I am trying to parse wavefront .obj file. From wikipedia I learned it's format specification. I am interested in volume analysis. Before that i worked in .stl files. I can compute volume of tetrahedron by using dot and cross product. In .stl files triangular faces' coordinates are given
i.e.
point1(x1, y1, z1),
point2(x2, y2, z2),
point3(x3, y3, z3).
But in wavefront .obj files :
Each face can contain three or more vertices.
f v1 v2 v3 v4 ..... like this. I do not know how to calculate the volume now. Because my understanding is, it will produce a polyhedron. Any idea will be extremely helpful. Thanks!
Each polygonal face can be subdivided in triangles and therefore used for tetrahedrons. If you are lucky enough, you can find polygons with 4 vertices only (easy to be decomposed in two triangles on the fly). For polygons with more than 4 vertices you need a triangulation algorithm to decompose the planar polygon in triangles. You can check Ear Clipping for example.
Related
I have a 3-D object described by surface facets. It is saved as OFF file. However, some of the vertices in a surface facet (polygon) are not exactly coplannar. Is there any good idea to fix these non-coplannar polygons to let them become coplannar?
My idea is
1) for every polygon, find the base plane of the polygon that most vertices fall into the plane, for example, the distance of vertex to the plane is less than machine precision (but how to find this base plane?);
2) for these vertices that are not fall into the base plane of the polygon, find the intersection of vertex's associated polygons' base planes and take it as the new location of this vertex.
Is there any problem? Do you have any better idea?
Thanks,
Tang Laoya
Maybe you should concentrate on fixing quadrilaterals, and repeat that over and over. Say two adjacent triangles have vertices abc and cbd, sharing edge bd.
Compute the volume of the tetrahedron abcd, and if it is small, decided to flatten. Or, maybe better, compute the dihedral angle at bd and flatten if
near pi; yes, that is probably a better measure.
Once you decide you would like to repair abcd, project d to the plane determined by abc, and replace d with that projected point. Now abcd are coplanar. You could do this for all the permutations and replace the point that moves the least.
Unfortunately, it seems this is order-dependent and could lead to cycles when
repeated over and over on all pairs of triangles.
I'm writing a data analysis program and part of it requires finding the volume of a shape. The shape information comes in the form of a lost of points, giving the radius and the angular coordinates of the point.
If the data points were uniformly distributed in coordinate space I would be able to perform the integral, but unfortunately the data points are basically randomly distributed.
My inefficient approach would be to find the nearest neighbours to each point and stitch the shape together like that, finding the volume of the stitched together parts.
Does anyone have a better approach to take?
Thanks.
IF those are surface points, one good way to do it would be to discretize the surface as triangles and convert the volume integral to a surface integral using Green's Theorem. Then you can use simple Gauss quadrature over the triangles.
Ok, here it is, along duffymo's lines I think.
First, triangulate the surface, and make sure you have consistent orientation of the triangles. Meaning that orientation of neighbouring triangle is such that the common edge is traversed in opposite directions.
Second, for each triangle ABC compute this expression: H*cross2D(B-A,C-A), where cross2D computes cross product using coordinates X and Y only, ignoring the Z coordinates, and H is the Z-coordinate of any convenient point in the triangle (although the barycentre would improve precision).
Third, sum up all the above expressions. The result would be the signed volume inside the surface (plus or minus depending on the choice of orientation).
Sounds like you want the convex hull of a point cloud. Fortunately, there are efficient ways of getting you there. Check out scipy.spatial.ConvexHull.
I have a map of a mountainous landscape, http://skimap.org/data/989/60/1218033025.jpg. It contains a number of known points, the lat-longs of which can be easily found out using Google maps. I wish to be able to pin any latitude longitude coordinate on the map, of course within the bounds of the landscape.
For this, I tried an approach that seems to be largely failing. I assumed the map to be equivalent to an aerial photograph of the Swiss landscape, without any info about the altitude or other coordinates of the camera. So, I assumed the plane perpendicular to the camera lens normal to be Ax+By+Cz-d=0.
I attempt to find the plane constants, using the known points. I fix my origin at a point, with z=0 at the sea level. I take two known points in the landscape, and using the equation for a line in 3D, I find the length of the projection of this line segment joining the two known points, on the plane. I multiply it by another constant K to account for the resizing of this length on a static 2d representation of this 3D image. The length between the two points on a 2d static representation of this image on this screen can be easily found in pixels, and the actual length of the line joining the two points, can be easily found, since I can calculate the distance between the two points with their lat-longs, and their heights above sea level.
So, I end up with an equation directly relating the distance between the two points on the screen 2d representation, lets call it Ls, and the actual length in the landscape, L. I have many other known points, so plugging them into the equation should give me values of the 4 constants. For this, I needed 8 known points (known parameters being their name, lat-long, and heights above sea level), one being my orogin, and the second being a fixed reference point. The rest 6 points generate a system of 6 linear equations in A^2, B^2, C^2, AB, BC and CA. Solving the system using a online tool, I get the result that the system has a unique solution with all 6 constants being 0.
So, it seems that the assumption that the map is equivalent to an aerial photograph taken from an aircraft, is faulty. Can someone please give me some pointers or any other ideas to get this to work? Do open street maps have a Mercator projection?
I would say that this impossible to do in an automatic way. The skimap should be considered as an image rather than a map, a map is an projection of the real world into one plane, since this doesn't fit skimaps very well they are drawn instead.
The best way is probably to manually define a lot of points in the skimap with known or estimated coordinates and use them to estimate the points betwween. To get an acceptable result you probably have to assign coordinates to each pixel in the skimap.
You could do something like the following: http://magazin.unic.com/en/2012/02/16/making-of-interactive-mobile-piste-map-by-laax/
I am solving the exact same issue. It is pretty hard and lots of maths. Taking me a few weeks to solve it. Interpolation is the key as well with lots of manual mapping. I would say that for a ski mountain it will take at least 1000/1500 points to be able to get the very basic. So, not a trivial task unless you can automate the collection of these points (what I am doing!) ;)
This seems like a question for which an answer should readily available on the web or books but my quest for an answer has led me so far only to blind alleys that turned out to be dead ends.
I'm trying to draw 3D lines in real-time with hidden surface removal (the lines are edges of solid objects).
So I have two 3D points that were projected to 2D points using perspective projection. For each point I have computed the depth of the point. Now I want to draw the line segment that joins the 2 points, and for hidden surface removal to work I have to compute, for each intermediary 2D point on the 2D line (that results from the projection) the depth of the corresponding 3D point (the 3D point that is projected on that intermediary 2D point).
My problem is that, since the depth function isn't linear when you do perspective projection, I can't interpolate the depth of the 2 original 3D points to compute the depth of the intermediary point.
So how do I compute the depth of each point on the line with a method that's compatible with the constraints of real-time rendering?
Thanks in advance for any help.
Use homogeneous coordinates, which can be linearly interpolated in screen space: http://www.cs.unc.edu/~olano/papers/2dh-tri/
I need to convert arbitrary triangulated 3D mesh to cloud of particles that are uniformly spaced.
First thought was to try find a way to fill one 3D triangle. And then fill each triangle of mesh, removing duplicated particles on edges, but that's just hard and too much work. I was hoping for some more-math way.
Can anyone point me to an algorithm which can help me do my task correctly... well, at least approximatively?
Thanks
There are two main options:
Voxelization of mesh. Easy to implement the conversion of mesh to voxels, but it's inaccurate since uniform spacing cannot be achieved: distance between cubes can be x, x*sqrt(2) or x*sqrt(3) depending if neighbor cubes are in same plane and adjacent.
Poisson disk sampling on surface. Hard to implement and lack of research material and code, but mathematically very correct. Some links:
http://research.microsoft.com/apps/pubs/default.aspx?id=135760
http://web.mysites.ntu.edu.sg/cwfu/public/Shared%20Documents/dualtiling/index.html
You could convert the TIN to raster using a GIS package or software such as R, then retrieve one point at the center of each pixel representing the value. (Example in ArcGIS)
EDIT: If the irregular 3D mesh has multiple heights per {x, y} a similar approach would be to sample the mesh using a voxel "grid" and keep one value per voxel. GRASS GIS has the functionality to take the vertices of the TIN (3d mesh) and convert them to voxels, then back to a regular 3d cloud.