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A spherical object is photographed from 6 different sides (cube faces). Since the radius and camera distance is known, the z coordinate of every pixel in the images can be calculated.
So I have multiple point clouds (nearly half spheres) of the same physical object as pcl::PointCloud<pcl::PointXYZRGB>.
I know the rough rotational relationship between the models (90 deg rotations), but to stitch them together to a full sphere correctly I need to know the rigid transform more precisely. How can I achieve this? The shapes have no significance in this case, stitching by color matching would be good. But the examples in the documentation all seem to only consider shapes, not colors.
The overlap of the partial models is about 40 degrees.
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!) ;)
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.
It's been a while since my math in university, and now I've come to need it like I never thought i would.
So, this is what I want to achieve:
Having a set of 3D points (geographical points, latitude and longitude, altitude doesn't matter), I want to display them on a screen, considering the direction I want to take into account.
This is going to be used along with a camera and a compass , so when I point the camera to the North, I want to display on my computer the points that the camera should "see". It's a kind of Augmented Reality.
Basically what (i think) i need is a way of transforming the 3D points viewed from above (like viewing the points on google maps) into a set of 3d Points viewed from a side.
The conversion of Latitude and longitude to 3-D cartesian (x,y,z) coordinates can be accomplished with the following (Java) code snippet. Hopefully it's easily converted to your language of choice. lat and lng are initially the latitude and longitude in degrees:
lat*=Math.PI/180.0;
lng*=Math.PI/180.0;
z=Math.sin(-lat);
x=Math.cos(lat)*Math.sin(-lng);
y=Math.cos(lat)*Math.cos(-lng);
The vector (x,y,z) will always lie on a sphere of radius 1 (i.e. the Earth's radius has been scaled to 1).
From there, a 3D perspective projection is required to convert the (x,y,z) into (X,Y) screen coordinates, given a camera position and angle. See, for example, http://en.wikipedia.org/wiki/3D_projection
It really depends on the degree of precision you require. If you're working on a high-precision, close-in view of points anywhere on the globe you will need to take the ellipsoidal shape of the earth into account. This is usually done using an algorithm similar to the one descibed here, on page 38 under 'Conversion between Geographical and Cartesian Coordinates':
http://www.icsm.gov.au/gda/gdatm/gdav2.3.pdf
If you don't need high precision the techniques mentioned above work just fine.
could anyone explain me exactly what these params mean ?
I've tried and the results where very weird so i guess i am missunderstanding some of the params for the perspective projection
* {a}_{x,y,z} - the point in 3D space that is to be projected.
* {c}_{x,y,z} - the location of the camera.
* {\theta}_{x,y,z} - The rotation of the camera. When {c}_{x,y,z}=<0,0,0>, and {\theta}_{x,y,z}=<0,0,0>, the 3D vector <1,2,0> is projected to the 2D vector <1,2>.
* {e}_{x,y,z} - the viewer's position relative to the display surface. [1]
Well, you'll want some 3D vector arithmetic to move your origin, and probably some quaternion-based rotation functions to rotate the vectors to match your direction. There are any number of good tutorials on using quaternions to rotate 3D vectors (since they're used a lot for rendering and such), and the 3D vector stuff is pretty simple if you can remember how vectors are represented.
well, just a pice ov advice, you can plot this points into a 3d space (you can do easily this using openGL).
You have to transforrm the lat/long into another system for example polar or cartesian.
So starting from lat/longyou put the origin of your space into the center of the heart, than you have to transform your data in cartesian coord:
z= R * sin(long)
x= R * cos(long) * sin(lat)
y= R * cos(long) * cos(lat)
R is the radius of the world, you can put it at 1 if you need only to cath the direction between yoour point of view anthe points you need "to see"
than put the Virtual camera in a point of the space you've created, and link data from your real camera (simply a vector) to the data of the virtual one.
The next stemp to gain what you want to do is to try to plot timages for your camera overlapped with your "virtual space", definitevly you should have a real camera that is a control to move the virtual one in a virtual space.