I wonder if someone can help me to find the area of a 2-D object in Cartesian plane , when we know coordinates of every points.
Eg : I want to calculate the area of a triangular. A(12,34) B(45,89) C(25,35)
I want a common algorithm to find any 2-D object's area.
Thank you.
Here you go, uses triangulation. This was literally the top result off Google when I searched "area of polygon given set of points". Please do your research before posting.
If your object is a simple polygon, there's no need to triangulate it to compute its area. There's a simple formula that depends only on the coordinates of the vertices. See http://en.wikipedia.org/wiki/Polygon#Area_and_centroid
Related
I understand how to use delaunay triangulation in 2d points?
But how to use delaunay triangulation in 3d points?
I mean I want to generate surface triangle mesh not tetrahedron mesh, so how can I use delaunay triangulation to generate 3d surface mesh?
Please give me some hint.
To triangulate a 3D point cloud you need the BallPivoting algorithm: https://vgc.poly.edu/~csilva/papers/tvcg99.pdf
There are two meanings of a 3D triangulation. One is when the whole space is filled, likely with tetrahedra (hexahedra and others may be also used). The other is called 2.5D, typically for terrains where the z is a property as the color or whatever, which doesn't influence the resulting triangulation.
If you use Shewchuk's triangle you can get the result.
If you are curious enough, you'll be able to select those tetrahedra that have one face not shared with other tetrahedra. These are the same tetrahedra "joined" with infinite/enclosing points. Extract those faces and you have your 3D surface triangulation.
If you want "direct" surface reconstruction then you undoubtly need to know in advance which vertices among the total given are in the surface. If you don't know them, perhaps the "maxima method" allows to find them out.
One your points cloud consists only of surface vertices, the triangulation method can be any one you like, from (adapted) incremental Chew's, Ruppert, etc to "ball-pivoting" method and "marching cubes" method.
The Delaunay tetrahedrization doesn't fit for two reasons
it fills a volume with tetrahedra, instead of defining a surface,
it fills the convex hull of the points, which is probably not what you expect.
To address the second problem, you need to accept concavities, and this implies that you need to specify a reference scale that tells what level of detail you want. This leads to the concept of Alpha Shapes, which are obtained as a subset of the faces.
Lookup "Alpha Shape" in an image search engine.
i'm working on a project where i have a cloud of points in space as input data, my goal is to create a surface.
I started by computing a regression plan for the cloud, then i projected my points on the plane using dot products :
My plane is represented by a point and a normal , i construct the axis of the plane's space using cross products then project each point on these axis.
then i triangulate in 2D (that's the point of the whole operation).
My problem is that my points now are in the plane space and i want to get them back to their inital position (inverse the transformation) to have my surface ON my points.
thank you :)
the best way is to keep the original positions and make the triangulation give you the indices rather than the positions , i hope it will help !
I have some dinamically created walls on the canvas, and I know the start end the end positions for every wall. The positions consist of the X and Z coordinate (y coordinate is constant, because the wall's height is predefined).
So, technically, I have some sections on the 2D plane (with start and end points).
The thing what I need is the polygon, wrap exactly the polygon which made by walls from plan view. Can anyone describe or link me an algorythm to do this?
Thanks in advance,
R
I found this: https://gis.stackexchange.com/questions/1200/concave-hull-definition-algorithms-and-practical-solutions
It's visualize my problem, I need an algorithm for get the concave hull!
R
I have an array of points. I want to know if this array of point represents a circle, a square or a triangle.
Where should i begin? (i use C#)
Thanks
Jon
Depending on your problem, a good approach for this problem may be to use the Hough transform and all its derived algorithm
It consists in a transformation of the image space to an other space where the coordinate represents the objects parameters (angle and initial point for a line, coordinates of the center and radius for a circle)
The algorithm transforms each point of your array of points in points in the other space. Then you have to search in the new space if some points are prevailing. From these points, you will get the parameters of your object.
Of course, you need to do it once to recognize the lines (so you will know how many lines are in your bitmap and where they are) and to it to recognize the circles (it is not exactly the same algorithm)
You may have a look to this lecture (for Hough Circle Transform), but you could easily find the algorithm for line
EDIT: you can also have a look to these answers
Shape recognition algorithm(s)
Detecting an object on the image based on geometrical form
imagine it is each of these one-by-one and try to fit each of these shapes on the data.. for a square, you could find the four extreme points, and try charting out a square that goes through all of them..
Once you have got a shape in place.. you could measure the distance between each of the points and the part of the shape that is nearest to it.. then square these distances and add them up.. the shape which has the smallest sum-of-squares is probably your best bet
Use the Hough Transform.
I'm going to take a wild stab and say if you have 3 points the shape represents a triangle, 4 points is some kind of quadrilateral, any more than that is a circle.
Perhaps there's more information to your problem you could provide.
I'm writing a mapping application that I am writing in python and I need to get the lat/lon centroid of N points.
Say I have two locations
a.lat = 101
a.lon = 230
b.lat = 146
b.lon = 200
Getting the center of two points is fairly easy using a euclidean formula. I would like
to be able to do it for more then two points.
Fundamentally I'm looking to do something like http://a.placebetween.us/ where one can enter multiple addresses and find a the spot that is equidistant for everyone.
Have a look at the pdf document linked below. It explains how to apply the plane figure algorithm that Bill the Lizard mentions, but on the surface of a sphere.
poster thumbnail and some details http://img51.imageshack.us/img51/4093/centroidspostersummary.jpg
Source: http://www.jennessent.com/arcgis/shapes_poster.htm
There is also a 25 MB full-size PDF available for download.
Credit goes to mixdev for finding the link to the original source, and of course to Jenness Enterprises for making the information available. Note: I am in no way affiliated with the author of this material.
Adding to Andrew Rollings' answer.
You will also need to make sure that if you have points on either side of the 0/360 longitude line that you are measuring in the "right direction"
Is the center of (0,359) and (0, 1) at (0,0) or (0,180)?
If you are averaging angles and have to deal with them crossing the 0/360 then it is safer to sum the sin and cos of each value and then Average = atan2(sum of sines,sum of cosines)
(be careful of the argument order in your atan2 function)
The math is pretty simple if the points form a plane figure. There's no guarantee, however, that a set of latitudes and longitudes are that simple, so it may first be necessary to find the convex hull of the points.
EDIT: As eJames points out, you have to make corrections for the surface of a sphere. My fault for assuming (without thinking) that this was understood. +1 to him.
The below PDF has a bit more detail than the poster from Jenness Enterprises. It also handles conversion in both directions and for a spheroid (such as the Earth) rather than a perfect sphere.
Converting between 3-D Cartesian and ellipsoidal latitude, longitude and height coordinates
Separately average the latitudes and longitudes.