Back to News
Advertisement
Advertisement

⚡ Community Insights

Discussion Sentiment

100% Positive

Analyzed from 409 words in the discussion.

Trending Topics

#triangle#sign#formula#area#areas#divergence#summing#geometric#same#projection

Discussion (10 Comments)Read Original on HackerNews

gurkwart•about 1 hour ago
There's a really elegant solution using Geometric Algebra, that to this day is one of the most satisfying things I've ever learnt. Steven de Keninck outlines it in his 2019 Siggraph talk [1].

[1] https://youtu.be/tX4H_ctggYo?t=4795

eterevsky•about 2 hours ago
Isn't the same as just taking every triangle from the mesh, calculating the volume of a prism-like polytope between it and its projection on one the planes, and then taking it with a + sign if its projection is oriented in one direction, and with a - sign if it's oriented in another? This kind of formula works based on the basic geometry.
Sharlin•41 minutes ago
Yes, this is essentially what the author derived (by means of calculus rather than geometric argument but the result is unsurprisingly the same). The 2D analog is easy to grok: to compute the area of a polygon, find the sum of the signed areas of each of the trapezoids formed by an edge and its projection on the x-axis. Turns out the negative areas of the right-to-left trapezoids cancel precisely out any excess area of the left-to-right trapezoids (or in the case of edges below the x-axis, add precisely the "missing" area).
cgadski•31 minutes ago
Yep. Using the same kind of calculus ideas, I can also think about a vector field that has a Dirac mass of divergence at some point and zero divergence everywhere else. Then you get an expression that you can sum over faces to determine if a polyhedron contains some point. Again, for the right vector field there is a simple geometric interpretation, namely the solid angle that a face makes with respect to the point.
aaa_aaa•about 1 hour ago
Yes I remember doing something like that in 90s for a survey/map engineering cad application. After delaunay triangulation, calculating approximate voulume is easy. But this probably is a more general solution
xigoi•about 1 hour ago
I wonder if this could be reversed to give an intuitive “proof” of the divergence theorem.
meindnoch•13 minutes ago
The divergence theorem can be intuitively summarized in one sentence: "what comes out is whatever went in, plus whatever was produced inside"
elikoga•about 2 hours ago
My belly says the naive formula is summing the triangle pyramid volumes to the origin with sign in orientation. It looks like that's what they derived. Which is a generalization of 2d polygon area calculated by summing triangle areas for each edge, I was taught this in a math camp where we calculated map polygon areas on gis data. I remember math knowledge being hard to get pre AI era but I didn't remember it being this hard.

No idea what the author means by "which are equivalent to rendering the mesh and then sampling the render".

less_less•15 minutes ago
> My belly says the naive formula is summing the triangle pyramid volumes to the origin with sign in orientation.

Yeah, that would also work but it's a slightly slower formula, sum(det(v1,v2,v3))/6. This one is summing sort of prism+pyramid shapes made by projecting each triangle to the yz plane.

arn3n•about 2 hours ago
I love these kinds of posts. Simple, fast, AI-free, and I learn something new.
N_Lens•about 2 hours ago
I'll accept any kind of jocularity in the current climate!
gigatexal•about 2 hours ago
Did they also work on the graphics stack for the Asahi project?
StilesCrisis•about 1 hour ago
Yup!
unkeen•about 1 hour ago
Sadly, there is no way to find out, f.ex. by a quick Google search.