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Mathematics and the art of linearizing the circle

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4

1

$begingroup$

One of the most prominent problems of ancient mathematics was the squaring of the circle: to construct the square with the same area as a given circle.

A related problem is linearizing the circle: to find a natural transition between a given line segment of length $L$ and the circle with circumference $U = 2pi R = L$ (which presupposes to find the radius $R = L/2pi$).

The main "problem" is: Along which paths are the points of the line segment to be moved to finally yield the circle such that the transition appears "natural".

By natural I mean this transition:

The points of the line segments follow these paths:

as can be seen here:

To be honest: Even though these paths look very much like circle segments, I'm not quite sure and I didn't define them by an explicit formula (which I didn't have at hand) but heuristically using some support points and splining.

My questions are:

• Are these paths really circle segments?




• If so: How to parametrize them?




• If not so: What kind of curves are they otherwise?

---

Please allow me – freely associating – to compare the pictures above with this (artificially symmetrized) picture of The Great Wave off Kanagawa

modular-arithmetic euclidean-geometry projective-geometry visualization art

share|cite|improve this question

edited 2 hours ago

Hans Stricker

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do we have an actual definition of what it means for this transition to "appear natural"?
  $endgroup$
  – Morgan Rodgers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think it would be "natural" to think of a family of curves $gamma_t$, $tin[0,1]$, of constant curvature such that $gamma_0$ is the flat line (curvature 0), $gamma_1$ is the circle (curvature 1/R) and each $gamma_t$ has curvature $tcdot1/R$.
  $endgroup$
  – Mars Plastic
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you could probably make a map that "appears natural" using circles for the paths, and you could also probably make a different map without using circles that also "appears natural". So without a definition of what that term means, I don't know what answer you are looking for.
  $endgroup$
  – Morgan Rodgers
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just for the composition, +1.
  $endgroup$
  – Allawonder
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe I'm missing something, but this 'transition' seems quite arbitrary to me.
  $endgroup$
  – rafa11111
  1 hour ago
  

  
  
  
  

 | 
show 8 more comments

4

1

$begingroup$

One of the most prominent problems of ancient mathematics was the squaring of the circle: to construct the square with the same area as a given circle.

A related problem is linearizing the circle: to find a natural transition between a given line segment of length $L$ and the circle with circumference $U = 2pi R = L$ (which presupposes to find the radius $R = L/2pi$).

The main "problem" is: Along which paths are the points of the line segment to be moved to finally yield the circle such that the transition appears "natural".

By natural I mean this transition:

The points of the line segments follow these paths:

as can be seen here:

To be honest: Even though these paths look very much like circle segments, I'm not quite sure and I didn't define them by an explicit formula (which I didn't have at hand) but heuristically using some support points and splining.

My questions are:

• Are these paths really circle segments?




• If so: How to parametrize them?




• If not so: What kind of curves are they otherwise?

---

Please allow me – freely associating – to compare the pictures above with this (artificially symmetrized) picture of The Great Wave off Kanagawa

modular-arithmetic euclidean-geometry projective-geometry visualization art

share|cite|improve this question

edited 2 hours ago

Hans Stricker

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do we have an actual definition of what it means for this transition to "appear natural"?
  $endgroup$
  – Morgan Rodgers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think it would be "natural" to think of a family of curves $gamma_t$, $tin[0,1]$, of constant curvature such that $gamma_0$ is the flat line (curvature 0), $gamma_1$ is the circle (curvature 1/R) and each $gamma_t$ has curvature $tcdot1/R$.
  $endgroup$
  – Mars Plastic
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you could probably make a map that "appears natural" using circles for the paths, and you could also probably make a different map without using circles that also "appears natural". So without a definition of what that term means, I don't know what answer you are looking for.
  $endgroup$
  – Morgan Rodgers
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just for the composition, +1.
  $endgroup$
  – Allawonder
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe I'm missing something, but this 'transition' seems quite arbitrary to me.
  $endgroup$
  – rafa11111
  1 hour ago
  

  
  
  
  

 | 
show 8 more comments

$begingroup$

One of the most prominent problems of ancient mathematics was the squaring of the circle: to construct the square with the same area as a given circle.

A related problem is linearizing the circle: to find a natural transition between a given line segment of length $L$ and the circle with circumference $U = 2pi R = L$ (which presupposes to find the radius $R = L/2pi$).

The main "problem" is: Along which paths are the points of the line segment to be moved to finally yield the circle such that the transition appears "natural".

By natural I mean this transition:

The points of the line segments follow these paths:

as can be seen here:

To be honest: Even though these paths look very much like circle segments, I'm not quite sure and I didn't define them by an explicit formula (which I didn't have at hand) but heuristically using some support points and splining.

My questions are:

• Are these paths really circle segments?




• If so: How to parametrize them?




• If not so: What kind of curves are they otherwise?

---

Please allow me – freely associating – to compare the pictures above with this (artificially symmetrized) picture of The Great Wave off Kanagawa

modular-arithmetic euclidean-geometry projective-geometry visualization art

share|cite|improve this question

edited 2 hours ago

Hans Stricker

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

$endgroup$

share|cite|improve this question

edited 2 hours ago

Hans Stricker

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

share|cite|improve this question

edited 2 hours ago

Hans Stricker

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

asked 2 hours ago

Hans StrickerHans Stricker

6,40143992

2 Answers
2

active

oldest

votes

3

$begingroup$

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2pi R/r$, with $r$ varying between $1$ and $+infty$. But I don't know if that is "natural" or not. Here's how it looks:

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

AretinoAretino

24.4k21443

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's essentially my approach. Good to know, that an expert like you came to the same conclusion. Concerning "natural": which approach could be more "natural"?
  $endgroup$
  – Hans Stricker
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How (and with which tools) did you create your elegant animated gif? (I had a hard time with mine.)
  $endgroup$
  – Hans Stricker
  10 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $C_r$ be $r$ and centre be $(-r,0)$, such that $(0,0)$ is on $C_r$ for all $r$. The coordinates of the point at arclength $s$ from $(0,0)$ are then given by $(r (cos(s/r) - 1), r sin(s/r))$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $f_t:[-pi, pi]tomathbb{R}^2$ where $tin[0,1]$ by
begin{align}
f_t(s) &= left(frac{cos(s(1-t))-1}{1-t}, frac{sin(s(1-t))}{1-t}right)&t<1\
f_1(s) &= (0, s)&
end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).

share|cite|improve this answer

answered 1 hour ago

KhorossKhoross

711

$endgroup$

add a comment | 

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3

$begingroup$

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2pi R/r$, with $r$ varying between $1$ and $+infty$. But I don't know if that is "natural" or not. Here's how it looks:

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

AretinoAretino

24.4k21443

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's essentially my approach. Good to know, that an expert like you came to the same conclusion. Concerning "natural": which approach could be more "natural"?
  $endgroup$
  – Hans Stricker
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How (and with which tools) did you create your elegant animated gif? (I had a hard time with mine.)
  $endgroup$
  – Hans Stricker
  10 mins ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2pi R/r$, with $r$ varying between $1$ and $+infty$. But I don't know if that is "natural" or not. Here's how it looks:

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

AretinoAretino

24.4k21443

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's essentially my approach. Good to know, that an expert like you came to the same conclusion. Concerning "natural": which approach could be more "natural"?
  $endgroup$
  – Hans Stricker
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How (and with which tools) did you create your elegant animated gif? (I had a hard time with mine.)
  $endgroup$
  – Hans Stricker
  10 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2pi R/r$, with $r$ varying between $1$ and $+infty$. But I don't know if that is "natural" or not. Here's how it looks:

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

AretinoAretino

24.4k21443

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

AretinoAretino

24.4k21443

answered 1 hour ago

AretinoAretino

24.4k21443

answered 1 hour ago

AretinoAretino

24.4k21443

2

$begingroup$

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $C_r$ be $r$ and centre be $(-r,0)$, such that $(0,0)$ is on $C_r$ for all $r$. The coordinates of the point at arclength $s$ from $(0,0)$ are then given by $(r (cos(s/r) - 1), r sin(s/r))$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $f_t:[-pi, pi]tomathbb{R}^2$ where $tin[0,1]$ by
begin{align}
f_t(s) &= left(frac{cos(s(1-t))-1}{1-t}, frac{sin(s(1-t))}{1-t}right)&t<1\
f_1(s) &= (0, s)&
end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).

share|cite|improve this answer

answered 1 hour ago

KhorossKhoross

711

$endgroup$

add a comment | 

2

$begingroup$

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $C_r$ be $r$ and centre be $(-r,0)$, such that $(0,0)$ is on $C_r$ for all $r$. The coordinates of the point at arclength $s$ from $(0,0)$ are then given by $(r (cos(s/r) - 1), r sin(s/r))$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $f_t:[-pi, pi]tomathbb{R}^2$ where $tin[0,1]$ by
begin{align}
f_t(s) &= left(frac{cos(s(1-t))-1}{1-t}, frac{sin(s(1-t))}{1-t}right)&t<1\
f_1(s) &= (0, s)&
end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).

share|cite|improve this answer

answered 1 hour ago

KhorossKhoross

711

$endgroup$

add a comment | 

$begingroup$

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $C_r$ be $r$ and centre be $(-r,0)$, such that $(0,0)$ is on $C_r$ for all $r$. The coordinates of the point at arclength $s$ from $(0,0)$ are then given by $(r (cos(s/r) - 1), r sin(s/r))$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $f_t:[-pi, pi]tomathbb{R}^2$ where $tin[0,1]$ by
begin{align}
f_t(s) &= left(frac{cos(s(1-t))-1}{1-t}, frac{sin(s(1-t))}{1-t}right)&t<1\
f_1(s) &= (0, s)&
end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).

share|cite|improve this answer

answered 1 hour ago

KhorossKhoross

711

$endgroup$

share|cite|improve this answer

answered 1 hour ago

KhorossKhoross

711

answered 1 hour ago

KhorossKhoross

711

answered 1 hour ago

KhorossKhoross

711