Mrthfx

What's a good word to describe a public place that looks like it wouldn't be rough?

Can a hotel cancel a confirmed reservation?

Can we use the stored gravitational potential energy of a building to produce power?

The vanishing of sum of coefficients: symmetric polynomials

"On one hand" vs "on the one hand."

Can you earn endless XP using a Flameskull and its self-revival feature?

How is the Incom shipyard still in business?

What are the advantages of using `make` for small projects?

Number of FLOP (Floating Point Operations) for exponentiation

En Passant For Beginners

What makes the Forgotten Realms "forgotten"?

Why did the villain in the first Men in Black movie care about Earth's Cockroaches?

What is the time complexity of enqueue and dequeue of a queue implemented with a singly linked list?

Does Windows 10's telemetry include sending *.doc files if Word crashed?

Dilemma of explaining to interviewer that he is the reason for declining second interview

Why zero tolerance on nudity in space?

Do authors have to be politically correct in article-writing?

Unwarranted claim of higher degree of accuracy in zircon geochronology

How to prove teleportation does not violate non-cloning theorem?

page split between longtable caption and table

Can polymorphing monsters spam their ability to effectively give themselves a massive health pool?

Does my logo design convey the right feelings for a University Student's Council?

A starship is travelling at 0.9c and collides with a small rock. Will it leave a clean hole through, or will more happen?

Avoiding morning and evening handshakes

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:

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 5 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 5 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 5 hours ago

AybeAybe

1536

$endgroup$

share|cite|improve this question

asked 5 hours ago

AybeAybe

1536

share|cite|improve this question

asked 5 hours ago

AybeAybe

1536

asked 5 hours ago

AybeAybe

1536

asked 5 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 (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...

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 11 mins ago

answered 38 mins ago

Jean MarieJean Marie

30.3k42153

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3132764%2fwhat-formula-could-mimic-the-following-curve%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 (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...

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 11 mins ago

answered 38 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 (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...

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 11 mins ago

answered 38 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 (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...

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 11 mins ago

answered 38 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 11 mins ago

answered 38 mins ago

Jean MarieJean Marie

30.3k42153

answered 38 mins ago

Jean MarieJean Marie

30.3k42153

answered 38 mins ago

Jean MarieJean Marie

30.3k42153