> On Jul 17, 2017, at 4:24 PM, Mario Meissner <mr.rash....@gmail.com> wrote: > > Okay, so let's make sure I understand the current code before continuing with > the new specifications. > > About the blade, yes I tried to just mix both because that made it simpler. > > Tomorrow I'll make the new blade (without the curve) and get going with the > code. It seems like I'm missing some basic concepts on what's going on > inside. I'll try to understand it as good as I can and then ask some > questions on what isn't clear to me. Is there any documentation on what > exactly happens during geometry evaluation and ray tracing in general?
This example is about as simple as it gets code-wise: http://brlcad.org/wiki/Example_Application For the abstract conceptual concepts of ray tracing, anything that talks about “solid” or “full path” ray tracing is generally relevant (as opposed to ray tracing for image synthesis, rendering pictures). Coincidentally relevant: http://slideplayer.com/slide/9271245/ What usually gets in the way are preconceptions. Consider for example an object, some unknown object. You don’t (yet) know what it looks like, how big it is, etc. All you have is a special ray/gun that shoots an infinitely thin ray through the object and tells you when you entered and exited the object. So you shoot a ray and hit it: shoot entry exit o-> ray . . . . . hit o–-–––o hit 6mm All you know at this point is that you hit it and 6mm later, the ray exited the object. That by itself isn’t very useful, but we can shoot a whole set of rays, each from a different position, and we start to get a picture of what the object looks like: o-> ray1 o–o o-> ray2 o–––––o o—> ray3 o––––––––—o o-> ray4 o–––––––––––——o o—> ray5 o—––––––——o o-> ray6 o–––––o o-> ray7 o–o We start to see it looks a bit like a rotated cube. Moreover, knowing the length of each segment and the spacing of the rays, we could actually get an estimate of area: o-> ray1 o–o 2mm o-> ray2 o–––––o 6mm o—> ray3 o––––––––—o 10mm o-> ray4 o–––––––––––——o 14mm o—> ray5 o—––––––——o 10mm o-> ray6 o–––––o 6mm o-> ray7 o–o 2mm –––– with 2mm spacing between rays: 50mm x 2mm = 100mm^2 area It’s probably a 10mm x 10mm square. Extend that example into layer after layer of samples and you don’t just have an area estimate, but now you have a (3D) volume estimate because each ray represents a 2D cross section (e.g., 2mm^2) which can be multiplied by each segment length: with 2mm spacing between rays: 50mm x 2mm x 2mm = 200mm^3 volume With a volume estimate and knowing density, you can obviously calculate mass too (mass = volume x density). And all of that was done without knowing anything about the geometry. That’s essentially what rtweight is doing. With your project, we’re changing the last step. In mathematical terms, we were doing a summation over all the individual volume bits, multiplying each by a constant density. New form, everything stays the same except we’re either going to multiply the volume by a different (e.g., average) density based on a function or we’re going to further parameterize the volume so an volume/mass integral can be estimated. Cheers! Sean ------------------------------------------------------------------------------ Check out the vibrant tech community on one of the world's most engaging tech sites, Slashdot.org! http://sdm.link/slashdot _______________________________________________ BRL-CAD Developer mailing list brlcad-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/brlcad-devel
