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Re: Shading Functions


Hi Des,
if you use ColorFunction the values of your function are normalized by
Mathematica to values between 0 and one. (See Help on ColorFunction,
the line reads
"The arguments provided for the function specified by ColorFunction are
always in the range 0 to 1.")
So you get other colors than in you self-made color function where
Mathematica allows you to go cyclically around if you want. (Try 

p2=Plot3D[{Sin[x y], colorFun[3Sin[x y]]},{x,-1,1},{y,-2,2},
AxesLabel->{X,Y,Z}];

to see what happens.)

Jurgen

Des Penny wrote:
> 
> Hi Folks:
> 
> I'm having trouble invoking the shading function in one form.
> 
> The following works ok:
> 
> In[1]:=
> p1=Plot3D[Sin[x y],{x,-1,1},{y,-2,2}, AxesLabel->{X,Y,Z}];
> 
> If we map the range of the plot into {0,0.8}, we will now be able to
> color the heights of a plot so that the lowest points are red and the
> highest points are Blue/Purple.
> 
> In[2]:=
> PlotRange[p1]
> Out[2]=
> {{-1.,1.},{-2.,2.},{-1.04461,1.04461}}
> 
> In[3]:=
> zrange=PlotRange[p1][[3]]
> Out[3]=
> {-1.04461,1.04461}
> 
> The max and min values of Sin are {-1,1}, so this checks.
> 
> We now want to map the heights, z, into the range {0,0.8}. Using the
> linear map, c=a z+b we find c=0.40 z+0.40 will work.
> 
> In[4]:=
> (0.4 zrange+0.4)
> Out[4]=
> {-0.0178423,0.817842}
> 
> If we now operate on each z value with Hue we will display this z value
> in the correct color:
> 
> In[5]:=
> Clear[colorFun];
> colorFun[z_]:=Hue[0.4(z+1)]
> 
> In[6]:=
> p2=Plot3D[{Sin[x y], colorFun[Sin[x y]]},{x,-1,1},{y,-2,2},
> AxesLabel->{X,Y,Z}];
> 
> This works fine.
> 
> However the following does not work correctly.  I'm not sure why.
> 
> In[7]:=
> Show[p2, ColorFunction->colorFun];
> 
> I was under the impression that the above statement caused colorFun to
> operate on the z values of the function.  We should get the same plot
> as the preceeding Plot3D but we don't.  The output of the above clearly
> shows that colorFun is operating, but not correctly.
> 
> Can anyone help explain what's going on?
> 
> Cheers,
> 
> Des Penny



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