$r(t)$ is a unit 4-vector. The derivatives of $r$ are known and well-behaved.

I'm interested in images of $r$ in stereographic projection – but (for purposes of this question) I don't yet know where the projection centre will be. Is there a convenient way to know what is the maximum possible curvature in the image of $r$ (around a given $t$), over all possible projection centres?

Context: some of my 3D prints, like this Klein bottle, are such projections. I prefer to adapt the spacing of the sample-points to the local curvature. To save time, I'd like to generate a ‘master’ design that can be projected from different centres, with the sample-points dense enough everywhere for each projection.


On second thought ...

The quantity that concerns me is not exactly the curvature, but the magnitude of the vector rejection of $r''$ on $r'$. Because of conformality, this number varies only with the scale of the mapping, which is inversely proportional to $1-\cos(d)$ where $d$ is distance from the projection centre. To keep my models finite, I generally cut them off at a fixed distance from the centre; that boundary is where I'll find my worst case.


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