FA-70239 / Map projection transforms / Member archive

Mollweide forward with Newton solve: iteration budget · case 04

High-latitude points are misplaced by kilometres.

Member previewVariant 4 · 3 implementations · 8 checks per implementation

Case contract

Input [lon, lat, lon0]; sphere R = 6371000, dlon wrapped to [-180, 180). Solve 2t + sin 2t = pi sin(phi) by Newton from t = phi, at most 50 steps, stopping when the residual is below 1e-12, with derivative 2 + 2 cos 2t. x = R (2 sqrt 2 / pi) dlon cos t, y = R sqrt 2 sin t, rounded to 2 decimals.

Why this case matters

Global thematic maps use Mollweide for equal area; an under-converged auxiliary angle skews high latitudes.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixture	Actual	Expected	Outcome
control #5	`[10594134.8, 6828427.27]`	`[10593963.16, 6828509.25]`	Failed

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