See the straightedge-and-compass operations and the exact link to the paper
For a parameter a > 0, the neutral step of the model gives t = (2 + 1/a)/3. Construct 1/a using similar triangles, add two units and divide the segment into three. Parallels then construct a, at, at². The last length is L₃(a). Here the index 3 denotes the root degree, not point L₃ of the rectangle.
The direct branch gives L = L₃(R); the reciprocal branch gives L₃(1/R), whose reciprocal is D. Multiply these lengths by u: ℓ = uD, h = uL. Place A–O = ℓ and O–B = h end to end on a line. The intersection H of the perpendicular at O with the semicircle of diameter AB satisfies OH² = ℓh.
L = (2R + 1)²/(9R), D = 9R/(R + 2)²
u⁺ = √(ℓh) = u(2R + 1)/(R + 2).
The chain panels are scaled independently for readability; use their numerical values to compare lengths. Successive bases are proportional and the red segments are parallel. If t > 1, the triangles grow and extend below their base.
The neutral start s = 1 has a degenerate telescoping drawing: the horizontals coincide. Here its step is constructed using the arithmetic mean (2 + 1/a)/3 prescribed by the model, then its chain is rebuilt. This is a geometric realization of the modern closure in the paper, not an attribution of that closure to ancient Pandrosion.
This page illustrates center G₃ from section 3 and “Pandrosion corrections drawn step by step” in V14. The B₅…B₁₁ family in section 6 adds a further correction of the center defect; it differs from the step illustrated here. For p = 3, G₃ coincides with Halley, or Padé [1/1].
Full protocol and scope of the checks · GeoGebra file of the two-sided center (import not verified)