I want to find a set of equally spaced points (blue points on the image) on a disk of radius $r$. Let $\mathbf{x}$ and $\mathbf{z}$ be the orthonormal basis vectors with origin at $O$, the coordinates of a point with respect to the basis is given as $(x,z)$. The radius $r$ is given as

$$\sqrt{x^2 + z^2} = r$$

I also make use of another basis which I call the integer coordinate system with the set of coordinates $(x_n,z_n) \in \mathbb{Z}^* \times \mathbb{Z}^*$, where $\mathbb{Z}^*$ represents the set of nonzero integers.

The coordinates $(x,z)$ is given as

$$x= l\ x_n \left(1-\frac{1}{2|x_n|}\right) \tag{1}$$ $$z= l \ z_n \left(1-\frac{1}{2|z_n|}\right) \tag{2}$$

where $l$ is the spacing between the points with respect to the basis {$\mathbf{x},\mathbf{z}$}.

enter image description here

How can I find the set of points $(x,z)$ that lie within and on this disk of radius $r$ using $(x_n,z_n)$?.

My guess: I think the answer could be:

$$\sqrt{\left[ l\ x_n \left(1-\frac{1}{2|x_n|}\right) \right]^2 + \left[ l \ z_n \left(1-\frac{1}{2|z_n|} \right) \right]^2} \leq r$$

  • $\begingroup$ For a start $\sqrt{x_n^2 + z_n^2} < r^2$ $\endgroup$ – Phil H Sep 12 '18 at 15:49
  • $\begingroup$ The points $(x_n,y_n)$ represents the blue points but in a coordinate system with nonzero integers. For example $(1,1)$, $(1,-1)$, $(-1,1)$ and $(-1,-1)$ are the coordinates for the four blue points around the origin in the nonzero integer coordinate system. The radius $r$ and $l$ is with respect to the basis ${\mathbf{x},\mathbf{z}}$ and not with respect to the integer coordinate system. Then how should I use $\sqrt{x_n^2 + z_n^2} < r^2$?. $\endgroup$ – dykes Sep 12 '18 at 16:32
  • 1
    $\begingroup$ This is an example of circle packing in a circle, where your desired points are just the centers of the interior circles. See en.wikipedia.org/wiki/Circle_packing_in_a_circle $\endgroup$ – David G. Stork Sep 12 '18 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.