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What formula could mimic the following curve?


Formula for area under the curveWhat do I call this curveFitting a curve - trending line formulaWhose curvature is smoother: Clothoid or Hypocycloid? Benefits of Hypocycloid in general for making smooth path between two points.Create a formula that creates a curve between two pointsderivative curve formulaWhat is the equation of the following polar curve?The pattern of Mills Mess juggling and a link invariant in 3D?Could the length of a curve be $0$?how is the following curve not simple curve?













2












$begingroup$


For the purpose of deforming a 3D mesh, I am looking for a formula to generate a curve I could evaluate like the following:



enter image description here



Its shape would be more or less a simplified version of wind waves over an ocean, where it starts slowly and ends more abruptly.



Which formula, if any, could allow me to draw such curve ?










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    For the purpose of deforming a 3D mesh, I am looking for a formula to generate a curve I could evaluate like the following:



    enter image description here



    Its shape would be more or less a simplified version of wind waves over an ocean, where it starts slowly and ends more abruptly.



    Which formula, if any, could allow me to draw such curve ?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      For the purpose of deforming a 3D mesh, I am looking for a formula to generate a curve I could evaluate like the following:



      enter image description here



      Its shape would be more or less a simplified version of wind waves over an ocean, where it starts slowly and ends more abruptly.



      Which formula, if any, could allow me to draw such curve ?










      share|cite|improve this question









      $endgroup$




      For the purpose of deforming a 3D mesh, I am looking for a formula to generate a curve I could evaluate like the following:



      enter image description here



      Its shape would be more or less a simplified version of wind waves over an ocean, where it starts slowly and ends more abruptly.



      Which formula, if any, could allow me to draw such curve ?







      curves






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      AybeAybe

      1536




      1536






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          Try the function



          $$f(x)=arctanleft(frac{asin(x-c)}{b+acos x}right) + d$$



          Also try $f(f(x))$ and other compositions of $f$ with itself.



          Screenshot
          The image shows the function $f(f(x))$, with $a=0.9$, $b=1$, $c=0.7$, $d=0.4$.



          I recommend that you use desmos to preview the function. For your convenience, here is a template that I have created. Just change the sliders to adjust the constants to your liking. You can also scale the $x$-axis if the peaks are spread out too much.



          I hope this helps.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you, exactly what I was looking for :)
            $endgroup$
            – Aybe
            3 hours ago



















          0












          $begingroup$

          @Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.



          I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.



          1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by



          $$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$



          Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:



          $$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$



          (this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).



          Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...



          enter image description here



          Fig. 1. A linear algebra solution : the blue curve as a "shadow" of the red curve.



          2) A "numerical analysis" method using quadratic splines.



          I will not enter into the details because it is not sure at all that you are acquainted with such curves, which are made of parabolas connected in a "smooth" way.



          Here is the Matlab program that explains all (notice the red, magenta and plus parabola with right translation variable $k$) and the figure it generates :



          clear all;close all;hold on;
          t=0:0.01:1;
          for k=0:5:15
          plot(2*t+k,t.^2,'r');
          plot(-2*t.^2+4*t+2+k,-4*t.^2+4*t+1,'m');
          plot(4+t.^2+k,(1-t).^2,'b');
          end;


          enter image description here



          Fig. 2 : A quadratic spline solution based on 3 parabolas (red, magenta, blue) connected in a smooth way, repeated "ad libidum".






          share|cite|improve this answer











          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Try the function



            $$f(x)=arctanleft(frac{asin(x-c)}{b+acos x}right) + d$$



            Also try $f(f(x))$ and other compositions of $f$ with itself.



            Screenshot
            The image shows the function $f(f(x))$, with $a=0.9$, $b=1$, $c=0.7$, $d=0.4$.



            I recommend that you use desmos to preview the function. For your convenience, here is a template that I have created. Just change the sliders to adjust the constants to your liking. You can also scale the $x$-axis if the peaks are spread out too much.



            I hope this helps.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Thank you, exactly what I was looking for :)
              $endgroup$
              – Aybe
              3 hours ago
















            5












            $begingroup$

            Try the function



            $$f(x)=arctanleft(frac{asin(x-c)}{b+acos x}right) + d$$



            Also try $f(f(x))$ and other compositions of $f$ with itself.



            Screenshot
            The image shows the function $f(f(x))$, with $a=0.9$, $b=1$, $c=0.7$, $d=0.4$.



            I recommend that you use desmos to preview the function. For your convenience, here is a template that I have created. Just change the sliders to adjust the constants to your liking. You can also scale the $x$-axis if the peaks are spread out too much.



            I hope this helps.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Thank you, exactly what I was looking for :)
              $endgroup$
              – Aybe
              3 hours ago














            5












            5








            5





            $begingroup$

            Try the function



            $$f(x)=arctanleft(frac{asin(x-c)}{b+acos x}right) + d$$



            Also try $f(f(x))$ and other compositions of $f$ with itself.



            Screenshot
            The image shows the function $f(f(x))$, with $a=0.9$, $b=1$, $c=0.7$, $d=0.4$.



            I recommend that you use desmos to preview the function. For your convenience, here is a template that I have created. Just change the sliders to adjust the constants to your liking. You can also scale the $x$-axis if the peaks are spread out too much.



            I hope this helps.






            share|cite|improve this answer











            $endgroup$



            Try the function



            $$f(x)=arctanleft(frac{asin(x-c)}{b+acos x}right) + d$$



            Also try $f(f(x))$ and other compositions of $f$ with itself.



            Screenshot
            The image shows the function $f(f(x))$, with $a=0.9$, $b=1$, $c=0.7$, $d=0.4$.



            I recommend that you use desmos to preview the function. For your convenience, here is a template that I have created. Just change the sliders to adjust the constants to your liking. You can also scale the $x$-axis if the peaks are spread out too much.



            I hope this helps.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 4 hours ago









            Haris GusicHaris Gusic

            2,168120




            2,168120








            • 1




              $begingroup$
              Thank you, exactly what I was looking for :)
              $endgroup$
              – Aybe
              3 hours ago














            • 1




              $begingroup$
              Thank you, exactly what I was looking for :)
              $endgroup$
              – Aybe
              3 hours ago








            1




            1




            $begingroup$
            Thank you, exactly what I was looking for :)
            $endgroup$
            – Aybe
            3 hours ago




            $begingroup$
            Thank you, exactly what I was looking for :)
            $endgroup$
            – Aybe
            3 hours ago











            0












            $begingroup$

            @Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.



            I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.



            1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by



            $$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$



            Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:



            $$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$



            (this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).



            Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...



            enter image description here



            Fig. 1. A linear algebra solution : the blue curve as a "shadow" of the red curve.



            2) A "numerical analysis" method using quadratic splines.



            I will not enter into the details because it is not sure at all that you are acquainted with such curves, which are made of parabolas connected in a "smooth" way.



            Here is the Matlab program that explains all (notice the red, magenta and plus parabola with right translation variable $k$) and the figure it generates :



            clear all;close all;hold on;
            t=0:0.01:1;
            for k=0:5:15
            plot(2*t+k,t.^2,'r');
            plot(-2*t.^2+4*t+2+k,-4*t.^2+4*t+1,'m');
            plot(4+t.^2+k,(1-t).^2,'b');
            end;


            enter image description here



            Fig. 2 : A quadratic spline solution based on 3 parabolas (red, magenta, blue) connected in a smooth way, repeated "ad libidum".






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              @Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.



              I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.



              1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by



              $$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$



              Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:



              $$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$



              (this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).



              Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...



              enter image description here



              Fig. 1. A linear algebra solution : the blue curve as a "shadow" of the red curve.



              2) A "numerical analysis" method using quadratic splines.



              I will not enter into the details because it is not sure at all that you are acquainted with such curves, which are made of parabolas connected in a "smooth" way.



              Here is the Matlab program that explains all (notice the red, magenta and plus parabola with right translation variable $k$) and the figure it generates :



              clear all;close all;hold on;
              t=0:0.01:1;
              for k=0:5:15
              plot(2*t+k,t.^2,'r');
              plot(-2*t.^2+4*t+2+k,-4*t.^2+4*t+1,'m');
              plot(4+t.^2+k,(1-t).^2,'b');
              end;


              enter image description here



              Fig. 2 : A quadratic spline solution based on 3 parabolas (red, magenta, blue) connected in a smooth way, repeated "ad libidum".






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                @Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.



                I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.



                1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by



                $$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$



                Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:



                $$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$



                (this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).



                Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...



                enter image description here



                Fig. 1. A linear algebra solution : the blue curve as a "shadow" of the red curve.



                2) A "numerical analysis" method using quadratic splines.



                I will not enter into the details because it is not sure at all that you are acquainted with such curves, which are made of parabolas connected in a "smooth" way.



                Here is the Matlab program that explains all (notice the red, magenta and plus parabola with right translation variable $k$) and the figure it generates :



                clear all;close all;hold on;
                t=0:0.01:1;
                for k=0:5:15
                plot(2*t+k,t.^2,'r');
                plot(-2*t.^2+4*t+2+k,-4*t.^2+4*t+1,'m');
                plot(4+t.^2+k,(1-t).^2,'b');
                end;


                enter image description here



                Fig. 2 : A quadratic spline solution based on 3 parabolas (red, magenta, blue) connected in a smooth way, repeated "ad libidum".






                share|cite|improve this answer











                $endgroup$



                @Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.



                I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.



                1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by



                $$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$



                Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:



                $$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$



                (this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).



                Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...



                enter image description here



                Fig. 1. A linear algebra solution : the blue curve as a "shadow" of the red curve.



                2) A "numerical analysis" method using quadratic splines.



                I will not enter into the details because it is not sure at all that you are acquainted with such curves, which are made of parabolas connected in a "smooth" way.



                Here is the Matlab program that explains all (notice the red, magenta and plus parabola with right translation variable $k$) and the figure it generates :



                clear all;close all;hold on;
                t=0:0.01:1;
                for k=0:5:15
                plot(2*t+k,t.^2,'r');
                plot(-2*t.^2+4*t+2+k,-4*t.^2+4*t+1,'m');
                plot(4+t.^2+k,(1-t).^2,'b');
                end;


                enter image description here



                Fig. 2 : A quadratic spline solution based on 3 parabolas (red, magenta, blue) connected in a smooth way, repeated "ad libidum".







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 7 mins ago

























                answered 34 mins ago









                Jean MarieJean Marie

                30.3k42153




                30.3k42153






























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