How can I derive a set of equations from this figure that will give me the maximum radius for a circle at any...
$begingroup$
Consider the following figure:
It is a circle with a number of points around the circumference. They are equidistant in the figure, but they do not necessarily have to be. Each point on the circle is the center of a smaller circle. Each point on that circle has a circle centered on it, and so on. You can keep iterating and adding additional circles on the points to infinity.
What I am trying to come up with is a way to determine a maximum radius for the point-centered circles for each iteration ($r_1$, $r_2$, $r_3$ in the figure) so that that the circles do not overlap with any circles in the current or previous iterations. (The red circles in the figure demonstrate the condition I am trying to avoid.)
Also, just like biggest circle does not have to have equidistant points, the subsequent iteration's points do not have to be equidistant either, and every iteration and every set of circles can be different.
Each circle can also have a different number of points. (I have just made them the same in the figure.)
geometry circle
edited Dec 18 '18 at 6:56
Saad
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
$endgroup$
add a comment |
$begingroup$
Consider the following figure:
It is a circle with a number of points around the circumference. They are equidistant in the figure, but they do not necessarily have to be. Each point on the circle is the center of a smaller circle. Each point on that circle has a circle centered on it, and so on. You can keep iterating and adding additional circles on the points to infinity.
What I am trying to come up with is a way to determine a maximum radius for the point-centered circles for each iteration ($r_1$, $r_2$, $r_3$ in the figure) so that that the circles do not overlap with any circles in the current or previous iterations. (The red circles in the figure demonstrate the condition I am trying to avoid.)
Also, just like biggest circle does not have to have equidistant points, the subsequent iteration's points do not have to be equidistant either, and every iteration and every set of circles can be different.
Each circle can also have a different number of points. (I have just made them the same in the figure.)
geometry circle
edited Dec 18 '18 at 6:56
Saad
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
$endgroup$
$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52
add a comment |
$begingroup$
Consider the following figure:
It is a circle with a number of points around the circumference. They are equidistant in the figure, but they do not necessarily have to be. Each point on the circle is the center of a smaller circle. Each point on that circle has a circle centered on it, and so on. You can keep iterating and adding additional circles on the points to infinity.
What I am trying to come up with is a way to determine a maximum radius for the point-centered circles for each iteration ($r_1$, $r_2$, $r_3$ in the figure) so that that the circles do not overlap with any circles in the current or previous iterations. (The red circles in the figure demonstrate the condition I am trying to avoid.)
Also, just like biggest circle does not have to have equidistant points, the subsequent iteration's points do not have to be equidistant either, and every iteration and every set of circles can be different.
Each circle can also have a different number of points. (I have just made them the same in the figure.)
geometry circle
edited Dec 18 '18 at 6:56
Saad
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
$endgroup$
Consider the following figure:
It is a circle with a number of points around the circumference. They are equidistant in the figure, but they do not necessarily have to be. Each point on the circle is the center of a smaller circle. Each point on that circle has a circle centered on it, and so on. You can keep iterating and adding additional circles on the points to infinity.
What I am trying to come up with is a way to determine a maximum radius for the point-centered circles for each iteration ($r_1$, $r_2$, $r_3$ in the figure) so that that the circles do not overlap with any circles in the current or previous iterations. (The red circles in the figure demonstrate the condition I am trying to avoid.)
Also, just like biggest circle does not have to have equidistant points, the subsequent iteration's points do not have to be equidistant either, and every iteration and every set of circles can be different.
Each circle can also have a different number of points. (I have just made them the same in the figure.)
geometry circle
geometry circle
edited Dec 18 '18 at 6:56
Saad
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
edited Dec 18 '18 at 6:56
Saad
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
edited Dec 18 '18 at 6:56
Saad
19.7k92352
edited Dec 18 '18 at 6:56
Saad
19.7k92352
edited Dec 18 '18 at 6:56
Saad
19.7k92352
19.7k92352
asked Dec 18 '18 at 6:41
JaybillJaybill
11
asked Dec 18 '18 at 6:41
JaybillJaybill
11
asked Dec 18 '18 at 6:41
JaybillJaybill
11
11
$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52
add a comment |
$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52
$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52
add a comment |
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$begingroup$
Must you have at least two points on each circle as centers of new circles?
$endgroup$
– John Wayland Bales
Dec 18 '18 at 7:23
$begingroup$
Are you interested in the maximal $r_n$ given that all of the $n-1$ previous ones are also the maximal ?
$endgroup$
– Sar
Dec 18 '18 at 7:28
$begingroup$
@JohnWaylandBales Yes, each circle will have at least 2 points.
$endgroup$
– Jaybill
Dec 18 '18 at 7:51
$begingroup$
@Sar I'm sorry, I don't quite understand the question.
$endgroup$
– Jaybill
Dec 18 '18 at 7:52