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How can an equation having ColorDistance function be solved?



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Can't solve equation having complex coefficientsRecursive functions taking list argumentsThe system can not be solvedHow to interpolate colors on scattered points for a smooth visualizationHow to add a common color-bar to a set of 2d maps produced from evaluating a function along three projections in Mathematica 8.0?How to imitate the opacity plot in EditColorFunctionHandling oscilloscope bitmap dataHow can I combine the ChromaticityPlot3D with a Sphere?Can this be solved even faster?










6












$begingroup$


I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
DistanceFunction -> "CIE2000"] == 0.01, l]


I got an error message.



How can I automate this process to complete the series.










share|improve this question









$endgroup$
















    6












    $begingroup$


    I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



    Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
    DistanceFunction -> "CIE2000"] == 0.01, l]


    I got an error message.



    How can I automate this process to complete the series.










    share|improve this question









    $endgroup$














      6












      6








      6


      1



      $begingroup$


      I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



      Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
      DistanceFunction -> "CIE2000"] == 0.01, l]


      I got an error message.



      How can I automate this process to complete the series.










      share|improve this question









      $endgroup$




      I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



      Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
      DistanceFunction -> "CIE2000"] == 0.01, l]


      I got an error message.



      How can I automate this process to complete the series.







      list-manipulation equation-solving color






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Apr 5 at 8:13









      MajisMajis

      1,477415




      1,477415




















          1 Answer
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          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45















          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45













          8












          8








          8





          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$



          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 5 at 8:26









          SzabolcsSzabolcs

          165k15452955




          165k15452955











          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45
















          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45















          $begingroup$
          Perfect. Now I can get the complete list. Thanks.
          $endgroup$
          – Majis
          Apr 5 at 8:45




          $begingroup$
          Perfect. Now I can get the complete list. Thanks.
          $endgroup$
          – Majis
          Apr 5 at 8:45

















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