I would like to make a file including polygon or lines from boundary points extracted from point cloud by concave hull method as shown in below page.
http://ait-survey.com/wp-content/uploads/2015/08/concave_hull_polygon1.png
Boundary points are 3D coordinates.
Also I would like to make polygon file import in AutoCAD.
Let me know how to make it.
That's a interesting problem. I would start with:
Identify the outer circle around the points, as shown here: http://through-the-interface.typepad.com/through_the_interface/2011/02/creating-the-smallest-possible-circle-around-2d-autocad-geometry-using-net.html
Run through the circle degrees (0 to 2PI), find the closest point. That should give you the order of the points, counter-clockwise.
Draw the polyline
I do not have a code 'ready-to-use'... may have some time late this week. But what do you think about the approach?
i haven’t been entirely sure what to google or search for to help solve my problem, really hoping someone here can help a little…
currently i have a 3d scene, it has a massive sphere with a texture mapped to it and the camera at the center of the sphere, so it’s much like a qtvr viewer.
i’d like a way to click on the polygons within the sphere and update the texture at that position with something and dot etc..
the only part of the process where i need help is converting the 2d mouse position to a point on the inside of the sphere.
hope this makes sense…
fyi, im only looking for a pure math solution..
The first thing you need to do is convert the screen coordinate into a line in 3d space. This will pass through the point you click and your eyepoint.
Once you have this line you can then intersect this line with your sphere to find the intersection point on the sphere.
You may get 2d coordinates of the polygons (triangles?) that are making up the sphere and then find the one that contains the mouse pointer point.
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 want to create a Voronoi diagram on several pairs of
latitudes/longitudes, but want to use the great circle distance
between them, not the (inaccurate) Pythagorean distance.
Can I make qhull/qvoronoi or some other Linux program do this?
I considered mapping the points to 3D, having qvoronoi create a 3D
Voronoi diagram[1], and intersecting the result with the unit sphere, but
I'm not sure that's easy.
[1] I realize the 3D distance between two latitudes/longitudes (the
"through the Earth" path) isn't the same as the great circle distance,
but it's easy to prove that this transformation preserves relative
distances, which is all that matters for a Voronoi diagram.
I assume you've found this article. From that, it seems like you have the right idea by using a 3D embedding. Your question is then how to intersect the result with the sphere.
First of all you need to consider how you're going to represent the voronoi diagram. If you want to work in lat/long coordinates in a 2D plane, then your voronoi diagram will contain curved edges, so maybe it is best to just use a 3D representation.
If you use a program like qvoronoi, you should in theory only need the inifinite hyperplane data (generated by Fo). This gives you the equation of the plane and the two points it corresponds to. Usually you only need to use the voronoi diagram to test for inclusion within regions, and the hyperplanes should be enough for that.
See also this question: Algorithm to compute a Voronoi diagram on a sphere?
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.