translate linear coordinates to circular space

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I have $5$ points ($P_0 rightarrow P_4$) on a straight line, and $4$ control points ($C_1 rightarrow C_4$) that I use to connect the points with quadratic bezier curves (as seen in the top part of the image below).

What I want to do is bend this 'linear' layout into 'circular' space, as seen in the bottom part. Is there a formula I can apply to my points' $xy$ coordinates to achieve this? If so I would like to have a variable with which I can control the curvature (e.g. $0$ for straight, $0.5$ for semi-circle, $1$ for full circle.

(apologies if I'm not expressing myself well - I'm a coder - so please feel free to edit the question)

coordinate-systems transformation

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edited Jul 26 at 9:54

Kenta S

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asked Jul 26 at 9:47

Hoff

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up vote
0
down vote

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I have $5$ points ($P_0 rightarrow P_4$) on a straight line, and $4$ control points ($C_1 rightarrow C_4$) that I use to connect the points with quadratic bezier curves (as seen in the top part of the image below).

What I want to do is bend this 'linear' layout into 'circular' space, as seen in the bottom part. Is there a formula I can apply to my points' $xy$ coordinates to achieve this? If so I would like to have a variable with which I can control the curvature (e.g. $0$ for straight, $0.5$ for semi-circle, $1$ for full circle.

(apologies if I'm not expressing myself well - I'm a coder - so please feel free to edit the question)

coordinate-systems transformation

share|cite|improve this question

edited Jul 26 at 9:54

Kenta S

1,1371418

asked Jul 26 at 9:47

Hoff

1012

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I have $5$ points ($P_0 rightarrow P_4$) on a straight line, and $4$ control points ($C_1 rightarrow C_4$) that I use to connect the points with quadratic bezier curves (as seen in the top part of the image below).

What I want to do is bend this 'linear' layout into 'circular' space, as seen in the bottom part. Is there a formula I can apply to my points' $xy$ coordinates to achieve this? If so I would like to have a variable with which I can control the curvature (e.g. $0$ for straight, $0.5$ for semi-circle, $1$ for full circle.

(apologies if I'm not expressing myself well - I'm a coder - so please feel free to edit the question)

coordinate-systems transformation

share|cite|improve this question

edited Jul 26 at 9:54

Kenta S

1,1371418

asked Jul 26 at 9:47

Hoff

1012

I have $5$ points ($P_0 rightarrow P_4$) on a straight line, and $4$ control points ($C_1 rightarrow C_4$) that I use to connect the points with quadratic bezier curves (as seen in the top part of the image below).

What I want to do is bend this 'linear' layout into 'circular' space, as seen in the bottom part. Is there a formula I can apply to my points' $xy$ coordinates to achieve this? If so I would like to have a variable with which I can control the curvature (e.g. $0$ for straight, $0.5$ for semi-circle, $1$ for full circle.

(apologies if I'm not expressing myself well - I'm a coder - so please feel free to edit the question)

coordinate-systems transformation

share|cite|improve this question

edited Jul 26 at 9:54

Kenta S

1,1371418

asked Jul 26 at 9:47

Hoff

1012

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 9:54

Kenta S

1,1371418

edited Jul 26 at 9:54

Kenta S

1,1371418

edited Jul 26 at 9:54

Kenta S

1,1371418

1,1371418

asked Jul 26 at 9:47

Hoff

1012

asked Jul 26 at 9:47

Hoff

1012

asked Jul 26 at 9:47

Hoff

1012

1012

add a comment | 

add a comment | 

1 Answer
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WLOG I assume that $P_0=(-1,0)$ and $P_4=(1,0)$ (if not, a translation+rotation+scaling will achieve that).

Now consider a circle of radius $r$ passing by the two given points. By Pythagoras, its center is at $(0,-h)$ with $h:=sqrtr^2-1$, and the half-aperture angle is $alpha=arcsindfrac1r$.

Using polar coordinates with the pole at $(0,-h)$, you will map the abscissas to angles ($xtotheta=dfracpi2+xalpha$) and the ordinates to moduli ($ytorho=r+y$), and

$$X=(r+y)cos(fracpi2+xalpha)=(r+y)sin(xalpha),\Y=(r+y)sin(fracpi2+xalpha)-h=(r+y)cos(xalpha)-h.$$

This way the origin $(0, 0)$ maps to the Cartesian coordinates $(0,r-h)$ and the endpoints $(pm1,0)$ to themselves.

---

Note that for very large $r$, $alphaapproxdfrac1r$ and $happrox r$ so that

$$Xapprox(r+y)frac xrapprox x,$$

$$Yapprox (r+y)cdot1-happrox y.$$

share|cite|improve this answer

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

add a comment | 

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1 Answer
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active

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1 Answer
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active

oldest

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votes

up vote
0
down vote

WLOG I assume that $P_0=(-1,0)$ and $P_4=(1,0)$ (if not, a translation+rotation+scaling will achieve that).

Now consider a circle of radius $r$ passing by the two given points. By Pythagoras, its center is at $(0,-h)$ with $h:=sqrtr^2-1$, and the half-aperture angle is $alpha=arcsindfrac1r$.

Using polar coordinates with the pole at $(0,-h)$, you will map the abscissas to angles ($xtotheta=dfracpi2+xalpha$) and the ordinates to moduli ($ytorho=r+y$), and

$$X=(r+y)cos(fracpi2+xalpha)=(r+y)sin(xalpha),\Y=(r+y)sin(fracpi2+xalpha)-h=(r+y)cos(xalpha)-h.$$

This way the origin $(0, 0)$ maps to the Cartesian coordinates $(0,r-h)$ and the endpoints $(pm1,0)$ to themselves.

---

Note that for very large $r$, $alphaapproxdfrac1r$ and $happrox r$ so that

$$Xapprox(r+y)frac xrapprox x,$$

$$Yapprox (r+y)cdot1-happrox y.$$

share|cite|improve this answer

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

add a comment | 

up vote
0
down vote

WLOG I assume that $P_0=(-1,0)$ and $P_4=(1,0)$ (if not, a translation+rotation+scaling will achieve that).

Now consider a circle of radius $r$ passing by the two given points. By Pythagoras, its center is at $(0,-h)$ with $h:=sqrtr^2-1$, and the half-aperture angle is $alpha=arcsindfrac1r$.

Using polar coordinates with the pole at $(0,-h)$, you will map the abscissas to angles ($xtotheta=dfracpi2+xalpha$) and the ordinates to moduli ($ytorho=r+y$), and

$$X=(r+y)cos(fracpi2+xalpha)=(r+y)sin(xalpha),\Y=(r+y)sin(fracpi2+xalpha)-h=(r+y)cos(xalpha)-h.$$

This way the origin $(0, 0)$ maps to the Cartesian coordinates $(0,r-h)$ and the endpoints $(pm1,0)$ to themselves.

---

Note that for very large $r$, $alphaapproxdfrac1r$ and $happrox r$ so that

$$Xapprox(r+y)frac xrapprox x,$$

$$Yapprox (r+y)cdot1-happrox y.$$

share|cite|improve this answer

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

add a comment | 

up vote
0
down vote

up vote
0
down vote

WLOG I assume that $P_0=(-1,0)$ and $P_4=(1,0)$ (if not, a translation+rotation+scaling will achieve that).

Now consider a circle of radius $r$ passing by the two given points. By Pythagoras, its center is at $(0,-h)$ with $h:=sqrtr^2-1$, and the half-aperture angle is $alpha=arcsindfrac1r$.

Using polar coordinates with the pole at $(0,-h)$, you will map the abscissas to angles ($xtotheta=dfracpi2+xalpha$) and the ordinates to moduli ($ytorho=r+y$), and

$$X=(r+y)cos(fracpi2+xalpha)=(r+y)sin(xalpha),\Y=(r+y)sin(fracpi2+xalpha)-h=(r+y)cos(xalpha)-h.$$

This way the origin $(0, 0)$ maps to the Cartesian coordinates $(0,r-h)$ and the endpoints $(pm1,0)$ to themselves.

---

Note that for very large $r$, $alphaapproxdfrac1r$ and $happrox r$ so that

$$Xapprox(r+y)frac xrapprox x,$$

$$Yapprox (r+y)cdot1-happrox y.$$

share|cite|improve this answer

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

WLOG I assume that $P_0=(-1,0)$ and $P_4=(1,0)$ (if not, a translation+rotation+scaling will achieve that).

Now consider a circle of radius $r$ passing by the two given points. By Pythagoras, its center is at $(0,-h)$ with $h:=sqrtr^2-1$, and the half-aperture angle is $alpha=arcsindfrac1r$.

Using polar coordinates with the pole at $(0,-h)$, you will map the abscissas to angles ($xtotheta=dfracpi2+xalpha$) and the ordinates to moduli ($ytorho=r+y$), and

$$X=(r+y)cos(fracpi2+xalpha)=(r+y)sin(xalpha),\Y=(r+y)sin(fracpi2+xalpha)-h=(r+y)cos(xalpha)-h.$$

This way the origin $(0, 0)$ maps to the Cartesian coordinates $(0,r-h)$ and the endpoints $(pm1,0)$ to themselves.

---

Note that for very large $r$, $alphaapproxdfrac1r$ and $happrox r$ so that

$$Xapprox(r+y)frac xrapprox x,$$

$$Yapprox (r+y)cdot1-happrox y.$$

share|cite|improve this answer

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 10:19

edited Jul 26 at 10:19

edited Jul 26 at 10:19

answered Jul 26 at 10:14

Yves Daoust

111k665203

answered Jul 26 at 10:14

Yves Daoust

111k665203

answered Jul 26 at 10:14

Yves Daoust

111k665203

111k665203

add a comment | 

add a comment | 

 

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