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enter image description hereHello i need to make a circle out of 20 smaller ones. The smaller circles radius is r=9.3cm heres what i wanna do:

user145433
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  • http://math.stackexchange.com/questions/429878/1-big-circle-formed-by-27-smaller-circles – lab bhattacharjee Apr 24 '14 at 15:38
  • im not very good at math so thats why i poseted here.What i need is someone to calculate the R of the bigger circle with only the r=9.3 of the 20 smaller circles.This image is what i need to do,is it posible? – user145433 Apr 24 '14 at 15:43
  • Yes it is possible. Write out the equations and you are done. – Calvin Lin Apr 24 '14 at 15:46

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If you take the centres of the smaller circles and draw lines between them, you get a 20 sided polygon (icosagon) with each side being twice the radius of a small circle. If you draw a line from the centre of each small circle to the centre of your large circle, it's $360^\circ/20 = 18^\circ$ for each slice.

If you take the point where two of the small circles touch $ = A$, the centre of one of those small circles $ = B$, and the centre of the large circle $ = C$, you get a triangle $ABC$.

$\overline{AB} = 9.3cm$

$\angle{ACB} = 9^\circ$

$\sin{9^\circ} = \frac{9cm}{h}$

$h \approx 57.53 cm$

So that's the radius of a circle made from the centres of the small circles. The radius you desire is $$\frac{9cm}{\sin{9^\circ}} - 9.3cm \approx 48.23 cm$$

RandomUser
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Draw a triangle between the center of the big circle and the centers of two adjacent small circles, and bisect it to make two narrow right triangles. (Sorry about the bad picture, but you get the idea.)

If you take one of the narrow right triangles, its small angle is $9^\circ$ (one-half of one-twentieth of $360^\circ$), the short side is $9.3$cm, and the hypotenuse is $R+9.3$cm, where $R$ is the radius of the large circle.

So $\sin(9^\circ)=\frac{9.3}{R+9.3}$. And $\sin 9^\circ\approx0.1564$ and you can solve for $R$.

enter image description here

Jeppe Stig Nielsen
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Steve Kass
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Let the radius of the large circle be $R$ CM

So, the perimeter of the concentric Circle with radius $=R+9.3 $ CM

will be $$2\pi(R+9.3)$$ which will be same as $$20\cdot2\cdot9.3$$

lab bhattacharjee
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  • so the outcome will be 372cm? I cant belive its as simple as that – user145433 Apr 24 '14 at 15:54
  • Are you approximating some dodecagon of small diameters with a circle, or similar? – Jeppe Stig Nielsen Apr 24 '14 at 15:54
  • i got 20 cilindars that have R=9.3 and wanna make a decorating well so i needed the calculation from someone good at math cause i wasted a whole hour trying to place them in a circle :( – user145433 Apr 24 '14 at 15:56
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    @user145433: No, it won't be 372cm! You have to solve the equation $2\pi(R+9.3) = 20 \times 2 \times 9.3$ for the variable $R$. Can you do this? (Also, the solution will not be exact, as hinted at in Jeppe's comment, but it will be close enough for a decorative well.) – TonyK Apr 24 '14 at 16:06
  • Your formula can't be exact. It seems to give $R=\frac{n-\pi}{\pi}r$ where $n=20$ in the image. It might be a good approximation for big $n$, but is not exact (consider $n=3$ for example). – Jeppe Stig Nielsen Apr 24 '14 at 16:07