Conference rusure::math

    For 1 circle, R = r
    	2	  R = 2r
    	3	  R = r(2+sqrt(4/5))
    	4	  R = r(1+sqrt(2))
    	5,6,7	  R = 3r
    	8	  ? now it begins to get interesting...
    		  The optimum arrangement appears to be
    			O O O
    			 O O
    			O O O
    	9	 Slightly larger...
    	10,11,12  	O O
    		       O O O
    		      O O O O
    		       O O O
    	13		?

Problems:

Prove that the the true formula *cannot* be expressed as [f(R/r)],
where f(x) is a polynomial function of x, and [x] is the floor function.

For very large R/r, what is a good approximate formula?


Aside:  I think the number for 3 circles (2+sqrt(4/5)) is wrong.

    
    re: .2
    
    > For very large R/r, what is a good approximate formula?
    
    How good is
    
    	pi/[ pi/2 + sqrt(3) ] * (R/r)^2   ?
    
    which is approximately 0.951177*(R/r)^2.
    

   I'd guess the the asymptotic ratio would be pi/(2*sqrt(3)),
   by comparing areas of closest packed circles to the region they lie in.

   I think the correct value 3 circles would be R = r*(1+2/sqrt(3)).

   - Jim

    I agree with Jim in .4 that 3 circles would be R/r = 1+2/sqrt(3).
    
    My table, so far is:
    
    minimum		maximum
    R/r			circles
    ==========		=======
    1			1
    2			2
    1+2/sqrt(3)		3
    1+sqrt(2)		4
    3			7
    2+sqrt(3)		8
    1+sqrt(8+2*sqrt(3))	12
    

    Jim, I can't seem to figure out how you got pi/(2*sqrt(3)) as the
    asymptotic ratio in note 553.4.
    
    My best estimate of the maximum number of circles where S=R/r for
    large R/r is:
    
            [ S^2 - ( 2 - pi/2 )*S - 2 ]/( 1/2 + sqrt(3)/pi ) + 2
    
    If this expression is accurate, then $1,000,000 in pennies could be
    laid out in a circle of radius approximately 320 feet, 5 inches.
    (I'm going to lay out my circle today and start filling it with
    pennies just to see how close the expression is.  Anyone want to
    help by sending pennies?)
     

re: the asymptotic ratio

    We can pack M*N circles of radius r into a rectangle with sides
    2*M*r+r and (N-1)*sqrt(3)*r+r.  So we can fit M*N circles into
    an area of size (2*M+1)*((N-1)*sqrt(3)+1)*r^2.  If we assume
    that M,N >> 1, the rectangle's area is roughly 2*M*N*sqrt(3)*r^2,
    and the number of small circles is roughly given by:

	{number of small circles} = {size of large area} / (2*sqrt(3)*r^2)

    Now we would expect the same approximation to hold for large circles.
    Thus, the number of small circles that fit into a circle of radius R,
    where R >> r, is roughly given by:

	{number of small circles} = pi*R^2 / (2*sqrt(3)*r^2)
    or
				       pi          2
	{number of small circles} = --------- (R/r)
				    2 sqrt(3)

     re: .-1  I believe I understand the argument given.  I'll spend
    some time examining it.
    
    In the meantime, please criticize the following discussion about the
    same topic: 
    
    
    Consider three tightly packed circles of radius r.  Using the centers
    of each circle as vertices, construct an equilateral triangle. 
    The area of this triangle is easily shown to be sqrt(3)*r^2.
    
    Now, consider the section within the triangle which is not part
    of any circle.  It's area is the area of the triangle less the area
    of the three one-sixths of a circle:  sqrt(3)*r^2 - (3/6)*pi*r^2,
    or
                           [ sqrt(3) - pi/2 ] * r^2
    
    Thus, one circle plus one "free area" consume
    
                      pi*r^2 + [ sqrt(3) - pi/2 ] * r^2, or
    
                          [ sqrt(3) + pi/2 ] * r^2.
    
    Next, note that for (n-2) circles added after the initial two, the
    total area consumed (including the first two circles) is 
    
              2*pi*r^2 + ( n - 2 )*[ sqrt(3) + pi/2 ] * r^2
    
    since for every circle added one "free area" is also created.
    
    Finally, consider the outermost circles that touch the enclosing
    circle's circumfrence.  To simplify calculations, yet lose little
    accuracy, think of the large circle not as a circle, but as a
    pi*R/r-sided polygon with one small circle at each vertex tangent to
    the small circle at the next vertex.  (There are pi*R/r sides because
    the circumfrence of the large circle is 2*pi*R and the distance between
    the centers of the small circles is 2*r: (2*pi*R)/(2*r) = pi*R/r.) The
    area between the perimeter of the polygon and the outermost circles is 
    
                       ( pi*R/r ) * ( 2 - pi/2 ) * r^2.
    
    The area of the n circles plus all free area (including the free
    area at the periphery) is
    
               2*pi*r^2 + ( n - 2 )*[ sqrt(3) + pi/2 ] * r^2 +
                       ( pi*R/r ) * ( 2 - pi/2 ) * r^2
    
    which must be less than pi*R^2, the area of the large circle.
    Substituting S for R/r and solving for n,
    
          n < [ S^2 - ( 2 - pi/2 )*S - 2 ]/( sqrt(3)/pi + 1/2 ) + 2.
    
    n circles of radius pi*r^2 would consume an area less than
    
    pi*{ [ R^2 - ( 2 - pi/2 )*r*R - 2*r^2 ]/( sqrt(3)/pi + 1/2 ) + 2*r^2}
    
    and, as a proportion of the larger circle,
    
         [ 1 - ( 2 - pi/2 )/S - 2/S^2 ]/( sqrt(3)/pi + 1/2 ) + 2/S^2
    
    The limit of this expression as S approaches infinity is simply
    
                            1/( sqrt(3)/pi + 1/2 )
    
    or,
    
                            pi/( sqrt(3) + pi/2 )
    

    You've assumed that there's one "hole" or "free area" per circle.
    I believe there should be two; your triangle having centers of three
    circles as vertices holds 1/2 circle and one "hole" -- replicated,
    this gives a ratio of two "holes" per circle.

    
    Yup, I'm sure that my problem stems from the assumption that there is
    only one hole per circle.  I'm convinced that there truly ARE two holes
    per circle.  What I'm having trouble understanding is that whenever I
    add a circle to a group of tightly packed circles, only one hole
    is created, not two.
    
    What's going on?
    

Lay out your pennies at the Xs below.  Now place another penny at the Y.
How many "holes" are created?

          X   Y
	X   X   X   X

Or another way:

The 'circumference' of a hole is 3/6 the circumference of a circle.
Setting the total sum of lengths equal (and ignoring the error caused by
the boundary of the large region we're filling, since most of the circles
are in the interior), we have: 

	{number of holes} * 3/6 = {number of circles} * 6/6