Let's consider there is a polygon with near orthogonal edges that is in the range of (70, 89)° or (91, 110)°. We know the sum of angles in a polygon is multiple of 180°.
The question is how to convert a polygon with customized shape to a minimum polygon with right angles.
As a very simple case, let's consider the example below:
What we want to achieve is, the polygon below:
The number of edges can be more. The example above is just a simple example.
I need a 3D triangular sphere mesh with uniformly sampled vertices, say V, with a predefined adjacency. Is there a specific way to achieve that?
I am having a Three OBJ file, Cone, Sphere and Cube. How can I find the dimension of these objects, so that I can use it in my collision detection class?
i.e How can I find Radius, Length of Cube and Radius and Height of cone? Or is there any better way for collision detection, I have hundreds of random particle in my game which may or may not collide with these objects.
Length of the cube would be the distance of two consecutive points in one face.
If by radius of a cube you mean its diagonal it would be sqrt{3} of its length.
For a sphere, you can find its center by averaging all the vertices.
Its radius would be the distance between any vertex and the center.
It works fine if the sphere is not high resolution.
Otherwise, you have to solve a system of equations to find the sphere that passes through four points. You can take a look at this:
https://www.quora.com/How-do-you-find-the-center-and-radius-of-a-sphere-given-any-four-arbitrary-points-x_1-y_1-z_1-x_2-y_2-z_2-x_3-y_3-z_3-x_4-y_4-z_4
For the cone: there is probably one vertex that is connected to all other vertices. This vertex is probably easy to find for you. Let's call that p.
Take three vertex other than p. Find the circle passing through those vertices. Call that c. The distance between p and c is the height. The distance between c and any vertex other than p would be the radius. To find the circle passing through three points, you again need to solve a system of equations with three unknowns. As the equation of a circle is (x-a)^2+(y-b)^2=r^2. You need to put the values of your three points in the equation and find a,b, and r. Note that this equation assumes that the circle is in 2D. To use it for 3D, you need to first find the plane passing through these three points. If you do not want to go through all these. you can again average all vertices except p and find the center of the circle. the radius would be the distance between the center and any point. I actually assumed that the circles and spheres in your input are uniformly sampled which is the case for most of the available Obj files for these shapes.
I would like to calculate an arbitrarily oriented minimum (area) bounding box of a polygon on a sphere (as a simplification of Earth).
For an axis-aligned version I found the great example from Jason Davies.
The idea is to have as input a list of lon/lat coordinates for the points of the polygon and as output the coordinates of the 4 points (lon/lat) of the bounding box.
Bonus: consider the (very likely) cases where the polygon crosses the antimeridian, contains one pole and/or spans more than one hemisphere.
In my current project I need to calculate the intersection area of triangles and the unit squares in an infinite grid.
For every triangle (given by three pairs of floating point numbers) I need to know the area (in the interval (0,1]) it has in common with every square it intersects.
Right now I convert both (the triangle and the square) to polygons and use Sutherland-Hodgman polygon clipping to calculate the intersection polygon, which I then use to calculate its area.
This approach now shows to be a performance bottleneck in my application. I guess a more specialized (analytical) algorithm would be much faster. Is there a standard solution for this problem, or do you have any idea? I only need the areas, not the shape of the intersections.
Your polygon are convex. There are some algorithms for convex polygons faster than general ones. I've used O'Rourke algorithm with success (code from his book here, I believe that good description exists). Note that some values may be precomputed for your squares.
If your polygons not always intersect, then you may at first check the fact of intersection with separating axes method.
Another option to try- Liang-Barski algorithm for clipping every triangle edge by square.
Edit: You can quickly find all intersections of triangle edges with grid using algorthm of Amanatides and Woo (example in grid traversal section here)
To process this task with hi performance , i suggest some modifications of
Vatti line sweep clipping.
http://en.wikipedia.org/wiki/Vatti_clipping_algorithm
Stepping from minimal Y vertex of your Triangle make such steps:
sort vertexes by Y coordinate
step Y higher to MIN(nextVertex.Y, nextGridBottom)
Calculate points of intersection of grid with edges.
Collect current trapezoid
repeat from step2 until vertex with highest Y coordinate.
Split trapezoids by X coordinate if required.
here is example of Trapezoidalization in X direction
http://www.personal.kent.edu/~rmuhamma/Compgeometry/MyCG/PolyPart/polyPartition.htm
It illustrate main idea of line sweep algorithm. Good luck.
You are not mentioning what precision you are looking for. In case you are looking for a analytical method, disregard this answer, but if you just want to do antialiasing I suggest a scanline edge-flag algorithm by Kiia Kallio. I have used it a few times and it is quite fast and can be set up for very high precision. I have a java implementation if you are interested.
You can take advantage of the regular pattern of squares.
I'm assuming the reason this is a bottleneck is because you have to wait while your algorithm finds all squares intersecting any of the triangles and computes all the areas of intersection. So we'll compute all the areas, but in batches for each triangle in order to get the most information from the fewest calculations.
First, as explained by others, for each edge of the triangle, you can find the sequence of squares that edge passes through, as well as the points at which it crosses each vertical or horizontal edge of a square.
Do this for all three sides, keeping a list of all the squares you encounter, but keep only one copy of each square. It may be useful to store the squares in multiple lists, so that all squares on a given row are all kept in the same list.
When you've found all squares the triangle's edges pass through, if two of those squares were on the same row, any squares between those two that are not in the list are completely inside the triangle, so 100% of each of those squares is covered.
For the other squares, the calculation of area can depend on how many vertices of the triangle are in the square (0, 1, 2, or 3) and where the edges of the triangle intersect the sides of the square. You can summarize all the cases in a few pencil-and-paper drawings, and come up with calculations for each one. For example, when an edge of triangle crosses two sides of the square, with one corner of the square on the "outside" side of the edge, that corner is one angle of a small triangle "cut off" by that edge of the larger triangle; use the points of intersection on the square's sides to compute the area of the small triangle and deduct it from the area of the square. If two points instead of one are "outside", you have a trapezoid whose two base lengths are found from the points of intersection, and whose height is the width of the square; deduct its area from the square. If three points are outside, deduct the entire area of the square and then add the area of the small triangle.
One vertex of the large triangle inside the square, three corners of the square outside that angle: draw a line from the remaining corner to the triangle's vertex, so you have two small triangles, deduct the entire square and add those triangles' areas. Two corners of the square outside the angle, draw lines to the vertex to get three small triangles, etc.
I'm phrasing this so that you always assume you start with the entire area of the square and reduce the area by some amount depending on how the edge of the triangle intersects the square. That way, in the case where the edges of the triangle intersect the square more than twice--such as one edge cuts across one corner of the square and another edge cuts across a different corner, you can just deduct the area cut off by the first edge, then deduct the area cut off by the second edge.
This will be a considerable number of special cases, though you can take advantage of symmetry; for example, you don't have to write the complete calculation for "cut off a triangle in one corner" four times.
You'll write a lot more code than if you just took someone's convex-polygon library off the shelf, and you will want to test the living daylights out of it to make sure you didn't forget to code any cases, but once you get it working, it shouldn't take much more effort to make it reasonably fast.