Search Results

Search found 3349 results on 134 pages for 'geometry conversion'.

Page 29/134 | < Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >

  • Converting from Latitude/Longitude to Cartesian Coordinates with a World File and map image.

    - by Heath
    I have a java applet that allows users to import a jpeg and world file from the local system. The user can then "click" draw lines on the image that was imported. Each endpoint of each line contains a set of X/Y and Lat/Long values. The XY is standard java coordinate space, the applet uses an affine transform calculation with the world file to determine the lat/long for every point on the canvas. I have a requirement that allows a user to type a distance into a text field and use the arrow key to draw a line in a certain direction (Up, Down, Left, Right) from a single selected point on the screen. I know how to determine the lat/long of a point given a source lat/long, distance, and bearing. So a user types "100" in the text field and presses the Right arrow key a line should be drawn 100 feet to the right from the currently selected point. My issue is I don't know how to convert the distance( which is in feet ) into the distance in pixels. This would then tell my where to plot the point.

    Read the article

  • Implementation of any Hamiltonian Path Problem algorithm

    - by Julien
    Hi all ! Here is my problem : I have an array of points, the points have three properties : the "x" and "y" coordinates, and a sequence number "n". The "x" and "y" are defined for all the points, the "n" are not. You can access and write them calling points[i]-x, points[i]-y, points[i]-n. i.e. : points[i]->n = var var = points[i]->n So the title maybe ruined the surprise, but I'm looking for a possible implementation of a solution to the Hamiltonian path problem : I need to set the "n" number of each point, so that the sequence is the shortest path (not cycle, the edges have to be disjoint) that goes exactly once through each point. I looked for a solution and I found The Bellman Ford Algorithm but I think it doesn't work since the problem doesn't specify that it has to go through all of the points is it correct ? If it is, does somebody has another algorithm and the implementation ? If the Bellman Ford Algorithm works, how would I implement it ? Thanks a lot, Julien

    Read the article

  • How to detect if an ellipse intersects(collides with) a circle

    - by php html
    I want to improve a collision system. Right now I detect if 2 irregular objects collide if their bounding rectangles collide. I want to obtain the for rectangle the corresponding ellipse while for the other one to use a circle. I found a method to obtain the ellipse coordinates but I have a problem when I try to detect if it intersects the circle. Do you know a algorithm to test if a circle intersects an ellipse?

    Read the article

  • How to calculate the length of a Path2D in Java?

    - by Sanoj
    I have some paths represented by Path2D. The Path consist of multiple CubicCurve2D or Line2D segments that are connected to each other. I would like to calculate or get the length from the start to the end of a Path. How can I calculate it or get it? Is it possible? I have checked the API documentation, but couldn't find any useful methods.

    Read the article

  • Bounding Boxes for Circle and Arcs in 3D

    - by David Rutten
    Given curves of type Circle and Circular-Arc in 3D space, what is a good way to compute accurate bounding boxes (world axis aligned)? Edit: found solution for circles, still need help with Arcs. C# snippet for solving BoundingBoxes for Circles: public static BoundingBox CircleBBox(Circle circle) { Point3d O = circle.Center; Vector3d N = circle.Normal; double ax = Angle(N, new Vector3d(1,0,0)); double ay = Angle(N, new Vector3d(0,1,0)); double az = Angle(N, new Vector3d(0,0,1)); Vector3d R = new Vector3d(Math.Sin(ax), Math.Sin(ay), Math.Sin(az)); R *= circle.Radius; return new BoundingBox(O - R, O + R); } private static double Angle(Vector3d A, Vector3d B) { double dP = A * B; if (dP <= -1.0) { return Math.PI; } if (dP >= +1.0) { return 0.0; } return Math.Acos(dP); }

    Read the article

  • formula for best approximation for center of 2D rotation with small angles

    - by RocketSurgeon
    This is not a homework. I am asking to see if problem is classical (trivial) or non-trivial. It looks simple on a surface, and I hope it is truly a simple problem. Have N points (N = 2) with coordinates Xn, Yn on a surface of 2D solid body. Solid body has some small rotation (below Pi/180) combined with small shifts (below 1% of distance between any 2 points of N). Possibly some small deformation too (<<0.001%) Same N points have new coordinates named XXn, YYn Calculate with best approximation the location of center of rotation as point C with coordinates XXX, YYY. Thank you

    Read the article

  • WPF: Combine Geometries / Canvas for Geometries?

    - by stefan.at.wpf
    Hello, I have 2 geometries A and B which I'd like to combine like shown at the end of the following drawing: http://www.imagebanana.com/view/9vm6zoy/geocombine.png How could one do this? Is there something like a "virtual canvas for geometries" where I can place A and B, move B accordingly and then get a geometrie from this? Thanks for any hint!

    Read the article

  • Parallel curve like algorithm for graphs

    - by skrat
    Is there a well know algorithm for calculating "parallel graph"? where by parallel graph I mean the same as parallel curve, vaguely called "offset curve", but with a graph instead of a curve. Given this picture how can I calculate points of black outlined polygons?

    Read the article

  • comparing two angles

    - by Elazar Leibovich
    Given four points in the plane, A,B,X,Y, I wish to determine which of the following two angles is smaller ?ABX or ?ABY. I'd rather not use cos or sqrt, in order to preserve accuracy. In the case where A=(-1,0),B=(0,0), I can compare the two angles ?ABX and ?ABY, by calculating the dot product of the vectors X,Y, and watch it's sign. What I can do in this case is: Determine whether or not ABX turns right or left If ABX turns left check whether or not Y and A are on the same side of the line on segment BX. If they are - ?ABX is a smaller than ABY. If ABX turns right, then Y and A on the same side of BX means that ?ABX is larger than ?ABY. But this seems too complicated to me. Any simpler approach?

    Read the article

  • Union of complex polygons

    - by grenade
    Given two polygons: POLYGON((1 0, 1 8, 6 4, 1 0)) POLYGON((4 1, 3 5, 4 9, 9 5, 4 1),(4 5, 5 7, 6 7, 4 4, 4 5)) How can I calculate the union (combined polygon)? Dave's example uses SQL server to produce the union, but I need to accomplish the same in code. I'm looking for a mathematical formula or code example in any language that exposes the actual math. I am attempting to produce maps that combine countries dynamically into regions. I asked a related question here: http://stackoverflow.com/questions/2653812/grouping-geographical-shapes

    Read the article

  • Algorithm to merge adjacent rectangles into polygon

    - by Glitch
    I guess that my problem is related to "convex hull", but no the same. All shapes in the drawing are rectangles with same width and height. Many are adjacent to each other. I want to combine those adjacent rectangles into polygons. Unlike "convex hull", the resuled polygons could be "hollow" inside. Is there any open source algorithm available?

    Read the article

  • How do I arbitrarily distort a textured polygon?

    - by Archagon
    I'd like to write a program that lets me arbitrarily distort a textured polygon by dragging its vertices. I want the texture to distort fluidly and without overlap, assuming the new polygon doesn't intersect itself. I should also be able to repeat the process with the new shape, and with a minimum amount of loss. Are there any algorithms for doing this?

    Read the article

  • How to find two most distant points?

    - by depesz
    This is a question that I was asked on a job interview some time ago. And I still can't figure out sensible answer. Question is: you are given set of points (x,y). Find 2 most distant points. Distant from each other. For example, for points: (0,0), (1,1), (-8, 5) - the most distant are: (1,1) and (-8,5) because the distance between them is larger from both (0,0)-(1,1) and (0,0)-(-8,5). The obvious approach is to calculate all distances between all points, and find maximum. The problem is that it is O(n^2), which makes it prohibitively expensive for large datasets. There is approach with first tracking points that are on the boundary, and then calculating distances for them, on the premise that there will be less points on boundary than "inside", but it's still expensive, and will fail in worst case scenario. Tried to search the web, but didn't find any sensible answer - although this might be simply my lack of search skills.

    Read the article

  • Finding intersection of two spheres

    - by Onkar Deshpande
    Hi, Consider the following problem - I am given 2 links of length L0 and L1. P0 is the point that the first link starts at and P1 is the point that I want the end of second link to be at in 3-D space. I am supposed to write a function that should take in these 3-D points (P0 and P1) as inputs and should find all configurations of the links that put the second link's end point at P1. My understanding of how to go about it is - Each link L0 and L1 will create a sphere S0 and S1 around itself. I should find out the intersection of those two spheres (which will be a circle) and print all points that are on the circumference of that circle. I saw gmatt's first reply on the http://stackoverflow.com/questions/1406375/finding-intersection-points-between-3-spheres but could not understand it properly since the images did not show up. I also saw a formula for finding out the intersection at mathworld[dot]wolfram[dot]com/Sphere-SphereIntersection[dot]html . I could find the radius of intersection by the method given on mathworld. Also I can find the center of that circle and then use the parametric equation of circle to find the points. The only doubt that I have is will this method work for the points P0 and P1 mentioned above ? Please comment and let me know your thoughts.

    Read the article

  • uniform generation of points on 3D box

    - by Myx
    Hello: I would like to generate random points on a 3D box defined by its (minx, miny, minz) and (maxx, maxy, maxz) corners. I was thinking of generating a random point inside of the box and then somehow projecting it onto one of the box sides. However, I don't have explicit plane information for the box sides and this seems like it will not produce a uniform distribution of points since if some sides of the box are bigger than others, those sides should have more points generated on them. Any suggestions are appreciated. Thanks.

    Read the article

  • Get Angle to Tangent that Intersects Point

    - by Christian Stewart
    I have a circle around a given point, call this point (x1, y1). I know the radius of the circle around this point. I also have a second point (x2, y2), that is a distance away, outside the radius of the circle. I need a algebraic way through code to calculate the heading (angle from vertical) needed to intersect the circle at 90* to the center point (I.E. get the angle of the tangent intersecting line 2) around the point (x1, y1) from the second point (x2, y2) A bit of background: Essentially the two points are GPS coordinates on a 2D map, I need to know the target heading to intersect the circle in order to follow its path around the center point. Thanks! Christian

    Read the article

  • What is the name of this geometrical function?

    - by Spike
    In a two dimentional integer space, you have two points, A and B. This function returns an enumeration of the points in the quadrilateral subset bounded by A and B. A = {1,1} B = {2,3} Fn(A,B) = {{1,1},{1,2},{1,3},{2,1},{2,2},{2,3}} I can implement it in a few lines of LINQ. private void UnknownFunction(Point to, Point from, List<Point> list) { var vectorX = Enumerable.Range(Math.Min(to.X, from.X), Math.Abs(to.X - from.Y) + 1); var vectorY = Enumerable.Range(Math.Min(to.Y, from.Y), Math.Abs(to.Y - from.Y) + 1); foreach (var x in vectorX) foreach (var y in vectorY) list.Add(new Point(x, y)); } I'm fairly sure that this is a standard mathematical operation, but I can't think what it is. Feel free to tell me that it's one line of code in your language of choice. Or to give me a cunning implementation with lambdas or some such. But mostly I just want to know what it's called. It's driving me nuts. It feels a little like a convolution, but it's been too long since I was at school for me to be sure.

    Read the article

  • correcting fisheye distortion programmatically

    - by Will
    I have some points that describe positions in a picture taken with a fisheye lens. I've found this description of how to generate a fisheye effect, but not how to reverse it. How do you calculate the radial distance from the centre to go from fisheye to rectilinear? My function stub looks like this: Point correct_fisheye(const Point& p,const Size& img) { // to polar const Point centre = {img.width/2,img.height/2}; const Point rel = {p.x-centre.x,p.y-centre.y}; const double theta = atan2(rel.y,rel.x); double R = sqrt((rel.x*rel.x)+(rel.y*rel.y)); // fisheye undistortion in here please //... change R ... // back to rectangular const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta)); fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y); return ret; } Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?

    Read the article

  • uniform generation of 3D points on cylinder/cone

    - by Myx
    Hello: I wish to randomly and uniformly generate points on a cylinder and a cone (separately). The cylinder is defined by its center, its radius and height. Same specifications for the cone. I am able to get the bounding box for each shape so I was thinking of generating points within the bounding box. However, I'm not sure how to project them onto the cylinder/cone or if this is the best idea. Any suggestions? Thanks.

    Read the article

  • Calculate the Hilbert value of a point for use in a Hilbert R-Tree?

    - by wrt
    I have an application where a Hilbert R-Tree (wikipedia) (citeseer) would seem to be an appropriate data structure. Specifically, it requires reasonably fast spatial queries over a data set that will experience a lot of updates. However, as far as I can see, none of the descriptions of the algorithms for this data structure even mention how to actually calculate the requisite Hilbert Value; which is the distance along a Hilbert Curve to the point. So any suggestions for how to go about calculating this?

    Read the article

  • Coordinates in distorted grid

    - by Carsten
    I have a grid in a 2D system like the one in the before image where all points A,B,C,D,A',B',C',D' are given (meaning I know the respective x- and y-coordinates). I need to calculate the x- and y-coordinates of A(new), B(new), C(new) and D(new) when the grid is distorted (so that A' is moved to A'(new), B' is moved to B'(new), C' is moved to C'(new) and D' is moved to D'(new)). The distortion happens in a way in which the lines of the grid are each divided into sub-lines of equal length (meaning for example that AB is divided into 5 parts of the equal length |AB|/5 and A(new)B(new) is divided into 5 parts of the equal length |A(new)B(new)|/5). The distortion is done with the DistortImage class of the Sandy 3D Flash engine. (My practical task is to distort an image using this class where the handles are not positioned at the corners of the image like in this demo but somewhere within it).

    Read the article

  • Grouping geographical shapes

    - by grenade
    I am using Dundas Maps and attempting to draw a map of the world where countries are grouped into regions that are specific to a business implementation. I have shape data (points and segments) for each country in the world. I can combine countries into regions by adding all points and segments for countries within a region to a new region shape. foreach(var region in GetAllRegions()){ var regionShape = new Shape { Name = region.Name }; foreach(var country in GetCountriesInRegion(region.Id)){ var countryShape = GetCountryShape(country.Id); regionShape.AddSegments(countryShape.ShapeData.Points, countryShape.ShapeData.Segments); } map.Shapes.Add(regionShape); } The problem is that the country border lines still show up within a region and I want to remove them so that only regional borders show up. Dundas polygons must start and end at the same point. This is the case for all the country shapes. Now I need an algorithm that can: Determine where country borders intersect at a regional border, so that I can join the regional border segments. Determine which country borders are not regional borders so that I can discard them. Sort the resulting regional points so that they sequentialy describe the shape boundaries. Below is where I have gotten to so far with the map. You can see that the country borders still need to be removed. For example, the border between Mongolia and China should be discarded whereas the border between Mongolia and Russia should be retained. The reason I need to retain a regional border is that the region colors will be significant in conveying information but adjacent regions may be the same color. The regions can change to include or exclude countries and this is why the regional shaping must be dynamic. EDIT: I now know that I what I am looking for is a UNION of polygons. David Lean explains how to do it using the spatial functions in SQL Server 2008 which might be an option but my efforts have come to a halt because the resulting polygon union is so complex that SQL truncates it at 43,680 characters. I'm now trying to either find a workaround for that or find a way of doing the union in code.

    Read the article

  • What is the maximum distance from an anchor point to a bezier curve?

    - by drawnonward
    Given a cubic bezier curve P0,P1,P2,P3 with the following properties: • Both P1 and P2 are on the same side of the line formed by P0 and P3. • P2 can be projected onto the line segment formed by P0 and P3 but P1 cannot. What is the T value for the point on the curve farthest from P3? Here is an image with an example curve. The curve bulges on the left, so there is a point on the curve farther from P3 than P0. I found this reference for finding the minimum distance from an arbitrary point to a curve. Is trial and error the only way to solve for maximum distance as well? Does it make any difference that the point is an anchor on the curve? Thanks

    Read the article

  • Choosing circle radius to fully fill a rectangle

    - by Andy
    Hi, the pixman image library can draw radial color gradients between two circles. I'd like the radial gradient to fill a rectangular area defined by "width" and "height" completely. Now my question, how should I choose the radius of the outer circle? My current parameters are the following: A) inner circle (start of gradient) center pointer of inner circle: (width*0.5|height*0.5) radius of inner circle: 1 color: black B) outer circle (end of gradient) center pointer of outer circle: (width*0.5|height*0.5) radius of outer circle: ??? color: white How should I choose the radius of the outer circle to make sure that the outer circle will entirely fill my bounding rectangle defined by width*height. There shall be no empty areas in the corners, the area shall be completely covered by the circle. In other words, the bounding rectangle width,height must fit entirely into the outer circle. Choosing outer_radius = max(width, height) * 0.5 as the radius for the outer circle is obviously not enough. It must be bigger, but how much bigger? Thanks!

    Read the article

< Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >