Looking for a sine curve with custom midpoint












1












$begingroup$


For a signal generating software, I'm looking for a function that generates a sine-based curve but with a shifted midpoint. Usually, a sine curve looks like this:



in       out  slope

-------------------

0.0 pi    0       1

0.5 pi    1       0

1.0 pi    0      -1

1.5 pi   -1       0

2.0 pi    0       1


I need to move the midpoint from pi to an arbitrary other value, like 0.7 pi or 1.4 pi. The quarter points should still be in the middle of their side, but if they're slightly off to make the curve "look better", that's fine, too. Example:



in       out  slope

-------------------

0.0 pi    0       1

0.7 pi    1       0

1.4 pi    0      -1

1.7 pi   -1       0

2.0 pi    0       1


What function gives me these values?






curves







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    For a signal generating software, I'm looking for a function that generates a sine-based curve but with a shifted midpoint. Usually, a sine curve looks like this:



    in       out  slope
    
-------------------
    
0.0 pi    0       1
    
0.5 pi    1       0
    
1.0 pi    0      -1
    
1.5 pi   -1       0
    
2.0 pi    0       1


    I need to move the midpoint from pi to an arbitrary other value, like 0.7 pi or 1.4 pi. The quarter points should still be in the middle of their side, but if they're slightly off to make the curve "look better", that's fine, too. Example:



    in       out  slope
    
-------------------
    
0.0 pi    0       1
    
0.7 pi    1       0
    
1.4 pi    0      -1
    
1.7 pi   -1       0
    
2.0 pi    0       1


    What function gives me these values?






    curves







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      For a signal generating software, I'm looking for a function that generates a sine-based curve but with a shifted midpoint. Usually, a sine curve looks like this:



      in       out  slope
      
-------------------
      
0.0 pi    0       1
      
0.5 pi    1       0
      
1.0 pi    0      -1
      
1.5 pi   -1       0
      
2.0 pi    0       1


      I need to move the midpoint from pi to an arbitrary other value, like 0.7 pi or 1.4 pi. The quarter points should still be in the middle of their side, but if they're slightly off to make the curve "look better", that's fine, too. Example:



      in       out  slope
      
-------------------
      
0.0 pi    0       1
      
0.7 pi    1       0
      
1.4 pi    0      -1
      
1.7 pi   -1       0
      
2.0 pi    0       1


      What function gives me these values?






      curves







      share|cite|improve this question









      $endgroup$




      For a signal generating software, I'm looking for a function that generates a sine-based curve but with a shifted midpoint. Usually, a sine curve looks like this:



      in       out  slope
      
-------------------
      
0.0 pi    0       1
      
0.5 pi    1       0
      
1.0 pi    0      -1
      
1.5 pi   -1       0
      
2.0 pi    0       1


      I need to move the midpoint from pi to an arbitrary other value, like 0.7 pi or 1.4 pi. The quarter points should still be in the middle of their side, but if they're slightly off to make the curve "look better", that's fine, too. Example:



      in       out  slope
      
-------------------
      
0.0 pi    0       1
      
0.7 pi    1       0
      
1.4 pi    0      -1
      
1.7 pi   -1       0
      
2.0 pi    0       1


      What function gives me these values?






      curves




      curves






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 20:33









      ygoeygoe

      1062




      1062






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          The simplest is to define a piecewise linear function that takes the new $x$ values to the old ones. So let $$y=begin {cases} frac x{1.4} & x le 1.4pi\pi+frac {x-1.4 pi}{0.6}&1.4pi lt x le 2pi end {cases}$$ and use $sin (y)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:23












          • $begingroup$
            True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
            $endgroup$
            – Ross Millikan
            Dec 9 '18 at 22:47



















          0












          $begingroup$

          Not exactly what you asked for but a derivable periodic function ($a$ is the "control parameter") :
          $$f_a(x):=sinleft(x-a,sin(x/2)^2right)$$
          $f_1(x);$ would produce the smooth :



          f_1(x)



          The idea is to replace the 'input' identity function $;xmapsto x;$ with something growing more slowly in $(0,2pi)$ and with derivative $1$ at $2kpi$ as illustrated here ( $xmapsto x-a,sin(x/2)^2$ ):



          modified input




          If you want the input function to be linear at the left you may indeed use a cubic curve to come back to the next linear part. I preferred the $sin$ function in the map :



          $$xmapsto begin{cases} x-a,x & x < pi\ x-a,xdfrac{1-sin(x-3pi/2)}2&x ge pi end{cases}$$
          half linear



          If the first part of the $sin$ function has to be preserved (stretched only) you'll have to replace the $,x < pi,$ by $,x < frac{pi}{1-a},$ with a somewhat more complicated function :
          $$xmapsto begin{cases} x-a,x & x < frac{pi}{1-a}\ x-dfrac{a,x}2left(1-sinleft(dfrac {x-pi-frac{pi}{2(1-a)}}{2-frac 1{1-a}}right)right)&x ge frac{pi}{1-a} end{cases}$$
          In this case we need $;0le ale dfrac 12;$ and the second part of the $sin$ function will collapse entirely for $a=dfrac 12$.



          Final result for $a=0.3$ :



          first sin






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
            $endgroup$
            – Raymond Manzoni
            Dec 9 '18 at 21:34










          • $begingroup$
            Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:22










          • $begingroup$
            I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:47










          • $begingroup$
            The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
            $endgroup$
            – ygoe
            Dec 12 '18 at 20:41










          • $begingroup$
            Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
            $endgroup$
            – Raymond Manzoni
            Dec 12 '18 at 22:50











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          The simplest is to define a piecewise linear function that takes the new $x$ values to the old ones. So let $$y=begin {cases} frac x{1.4} & x le 1.4pi\pi+frac {x-1.4 pi}{0.6}&1.4pi lt x le 2pi end {cases}$$ and use $sin (y)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:23












          • $begingroup$
            True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
            $endgroup$
            – Ross Millikan
            Dec 9 '18 at 22:47
















          0












          $begingroup$

          The simplest is to define a piecewise linear function that takes the new $x$ values to the old ones. So let $$y=begin {cases} frac x{1.4} & x le 1.4pi\pi+frac {x-1.4 pi}{0.6}&1.4pi lt x le 2pi end {cases}$$ and use $sin (y)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:23












          • $begingroup$
            True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
            $endgroup$
            – Ross Millikan
            Dec 9 '18 at 22:47














          0












          0








          0





          $begingroup$

          The simplest is to define a piecewise linear function that takes the new $x$ values to the old ones. So let $$y=begin {cases} frac x{1.4} & x le 1.4pi\pi+frac {x-1.4 pi}{0.6}&1.4pi lt x le 2pi end {cases}$$ and use $sin (y)$






          share|cite|improve this answer









          $endgroup$



          The simplest is to define a piecewise linear function that takes the new $x$ values to the old ones. So let $$y=begin {cases} frac x{1.4} & x le 1.4pi\pi+frac {x-1.4 pi}{0.6}&1.4pi lt x le 2pi end {cases}$$ and use $sin (y)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 21:23









          Ross MillikanRoss Millikan

          295k23198371




          295k23198371












          • $begingroup$
            While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:23












          • $begingroup$
            True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
            $endgroup$
            – Ross Millikan
            Dec 9 '18 at 22:47


















          • $begingroup$
            While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:23












          • $begingroup$
            True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
            $endgroup$
            – Ross Millikan
            Dec 9 '18 at 22:47
















          $begingroup$
          While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:23






          $begingroup$
          While that's simple and comprehensible, there's a break at the midpoint where the curve suddenly changes direction. It's not smooth. (Forgive me my simple math words, I've learnt them in German only at school, long ago.)
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:23














          $begingroup$
          True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
          $endgroup$
          – Ross Millikan
          Dec 9 '18 at 22:47




          $begingroup$
          True, there is a break in the derivative. You have five points based on what you specified. You can put any curve you want through them. You could use a Bezier curve, which is continuous through the second derivative, but is harder to derive.
          $endgroup$
          – Ross Millikan
          Dec 9 '18 at 22:47











          0












          $begingroup$

          Not exactly what you asked for but a derivable periodic function ($a$ is the "control parameter") :
          $$f_a(x):=sinleft(x-a,sin(x/2)^2right)$$
          $f_1(x);$ would produce the smooth :



          f_1(x)



          The idea is to replace the 'input' identity function $;xmapsto x;$ with something growing more slowly in $(0,2pi)$ and with derivative $1$ at $2kpi$ as illustrated here ( $xmapsto x-a,sin(x/2)^2$ ):



          modified input




          If you want the input function to be linear at the left you may indeed use a cubic curve to come back to the next linear part. I preferred the $sin$ function in the map :



          $$xmapsto begin{cases} x-a,x & x < pi\ x-a,xdfrac{1-sin(x-3pi/2)}2&x ge pi end{cases}$$
          half linear



          If the first part of the $sin$ function has to be preserved (stretched only) you'll have to replace the $,x < pi,$ by $,x < frac{pi}{1-a},$ with a somewhat more complicated function :
          $$xmapsto begin{cases} x-a,x & x < frac{pi}{1-a}\ x-dfrac{a,x}2left(1-sinleft(dfrac {x-pi-frac{pi}{2(1-a)}}{2-frac 1{1-a}}right)right)&x ge frac{pi}{1-a} end{cases}$$
          In this case we need $;0le ale dfrac 12;$ and the second part of the $sin$ function will collapse entirely for $a=dfrac 12$.



          Final result for $a=0.3$ :



          first sin






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
            $endgroup$
            – Raymond Manzoni
            Dec 9 '18 at 21:34










          • $begingroup$
            Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:22










          • $begingroup$
            I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:47










          • $begingroup$
            The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
            $endgroup$
            – ygoe
            Dec 12 '18 at 20:41










          • $begingroup$
            Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
            $endgroup$
            – Raymond Manzoni
            Dec 12 '18 at 22:50
















          0












          $begingroup$

          Not exactly what you asked for but a derivable periodic function ($a$ is the "control parameter") :
          $$f_a(x):=sinleft(x-a,sin(x/2)^2right)$$
          $f_1(x);$ would produce the smooth :



          f_1(x)



          The idea is to replace the 'input' identity function $;xmapsto x;$ with something growing more slowly in $(0,2pi)$ and with derivative $1$ at $2kpi$ as illustrated here ( $xmapsto x-a,sin(x/2)^2$ ):



          modified input




          If you want the input function to be linear at the left you may indeed use a cubic curve to come back to the next linear part. I preferred the $sin$ function in the map :



          $$xmapsto begin{cases} x-a,x & x < pi\ x-a,xdfrac{1-sin(x-3pi/2)}2&x ge pi end{cases}$$
          half linear



          If the first part of the $sin$ function has to be preserved (stretched only) you'll have to replace the $,x < pi,$ by $,x < frac{pi}{1-a},$ with a somewhat more complicated function :
          $$xmapsto begin{cases} x-a,x & x < frac{pi}{1-a}\ x-dfrac{a,x}2left(1-sinleft(dfrac {x-pi-frac{pi}{2(1-a)}}{2-frac 1{1-a}}right)right)&x ge frac{pi}{1-a} end{cases}$$
          In this case we need $;0le ale dfrac 12;$ and the second part of the $sin$ function will collapse entirely for $a=dfrac 12$.



          Final result for $a=0.3$ :



          first sin






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
            $endgroup$
            – Raymond Manzoni
            Dec 9 '18 at 21:34










          • $begingroup$
            Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:22










          • $begingroup$
            I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:47










          • $begingroup$
            The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
            $endgroup$
            – ygoe
            Dec 12 '18 at 20:41










          • $begingroup$
            Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
            $endgroup$
            – Raymond Manzoni
            Dec 12 '18 at 22:50














          0












          0








          0





          $begingroup$

          Not exactly what you asked for but a derivable periodic function ($a$ is the "control parameter") :
          $$f_a(x):=sinleft(x-a,sin(x/2)^2right)$$
          $f_1(x);$ would produce the smooth :



          f_1(x)



          The idea is to replace the 'input' identity function $;xmapsto x;$ with something growing more slowly in $(0,2pi)$ and with derivative $1$ at $2kpi$ as illustrated here ( $xmapsto x-a,sin(x/2)^2$ ):



          modified input




          If you want the input function to be linear at the left you may indeed use a cubic curve to come back to the next linear part. I preferred the $sin$ function in the map :



          $$xmapsto begin{cases} x-a,x & x < pi\ x-a,xdfrac{1-sin(x-3pi/2)}2&x ge pi end{cases}$$
          half linear



          If the first part of the $sin$ function has to be preserved (stretched only) you'll have to replace the $,x < pi,$ by $,x < frac{pi}{1-a},$ with a somewhat more complicated function :
          $$xmapsto begin{cases} x-a,x & x < frac{pi}{1-a}\ x-dfrac{a,x}2left(1-sinleft(dfrac {x-pi-frac{pi}{2(1-a)}}{2-frac 1{1-a}}right)right)&x ge frac{pi}{1-a} end{cases}$$
          In this case we need $;0le ale dfrac 12;$ and the second part of the $sin$ function will collapse entirely for $a=dfrac 12$.



          Final result for $a=0.3$ :



          first sin






          share|cite|improve this answer











          $endgroup$



          Not exactly what you asked for but a derivable periodic function ($a$ is the "control parameter") :
          $$f_a(x):=sinleft(x-a,sin(x/2)^2right)$$
          $f_1(x);$ would produce the smooth :



          f_1(x)



          The idea is to replace the 'input' identity function $;xmapsto x;$ with something growing more slowly in $(0,2pi)$ and with derivative $1$ at $2kpi$ as illustrated here ( $xmapsto x-a,sin(x/2)^2$ ):



          modified input




          If you want the input function to be linear at the left you may indeed use a cubic curve to come back to the next linear part. I preferred the $sin$ function in the map :



          $$xmapsto begin{cases} x-a,x & x < pi\ x-a,xdfrac{1-sin(x-3pi/2)}2&x ge pi end{cases}$$
          half linear



          If the first part of the $sin$ function has to be preserved (stretched only) you'll have to replace the $,x < pi,$ by $,x < frac{pi}{1-a},$ with a somewhat more complicated function :
          $$xmapsto begin{cases} x-a,x & x < frac{pi}{1-a}\ x-dfrac{a,x}2left(1-sinleft(dfrac {x-pi-frac{pi}{2(1-a)}}{2-frac 1{1-a}}right)right)&x ge frac{pi}{1-a} end{cases}$$
          In this case we need $;0le ale dfrac 12;$ and the second part of the $sin$ function will collapse entirely for $a=dfrac 12$.



          Final result for $a=0.3$ :



          first sin







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:30

























          answered Dec 9 '18 at 21:24









          Raymond ManzoniRaymond Manzoni

          37.1k563117




          37.1k563117












          • $begingroup$
            ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
            $endgroup$
            – Raymond Manzoni
            Dec 9 '18 at 21:34










          • $begingroup$
            Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:22










          • $begingroup$
            I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:47










          • $begingroup$
            The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
            $endgroup$
            – ygoe
            Dec 12 '18 at 20:41










          • $begingroup$
            Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
            $endgroup$
            – Raymond Manzoni
            Dec 12 '18 at 22:50


















          • $begingroup$
            ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
            $endgroup$
            – Raymond Manzoni
            Dec 9 '18 at 21:34










          • $begingroup$
            Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:22










          • $begingroup$
            I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
            $endgroup$
            – ygoe
            Dec 9 '18 at 22:47










          • $begingroup$
            The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
            $endgroup$
            – ygoe
            Dec 12 '18 at 20:41










          • $begingroup$
            Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
            $endgroup$
            – Raymond Manzoni
            Dec 12 '18 at 22:50
















          $begingroup$
          ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
          $endgroup$
          – Raymond Manzoni
          Dec 9 '18 at 21:34




          $begingroup$
          ($xmapsto x-a,sin(x/2)$ appears nicer in $(0,2pi)$ but is reverted in the following interval)
          $endgroup$
          – Raymond Manzoni
          Dec 9 '18 at 21:34












          $begingroup$
          Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:22




          $begingroup$
          Hm, I've tried it but the result's didn't look too good. Especially the selection of the midpoint is very limited. But the idea of putting another function before the sin(x) is interesting. I'll have to play with that for a bit.
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:22












          $begingroup$
          I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:47




          $begingroup$
          I've found something interesting. A Bézier curve might be helpful to build a smooth connection from the first to the second section. Then apply sin on that. I'll look into that tomorrow.
          $endgroup$
          – ygoe
          Dec 9 '18 at 22:47












          $begingroup$
          The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
          $endgroup$
          – ygoe
          Dec 12 '18 at 20:41




          $begingroup$
          The Bézier function indeed did lead to a curve similar to what I had in mind. There's a JavaScript implementation of it which I ported to C# for my case. github.com/gre/bezier-easing I use control points that are symmetric so that there's no break when repeating the pattern. The resulting "easing" value is then put into sin().
          $endgroup$
          – ygoe
          Dec 12 '18 at 20:41












          $begingroup$
          Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
          $endgroup$
          – Raymond Manzoni
          Dec 12 '18 at 22:50




          $begingroup$
          Thanks for the follow up @ygoe! You may propose your solution as an answer here, excellent continuation,
          $endgroup$
          – Raymond Manzoni
          Dec 12 '18 at 22:50


















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