I first put the equation in inhomogenous form, assuming z5 neq 0.I would like to render these manifolds using raytracing (i.e., this only requires to numerically compute the intersection between a line and the manifold), and in this context, Id like to understand why I would need 2 algebraic equations (the equation of my line) to compute an (or several) intersection point(s).I didnt really understand how the projection in 3D worked in his paper as well:s Thanks.
Then he chooses three variables for the projection, in his case the two real parts and a linear combination of the imaginary parts. The fourth variable, a linearly independent combination of imaginary parts, is ignored. This is an orthogonal projection, like how the sun casts shadows when directly overhead. Provide details and share your research But avoid Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. MathJax reference. To learn more, see our tips on writing great answers. Not the answer youre looking for Browse other questions tagged algebraic-geometry 3d visualization or ask your own question.
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