Polar coordinates in terms of distance and arc length
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On the picture, we have a point on the circle $(a, b)$ parametrized using the arc length and distance to diameter. As you can see, $a = rtheta$. I want to write polar coordinates, $(r, theta)$ in terms of $(a, b)$. As you can see, if we palce this circle to origin, this is quite like transformation from cartesian to polar coordinates but with one ceveat: $a$ is not $x$ coordinate value but arc length.
geometry euclidean-geometry
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On the picture, we have a point on the circle $(a, b)$ parametrized using the arc length and distance to diameter. As you can see, $a = rtheta$. I want to write polar coordinates, $(r, theta)$ in terms of $(a, b)$. As you can see, if we palce this circle to origin, this is quite like transformation from cartesian to polar coordinates but with one ceveat: $a$ is not $x$ coordinate value but arc length.
geometry euclidean-geometry
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On the picture, we have a point on the circle $(a, b)$ parametrized using the arc length and distance to diameter. As you can see, $a = rtheta$. I want to write polar coordinates, $(r, theta)$ in terms of $(a, b)$. As you can see, if we palce this circle to origin, this is quite like transformation from cartesian to polar coordinates but with one ceveat: $a$ is not $x$ coordinate value but arc length.
geometry euclidean-geometry
On the picture, we have a point on the circle $(a, b)$ parametrized using the arc length and distance to diameter. As you can see, $a = rtheta$. I want to write polar coordinates, $(r, theta)$ in terms of $(a, b)$. As you can see, if we palce this circle to origin, this is quite like transformation from cartesian to polar coordinates but with one ceveat: $a$ is not $x$ coordinate value but arc length.
geometry euclidean-geometry
asked Jul 31 at 10:21
meguli
35829
35829
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1 Answer
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Probably the equation b/r = cos(a/r) cannot be solved for r with elementary methods/functions. And r must be a solution to this equation.
You could transform to (b/a)*x = cos(x) with x=a/r.
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Probably the equation b/r = cos(a/r) cannot be solved for r with elementary methods/functions. And r must be a solution to this equation.
You could transform to (b/a)*x = cos(x) with x=a/r.
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
add a comment |Â
up vote
1
down vote
Probably the equation b/r = cos(a/r) cannot be solved for r with elementary methods/functions. And r must be a solution to this equation.
You could transform to (b/a)*x = cos(x) with x=a/r.
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Probably the equation b/r = cos(a/r) cannot be solved for r with elementary methods/functions. And r must be a solution to this equation.
You could transform to (b/a)*x = cos(x) with x=a/r.
Probably the equation b/r = cos(a/r) cannot be solved for r with elementary methods/functions. And r must be a solution to this equation.
You could transform to (b/a)*x = cos(x) with x=a/r.
answered Jul 31 at 14:06
jtm
112
112
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
add a comment |Â
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
I am also looking into $cos(theta) / theta = b / a$ which is another form for what you said. Still, it does not help me write $(r, theta)$ in terms of $a, b$.
â meguli
Aug 1 at 5:09
add a comment |Â
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