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WANRU LIU

listcontourplot on a sector of a circle

WANRU LIU

Posted 8 years ago

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s = Import["Desktop/tension.xlsx"]; ListContourPlot[s, Contours -> 10, PlotLegends -> Automatic] In fact, it is a contour of a circular part from (2,0) (4,0) to (0,4) (0,2). However, I cannot get a contour on a sector of a circle. It has redundant part How can I delete that red part to make it a sector of a circle?? Thanks a lot!!!:) Attachments:

POSTED BY: WANRU LIU

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Udo Krause

Udo Krause, NOrganization

Posted 8 years ago

Draw a white disk above the disliked area as follows In[1]:= s = Import[FileNameJoin[{NotebookDirectory[], "..", "Abt_skrot", "tension.xlsx"}]]; In[8]:= Show[{ListContourPlot[s, Contours -> 10, PlotLegends -> Automatic, DisplayFunction -> Identity], Graphics[{White, Disk[{0, 0}, 2, {0, \[Pi]/2}]}, DisplayFunction -> Identity]}, DisplayFunction -> $DisplayFunction, AspectRatio -> 1] to see Ask yourself whether this graphics is cheating because the shown inner boundary does not fulfill any boundary condition, right?

Draw a white disk above the disliked area as follows

In[1]:= s = Import[FileNameJoin[{NotebookDirectory[], "..", "Abt_skrot", "tension.xlsx"}]];
In[8]:= Show[{ListContourPlot[s, Contours -> 10, PlotLegends -> Automatic, DisplayFunction -> Identity], 
              Graphics[{White, Disk[{0, 0}, 2, {0, \[Pi]/2}]}, DisplayFunction -> Identity]}, 
             DisplayFunction -> $DisplayFunction, AspectRatio -> 1]

to see

aWhiteDisk

Ask yourself whether this graphics is cheating because the shown inner boundary does not fulfill any boundary condition, right?

POSTED BY: Udo Krause

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