What formula could mimic the following curve?
$begingroup$
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
$endgroup$
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
$endgroup$
1
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago
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
$endgroup$
For the purpose of deforming a 3D mesh, I am looking for a formula to generate a curve I could evaluate like the following:
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
curves
asked 9 hours ago
AybeAybe
1636
1636
1
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago
add a comment |
1
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago
1
1
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour ago
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".
$endgroup$
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
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active
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votes
$begingroup$
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.
$endgroup$
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour 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.
$endgroup$
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour 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.
$endgroup$
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.
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.
edited 8 hours ago
answered 9 hours ago
Haris GusicHaris Gusic
2,228121
2,228121
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour ago
add a comment |
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour ago
1
1
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
$begingroup$
Thank you, exactly what I was looking for :)
$endgroup$
– Aybe
8 hours ago
2
2
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
@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
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour ago
$begingroup$
@Aybe: Replace $x$ with $2pi (x-x_0)$ for some $x_0$?
$endgroup$
– Mehrdad
1 hour ago
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".
$endgroup$
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
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".
$endgroup$
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
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".
$endgroup$
@Haris Gusic : I have seen your solution which fits nicely the objectives of the asker with its 3 different tunable parameters.
I propose here two alternatives, one which is more intuitive, using linear algebra, the other one which is more 'numerical analysis' oriented.
1) I have been striken by the fact that the curve desired by Aybe is a perspective (or shadow) of a sine curve (or a power of a sine curve). Let us see it on the (red) curve of $y=sin(x)^n$ and its (blue) perspective image, which is an alternate answer to the question, with parametric equations given by
$$begin{cases}x&=&t+0.8sin(t)^n\y&=&0.1sin(t)^nend{cases} text{here, with } n=4.$$
Why that ? This "shadowing effect" is rendered by a so-called "skewing" or "transvection" linear operation with an upper triangular matrix:
$$color{blue}{binom{x}{y}}=begin{pmatrix}1&0.8\0&0.1end{pmatrix}color{red}{binom{t}{sin(t)^n}}$$
(this matrix reflects the fact that the horizontal direction is preserved whereas the former vertical direction has been bent rightwards).
Remark : the second column entries of the matrix as well as the sine exponent are tunable... You can even, in this way, obtain breaking waves...
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".
edited 5 hours ago
answered 5 hours ago
Jean MarieJean Marie
30.4k42153
30.4k42153
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
add a comment |
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
$begingroup$
Thank you, this looks very interesting but I don't understand how I can draw it from the formulas you've posted :)
$endgroup$
– Aybe
1 hour ago
add a comment |
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1
$begingroup$
anyone for function golf on Area 51? (similar to code golf) ;-)
$endgroup$
– uhoh
2 hours ago