[dlghlxon] two moving rows of dots

consider a horizontal row of dots equally spaced 1 unit apart.  put another such row above it, also with horizontal spacings of 1 unit.  let the vertical spacing between rows be A = sqrt(15)/4 ~= 0.968 .

let the upper row slide horizontally at a constant velocity.  at some snapshot in time, consider a dot in the upper row, and the dot in the lower row that it is closest to.  if the upper dot is directly above the lower dot, that is, their relative horizontal offset is zero, then the distance between them is A.  if their relative horizontal offset is 0.25, then the distance between them is 1 by Pythagoras.  if their relative horizontal offset is 0.5, then the distance between them is sqrt(19)/4 ~= 1.090 .  this is the maximum possible separation: if the offset is greater, then it gets closer to the next dot in the row.  thus, half the time (0 to 0.25) the vertical distance is less than the horizontal distance between dots (namely 1), and the other half of the time (0.25 to 0.5), the vertical distance is greater.

motivation is dots in motion but staying well separated.  rows can be stacked.

previously: dots arranged in rings instead of rows.

if we want the average distance to the nearest dot in the other row to equal 1, then I don't think there is a closed form solution for the vertical space h between rows.  Mathematica:

N[Solve[ 2*Integrate[ Sqrt[h^2+x^2], {x, 0, 1/2}, Assumptions -> Element[h, Reals] && h > 0]==1, {h}, Reals], 50]

yields h ~= 0.95813624081219179188949156285292561053238539746725

(Inverse Symbolic Calculator finds nothing.)

if the rows of dots are not moving, then both the square lattice and equilateral triangular lattice achieve equal separation vertically and horizontally.