Search Results

Search found 779 results on 32 pages for 'coordinate'.

Page 2/32 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Inverse projection: question about w coordinate

    - by fayeWilly
    I have to perform in shader an inverse projection from a u/v of a render target. What I do is: Get NDC as 2*(u,v,depth) - 1 Then world space as tmp = (P*V)^-1 * (NDC,1.0); world space = tmp/tmp.w; This apparently works, but I am confused about the w division there. Why this work? Shouldn't be a multiplication by a w somewhere (as in the "forward" pipeline there is the perpsective division?) Thank you, Faye

    Read the article

  • Coordinate and positioning problem on iOS with cocos2d-x

    - by Vexille
    I'm using cocos2d-x alongside with Marmalade and running some tests and tutorials before starting an actual project with them. So far things are working reasonably well on the windows simulator, Android and even on Blackberry's Playbook, but on iOS devices (iPhone and iPad) the positioning seems to be off. To make things clearer, I put together a scene that just draws an image in the middle of the screen. It worked as expected on everything else, but this is the result I got on an iPhone: To get the coordinates for the center of the screen I'm using the VisibleRect class from the TestCpp sample. It just uses sharedOpenGLView to get the visible size and visible origin, and calculate the center from that. CCSprite* test = CCSprite::create("Ball.png", CCRectMake(0, 0, 80, 80) ); test->setPosition( ccp(VisibleRect::center().x, VisibleRect::center().y) ); this->addChild(test); Also I have a noBorder policy set on AppDelegate: CCEGLView::sharedOpenGLView()->setDesignResolutionSize(designSize.width, designSize.height, kResolutionNoBorder); One funny thing is that I tried to deploy the TestCpp sample project to some iOS devices and it worked reasonably well on the iPhone, but on the iPad the application was only being drawn on a small portion of the screen - just like what happened on the iPhone when I tried using the ShowAll policy.

    Read the article

  • Coordinate spaces and transformation matrices

    - by Belgin
    I'm trying to get an object from object space, into projected space using these intermediate matrices: The first matrix (I) is the one that transforms from object space into inertial space, but since my object is not rotated or translated in any way inside the object space, this matrix is the 4x4 identity matrix. The second matrix (W) is the one that transforms from inertial space into world space, which is just a scale transform matrix of factor a = 14.1 on all coordinates, since the inertial space origin coincides with the world space origin. /a 0 0 0\ W = |0 a 0 0| |0 0 a 0| \0 0 0 1/ The third matrix (C) is the one that transforms from world space, into camera space. This matrix is a translation matrix with a translation of (0, 0, 10), because I want the camera to be located behind the object, so the object must be positioned 10 units into the z axis. /1 0 0 0\ C = |0 1 0 0| |0 0 1 10| \0 0 0 1/ And finally, the fourth matrix is the projection matrix (P). Bearing in mind that the eye is at the origin of the world space and the projection plane is defined by z = 1, the projection matrix is: /1 0 0 0\ P = |0 1 0 0| |0 0 1 0| \0 0 1/d 0/ where d is the distance from the eye to the projection plane, so d = 1. I'm multiplying them like this: (((P x C) x W) x I) x V, where V is the vertex' coordinates in column vector form: /x\ V = |y| |z| \1/ After I get the result, I divide x and y coordinates by w to get the actual screen coordinates. Apparenly, I'm doing something wrong or missing something completely here, because it's not rendering properly. Here's a picture of what is supposed to be the bottom side of the Stanford Dragon: Also, I should add that this is a software renderer so no DirectX or OpenGL stuff here.

    Read the article

  • Converting Degrees to X and Y Coordinate change

    - by gopgop
    I am using a float positioning system in my game. IE float x,y,z now I want to get the location of the mouse, then to fire an arrow to it. X0 = the players X location X1 = the mouse X location Y0 = the players Y location Y1 = the mouse Y location I want to make a method that the parameters are Degrees and it sets my Yspeed and my Xspeed accordingly to get to the mouse xy starting at player xy How do would I accomplish this?

    Read the article

  • Why do we need normalized coordinate system? Options

    - by jcyang
    Hi, I have problem understand following sentences in my textbook Computer Graphics with OpenGL. "To make viewing process independent of the requirements of any output device,graphic system convert object descriptions to normalized coordinates and apply the clipping routines." Why normalized coordinates could make viewing process independent of the requirements of any output devices? Isn't the projection coordinates already independent of output device?We only need to first scale and then translate the projection coordinate then we will get device coordinate. So why do we need first convert the projection coordinate to normalized coordinate first? "Clipping is usually performed in normlized coordinates.This allows us to reduce computations by first concatenating the various transformation matrices" Why clipping is usually performed in normlized coordinates? What kind of transformation concatenated? thanks. jcyang.

    Read the article

  • Why / how does XNA's right-handed coordinate system effect anything if you can specify near/far Z values?

    - by vargonian
    I am told repeatedly that XNA Game Studio uses a right-handed coordinate system, and I understand the difference between a right-handed and left-handed coordinate system. But given that you can use a method like Matrix.CreateOrthographicOffCenter to create your own custom projection matrix, specifying the left, right, top, bottom, zNear and zFar values, when does XNA's coordinate system come into play? For example, I'm told that in a right-handed coordinate system, increasingly negative Z values go "into" the screen. But I can easily create my projection matrix like this: Matrix.CreateOrthographicOffCenter(left, right, bottom, top, 0.1f, 10000f); I've now specified a lower value for the near Z than the far Z, which, as I understand it, means that positive Z now goes into the screen. I can similarly tweak the values of left/right/top/bottom to achieve similar results. If specifying a lower zNear than zFar value doesn't affect the Z direction of the coordinate system, what does it do? And when is the right-handed coordinate system enforced? The reason I ask is that I'm trying to implement a 2.5D camera that supports zooming and rotation, and I've spent two full days encountering one unexpected result after another.

    Read the article

  • A Star Path finding endless loop

    - by PoeHaH
    I have implemented A* algorithm. Sometimes it works, sometimes it doesn't, and it goes through an endless loop. After days of debugging and googling, I hope you can come to the rescue. This is my code: The algorythm: public ArrayList<Coordinate> findClosestPathTo(Coordinate start, Coordinate goal) { ArrayList<Coordinate> closed = new ArrayList<Coordinate>(); ArrayList<Coordinate> open = new ArrayList<Coordinate>(); ArrayList<Coordinate> travelpath = new ArrayList<Coordinate>(); open.add(start); while(open.size()>0) { Coordinate current = searchCoordinateWithLowestF(open); if(current.equals(goal)) { return travelpath; } travelpath.add(current); open.remove(current); closed.add(current); ArrayList<Coordinate> neighbors = current.calculateCoordAdjacencies(true, rowbound, colbound); for(Coordinate n:neighbors) { if(closed.contains(n) || map.isWalkeable(n)) { continue; } int gScore = current.getGvalue() + 1; boolean gScoreIsBest = false; if(!open.contains(n)) { gScoreIsBest = true; n.setHvalue(manhattanHeuristic(n,goal)); open.add(n); } else { if(gScore<n.getGvalue()) { gScoreIsBest = true; } } if(gScoreIsBest) { n.setGvalue(gScore); n.setFvalue(n.getGvalue()+n.getHvalue()); } } } return null; } What I have found out is that it always fails whenever there's an obstacle in the path. If I'm running it on 'open terrain', it seems to work. It seems to be affected by this part: || map.isWalkeable(n) Though, the isWalkeable function seems to work fine. If additional code is needed, I will provide it. Your help is greatly appreciated, Thanks :)

    Read the article

  • How to get the coordinate of Gridlayout

    - by Jessy
    Hi, I set my JPanel to GridLayout (6,6), with dimension (600,600) Each cell of the grid will display one pictures with different widths and heights. The picture first add to a JLabel, and the JLabel then added to the cells. How can retrieved the coordinate of the pictures in the cells and not the coordinate of cells? So far the out give these coordinate which equal height and width even on screen the pictures showed in different sizes. e.g. java.awt.Rectangle[x=100,y=100,width=100,height=100] java.awt.Rectangle[x=200,y=100,width=100,height=100] java.awt.Rectangle[x=300,y=100,width=100,height=100] The reason why I used GridLayout instead of gridBagLayout is that, I want each pictures to have boundary. If I use GridBagLayout, the grid will expand according to the picture size. I want grid size to be in fix size. JPanel pDraw = new JPanel(new GridLayout(6,6)); pDraw.setPreferredSize(new Dimension(600,600)); for (int i =0; i<(6*6); i++) { //get random number for height and width of the image int x = rand.nextInt(40)+(50); int y = rand.nextInt(40)+(50); ImageIcon icon = createImageIcon("bird.jpg"); //rescale the image according to the size selected Image img = icon.getImage().getScaledInstance(x,y,img.SCALE_SMOOTH); icon.setImage(img ); JLabel label = new JLabel(icon); pDraw.add(label); } for(Component component:components) { //retrieve the coordinate System.out.println(component.getBounds()); }

    Read the article

  • Projecting a 3D point to 2D screen coordinate OpenTK

    - by sinsro
    Using Monotouch and OpenTK I am trying to get the screen coordinate of one 3D point. I have my world view projection matrix set up, and OpenGL makes sense of it and projects my 3D model perfectly, but how to use the same matrix to project just one point from 2D to 3D? I thought I could simply use: Vector3.Transform(ref input3Dpos, ref matWorldViewProjection, out projected2Dpos); Then have the projected screen coordinate in projected2DPos. But the resulting Vector4 does not seem to represent the proper projected screen coordinate. And I do not know how to calculate it from there on.

    Read the article

  • photoshop - How to show the cursor coordinate in selection and dropper

    - by q0987
    I am using Adobe Photoshop CS4 to edit a picture. The problems I have are that I don't know where I can see the mouse cursor coordinate while I move my mouse. Q1 For example, in Paint.Net, whenever I move mouse cursor, I can immediately see the feedback from bottom-right corner (x, y), how can I see such info in photoshop while I use eyedropper function? Q2 For example, when in crop mode, I drag-drop a rectangle but I cannot see the size of the created region of interest? Thank you

    Read the article

  • Why does DirectX use a left-handed coordinate system?

    - by greyfade
    I considered posting on Stack Overflow, but the question strikes me as being far too subjective since I can't think of a reasonable technical explanation for Microsoft's choice in this matter. But this question has bugged me for so long and the issue keeps coming up in one of my projects, and I have never actually seen an attempt at explaining this: OpenGL uses a right-handed coordinate system, where the +Z part of the world coordinate system extends toward the viewer. DirectX uses a left-handed system where the +Z part of the world coordinate extends into the screen, away from the viewer. I never used the Glide API, so I don't know how it worked, but from what I can gather, it uses a left-handed system as well. Is there a technical reason for this? And if not, is there some conceptual advantage to a particular handedness of a coordinate system? Why would one choose one over the other?

    Read the article

  • How to generate normal coordinate?

    - by rbchr
    Hi! I'm developping an game using the API opengl es 2.0. I need to know how to generate the normal coordinate because i need them to developp the lighting. I'm wondering if there is a software or an algorithm that generate normal coordinate. Great thanks!

    Read the article

  • Need a list of all countries in the world, with a longitude and latitude coordinate

    - by ptrn
    I need a list of all the countries in the world, with one lat/long coordinate for the country. I had a look at GeoNames, but all I can seem to find are lists that have the countries as well as cities in the same list. I guess I can just parse it and filter out the countries, but I was hoping to avoid that. Basically, what I need; Country name - Lat/long coordinate (some sort of center of the country) It doesn't need to say anything about continent, but I wouldn't mind if it did. _L

    Read the article

  • C++ library for Coordinate Transformation Matrices (CTM)?

    - by BastiBense
    I'm looking for a C++ library which allows for easy integration of Coordinate Transformation Matrices (CTM) in my application. You might know CTMs from PDF or PostScript. For one project we are using C++/Qt4 as a framework, which offers a QTransform class, which provides methods like .translate(double x, double y) or .rotate(double degrees). After doing some transformations, it would allow me to get all 6 CTM values, which I could feed into a PDF library or use a transformation matrix in export files. Qt's API also allows for arbitrary mapping of polygons (QPolygon), rectangles (QRect) and other primitive data structures into transformed coordinate systems. So basically I'm looking for something similar to what Qt provides, but without the need of using Qt. I know I could do the matrix multiplications myself, but I'm not really interested in doing so, as I'm very sure that someone already solved this problem, so please no links to books or other guides on how to multiply matrices. Thanks!

    Read the article

  • How to transform coordinate from WGS84 to a coordinate in a projection in PROJ.4?

    - by Sanoj
    I have a GPS-coordinate in WGS84 that I would like to transform to a map-projection coordinate in SWEREF99 TM using PROJ.4 in Java or proj4js in JavaScript. Its hard to find documentation for PROJ.4 and how to us it. If you have a good link, please post it as a comment. The PROJ.4 parameters for SWEREF99 TM is +proj=utm +zone=33 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=m +no_defs I have tried to use a PROJ.4 Java library and tried this code and values: String[] proj4_w = new String[] { "+proj=utm", "+zone=33", "+ellps=GRS80", "+towgs84=0,0,0,0,0,0,0", "+units=m", "+no_defs" }; Projection proj = ProjectionFactory.fromPROJ4Specification(proj4_w); Point2D.Double testLatLng = new Point2D.Double(55.0000, 12.7500); Point2D.Double testProjec = proj.transform(testLatLng, new Point2D.Double()); This give me the point Point2D.Double[5197915.86288144, 1822635.9083898761] but I should be N: 6097106.672, E: 356083.438 What am I doing wrong? and what method and parameters should I use instead? The correct values is taken from Lantmäteriet. I am not sure if proj.transform(testLatLng, new Point2D.Double()); is the right method to use.

    Read the article

  • Calculating bounding box a certain distance away from a lat/long coordinate in Java

    - by Bryce Thomas
    Given a coordinate (lat, long), I am trying to calculate a square bounding box that is a given distance (e.g. 50km) away from the coordinate. So as input I have lat, long and distance and as output I would like two coordinates; one being the south-west (bottom-left) corner and one being the north-east (top-right) corner. I have seen a couple of answers on here that try to address this question in Python, but I am looking for a Java implementation in particular. Just to be clear, I intend on using the algorithm on Earth only and so I don't need to accommodate a variable radius. It doesn't have to be hugely accurate (+/-20% is fine) and it'll only be used to calculate bounding boxes over small distances (no more than 150km). So I'm happy to sacrifice some accuracy for an efficient algorithm. Any help is much appreciated. Edit: I should have been clearer, I really am after a square, not a circle. I understand that the distance between the center of a square and various points along the square's perimeter is not a constant value like it is with a circle. I guess what I mean is a square where if you draw a line from the center to any one of the four points on the perimeter that results in a line perpendicular to a side of the perimeter, then those 4 lines have the same length.

    Read the article

  • php PHPExcel split Excel cell coordinate

    - by Shiro
    currently, I used PHPExcel to import excel file, there is a function $cell-getCoordinate(); I would like to ask any solution for split the cell coordinate alphabet and integer? e.g A1, A2, I need to know currently which row, and until which column. I do some research about split, but not luck for it. Any idea?

    Read the article

  • Adding distance to a GPS coordinate

    - by fgs
    Hi! I'm trying to generate some points at random distances away from a fixed point using GPS. How can I add distance in meters to a GPS coordinate? I've looked at UTM to GPS conversion but is there a simpler method to achieve this? I'm working on Android platform just in case. Cheers, fgs

    Read the article

  • OpenGL Coordinate system confusion

    - by user146780
    Maybe I set up GLUT wrong. Basically I want verticies to be reletive to their size in pixels. Ex:right now if I create a hexagon, it hakes up the whole screen even though the units are 6. #include <iostream> #include <stdlib.h> //Needed for "exit" function #include <cmath> //Include OpenGL header files, so that we can use OpenGL #ifdef __APPLE__ #include <OpenGL/OpenGL.h> #include <GLUT/glut.h> #else #include <GL/glut.h> #endif using namespace std; //Called when a key is pressed void handleKeypress(unsigned char key, //The key that was pressed int x, int y) { //The current mouse coordinates switch (key) { case 27: //Escape key exit(0); //Exit the program } } //Initializes 3D rendering void initRendering() { //Makes 3D drawing work when something is in front of something else glEnable(GL_DEPTH_TEST); } //Called when the window is resized void handleResize(int w, int h) { //Tell OpenGL how to convert from coordinates to pixel values glViewport(0, 0, w, h); glMatrixMode(GL_PROJECTION); //Switch to setting the camera perspective //Set the camera perspective glLoadIdentity(); //Reset the camera gluPerspective(45.0, //The camera angle (double)w / (double)h, //The width-to-height ratio 1.0, //The near z clipping coordinate 200.0); //The far z clipping coordinate } //Draws the 3D scene void drawScene() { //Clear information from last draw glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); glLoadIdentity(); //Reset the drawing perspective glPolygonMode(GL_FRONT_AND_BACK, GL_FILL); glBegin(GL_POLYGON); //Begin quadrilateral coordinates //Trapezoid glColor3f(255,0,0); for(int i = 0; i < 6; ++i) { glVertex2d(sin(i/6.0*2* 3.1415), cos(i/6.0*2* 3.1415)); } glEnd(); //End quadrilateral coordinates glutSwapBuffers(); //Send the 3D scene to the screen } int main(int argc, char** argv) { //Initialize GLUT glutInit(&argc, argv); glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGBA | GLUT_DEPTH); glutInitWindowSize(400, 400); //Set the window size //Create the window glutCreateWindow("Basic Shapes - videotutorialsrock.com"); initRendering(); //Initialize rendering //Set handler functions for drawing, keypresses, and window resizes glutDisplayFunc(drawScene); glutKeyboardFunc(handleKeypress); glutReshapeFunc(handleResize); glutMainLoop(); //Start the main loop. glutMainLoop doesn't return. return 0; //This line is never reached } How can I make it so that a polygon of 0,0 10,0 10,10 0,10 defines a polygon starting at the top left of the screen and is a width and height of 10 pixels? Thanks

    Read the article

  • Actionscript change default coordinate

    - by Comma
    I have a sprite inside another sprite SpriteB is inside SpriteA I would like to change the default coordinate(top left corner) to say 250,10 When SpriteB.x = 0, SpriteB.y = 0 puts SpriteB to 250,10 in SpriteA Is this possible?

    Read the article

  • Mismatched coordinate systems in LWJGL with Mouse and Textures

    - by Braains
    I'm not really sure how to expand on this other than to say that it appears that my LWJGL seems to have different coordinate systems for the Mouse and for painting Textures. It seems that Textures have the usual Java2D way of putting (0, 0) in the upper-left corner, while the Mouse goes by the more sensible way of having the origin in the lower-left corner. I've checked my code a bunch but I don't see anything modifying the values between where I read them and where I use them. It's thrown me for a loop and I can't quite figure it out. I'll post all the code that involves the Mouse input and Texture painting for you guys to look at. private static void pollHelpers() { while(Mouse.next()) { InputHelper.acceptMouseInput(Mouse.getEventButton(), Mouse.getEventX(), Mouse.getEventY()); } while (Keyboard.next()) { if (Keyboard.getEventKeyState()) { InputHelper.acceptKeyboardInput(Keyboard.getEventKey(), true); } else { InputHelper.acceptKeyboardInput(Keyboard.getEventKey(), false); } } } public static void acceptMouseInput(int mouseButton, int x, int y) { for(InputHelper ie: InputHelper.instances) { if(ie.checkRectangle(x, y)) { ie.sendMouseInputToParent(mouseButton); } } } private void sendMouseInputToParent(int mouseButton) { parent.onClick(mouseButton); } public boolean checkRectangle(int x, int y) { //y = InputManager.HEIGHT - y; See below for explanation return x > parent.getX() && x < parent.getX() + parent.getWidth() && y > parent.getY() && y < parent.getY() + parent.getHeight(); } I put this line of code in because it temporarily fixed the coordinate system problem. However, I want to have my code to be as independent as possible so I want to remove as much reliance on other classes as possible. These are the only methods that touch the Mouse input, and as far as I can tell none of them change anything so all is good here. Now for the Texture methods: public void draw() { if(!clicked || deactivatedImage == null) { activatedImage.bind(); glBegin(GL_QUADS); { DrawHelper.drawGLQuad(activatedImage, x, y, width, height); } glEnd(); } else { deactivatedImage.bind(); glBegin(GL_QUADS); { DrawHelper.drawGLQuad(deactivatedImage, x, y, width, height); } glEnd(); } } public static void drawGLQuad(Texture texture, float x, float y, float width, float height) { glTexCoord2f(x, y); glVertex2f(x, y); glTexCoord2f(x, y + texture.getHeight()); glVertex2f(x, y + height); glTexCoord2f(x + texture.getWidth(), y + texture.getHeight()); glVertex2f(x + width, y +height); glTexCoord2f(x + texture.getWidth(), y); glVertex2f(x + width, y); } I'll be honest, I do not have the faintest clue as to what is really happening in this method but I was able to put it together on my own and get it to work. my guess is that this is where the problem is. Thanks for any help!

    Read the article

  • Drop shadow coordinate different from iPhone OS 3.2?

    - by Axel
    I'm trying to draw a custom view with a drop shadow. I'm having different results from iPhone OS 3.2. Has the coordinate system changed for CGContextSetShadowWithColor from 3.2? Here is the code I'm using: CGContextRef graphicContext = UIGraphicsGetCurrentContext(); CGColorSpaceRef colorspace = CGColorSpaceCreateDeviceRGB(); CGFloat values1[4] = { 0, 0, 0, 1}; CGColorRef blackColor = CGColorCreate (colorspace, values1); CGContextSetShadowWithColor(graphicContext, CGSizeMake(0, -1), 3, blackColor); CGColorRelease(blackColor); CGFloat values2[4] = { .5, .5, 1, 1}; CGColorRef fillColor = CGColorCreate (colorspace, values2); CGContextSetFillColorWithColor(graphicContext, fillColor); CGColorRelease(fillColor); CGContextFillRect(graphicContext,CGRectMake(40, 40, 100, 100)); Here is how it looks on 3.1 and from 3.2. All my views using drop shadow look different depending on the version of iPhone OS. Am I missing something?

    Read the article

  • Getting x/y coordinate of a UITouch...

    - by Tarek
    HI, I have been trying to get the x/y coordinates from a touch on any iDevice. When getting the touch locations, everything looks ok if the touch is in the middle of the screen. But if I drag my finger to the bottom of the screen, I can only get a y coordinate of 1015. It should be getting to 1023. Same thing for dragging my finger to the top of the screen. I get -6. It should be 0. I have explicitly set the window and views to an origin of 0,0 and the width, height of the device's screen. Still nothing. I am really lost on what might be going on. Is something shifted? Am I not reading the x/y coordinates properly. Does something need to be transformed or converted? Any help would be much appreciated. T

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >