i, -\[Pi], \[Pi], .25}, {j, -\[Pi], \[Pi], .25}], 1]; > > If I plot this, everything is fine. To get the derivatives I interpolated > the set of data > > f = Interpolation[data]; > > If I then plot the absolute value of the gradient in the same range it > also looks correct. > > ListPlot3D[ > Flatten[Table[{i, j, > Sqrt[D[f[x, y], x]^2 + D[f[x, y], y]^2] /. {x -> i, > y -> j}}, {i, -\[Pi], \[Pi], .25}, {j, -\[Pi], \[Pi], .25}], 1]] > > Now I want to plot the original data with the surface colored according to > the values of the last plot. I tried > > ListPlot3D[data, > ColorFunction -> (ColorData["Rainbow"][ > Sqrt[D[f[x, #2], x]^2 + D[f[#1, y], y]^2] /. {x -> #1, > y -> #2}] &)] > > but the result is definitely off. What am I making wrong? > > Thanks in advance, > Conrad > >

**References**:**color surface according to absolut value of the gradient***From:*conrad.clausz@gmail.com