What formula could mimic the following curve?












8












$begingroup$


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











$endgroup$








  • 6




    $begingroup$
    anyone for function golf on Area 51? (similar to code golf) ;-)
    $endgroup$
    – uhoh
    Mar 3 at 2:56








  • 1




    $begingroup$
    I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
    $endgroup$
    – Jean Marie
    Mar 3 at 8:27


















8












$begingroup$


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











$endgroup$








  • 6




    $begingroup$
    anyone for function golf on Area 51? (similar to code golf) ;-)
    $endgroup$
    – uhoh
    Mar 3 at 2:56








  • 1




    $begingroup$
    I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
    $endgroup$
    – Jean Marie
    Mar 3 at 8:27
















8












8








8


0



$begingroup$


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 ?







geometry curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 3 at 8:27









Jean Marie

30.6k42154




30.6k42154










asked Mar 2 at 19:43









AybeAybe

1836




1836








  • 6




    $begingroup$
    anyone for function golf on Area 51? (similar to code golf) ;-)
    $endgroup$
    – uhoh
    Mar 3 at 2:56








  • 1




    $begingroup$
    I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
    $endgroup$
    – Jean Marie
    Mar 3 at 8:27
















  • 6




    $begingroup$
    anyone for function golf on Area 51? (similar to code golf) ;-)
    $endgroup$
    – uhoh
    Mar 3 at 2:56








  • 1




    $begingroup$
    I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
    $endgroup$
    – Jean Marie
    Mar 3 at 8:27










6




6




$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
Mar 3 at 2:56






$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
Mar 3 at 2:56






1




1




$begingroup$
I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
$endgroup$
– Jean Marie
Mar 3 at 8:27






$begingroup$
I have taken the liberty to add the tag "geometry" to the tag "curves" ( a tag "shape" would have been the most accurate)
$endgroup$
– Jean Marie
Mar 3 at 8:27












2 Answers
2






active

oldest

votes


















16












$begingroup$

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.



EDIT: As per the suggestion by @J. M. is not a mathematician, you can replace $arctan$ with the function $$g(x) = frac{px}{sqrt{q+(px)^2}}$$
if you need a greater variety of waves.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, exactly what I was looking for :)
    $endgroup$
    – Aybe
    Mar 2 at 21:38






  • 6




    $begingroup$
    What a lovely function. Can you perhaps say a few words on how you came up with it?
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 3:35






  • 1




    $begingroup$
    @Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
    $endgroup$
    – Aybe
    Mar 3 at 4:32








  • 2




    $begingroup$
    @Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
    $endgroup$
    – Mehrdad
    Mar 3 at 4:37








  • 1




    $begingroup$
    @Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
    $endgroup$
    – Aybe
    Mar 3 at 5:47





















10












$begingroup$

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



I propose here two alternatives, an intuitive one, using linear algebra, and another one more 'numerical analysis' oriented.



1) I have been striken by the fact that the curve desired by Aybe can be considered as a perspective view (or shadow) of a sine curve (or a power of a sine curve) : see Fig. 1 displaying the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, with parametric equations given by



$$begin{cases}x&=&t+asin(t)^n\y&=&bsin(t)^nend{cases} text{here, with } begin{cases}n&=&4\a&=&0.8\b&=&0.1end{cases}$$



enter image description here



Fig. 1. The blue curve as a "shadow" of the red curve.



Why that ? This "shadow effect" is rendered by a so-called horizontal "shear mapping" (https://en.wikipedia.org/wiki/Shear_mapping) or "transvection", a linear operation with an upper triangular matrix:



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



(The first column of this matrix reflects the fact that the horizontal direction is preserved whereas the second column with $a,b>0$ gives to the former vertical direction a certain leaning to the right).



enter image description here



Fig. 2 : Graphical representation using Desmos. Please note that $sin$ function has been placed between absolute value signs in order to allow non-integer exponents. A supplementary parameter $m$ has also been introduced. This is a very tunable solution : one can even, in this way, obtain breaking waves...



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 arcs of parabolas connected in a "smooth" way (https://wordsandbuttons.online/quadric_splines_are_useful_too.html).



enter image description here



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



Here is the Matlab program that has generated Figure 3 (please note the three plotting operations for the red, magenta and blue parabolas with right translation variable $k$) :



clear all;close all;hold on;
t=0:0.01:1;
for k=0:6:18
plot(t.^2 + 2*t+k, t.^2,'r');
plot(-2*t.^2+4*t+3+k,-2*t.^2+2*t+1,'m');
plot(t.^2+5+k,(1-t).^2,'b');
end;


If you want to do the same with Desmos, here is a way to do it (it can be very instructive to enlarge a little the domain of parameter $t$ by taking for example $-0.5 leq t leq 1.5$ in order to understand what are these parabolas):



enter image description here



Fig. 4.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
    $endgroup$
    – Aybe
    Mar 3 at 4:13










  • $begingroup$
    @Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
    $endgroup$
    – Jean Marie
    Mar 3 at 7:11






  • 1




    $begingroup$
    @Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 8:36










  • $begingroup$
    Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
    $endgroup$
    – Aybe
    Mar 3 at 10:01










  • $begingroup$
    This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
    $endgroup$
    – Aybe
    Mar 4 at 4:35













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









16












$begingroup$

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.



EDIT: As per the suggestion by @J. M. is not a mathematician, you can replace $arctan$ with the function $$g(x) = frac{px}{sqrt{q+(px)^2}}$$
if you need a greater variety of waves.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, exactly what I was looking for :)
    $endgroup$
    – Aybe
    Mar 2 at 21:38






  • 6




    $begingroup$
    What a lovely function. Can you perhaps say a few words on how you came up with it?
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 3:35






  • 1




    $begingroup$
    @Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
    $endgroup$
    – Aybe
    Mar 3 at 4:32








  • 2




    $begingroup$
    @Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
    $endgroup$
    – Mehrdad
    Mar 3 at 4:37








  • 1




    $begingroup$
    @Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
    $endgroup$
    – Aybe
    Mar 3 at 5:47


















16












$begingroup$

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.



EDIT: As per the suggestion by @J. M. is not a mathematician, you can replace $arctan$ with the function $$g(x) = frac{px}{sqrt{q+(px)^2}}$$
if you need a greater variety of waves.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, exactly what I was looking for :)
    $endgroup$
    – Aybe
    Mar 2 at 21:38






  • 6




    $begingroup$
    What a lovely function. Can you perhaps say a few words on how you came up with it?
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 3:35






  • 1




    $begingroup$
    @Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
    $endgroup$
    – Aybe
    Mar 3 at 4:32








  • 2




    $begingroup$
    @Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
    $endgroup$
    – Mehrdad
    Mar 3 at 4:37








  • 1




    $begingroup$
    @Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
    $endgroup$
    – Aybe
    Mar 3 at 5:47
















16












16








16





$begingroup$

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.



EDIT: As per the suggestion by @J. M. is not a mathematician, you can replace $arctan$ with the function $$g(x) = frac{px}{sqrt{q+(px)^2}}$$
if you need a greater variety of waves.






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.



EDIT: As per the suggestion by @J. M. is not a mathematician, you can replace $arctan$ with the function $$g(x) = frac{px}{sqrt{q+(px)^2}}$$
if you need a greater variety of waves.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 3 at 10:56

























answered Mar 2 at 20:18









Haris GusicHaris Gusic

2,603322




2,603322








  • 1




    $begingroup$
    Thank you, exactly what I was looking for :)
    $endgroup$
    – Aybe
    Mar 2 at 21:38






  • 6




    $begingroup$
    What a lovely function. Can you perhaps say a few words on how you came up with it?
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 3:35






  • 1




    $begingroup$
    @Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
    $endgroup$
    – Aybe
    Mar 3 at 4:32








  • 2




    $begingroup$
    @Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
    $endgroup$
    – Mehrdad
    Mar 3 at 4:37








  • 1




    $begingroup$
    @Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
    $endgroup$
    – Aybe
    Mar 3 at 5:47
















  • 1




    $begingroup$
    Thank you, exactly what I was looking for :)
    $endgroup$
    – Aybe
    Mar 2 at 21:38






  • 6




    $begingroup$
    What a lovely function. Can you perhaps say a few words on how you came up with it?
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 3:35






  • 1




    $begingroup$
    @Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
    $endgroup$
    – Aybe
    Mar 3 at 4:32








  • 2




    $begingroup$
    @Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
    $endgroup$
    – Mehrdad
    Mar 3 at 4:37








  • 1




    $begingroup$
    @Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
    $endgroup$
    – Aybe
    Mar 3 at 5:47










1




1




$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
Mar 2 at 21:38




$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
Mar 2 at 21:38




6




6




$begingroup$
What a lovely function. Can you perhaps say a few words on how you came up with it?
$endgroup$
– J. M. is not a mathematician
Mar 3 at 3:35




$begingroup$
What a lovely function. Can you perhaps say a few words on how you came up with it?
$endgroup$
– J. M. is not a mathematician
Mar 3 at 3:35




1




1




$begingroup$
@Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
$endgroup$
– Aybe
Mar 3 at 4:32






$begingroup$
@Haris Gusic I am trying to 'map' the interesting range to the 0 to 1 range but I am struggling, if you have an idea it's welcome!
$endgroup$
– Aybe
Mar 3 at 4:32






2




2




$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
Mar 3 at 4:37






$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
Mar 3 at 4:37






1




1




$begingroup$
@Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
$endgroup$
– Aybe
Mar 3 at 5:47






$begingroup$
@Mehrdad It works except that it doesn't start at (0, 0). Desmos tells when sign changes so I know it starts at (-1.63, -0.204) and scales to (2PI, 2.0), I could just go on from there but still ... I was hoping to fix it directly in the formula but failed miserably :)
$endgroup$
– Aybe
Mar 3 at 5:47













10












$begingroup$

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



I propose here two alternatives, an intuitive one, using linear algebra, and another one more 'numerical analysis' oriented.



1) I have been striken by the fact that the curve desired by Aybe can be considered as a perspective view (or shadow) of a sine curve (or a power of a sine curve) : see Fig. 1 displaying the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, with parametric equations given by



$$begin{cases}x&=&t+asin(t)^n\y&=&bsin(t)^nend{cases} text{here, with } begin{cases}n&=&4\a&=&0.8\b&=&0.1end{cases}$$



enter image description here



Fig. 1. The blue curve as a "shadow" of the red curve.



Why that ? This "shadow effect" is rendered by a so-called horizontal "shear mapping" (https://en.wikipedia.org/wiki/Shear_mapping) or "transvection", a linear operation with an upper triangular matrix:



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



(The first column of this matrix reflects the fact that the horizontal direction is preserved whereas the second column with $a,b>0$ gives to the former vertical direction a certain leaning to the right).



enter image description here



Fig. 2 : Graphical representation using Desmos. Please note that $sin$ function has been placed between absolute value signs in order to allow non-integer exponents. A supplementary parameter $m$ has also been introduced. This is a very tunable solution : one can even, in this way, obtain breaking waves...



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 arcs of parabolas connected in a "smooth" way (https://wordsandbuttons.online/quadric_splines_are_useful_too.html).



enter image description here



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



Here is the Matlab program that has generated Figure 3 (please note the three plotting operations for the red, magenta and blue parabolas with right translation variable $k$) :



clear all;close all;hold on;
t=0:0.01:1;
for k=0:6:18
plot(t.^2 + 2*t+k, t.^2,'r');
plot(-2*t.^2+4*t+3+k,-2*t.^2+2*t+1,'m');
plot(t.^2+5+k,(1-t).^2,'b');
end;


If you want to do the same with Desmos, here is a way to do it (it can be very instructive to enlarge a little the domain of parameter $t$ by taking for example $-0.5 leq t leq 1.5$ in order to understand what are these parabolas):



enter image description here



Fig. 4.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
    $endgroup$
    – Aybe
    Mar 3 at 4:13










  • $begingroup$
    @Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
    $endgroup$
    – Jean Marie
    Mar 3 at 7:11






  • 1




    $begingroup$
    @Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 8:36










  • $begingroup$
    Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
    $endgroup$
    – Aybe
    Mar 3 at 10:01










  • $begingroup$
    This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
    $endgroup$
    – Aybe
    Mar 4 at 4:35


















10












$begingroup$

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



I propose here two alternatives, an intuitive one, using linear algebra, and another one more 'numerical analysis' oriented.



1) I have been striken by the fact that the curve desired by Aybe can be considered as a perspective view (or shadow) of a sine curve (or a power of a sine curve) : see Fig. 1 displaying the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, with parametric equations given by



$$begin{cases}x&=&t+asin(t)^n\y&=&bsin(t)^nend{cases} text{here, with } begin{cases}n&=&4\a&=&0.8\b&=&0.1end{cases}$$



enter image description here



Fig. 1. The blue curve as a "shadow" of the red curve.



Why that ? This "shadow effect" is rendered by a so-called horizontal "shear mapping" (https://en.wikipedia.org/wiki/Shear_mapping) or "transvection", a linear operation with an upper triangular matrix:



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



(The first column of this matrix reflects the fact that the horizontal direction is preserved whereas the second column with $a,b>0$ gives to the former vertical direction a certain leaning to the right).



enter image description here



Fig. 2 : Graphical representation using Desmos. Please note that $sin$ function has been placed between absolute value signs in order to allow non-integer exponents. A supplementary parameter $m$ has also been introduced. This is a very tunable solution : one can even, in this way, obtain breaking waves...



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 arcs of parabolas connected in a "smooth" way (https://wordsandbuttons.online/quadric_splines_are_useful_too.html).



enter image description here



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



Here is the Matlab program that has generated Figure 3 (please note the three plotting operations for the red, magenta and blue parabolas with right translation variable $k$) :



clear all;close all;hold on;
t=0:0.01:1;
for k=0:6:18
plot(t.^2 + 2*t+k, t.^2,'r');
plot(-2*t.^2+4*t+3+k,-2*t.^2+2*t+1,'m');
plot(t.^2+5+k,(1-t).^2,'b');
end;


If you want to do the same with Desmos, here is a way to do it (it can be very instructive to enlarge a little the domain of parameter $t$ by taking for example $-0.5 leq t leq 1.5$ in order to understand what are these parabolas):



enter image description here



Fig. 4.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
    $endgroup$
    – Aybe
    Mar 3 at 4:13










  • $begingroup$
    @Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
    $endgroup$
    – Jean Marie
    Mar 3 at 7:11






  • 1




    $begingroup$
    @Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 8:36










  • $begingroup$
    Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
    $endgroup$
    – Aybe
    Mar 3 at 10:01










  • $begingroup$
    This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
    $endgroup$
    – Aybe
    Mar 4 at 4:35
















10












10








10





$begingroup$

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



I propose here two alternatives, an intuitive one, using linear algebra, and another one more 'numerical analysis' oriented.



1) I have been striken by the fact that the curve desired by Aybe can be considered as a perspective view (or shadow) of a sine curve (or a power of a sine curve) : see Fig. 1 displaying the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, with parametric equations given by



$$begin{cases}x&=&t+asin(t)^n\y&=&bsin(t)^nend{cases} text{here, with } begin{cases}n&=&4\a&=&0.8\b&=&0.1end{cases}$$



enter image description here



Fig. 1. The blue curve as a "shadow" of the red curve.



Why that ? This "shadow effect" is rendered by a so-called horizontal "shear mapping" (https://en.wikipedia.org/wiki/Shear_mapping) or "transvection", a linear operation with an upper triangular matrix:



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



(The first column of this matrix reflects the fact that the horizontal direction is preserved whereas the second column with $a,b>0$ gives to the former vertical direction a certain leaning to the right).



enter image description here



Fig. 2 : Graphical representation using Desmos. Please note that $sin$ function has been placed between absolute value signs in order to allow non-integer exponents. A supplementary parameter $m$ has also been introduced. This is a very tunable solution : one can even, in this way, obtain breaking waves...



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 arcs of parabolas connected in a "smooth" way (https://wordsandbuttons.online/quadric_splines_are_useful_too.html).



enter image description here



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



Here is the Matlab program that has generated Figure 3 (please note the three plotting operations for the red, magenta and blue parabolas with right translation variable $k$) :



clear all;close all;hold on;
t=0:0.01:1;
for k=0:6:18
plot(t.^2 + 2*t+k, t.^2,'r');
plot(-2*t.^2+4*t+3+k,-2*t.^2+2*t+1,'m');
plot(t.^2+5+k,(1-t).^2,'b');
end;


If you want to do the same with Desmos, here is a way to do it (it can be very instructive to enlarge a little the domain of parameter $t$ by taking for example $-0.5 leq t leq 1.5$ in order to understand what are these parabolas):



enter image description here



Fig. 4.






share|cite|improve this answer











$endgroup$



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



I propose here two alternatives, an intuitive one, using linear algebra, and another one more 'numerical analysis' oriented.



1) I have been striken by the fact that the curve desired by Aybe can be considered as a perspective view (or shadow) of a sine curve (or a power of a sine curve) : see Fig. 1 displaying the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, with parametric equations given by



$$begin{cases}x&=&t+asin(t)^n\y&=&bsin(t)^nend{cases} text{here, with } begin{cases}n&=&4\a&=&0.8\b&=&0.1end{cases}$$



enter image description here



Fig. 1. The blue curve as a "shadow" of the red curve.



Why that ? This "shadow effect" is rendered by a so-called horizontal "shear mapping" (https://en.wikipedia.org/wiki/Shear_mapping) or "transvection", a linear operation with an upper triangular matrix:



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



(The first column of this matrix reflects the fact that the horizontal direction is preserved whereas the second column with $a,b>0$ gives to the former vertical direction a certain leaning to the right).



enter image description here



Fig. 2 : Graphical representation using Desmos. Please note that $sin$ function has been placed between absolute value signs in order to allow non-integer exponents. A supplementary parameter $m$ has also been introduced. This is a very tunable solution : one can even, in this way, obtain breaking waves...



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 arcs of parabolas connected in a "smooth" way (https://wordsandbuttons.online/quadric_splines_are_useful_too.html).



enter image description here



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



Here is the Matlab program that has generated Figure 3 (please note the three plotting operations for the red, magenta and blue parabolas with right translation variable $k$) :



clear all;close all;hold on;
t=0:0.01:1;
for k=0:6:18
plot(t.^2 + 2*t+k, t.^2,'r');
plot(-2*t.^2+4*t+3+k,-2*t.^2+2*t+1,'m');
plot(t.^2+5+k,(1-t).^2,'b');
end;


If you want to do the same with Desmos, here is a way to do it (it can be very instructive to enlarge a little the domain of parameter $t$ by taking for example $-0.5 leq t leq 1.5$ in order to understand what are these parabolas):



enter image description here



Fig. 4.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 5 at 5:51

























answered Mar 3 at 0:13









Jean MarieJean Marie

30.6k42154




30.6k42154








  • 1




    $begingroup$
    Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
    $endgroup$
    – Aybe
    Mar 3 at 4:13










  • $begingroup$
    @Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
    $endgroup$
    – Jean Marie
    Mar 3 at 7:11






  • 1




    $begingroup$
    @Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 8:36










  • $begingroup$
    Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
    $endgroup$
    – Aybe
    Mar 3 at 10:01










  • $begingroup$
    This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
    $endgroup$
    – Aybe
    Mar 4 at 4:35
















  • 1




    $begingroup$
    Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
    $endgroup$
    – Aybe
    Mar 3 at 4:13










  • $begingroup$
    @Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
    $endgroup$
    – Jean Marie
    Mar 3 at 7:11






  • 1




    $begingroup$
    @Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
    $endgroup$
    – J. M. is not a mathematician
    Mar 3 at 8:36










  • $begingroup$
    Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
    $endgroup$
    – Aybe
    Mar 3 at 10:01










  • $begingroup$
    This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
    $endgroup$
    – Aybe
    Mar 4 at 4:35










1




1




$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
Mar 3 at 4:13




$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
Mar 3 at 4:13












$begingroup$
@Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
$endgroup$
– Jean Marie
Mar 3 at 7:11




$begingroup$
@Aybe : Desmos, for example, handles as well cartesian graphing ($y=f(x)$) and parametric plot graphing ($x=x(t),y=y(t)$). I just included a way to do it in my text.
$endgroup$
– Jean Marie
Mar 3 at 7:11




1




1




$begingroup$
@Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
$endgroup$
– J. M. is not a mathematician
Mar 3 at 8:36




$begingroup$
@Aybe, to get a periodic function from Jean's second construction, you can compose the piecewise-parabolic function he has with a sawtooth function, as in this answer.
$endgroup$
– J. M. is not a mathematician
Mar 3 at 8:36












$begingroup$
Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
$endgroup$
– Aybe
Mar 3 at 10:01




$begingroup$
Right, I need to do that in front of my computer because it's not exactly easy from a phone :)
$endgroup$
– Aybe
Mar 3 at 10:01












$begingroup$
This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
$endgroup$
– Aybe
Mar 4 at 4:35






$begingroup$
This is really interesting but I just don't understand how to transform it to C# code, usually I do evaluate say 100 steps between 0 and 1 so t or x if you prefer then a generate a vertex out of it; right now I'm puzzled on how to do it yet I'd like to try it!
$endgroup$
– Aybe
Mar 4 at 4:35




















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