I may be misunderstanding, Tony, but your calculation is "unfettered" whereas if you have a finite space to fill and you use larger rather than smaller "chunks" then you can't get as many in the finite space. I'll fit a pulverized sugar cube between center sills a lot more effectively than I will a solid sugar cube of the same volume. 

      Yes, obviously a space that can't take an integral number of large spheres will not fit this calculation. But if you choose any size sphere, for a space where they fit "end to end" as well as "side to side," the calculation of the empty volume is independent of sphere size.

       Obviously if your sugar cube won't fit between the sills at all, then your argument is easily successful <g>. Note also that mixing sizes of the "bits" means that little ones can fill between the large ones, and the calculation for uniform spheres doesn't apply.