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What formula could mimic the following curve?

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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:

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

asked 4 hours ago

AybeAybe

1536

$endgroup$

add a comment | 

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:

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

asked 4 hours ago

AybeAybe

1536

$endgroup$

add a comment | 

$begingroup$

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

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

asked 4 hours ago

AybeAybe

1536

$endgroup$

share|cite|improve this question

asked 4 hours ago

AybeAybe

1536

share|cite|improve this question

asked 4 hours ago

AybeAybe

1536

asked 4 hours ago

AybeAybe

1536

asked 4 hours ago

AybeAybe

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.

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

edited 3 hours ago

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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 curve of $y=sin(x)^n$ (in red) and its perspective image in blue 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...

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;

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

edited 8 mins ago

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

$endgroup$

add a comment | 

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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.

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

edited 3 hours ago

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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.

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

edited 3 hours ago

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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.

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

edited 3 hours ago

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

answered 4 hours ago

Haris GusicHaris Gusic

2,168120

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 curve of $y=sin(x)^n$ (in red) and its perspective image in blue 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...

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;

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

edited 8 mins ago

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

$endgroup$

add a comment | 

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 curve of $y=sin(x)^n$ (in red) and its perspective image in blue 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...

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;

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

edited 8 mins ago

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

$endgroup$

add a comment | 

$begingroup$

@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 curve of $y=sin(x)^n$ (in red) and its perspective image in blue 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...

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;

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

edited 8 mins ago

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

$endgroup$

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;

share|cite|improve this answer

edited 8 mins ago

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

answered 26 mins ago

Jean MarieJean Marie

30.3k42153

answered 26 mins ago

Jean MarieJean Marie

30.3k42153