Alright, so we have a hexagon assembled as follows;

19 spheres in a line. Two lines on either side 18 spheres long. Two lines 17 spheres long. 16,15,14,13,12,11,10,9 - so 10 lines on either side ranging from 17 spheres long to 9 spheres long.

18 + 9 = 27
17 + 10 = 27
16 + 11 = 27
15 + 12 = 27
14 + 13 = 27

On each side of the 19 sphere line...

10*27+19 = 289 spheres

in the first disk. So, if we had 275 spheres in this layer, we would have 14 spares - places for landing gear and other equipment.

Now we stack a disk of spheres on top of this one, between the spheres in hexagonal close pack fashion...

18 spheres in a line. Two lines on either side 17 spheres long. Two lines on either side 16 spheres long. 16,15,14,13,12,11,10,9 - so 9 lines on either side ranging from 17 to 9 spheres in length. This forms a hexagon 9 sphere diameters on a side.

17 + 9 = 26
16 + 10 = 26
15 + 11 = 26
14 + 12 = 26
13

On either side - so we have 9 * 26 + 18 = 252

Well, we have 164 + 56 + 18 = 230

So, there are plenty of locations... 22 locations to spare.

The last two stages are 56 + 18 = 74

So, a line of spheres 10 long, with five rows on either side, 9,8,7,6,5 total

2x 9+5=14, 8+6=14, 7 + 10 = 80

which is 6 more locations than needed.

This is both the third and fourth stages. The last stage is only 18. Here we have a line of 5 with 4 on either side and 3 on either side. A hexagon that's 3 units long on a side and 6 units in diameter.

5 + 2* (4+3) = 19

Which is 1 more space than needed.

Here's an earlier design using larger spheres

http://www.scribd.com/doc/40549127/Disk-Moonship

http://www.scribd.com/doc/40623446/Disk-Moonship-2

The current proposal uses 0.45 m diameter spheres. The project costs $22 million per person to the moon. The project time is five years - that's 2019 - the fiftieth anniversary of Apollo.

Budget (thousands)

$1,048 Test Complete
$2,095 Sphere on Orbit
$4,190 Person Ballistic
$8,381 Person on Orbit
$6,286 Person on Moon

$22,000