Packing circles? [Resolved]

[darrenl]

Take one Circle radius r

I want to fit n circles of the same radius (r1) along the inside edge of that circle so that each inner circle touches the next. (much like an old style dialing telephone).

How do i compute the radius r1 from the two values r and d.

[Acidic]

do you mean r and d or r and n?

[darrenl]

r and n of course, typo.

[darrenl]

Anybody?

[Acidic]

try this:
let's say the big circle has a diameter of 10cm.
and the little on have a radius of 1cm.

draw the big circle, then a smaller circle inside it with 4cm radius ( (10/2)-1 ).

This line will go through the centres of all the smaller circles. Then find the length of this circumferencde, on divide it by the diameter of the smaller circles. This should find the number of smaller circles taht can fit in.

[Acidic]

or since you want to find radius of little circle, do some of the above calculations backwards, I think this will work.

[darrenl]

I think you misunderstand,

One Big circle, radius R

I want to pack n circles around the inside edge.

I need to calculate the optimum radius of the smaller circles so that I can pack them inside the edge of the large circle.

If the radius of the large circle is 10cm and I have to pack 2 circles inside of it, then the radius of the little circles is 10/2=5, but this does not work for packing 3,4,5,6,7,8.....

is there an equation for this, or do I have to iteratively suck and see?

[darrenl]

In fact here is diagram to help explain what I want.

[Acidic]

must they always touch the outer circle?

[darrenl]

yes

[si_the_geek]

right then.. it took me a while to get around to it, but here's the code:

Private Sub Form_Load()
 
Dim n As Integer
Dim r As Double
Dim r1 As Double
 
  'n = 3   '(currently set in loop below)
  r = 2000
 
Dim centre_X As Double
Dim centre_Y As Double
   centre_X = r
   centre_Y = r
   Me.Show
   
 
Const pi = 3.14159265358979         '180°
Const pi2 = 3.14159265358979 * 2    '360°
Const pi_d2 = 3.14159265358979 / 2  ' 90°
Dim radians_per_circle As Double
Dim ang As Double
Dim i As Long
Dim s As Double
   
  For n = 1 To 20
    Me.Cls          'draw outer circle
    Me.Circle (centre_X, centre_Y), r, vbBlack
 
                  'find radians (of outer circle) per inner circle
    radians_per_circle = pi2 / n
 
                  'find radius of inner circle
    s = Sin(radians_per_circle / 2)
    r1 = (r * s) / (s + 1)
   
    For i = 0 To n
      ang = (radians_per_circle * i) - pi_d2
  
      Circle (centre_X + (Cos(ang) * (r - r1)), _
              centre_Y + (Sin(ang) * (r - r1))), _
              r1, _
              vbRed
    Next i
 
    MsgBox n
  Next n
 
End Sub

basic explanation:
In the outer circle there are 2*pi radians, this is divided equally for each of the inner circles (radians_per_circle), each basically forming a "triangle" (not strictly true for small values of n).

If you divide each of these triangles into 2, you can use standard trig. (treating r1 as the 'opposite', and r-r1 as the hypotenuese (sp?)) to get this formula:
r1 = sin(radians_per_circle/2) * (r - r1)
which can be rearranged...
=> r1 = sin(radians_per_circle/2) * r - sin(radians_per_circle/2) * r1
=> r1 + r1 * sin(radians_per_circle/2) = sin(radians_per_circle/2) * r
=> r1 * (1 + sin(radians_per_circle/2)) = sin(radians_per_circle/2) * r
..to this:
=> r1 = (sin(radians_per_circle/2) * r) / (sin(radians_per_circle/2) + 1)
which is what I used above.

It's not 100% accurate, but I think it's close enough

[darrenl]

Thanks very much, perfect. Just what the doctor ordered.