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  • 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

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  • 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.

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  • 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?

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  • 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.

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  • 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?

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  • 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).

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  • 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.

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  • 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!

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  • 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

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  • reflection paths between points in2d

    - by Chris H
    Just wondering if there was a nice (already implemented/documented) algorithm to do the following Given any shape (without crossing edges) and two points inside that shape, compute all the paths between the two points such that all reflections are perfect reflections. The path lengths should be limited to a certain length otherwise there are infinite solutions. I'm not interested in just shooting out rays to try to guess how close I can get, I'm interested in algorithms that can do it perfectly. Search based, not guess/improvement based.

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  • Calculating co-ordinate of a point on a path given a distance

    - by Alex
    I'm working on a project that surveys the condition of a road or highway using a calibrated trip computer connected to a rugged-PC. An operator keys in defect codes as they travel along a pre-defined route. I need to show an indicator on the map screen that shows the vehicles current position, taking into account the distance data from the trip computer. I know the exact lat lon co-ordinates at the starting point of each section of road, and the road is made up of a series of points. The question is: how can I calculate the lat lon co-ordinates of the vehicle assuming that it has continued on the route and traveled a certain distance (e.g. 1.4km). The co-ordinates would be 'locked onto' the road line, as shown in blue on the diagram below. Thanks, Alex

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  • fit a ellipse in Python given a set of points xi=(xi,yi)

    - by Gianni
    I am computing a series of index from a 2D points (x,y). One index is the ratio between minor and major axis. To fit the ellipse i am using the following post when i run these function the final results looks strange because the center and the axis length are not in scale with the 2D points center = [ 560415.53298363+0.j 6368878.84576771+0.j] angle of rotation = (-0.0528033467597-5.55111512313e-17j) axes = [0.00000000-557.21553487j 6817.76933256 +0.j] thanks in advance for help import numpy as np from numpy.linalg import eig, inv def fitEllipse(x,y): x = x[:,np.newaxis] y = y[:,np.newaxis] D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x))) S = np.dot(D.T,D) C = np.zeros([6,6]) C[0,2] = C[2,0] = 2; C[1,1] = -1 E, V = eig(np.dot(inv(S), C)) n = np.argmax(np.abs(E)) a = V[:,n] return a def ellipse_center(a): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] num = b*b-a*c x0=(c*d-b*f)/num y0=(a*f-b*d)/num return np.array([x0,y0]) def ellipse_angle_of_rotation( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] return 0.5*np.arctan(2*b/(a-c)) def ellipse_axis_length( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g) down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) res1=np.sqrt(up/down1) res2=np.sqrt(up/down2) return np.array([res1, res2]) if __name__ == '__main__': points = [(560036.4495758876, 6362071.890493258), (560036.4495758876, 6362070.890493258), (560036.9495758876, 6362070.890493258), (560036.9495758876, 6362070.390493258), (560037.4495758876, 6362070.390493258), (560037.4495758876, 6362064.890493258), (560036.4495758876, 6362064.890493258), (560036.4495758876, 6362063.390493258), (560035.4495758876, 6362063.390493258), (560035.4495758876, 6362062.390493258), (560034.9495758876, 6362062.390493258), (560034.9495758876, 6362061.390493258), (560032.9495758876, 6362061.390493258), (560032.9495758876, 6362061.890493258), (560030.4495758876, 6362061.890493258), (560030.4495758876, 6362061.390493258), (560029.9495758876, 6362061.390493258), (560029.9495758876, 6362060.390493258), (560029.4495758876, 6362060.390493258), (560029.4495758876, 6362059.890493258), (560028.9495758876, 6362059.890493258), (560028.9495758876, 6362059.390493258), (560028.4495758876, 6362059.390493258), (560028.4495758876, 6362058.890493258), (560027.4495758876, 6362058.890493258), (560027.4495758876, 6362058.390493258), (560026.9495758876, 6362058.390493258), (560026.9495758876, 6362057.890493258), (560025.4495758876, 6362057.890493258), (560025.4495758876, 6362057.390493258), (560023.4495758876, 6362057.390493258), (560023.4495758876, 6362060.390493258), (560023.9495758876, 6362060.390493258), (560023.9495758876, 6362061.890493258), (560024.4495758876, 6362061.890493258), (560024.4495758876, 6362063.390493258), (560024.9495758876, 6362063.390493258), (560024.9495758876, 6362064.390493258), (560025.4495758876, 6362064.390493258), (560025.4495758876, 6362065.390493258), (560025.9495758876, 6362065.390493258), (560025.9495758876, 6362065.890493258), (560026.4495758876, 6362065.890493258), (560026.4495758876, 6362066.890493258), (560026.9495758876, 6362066.890493258), (560026.9495758876, 6362068.390493258), (560027.4495758876, 6362068.390493258), (560027.4495758876, 6362068.890493258), (560027.9495758876, 6362068.890493258), (560027.9495758876, 6362069.390493258), (560028.4495758876, 6362069.390493258), (560028.4495758876, 6362069.890493258), (560033.4495758876, 6362069.890493258), (560033.4495758876, 6362070.390493258), (560033.9495758876, 6362070.390493258), (560033.9495758876, 6362070.890493258), (560034.4495758876, 6362070.890493258), (560034.4495758876, 6362071.390493258), (560034.9495758876, 6362071.390493258), (560034.9495758876, 6362071.890493258), (560036.4495758876, 6362071.890493258)] a_points = np.array(points) x = a_points[:, 0] y = a_points[:, 1] from pylab import * plot(x,y) show() a = fitEllipse(x,y) center = ellipse_center(a) phi = ellipse_angle_of_rotation(a) axes = ellipse_axis_length(a) print "center = ", center print "angle of rotation = ", phi print "axes = ", axes from pylab import * plot(x,y) plot(center[0:1],center[1:], color = 'red') show() each vertex is a xi,y,i point plot of 2D point and center of fit ellipse

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  • Divide a path into N sections using Java or PostgreSQL/PostGIS

    - by Guido
    Imagine a GPS tracking system that is following the position of several objects. The points are stored in a database (PostgreSQL + PostGIS). Each path is composed by a different number of points. That is the reason why, in order to compare a pair of paths, I need to divide every path in a set of 100 points. Do you know any PostGIS function that already implement this algorithm? I've not been able to find it. If not, I'd like to solve it using Java. In this case I'd like to know an efficient and easy to implement algorithm to divide a path into N points. The most simple example could be to divide this path into three points: position 1 : x=1, y=2 position 2 : x=1, y=3 And the result should be: position 1 : x=1, y=2 (starting point) position 2 : x=5, y=2.5 position 3 : x=9, y=3 (end point) Edit: By 'compare a pair of paths' I mean to calculate the distance between two paths. I plan to divide each path in 100 points, and sum the euclidean distance between each one of these points as the distance between the two paths.

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  • Merge overlapping triangles into a polygon

    - by nornagon
    I've got a bunch of overlapping triangles from a 3D model projected into a 2D plane. I need to merge each island of touching triangles into a closed, non-convex polygon. The resultant polygons shouldn't have any holes in them (since the source data doesn't). Many of the source triangles share (floating point identical) edges with other triangles in the source data. What's the easiest way to do this? Performance isn't particularly important, since this will be done at design time.

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  • Render only the segment/area of a circle that intersects the main circle

    - by Greenhouse Gases
    I absolutely love maths (or 'math' as most of you would say!) but I haven't done it to a level where I know the answer to this problem. I have a main circle which could have a centre point at any x and y on a display. Other circles will move around the display at will but at any given call to a render method I want to render not only those circles that intersect the main circle, but also only render the segment of that circle that is visible inside the main circle. An analogy would be a shadow cast on a real life object, and I only want to draw the part of that object that is 'illuminated'. I want to do this preferably in Java, but if you have a raw formula that would be appreciated. I wonder how one might draw the shape and fill it in Java, I'm sure there must be some variation on a polyline with arcs or something? Many thanks

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  • Convet from line points to shape points

    - by VOX
    I have an array of points that make up a line. However I need to draw the line with the width of n pixels. How can I transform that points for lines to points for polygon (or a shape) so I can directly draw it on canvas. I'm developing two program at the same time, one is j2me and another is .NET CF. j2me doesn't support drawing lines with width. Please take a look at the picture. link text

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  • Calculate cubic bezier T value where tangent is perpendicular to anchor line.

    - by drawnonward
    Project a cubic bezier p1,p2,p3,p4 onto the line p1,p4. When p2 or p3 does not project onto the line segment between p1 and p4, the curve will bulge out from the anchor points. Is there a way to calculate the T value where the tangent of the curve is perpendicular to the anchor line? This could also be stated as finding the T values where the projected curve is farthest from the center of the line segment p1,p4. When p2 and p3 project onto the line segment, then the solutions are 0 and 1 respectively. Is there an equation for solving the more interesting case? The T value seems to depend only on the distance of the mapped control points from the anchor line segment. I can determine the value by refining guesses, but I am hoping there is a better way.

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  • Grouping rectangles (getting the bounding boxes of rects)

    - by hyn
    What is a good, fast way to get the "final" bounding boxes of a set of random (up to about 40, not many) rectangles? By final I mean that all bounding boxes don't intersect with any other. Brute force way: in a double for loop, for each rect, test for intersection against every other rect. The intersecting rects become a new rect (replaced), indicating the bounding box. Start over and repeat until no intersection is detected. Because the rects are random every time, and the rect count is relatively small, collision detection using spatial hashing seems like overkill. Is there a way to do this more effectively?

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  • Method of transforming 3D vectors with a matrix

    - by Drew Noakes
    I've been doing some reading on transforming Vector3 with matrices, and am tossing up digging deeper into the math and coding this myself versus using existing code. For whatever reason my school curriculum never included matrices, so I'm filling a gap in my knowledge. Thankfully I only need a few simple things, I think. Context is that I'm programming a robot for the RoboCup 3D league. I'm coding it in C# but it'll have to run on Mono. Ideally I wouldn't use any existing graphics libraries for this (WinForms/WPF/XNA) as all I really need is a neat subset of matrix transformations. Specifically, I need translation and x/y/z rotations, and a way of combining multiple transformations into a single matrix. This will then be applied to my own Vector3 type to produce the transformed Vector3. I've read different advice about this. For example, some model the transformation with a 4x3 matrix, others with a 4x4 matrix. Also, some examples show that you need a forth value for the vector's matrix of 1. What happens to this value when it's included in the output? [1 0 0 0] [x y z 1] * [0 1 0 0] = [a b c d] [0 0 1 0] [2 4 6 1] The parts I'm missing are: What sizes my matrices should be Compositing transformations by multiplying the transformation matrices together Transforming 3D vectors with the resulting matrix As I mostly just want to get this running, any psuedo-code would be great. Information about what matrix values perform what transformations is quite clearly defined on many pages, so need not be discussed here unless you're very keen :)

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  • How do i pack multiple rectangles in a 2d box tetris style

    - by mglmnc
    I have a number of rectangles of various widths and heights. I have a larger rectangular platform to put them on. I want to pack them on one side of the platform so they spread in the lengthwise (X) dimension but keep the widthwise (Y) dimension to a minimal. That is to place them like a tetris game. There can be no overlaps but there can be gaps. Is there an algorithm out there to do this?

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  • Projection matrix + world plane ~> Homography from image plane to world plane

    - by B3ret
    I think I have my wires crossed on this, it should be quite easy. I have a projection matrix from world coordinates to image coordinates (4D homogeneous to 3D homgeneous), and therefore I also have the inverse projection matrix from image coordinates to world "rays". I want to project points of the image back onto a plane within the world (which is given of course as 4D homogeneous vector). The needed homography should be uniquely identified, yet I can not figure out how to compute it. Of course I could also intersect the back-projected rays with the world plane, but this seems not a good way, knowing that there MUST be a homography doing this for me. Thanks in advance, Ben

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