Where the origin lands
Take the origin H = (3/10, −1/10), where Holden's two lines cross in the unit square, and step it out: H, 2H, 3H, … Split the square of side n into its n × n unit squares. Then nH always lands exactly on an origin of the unit square it falls in, turned by a whole number of quarter turns, or, for the multiples of 5, exactly on a corner. The turn follows the same rhythm of five as the Pentametric areas, and at the Lucas numbers it turns one quarter turn per Fibonacci square. From the note Counting the Pentametric areas from an empty square, section 4.
1 · The landing
2 · Adding H, one step at a time
Nothing here needs more than adding and taking away. Each step adds H once more, and then the centre of the nearest unit square is taken away: what is left is the landing. Five steps make 5H = (3/2, −1/2), a corner of the grid, and from there everything repeats with the next five. Click a step to see it above.
Read the turn as a breath, with products and squares only: for the landing d = nH − c, the number 100[(d·H)² − (d×H)²] is +1 when the landing is not turned or turned half a turn, along the odd line, and −1 when it is turned a quarter, along the even line. That is the same +1, −1, −1, +1, rest as the blue square of 2 in the counting of the areas: 1, 4, 6, 9 breathe in, 2, 3, 7, 8 breathe out, 0 and 5 rest.
3 · A quarter turn per Fibonacci square
At the Lucas numbers 1, 3, 4, 7, 11, 18, … the landing turns exactly one quarter turn from each Fibonacci square to the next: clockwise. The centres of the Fibonacci squares themselves, seen from H in Holden's tiling, also turn one quarter turn per square, at the same rate, but anticlockwise. No Lucas number is a multiple of 5, so the landing never rests there, and the breath alternates.
The landing ↻ clockwise
Holden's centres ↺ anticlockwise
Both on one dial
Why: L₁ = 1, L₂ = 3, and each Lucas number is the sum of the two before, so modulo 5 they run 1, 3, 4, 2, 1, 3, … the powers of 3, which is exactly the order of the four turns. The arithmetic behind it is classical (10H = 3 − i in the Gaussian integers, 3² + 1² = 10, 5 = (2 + i)(2 − i)); the geometric reading is from the note above. Every value on this page is computed exactly. To turn the spirals yourself and watch where they cross, open the Spiral Workbench.