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New to the site, but I've run into a math problem, and I thought I'd ask the amazing internet for help. The problem is as follows: I want to find 8 plots in a 100x100 circle 2D plane, all 8 plots must have an equal distance between each other. Furthermore, I need to do the same with 7 plots, 6 plots.. all down to 3 plots. Further-furthermore - I need a way to slightly decrease the circle and then find those plots again with the newly created smaller circle. It all sounds very confusing to me, so I found my inner paint artists and tried to illustrate what I meant.

Click here to see the image. I hope it makes slightly sense. Thank you in advance.

gt6989b
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  • It's all separate circles really. The one with 7 plots would like something like this – Chrazini Apr 30 '18 at 14:50
  • I think you are interested in the Polar Coordinate system. Instead of plotting $(x,y)$ coordinates we plot $(r,\theta)$, where $r$ is distance from $(0,0)$ and $\theta$ is angle, usually in radians instead of degrees, measured from East going counterclockwise. For example, a point that is 3/8 of the way around a circle of radius 5 is $(5,\frac{3}{8}\cdot2\pi)$. Then use the relation $x=r\cdot\cos\theta$ and $y=r\cdot\sin\theta$ to convert to normal coordinates. –  Apr 30 '18 at 15:03

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If you want each of your circles to have the same number of points/plots on them, then just use the formula $$ (x, y) = \left( (50 - kr)\cdot\cos\left(n\cdot\frac{2\pi}{p}\right) + 50, (50 - kr)\cdot\sin\left(n\cdot\frac{2\pi}{p}\right) + 50 \right) $$ where $p$ is the number of plots that you want, $n$ ranges between $0, 1, \ldots, p-1$ to give you equally spaced points around the circle(s), $r$ is the distance between the different sized circles, and $k$ ranges from $0$ to the integer part of $50/r$ where $k = 0$ corresponds to the largest circle, $k=1$ corresponds to the next largest circle, and so on. Is this what you're after?


Example. Suppose you want eight equally spaced points, so that $p = 8$. Then you have the following cosine and sine values: $$ \begin{array}{r|c|c|c|c|c|c|c|c} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline \cos(2n\pi/8) & 1 & 1/\sqrt{2} & 0 & -1/\sqrt{2} & -1 & -1/\sqrt{2} & 0 & 1/\sqrt{2}\\ \sin(2n\pi/8) & 0 & 1/\sqrt{2} & 1 & 1/\sqrt{2} & 0 & -1/\sqrt{2} & -1 & -1/\sqrt{2}\\ \end{array} $$ These can be determined quickly by using Excel (or some variant), especially for other values of $p$. Thus, the coordinates on the largest circle are $$ \begin{array}{cc} (100, 50), & \big((50/\sqrt{2}) + 50, (50/\sqrt{2}) + 50\big),\\ (50, 100), & \big((-50/\sqrt{2}) + 50, (50/\sqrt{2}) + 50\big),\\ (0, 50), & \big((-50/\sqrt{2}) + 50, (-50/\sqrt{2}) + 50\big),\\ (50, 0), & \big((50/\sqrt{2}) + 50, (-50/\sqrt{2}) + 50\big),\\ \end{array} $$ or, in decimal, $$ \begin{array}{cc} (100, 50), & (85.3553, 85.3553),\\ (50, 100), & (14.6447, 85.3553),\\ (0, 50), & (14.6447, 14.6447),\\ (50, 0), & (85.3553, 14.6447).\\ \end{array} $$ If your distance to the next largest circle is $10$, then the coordinates are $$ \begin{array}{cc} (90, 50), & \big((40/\sqrt{2}) + 50, (40/\sqrt{2}) + 50\big),\\ (50, 90), & \big((-40/\sqrt{2}) + 50, (40/\sqrt{2}) + 50\big),\\ (10, 50), & \big((-40/\sqrt{2}) + 50, (-40/\sqrt{2}) + 50\big),\\ (50, 10), & \big((40/\sqrt{2}) + 50, (-40/\sqrt{2}) + 50\big),\\ \end{array} $$ or, in decimal, $$ \begin{array}{cc} (90, 50), & (78.2843, 78.2843),\\ (50, 90), & (21.7157, 78.2843),\\ (10, 50), & (21.7157, 21.7157),\\ (50, 10), & (78.2843, 21.7157).\\ \end{array} $$ All decimal values have been rounded to four significant figures. I hope it is clear now how one might continue, and do the same with different values for $p$.


Excel Spreadsheet. The following image gives a small example of what I would do to generate the points. I have included the equations for cells B$3$, B$4$, B$6$, and B$7$ at the bottom --- these can be dragged across to apply to all cells in the range that you require. Then you can change the values for $p, k, r$ to get the points on different circles according to which circle you want the coordinates for.

enter image description here

Bilbottom
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  • I can't seem to get the results I want. Can you try and give me an example? – Chrazini Apr 30 '18 at 16:33
  • @Chrazini Is this update sufficient? – Bilbottom Apr 30 '18 at 17:09
  • You have been so tremendously helpful. I appreciate the time you put into this very detailed response. I wish you a fantastic day. Thank you so much! – Chrazini Apr 30 '18 at 17:34
  • You mentioned it was easy to determine this quickly using Excel. I can't seem to figure out how to do this easily. Since I need a lot of values, it would save me a tremendous amount of time. I've been trying for a while now, but I can't seem to make it work. I can do the calculations one by one, but making it happen automatically is tough. You've helped me a lot already, so I almost don't want to ask for any more help, but if you have a few more minutes at some point, your assistance would be appreciated. If you don't just ignore this message. I'm still very grateful for all your help. – Chrazini May 02 '18 at 18:00
  • @Chrazini See update. – Bilbottom May 02 '18 at 20:10