Math Forum / Math Software / Mathematica / October 2008

Hi,

the legend here is complete right, try
Plot3D[2 Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}]

and so, simply remove the 2

Plot3D[Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2}]

and the problem is fixed.

Regards
  Jens

In addition to my previous reply I should have taken the automatic
ranging of ContourPlot into account as it scales values to give a 0..1
range.

So, if you want to make sure the legend maps GrayLevels 0..1 to the
-2..2 range (in case the ends of this range are not within the plot
region) you have to adjust the scaling of ContourPlot accordingly:

contourPlotFun =
 ContourPlot[
  2 Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2},
  ColorFunctionScaling -> False,
  ColorFunction -> (GrayLevel[1 - ((# + 2)/4)] &)];

ShowLegend[contourPlotFun, {GrayLevel[1 - #] &, 10, " 2", "-2",
 LegendPosition -> {1.1, -.4}, LegendShadow -> None}]

Cheers -- Sjoerd

Er,

In my version of Mathematica (6.03 WinXP) the ContourPlot is in color
by default. It doesn't make sense to add a grayscale legend. So either
you color the legend or you add a greyscale ColorFunction to the
ContourPlot:

contourPlotFun =
 ContourPlot[
  2 Sin[x y], {x, -\[Pi]/2, \[Pi]/2}, {y, -\[Pi]/2, \[Pi]/2},
  ColorFunction -> (GrayLevel[1 - #] &)];

ShowLegend[contourPlotFun, {GrayLevel[1 - #] &, 10, " 2", "-2",
 LegendPosition -> {1.1, -.4}}]

The legend doesn't adapt to the range of your plot. You have to
provide the range yourself. In the expression above I put in 2 and -2
and that does the trick.

Cheers,

Sjoerd