Determining sufficient area of a 2D annulus such that it can fit 48 circles each with radius of 35

Question

I'm helping my mother with a project she wishes to complete as part of her job as a kindergarten teacher. She wishes to create a $2D$ annulus (doughnut) and fill it with $48$ Circles as part of a practical activity with her students. Each of radius $35$mm. There will ideally be $2$ layers of circles. I've included a diagram for increased clarity.

I have concluded the the distance from the top of the inner circle, to the top of the outer circle is $140$mm, but aside from that have been largely unsuccessful in any mathematical attempt at this problem. I am asking for any more experienced mathematicians to solve this problem for me as the solution is needed ASAP. I understand that questions with little attempt at an answer violate site rules but I am really stuck on this problem, so many thanks to those who answer it.

I am seeking the diameter of the outer, and inner circle so that it can be cut out from cardboard, as well as how many circles will be on the inside and outside layer. I also understand that it is unlikely that $48$ circles will be the best fit for the determined circle, so the amount of circles is also flexible.

Answer 1

The naive arrangement that satisfies the two-row requirement involves placing half of the circles in one row, and half in the second, such that adjacent circles in the inner row will be tangent, and the circles in the outer row will be tangent to the circles in the inner row as shown in the diagram below.

In this way, we can see that if $r$ is the common radius of the small circles, then the annulus will have inner radius $$R_i = \left( -1 + \csc \frac{\pi}{24}\right) r,$$ and outer radius $$R_o = \left(1 + \sqrt{3} + \cot \frac{\pi}{24}\right) r.$$ For $r = 35$ mm, this gives $$R_i \approx 233.145 \text{ mm}, \quad R_o \approx 361.473 \text{ mm}.$$

However, this naive arrangement looks like there are spaces in between the circles of the outer row that could fit additional circles. The question is, how many? The distance between adjacent outer circles is exactly $$\left(-2 + \sqrt{5 - 2 \sqrt{6}} \csc \frac{\pi}{24}\right) r,$$ which means there is a "wastage" of about $5$ circles. This suggests moving at least one circle from the inner row to the outer. But if we do this, we must also adjust the spacing between the inner and outer rows. Some trial and error shows that having $21$ circles for the inner row, and $27$ in the outer row, will work best, as shown in the figure below, with the 'naive' annulus in gray, and the "improvement" shown outlined in black.

This improved annulus has inner and outer radii $$R_i = \left(-1 + \csc \frac{\pi}{21} \right) r, \\ R_o = \left(3 + \csc \frac{\pi}{21} \right) r,$$ and again for $r = 35$ mm, this gives $$R_i \approx 199.833 \text{ mm}, \quad R_o \approx 339.833 \text{ mm}.$$

Note that the outer row circles still are not mutually tangent: there is a small amount of space between them, $$\Delta = \left(-2 + 2 \left(2 + \csc \frac{\pi}{21}\right)\sin \frac{\pi}{27}\right) r \approx 0.0222239 r.$$ But it should be obvious that no more circles can fit in the outer row.

Although this arrangement also has waste in the form of potentially suboptimal placement of the outer row relative to the inner row, it is so small that any further improvement is not likely to make a substantial difference in the outer radius of the annulus. That is to say, the naive arrangement minimizes the "width" of the annular band $R_o - R_i$, but the "improvement" is very close to the minimum outer radius attainable.

Answer 2

This answer isn't very different from the first part of the excellent one given by @heropup. We consider necklaces made of 48 disks with alternating distances to the origin (see three cases on fig. 2) ; otherwise said, when the centers of touching circles are connected by a line segment, these line segments represent a starshaped polygon (with alternating spikes), with the limit case displayed on the right of Fig. 2.

The interest of this answer is to show, rather counter-intuitively, that one row (= necklace) with 48 touching disks (last case of Fig. 2) is more advantageous for the minimization of the area of the enclosing annulus than 2 rows (necklaces) with 24 disks on each row (first and second case of Fig. 2).

A first figure giving the area $A(r)$ of the enclosing annulus as a function of the internal radius $r$ :

Fig. 1: Annulus area $A(r)$ as a function of internal radius $r$.

The second figure below gives the configuration in three cases corresponding to the initial point, the maximal point, and the endpoint of Fig. 1, the third case being the most favorable :

Fig. 2 : 3 different cases for $r$ with progressive decrease of the spikes.

Instead of the calculation details, here is the Matlab program that has generated the different figures just seen on which one understands many aspects:

a=pi/24;s=sin(a);c=cos(a);
mini=35*(1/s-1);maxi=35*(2/s-1); % see solution by heropup.
r=mini:maxi; % range of inner radius r
R=@(r)((r+35)*c+35+sqrt(4900-s^2*(r+35).^2)); 
% R(r) = outer radius as a function of inner radius r
A=@(r)(pi*(R(r).^2-r.^2)); % annulus area
figure(1);plot(r,A(r));grid on
rn=[234,410,501]; % the 3 cases for r
uc=exp(i*(0:0.01:2*pi)); % unit circle 
for k=1:3
    r=rn(k);
    figure(k+1);hold on;axis equal off;
    plot(r*uc);plot(R(r)*uc);
    quiver(0,0,r,0,1);text(r/3,40,['r = ',num2str(r)]); % arrow
    quiver(0,0,0,R(r),1);text(20,r/2,['R = ',num2str(round(R(r)))]);
    text(-r/4,-r/2,['Area = ',num2str(round(A(r)))]);
    for k=1:24; 
       plot(exp(i*k*a/2)*(r+35+35*uc)); % internal disks
       plot(exp(i*(k+1/2)*a/2)*(R(r)-35+35*uc)); % external  ones
    end;
 end;