FA-70233 / Map projection transforms / Member archive

Mollweide forward with Newton solve: newton derivative · case 03

Mid-latitude points jitter between two wrong positions depending on iteration parity.

Member previewVariant 3 · 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 #2	`[-6033228.55, -7735267.54]`	`[-7888806.65, -6684729.63]`	Failed

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This record includes three runnable implementations, regression fixtures, execution results, and source hashes.

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