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  • How are DX and DY coordinates calculated in flash?

    - by Meganlee1984
    I'm trying to update a clients site and the original developer left almost no instructions. The code is all updated through XML. Here is a sample of the code enter code here<FOLDER NAME="COMMERCIAL"> <GALLERY NAME="LOCANDA VERDE: New York"> <IMG HEIGHT="500" CAPTION="Some photo" WIDTH="393" SRC="locanda1.jpg" DX="60" DY="40" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="300" CAPTION="Some photo" WIDTH="450" SRC="locanda2.jpg" DX="160" DY="80" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="500" CAPTION="Some photo" WIDTH="393" SRC="locanda3.jpg" DX="80" DY="260" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="500" CAPTION="Some photo" WIDTH="393" SRC="locanda4.jpg" DX="120" DY="60" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="393" CAPTION="Some photo" WIDTH="500" SRC="locanda5.jpg" DX="180" DY="100" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="500" CAPTION="Some photo" WIDTH="433" SRC="locanda6.jpg" DX="60" DY="140" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> <IMG HEIGHT="500" CAPTION="Some photo" WIDTH="393" SRC="locanda7.jpg" DX="100" DY="200" LINKTEXT="" LINKURL="" INFOTEXT="SOHO, NEW YORK"/> </GALLERY>`enter code here It relates to this page: http://meyerdavis.com/ Click Commercial Click Laconda Verde New York. The xml file pulls a jpg from 2 places, one is a thumb nail that are all 60 x 60 and then one is the bigger sized image. The issue that I'm having is that I can't figure out how the DX and DY coordinates are generated for each item. Any help would be much appreciated. ` Edit: Here's the code from the comment below. platformblock.expandspeed = 0.02; platformblock.h = 450 - platformblock.dy1; //platformblock.h = 402; platformblock.dy2 = 0; platformblock.onResize(); /**/ platformblock.onEnterFrame = function() { this.dy1 += (48 - this.dy1)*this.expandspeed; this.h = 450 - this.dy1; if(this.expandspeed<0.3) { this.expandspeed += 0.02; } if(Math.abs(this.dy1-48)<0.2) { this.dy1 = 48; } if(this.platform._height==402 && this.dy1==48){ this.h = null; this.onResize(); this.onEnterFrame = null; } } platformblock._resizeto(800, 402, _root.play, _root, 0.08); titleblockcontainer.play(); /**/ stop();

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  • Grid overlayed on image using javascript, need help getting grid coordinates.

    - by Alos
    Hi I am fairly new to javascript and could use some help, I am trying to overlay a grid on top of an image and then be able to have the user click on the grid and get the grid coordinate from the box that the user clicked. I have been working with the code from the following stackoverflow question: Creating a grid overlay over image. link text Here is the code that I have so far: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> </script> <script type="text/javascript"> var SetGrid = function(el, sz, nr, nc){ //get number of rows/columns according to the 'grid' size //numcols = el.getSize().x/sz; //numrows = el.getSize().y/sz; numcols = 48; numrows = 32; //create table element for injecting cols/rows var gridTable = new Element('table', { 'id' : 'gridTable', 'styles' : { 'width' : el.getSize().x, 'height' : el.getSize().y, 'top' : el.getCoordinates().top, 'left' : el.getCoordinates().left } }); //inject rows/cols into gridTable for (row = 1; row<=numrows; row++){ thisRow = new Element('tr', { 'id' : row, 'class' : 'gridRow' }); for(col = 1; col<=numcols; col++){ thisCol = new Element('td', { 'id' : col, 'class' : 'gridCol0' }); //each cell gets down/up over event... down starts dragging|up stops|over draws area if down was fired thisCol.addEvents({ 'mousedown' : function(){ dragFlag = true; startRow = this.getParent().get('id'); startCol = this.get('id'); }, 'mouseup' : function(){ dragFlag = false; }, 'mouseover' : function(){ if (dragFlag==true){ this.set('class', 'gridCol'+$$('#lvlSelect .on').get('value')); } }, 'click' : function(){ //this.set('class', 'gridCol'+$$('#lvlSelect .on').get('id').substr(3, 1) ); str = $$('#lvlSelect .on').get('id'); alert(str.substr(2, 3)); } }); thisCol.inject(thisRow, 'bottom'); }; thisRow.inject(gridTable, 'bottom'); }; gridTable.inject(el.getParent()); } //sens level selector func var SetSensitivitySelector = function(el, sz, nr, nc){ $$('#lvlSelect ul li').each(function(el){ el.addEvents({ 'click' : function(){ $$('#lvlSelect ul li').set('class', ''); this.set('class', 'on'); }, 'mouseover' : function(){ el.setStyle('cursor','pointer'); }, 'mouseout' : function(){ el.setStyle('cursor',''); } }); }); } //execute window.addEvent('load', function(){ SetGrid($('imagetomap'), 32); SetSensitivitySelector(); }); var gridSize = { x: 48, y: 32 }; var img = document.getElementById('imagetomap'); img.onclick = function(e) { if (!e) e = window.event; alert(Math.floor(e.offsetX/ gridSize.x) + ', ' + Math.floor(e.offsetY / gridSize.y)); } </script> <style> #imagetomapdiv { float:left; display: block; } #gridTable { border:1px solid red; border-collapse:collapse; position:absolute; z-index:5; } #gridTable td { opacity:0.2; filter:alpha(opacity=20); } #gridTable .gridCol0 { border:1px solid gray; background-color: none; } #gridTable .gridCol1 { border:1px solid gray; background-color: green; } #gridTable .gridCol2 { border:1px solid gray; background-color: blue; } #gridTable .gridCol3 { border:1px solid gray; background-color: yellow; } #gridTable .gridCol4 { border:1px solid gray; background-color: orange; } #gridTable .gridCol5 { border:1px solid gray; background-color: red; } #lvlSelect ul {float: left; display:block; position:relative; margin-left: 20px; padding: 10px; } #lvlSelect ul li { width:40px; text-align:center; display:block; border:1px solid black; position:relative; padding: 10px; list-style:none; opacity:0.2; filter:alpha(opacity=20); } #lvlSelect ul li.on { opacity:1; filter:alpha(opacity=100); } #lvlSelect ul #li0 { background-color: none; } #lvlSelect ul #li1 { background-color: green; } #lvlSelect ul #li2 { background-color: blue; } #lvlSelect ul #li3 { background-color: yellow; } #lvlSelect ul #li4 { background-color: orange; } #lvlSelect ul #li5 { background-color: red; } </style> </div> <div id="lvlSelect"> <ul> <li value="0" id="li0">0</li> <li value="1" id="li1">1</li> <li value="2" id="li2">2</li> <li value="3" id="li3">3</li> <li value="4" id="li4">4</li> <li value="5" id="li5" class="on">5</li> </ul> </div> In this example the grid box changes color when the image is grid box is clicked, but I would like to be able to have the coordinates of the box. Any help would be great. Thank you

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  • Finding the closest grid coordinate to the mouse position with javascript/jQuery

    - by Acorn
    What I'm trying to do is make a grid of invisible coordinates on the page equally spaced. I then want a <div> to be placed at whatever grid coordinate is closest to the pointer when onclick is triggered. Here's the rough idea: I have the tracking of the mouse coordinates and the placing of the <div> worked out fine. What I'm stuck with is how to approach the problem of the grid of coordinates. First of all, should I have all my coordinates in an array which I then compare my onclick coordinate to? Or seeing as my grid coordinates follow a rule, could I do something like finding out which coordinate that is a multiple of whatever my spacing is is closest to the onclick coordinate? And then, where do I start with working out which grid point coordinate is closest? What's the best way of going about it? Thanks!

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  • Finding the closest grid coordinate to the mouse onclick with javascript/jQuery

    - by Acorn
    What I'm trying to do is make a grid of invisible coordinates on the page equally spaced. I then want a <div> to be placed at whatever grid coordinate is closest to the pointer when onclick is triggered. Here's the rough idea: I have the tracking of the mouse coordinates and the placing of the <div> worked out fine. What I'm stuck with is how to approach the problem of the grid of coordinates. First of all, should I have all my coordinates in an array which I then compare my onclick coordinate to? Or seeing as my grid coordinates follow a rule, could I do something like finding out which coordinate that is a multiple of whatever my spacing is is closest to the onclick coordinate? And then, where do I start with working out which grid point coordinate is closest? What's the best way of going about it? Thanks!

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  • Any way to loop through FPDF code with proper XY coordinates?

    - by JM4
    At the end of a form collection, I provide the consumer a printable PDF with the information they just entered. I already run through a loop to store the variables themselves but am wondering if it is at all possible to build a loop that builds on itself for FPDF. The catch is this, each new variable (#1, #2, #3) will change location by a determined amount of space. For example: I print the Member #1 First name at coordinate at coordinate (95, 101). I print Member #2 First name at coordinate (95, 110)... and so on. Each known variable will be 9.5mm greater than its previous entry (therefor Member #9 will be 40mm higher than Member 6) My sample code for the FPDF itself is: $pdf->SetFont('Arial','', 7); $pdf->SetXY(8,76.5); $pdf->Cell(20,0,$f1name); $pdf->SetFont('Arial','', 5); $pdf->SetXY(50.5,76.5); $pdf->Cell(20,0,$f1address); $pdf->SetFont('Arial','', 7); $pdf->SetXY(95.7,76.5); $pdf->Cell(20,0,$f1city); $pdf->SetXY(129.5,76.5); $pdf->Cell(20,0,$f1state); $pdf->SetXY(139.1,76.5); $pdf->Cell(20,0,$f1zip); $pdf->SetXY(151,76.5); $pdf->Cell(20,0,$f1dob); $pdf->SetXY(168,76.5); $pdf->Cell(20,0,$f1ssn); $pdf->SetXY(186,76.5); $pdf->Cell(20,0,$f1phone); $pdf->SetXY(55,81.1); $pdf->Cell(20,0,$f1email); $pdf->SetXY(129,81.1); $pdf->Cell(20,0,$f1fednum); Ideally, all Y variables with $f2 would be 9.5mm greater than f1's Y values.

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  • 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.

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  • How to Load Bing Map using Coordinates from Database?

    - by Lukasz
    I have latitude and longitude saved inside a database. I have the bing map loading and I can set the VELatLong using regular values but can't seem to be able to load them from the database. Whatever I try the map just doesn't show at all. <script type="text/javascript"> var map = null; var selStyle = VEMapStyle.Road; var selMode = VEMapMode.Mode2D; var zoom = 14; var latLon = new VELatLong(40.67959657544238, -73.94073486328126); // NYC, NY var locationPin = null; function GetMap() { map = new VEMap("myMap"); map.onLoadMap = InitialPin; map.SetCredentials("--KEY HERE--"); map.LoadMap(latLon, zoom, selStyle, false, selMode, false); map.AttachEvent("onclick", OnClick_PinHandler); } window.onload = GetMap; window.onunload = DisposeMap; </script> Thanks for your help!

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  • Using PHP to extract data from an ATOM feed and using it reverse lookup coordinates to an actual add

    - by gbhall
    Basically I have a public feed: http://www.google.com/latitude/apps/badge/api?user=-1671995934285587708&type=atom If you go to my Google Profile you can see it says: "Gareth is in 6 Seaside Gardens, Mullaloo WA 6027, Australia (1 minute ago)" google.com/profiles/Gareth.B.Hall How can I, using PHP, display my location on a website the same way it's displayed on my Google Profile? Thanks

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  • What could make GetCursorPos return incorrect coordinates of {0,0} ?

    - by Dave Moore
    We are seeing bad behavior in an application when it runs on Server 2008 (not R2). This is a WinForms application, and Control.MousePosition is returning {0,0} no matter where the mouse is on the screen... Control.MousePosition just makes a P/Invoke call to Win32 api GetCursorPos(). There is a control in our library that calls SetWindowsHookEx to hook WH_CALLWNDPROCRET for our entire process. I'm suspicious of this code, but tracing statements show that we're getting in + out of that hook cleanly. What else should I be looking for? Thanks, Dave

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  • UIKeyboard slides from wrong coordinates when in landscape, why?

    - by Horatiu Paraschiv
    Hi I have an app that starts in Landscape Right mode. When I tap a textField the keyboard should slide up, however first time when the keyboard slides it slides from the right (where the home button is). After that it slides up and down as expected. So only the first time when I call the keyboard in it comes from where it would come if the app would be in portrait mode. Does anyone know how can I fix this?

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  • How would you sample a real-time stream of coordinates to create a Speed Graph?

    - by Andrew Johnson
    I have a GPS device, and I am receiving continuous points, which I store in an array. These points are time stamped. I would like to graph distance/time (speed) vs. distance in real-time; however, I can only plot 50 of the points because of hardware constraints. How would you select points from the array to graph? For example, one algorithm might be to select every Nth point from the array, where N results in 50 points total. Code: float indexModifier = 1; if (MIN(50,track.lastPointIndex) == 50) { indexModifier = track.lastPointIndex/50.0f; } index = ceil(index*indexModifier); Another algorithm might be to keep an array of 50 points, and throw out the point with the least speed change each time you get a new point.

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  • Incorrect logic flow? function that gets coordinates for a sudoku game

    - by igor
    This function of mine keeps on failing an autograder, I am trying to figure out if there is a problem with its logic flow? Any thoughts? Basically, if the row is wrong, "invalid row" should be printed, and clearInput(); called, and return false. When y is wrong, "invalid column" printed, and clearInput(); called and return false. When both are wrong, only "invalid row" is to be printed (and still clearInput and return false. Obviously when row and y are correct, print no error and return true. My function gets through most of the test cases, but fails towards the end, I'm a little lost as to why. bool getCoords(int & x, int & y) { char row; bool noError=true; cin>>row>>y; row=toupper(row); if(row>='A' && row<='I' && isalpha(row) && y>=1 && y<=9) { x=row-'A'; y=y-1; return true; } else if(!(row>='A' && row<='I')) { cout<<"Invalid row"<<endl; noError=false; clearInput(); return false; } else { if(noError) { cout<<"Invalid column"<<endl; } clearInput(); return false; } }

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  • which layout engine for finding coordinates of html elements on the web page?

    - by Mexx
    I am doing some web data classification task and was thinking if I could get the co-ordinates of html elements as they would appear on a web-browser without taking into consideration any css or javascript being referred in the web page. My language of programming is c++ and the need results for a couple million of pages, so it has to be fast. I know there is a Microsoft COM component which renders the page in a web browser control and then can be queried for position of different html tags. But this is not suitable in my case as it first renders the whole page which takes up a lot of time. So as I found out, there are open-source layout engines WebKit, Gecko that can probably be used for this. But that's a huge piece of code and I need someone to direct me to the right classes or right modules to look into or any previous/similar work someone has done previously. Also, please let me know what you guys think is a good choice if I want to customize the existing code for use with multiple threads to make it faster. Thanks

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  • how do I convert a php array of names and coordinates to a javascript array for google map points?

    - by princyp
    I have a php array that has a bunch of data that I need but specifically I need just the name and longitude and latitude from each item in the array so that I can display points on a google map. The google map array needs to look like this in the end var points = [ ['test name', 37.331689, -122.030731, 4] ['test name 2', 37.331689, -122.030731, 4] ]; What is the best way to put my php data into a js array?

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  • Android - determine specific locations (X,Y coordinates) on a Bitmap on different resolutions?

    - by Mike
    My app that I am trying to create is a board game. It will have one bitmap as the board and pieces that will move to different locations on the board. The general design of the board is square, has a certain number of columns and rows and has a border for looks. Think of a chess board or scrabble board. Before using bitmaps, I first created the board and boarder by manually drawing it - drawLine & drawRect. I decided how many pixels in width the border would be based on the screen width and height passed in on "onSizeChanged". The remaining screen I divided by the number of columns or rows I needed. For examples sake, let's say the screen dimensions are 102 x 102. I may have chosen to set the border at 1 and set the number of rows & columns at 10. That would leave 100 x 100 left (reduced by two to account for the top & bottom border, as well as left/right border). Then with columns and rows set to 10, that would leave 10 pixels left for both height and width. No matter what screen size is passed in, I store exactly how many pixels in width the boarder is and the height & width of each square on the board. I know exactly what location on the screen to move the pieces to based on a simple formula and I know exactly what cell a user touched to make a move. Now how does that work with bitmaps? Meaning, if I create 3 different background bitmaps, once for each density, won't they still be resized to fit each devices screen resolution, because from what I read there were not just 3 screen resolutions, but 5 and now with tablets - even more. If I or Android scales the bitmaps up or down to fit the current devices screen size, how will I know how wide the border is scaled to and the dimensions of each square in order to figure out where to move a piece or calculate where a player touched. So far the examples I have looked at just show how to scale the overall bitmap and get the overall bitmaps width and height. But, I don't see how to tell how many pixels wide or tall each part of the board would be after it was scaled. When I draw each line and rectangle myself based in the screen dimensions from onSizeChanged, I always know these dimensions. If anyone has any sample code or a URL to point me to that I can a read about this with bitmaps, I would appreciate it. Thanks, --Mike BTW, here is some sample code (very simplified) on how I know the dimensions of my game board (border and squares) no matter the screen size. Now I just need to know how to do this with the board as a bitmap that gets scaled to any screen size. @Override protected void onSizeChanged(int w, int h, int oldw, int oldh) { intScreenWidth = w; intScreenHeight = h; // Set Border width - my real code changes this value based on the dimensions of w // and h that are passed in. In other words bigger screens get a slightly larger // border. intOuterBorder = 1; /** Reserve part of the board for the boardgame and part for player controls & score My real code forces this to be square, but this is good enough to get the point across. **/ floatBoardHeight = intScreenHeight / 4 * 3; // My real code actually causes floatCellWidth and floatCellHeight to // be equal (Square). floatCellWidth = (intScreenWidth - intOuterBorder * 2 ) / intNumColumns; floatCellHeight = (floatBoardHeight - intOuterBorder * 2) / intNumRows; super.onSizeChanged(w, h, oldw, oldh); }

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  • LWJGL: Camera distance from image plane?

    - by Rogem
    Let me paste some code before I ask the question... public static void createWindow(int[] args) { try { Display.setFullscreen(false); DisplayMode d[] = Display.getAvailableDisplayModes(); for (int i = 0; i < d.length; i++) { if (d[i].getWidth() == args[0] && d[i].getHeight() == args[1] && d[i].getBitsPerPixel() == 32) { displayMode = d[i]; break; } } Display.setDisplayMode(displayMode); Display.create(); } catch (Exception e) { e.printStackTrace(); System.exit(0); } } public static void initGL() { GL11.glEnable(GL11.GL_TEXTURE_2D); GL11.glShadeModel(GL11.GL_SMOOTH); GL11.glClearColor(0.0f, 0.0f, 0.0f, 0.0f); GL11.glClearDepth(1.0); GL11.glEnable(GL11.GL_DEPTH_TEST); GL11.glDepthFunc(GL11.GL_LEQUAL); GL11.glMatrixMode(GL11.GL_PROJECTION); GL11.glLoadIdentity(); GLU.gluPerspective(45.0f, (float) displayMode.getWidth() / (float) displayMode.getHeight(), 0.1f, 100.0f); GL11.glMatrixMode(GL11.GL_MODELVIEW); GL11.glHint(GL11.GL_PERSPECTIVE_CORRECTION_HINT, GL11.GL_NICEST); } So, with the camera and screen setup out of the way, I can now ask the actual question: How do I know what the camera distance is from the image plane? I also would like to know what the angle between the image plane's center normal and a line drawn from the middle of one of the edges to the camera position is. This will be used to consequently draw a vector from the camera's position through the player's click-coordinates to determine the world coordinates they clicked (or could've clicked). Also, when I set the camera coordinates, do I set the coordinates of the camera or do I set the coordinates of the image plane? Thank you for your help. EDIT: So, I managed to solve how to calculate the distance of the camera... Here's the relevant code... private static float getScreenFOV(int dim) { if (dim == 0) { float dist = (float) Math.tan((Math.PI / 2 - Math.toRadians(FOV_Y))/2) * 0.5f; float FOV_X = 2 * (float) Math.atan(getScreenRatio() * 0.5f / dist); return FOV_X; } else if (dim == 1) { return FOV_Y; } return 0; } FOV_Y is the Field of View that one defines in gluPerspective (float fovy in javadoc). This seems to be (and would logically be) for the height of the screen. Now I just need to figure out how to calculate that vector.

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  • how to use kml file in my code..

    - by zjm1126
    i download a kml file : <?xml version="1.0" encoding="UTF-8"?> <kml xmlns="http://www.opengis.net/kml/2.2"> <Document> <Style id="transGreenPoly"> <LineStyle> <width>1.5</width> </LineStyle> <PolyStyle> <color>7d00ff00</color> </PolyStyle> </Style> <Style id="transYellowPoly"> <LineStyle> <width>1.5</width> </LineStyle> <PolyStyle> <color>7d00ffff</color> </PolyStyle> </Style> <Style id="transRedPoly"> <LineStyle> <width>1.5</width> </LineStyle> <PolyStyle> <color>7d0000ff</color> </PolyStyle> </Style> <Style id="transBluePoly"> <LineStyle> <width>1.5</width> </LineStyle> <PolyStyle> <color>7dff0000</color> </PolyStyle> </Style> <Folder> <name>Placemarks</name> <open>0</open> <Placemark> <name>Simple placemark</name> <description>Attached to the ground. Intelligently places itself at the height of the underlying terrain.</description> <Point> <coordinates>-122.0822035425683,37.42228990140251,0</coordinates> </Point> </Placemark> <Placemark> <name>Descriptive HTML</name> <description><![CDATA[Click on the blue link!<br/><br/> Placemark descriptions can be enriched by using many standard HTML tags.<br/> For example: <hr/> Styles:<br/> <i>Italics</i>, <b>Bold</b>, <u>Underlined</u>, <s>Strike Out</s>, subscript<sub>subscript</sub>, superscript<sup>superscript</sup>, <big>Big</big>, <small>Small</small>, <tt>Typewriter</tt>, <em>Emphasized</em>, <strong>Strong</strong>, <code>Code</code> <hr/> Fonts:<br/> <font color="red">red by name</font>, <font color="#408010">leaf green by hexadecimal RGB</font> <br/> <font size=1>size 1</font>, <font size=2>size 2</font>, <font size=3>size 3</font>, <font size=4>size 4</font>, <font size=5>size 5</font>, <font size=6>size 6</font>, <font size=7>size 7</font> <br/> <font face=times>Times</font>, <font face=verdana>Verdana</font>, <font face=arial>Arial</font><br/> <hr/> Links: <br/> <a href="http://earth.google.com/">Google Earth!</a> <br/> or: Check out our website at www.google.com <hr/> Alignment:<br/> <p align=left>left</p> <p align=center>center</p> <p align=right>right</p> <hr/> Ordered Lists:<br/> <ol><li>First</li><li>Second</li><li>Third</li></ol> <ol type="a"><li>First</li><li>Second</li><li>Third</li></ol> <ol type="A"><li>First</li><li>Second</li><li>Third</li></ol> <hr/> Unordered Lists:<br/> <ul><li>A</li><li>B</li><li>C</li></ul> <ul type="circle"><li>A</li><li>B</li><li>C</li></ul> <ul type="square"><li>A</li><li>B</li><li>C</li></ul> <hr/> Definitions:<br/> <dl> <dt>Google:</dt><dd>The best thing since sliced bread</dd> </dl> <hr/> Centered:<br/><center> Time present and time past<br/> Are both perhaps present in time future,<br/> And time future contained in time past.<br/> If all time is eternally present<br/> All time is unredeemable.<br/> </center> <hr/> Block Quote: <br/> <blockquote> We shall not cease from exploration<br/> And the end of all our exploring<br/> Will be to arrive where we started<br/> And know the place for the first time.<br/> <i>-- T.S. Eliot</i> </blockquote> <br/> <hr/> Headings:<br/> <h1>Header 1</h1> <h2>Header 2</h2> <h3>Header 3</h3> <h3>Header 4</h4> <h3>Header 5</h5> <hr/> Images:<br/> <i>Remote image</i><br/> <img src="http://code.google.com/apis/kml/documentation/googleSample.png"><br/> <i>Scaled image</i><br/> <img src="http://code.google.com/apis/kml/documentation/googleSample.png" width=100><br/> <hr/> Simple Tables:<br/> <table border="1" padding="1"> <tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr> <tr><td>a</td><td>b</td><td>c</td><td>d</td><td>e</td></tr> </table> <br/>]]></description> <Point> <coordinates>-122,37,0</coordinates> </Point> </Placemark> </Folder> <Folder> <name>Google Campus - Polygons</name> <open>0</open> <description>A collection showing how easy it is to create 3-dimensional buildings</description> <Placemark> <name>Building 40</name> <styleUrl>#transRedPoly</styleUrl> <Polygon> <extrude>1</extrude> <altitudeMode>relativeToGround</altitudeMode> <outerBoundaryIs> <LinearRing> <coordinates> -122.0848938459612,37.42257124044786,17 -122.0849580979198,37.42211922626856,17 -122.0847469573047,37.42207183952619,17 -122.0845725380962,37.42209006729676,17 -122.0845954886723,37.42215932700895,17 -122.0838521118269,37.42227278564371,17 -122.083792243335,37.42203539112084,17 -122.0835076656616,37.42209006957106,17 -122.0834709464152,37.42200987395161,17 -122.0831221085748,37.4221046494946,17 -122.0829247374572,37.42226503990386,17 -122.0829339169385,37.42231242843094,17 -122.0833837359737,37.42225046087618,17 -122.0833607854248,37.42234159228745,17 -122.0834204551642,37.42237075460644,17 -122.083659133885,37.42251292011001,17 -122.0839758438952,37.42265873093781,17 -122.0842374743331,37.42265143972521,17 -122.0845036949503,37.4226514386435,17 -122.0848020460801,37.42261133916315,17 -122.0847882750515,37.42256395055121,17 -122.0848938459612,37.42257124044786,17 </coordinates> </LinearRing> </outerBoundaryIs> </Polygon> </Placemark> <Placemark> <name>Building 41</name> <styleUrl>#transBluePoly</styleUrl> <Polygon> <extrude>1</extrude> <altitudeMode>relativeToGround</altitudeMode> <outerBoundaryIs> <LinearRing> <coordinates> -122.0857412771483,37.42227033155257,17 -122.0858169768481,37.42231408832346,17 -122.085852582875,37.42230337469744,17 -122.0858799945639,37.42225686138789,17 -122.0858860101409,37.4222311076138,17 -122.0858069157288,37.42220250173855,17 -122.0858379542653,37.42214027058678,17 -122.0856732640519,37.42208690214408,17 -122.0856022926407,37.42214885429042,17 -122.0855902778436,37.422128290487,17 -122.0855841672237,37.42208171967246,17 -122.0854852065741,37.42210455874995,17 -122.0855067264352,37.42214267949824,17 -122.0854430712915,37.42212783846172,17 -122.0850990714904,37.42251282407603,17 -122.0856769818632,37.42281815323651,17 -122.0860162273783,37.42244918858723,17 -122.0857260327004,37.42229239604253,17 -122.0857412771483,37.42227033155257,17 </coordinates> </LinearRing> </outerBoundaryIs> </Polygon> </Placemark> <Placemark> <name>Building 42</name> <styleUrl>#transGreenPoly</styleUrl> <Polygon> <extrude>1</extrude> <altitudeMode>relativeToGround</altitudeMode> <outerBoundaryIs> <LinearRing> <coordinates> -122.0857862287242,37.42136208886969,25 -122.0857312990603,37.42136935989481,25 -122.0857312992918,37.42140934910903,25 -122.0856077073679,37.42138390166565,25 -122.0855802426516,37.42137299550869,25 -122.0852186221971,37.42137299504316,25 -122.0852277765639,37.42161656508265,25 -122.0852598189347,37.42160565894403,25 -122.0852598185499,37.42168200156,25 -122.0852369311478,37.42170017860346,25 -122.0852643957828,37.42176197982575,25 -122.0853239032746,37.42176198013907,25 -122.0853559454324,37.421852864452,25 -122.0854108752463,37.42188921823734,25 -122.0854795379357,37.42189285337048,25 -122.0855436229819,37.42188921797546,25 -122.0856260178042,37.42186013499926,25 -122.085937287963,37.42186013453605,25 -122.0859428718666,37.42160898590042,25 -122.0859655469861,37.42157992759144,25 -122.0858640462341,37.42147115002957,25 -122.0858548911215,37.42140571326184,25 -122.0858091162768,37.4214057134039,25 -122.0857862287242,37.42136208886969,25 </coordinates> </LinearRing> </outerBoundaryIs> </Polygon> </Placemark> <Placemark> <name>Building 43</name> <styleUrl>#transYellowPoly</styleUrl> <Polygon> <extrude>1</extrude> <altitudeMode>relativeToGround</altitudeMode> <outerBoundaryIs> <LinearRing> <coordinates> -122.0844371128284,37.42177253003091,19 -122.0845118855746,37.42191111542896,19 -122.0850470999805,37.42178755121535,19 -122.0850719913391,37.42143663023161,19 -122.084916406232,37.42137237822116,19 -122.0842193868167,37.42137237801626,19 -122.08421938659,37.42147617161496,19 -122.0838086419991,37.4214613409357,19 -122.0837899728564,37.42131306410796,19 -122.0832796534698,37.42129328840593,19 -122.0832609819207,37.42139213944298,19 -122.0829373621737,37.42137236399876,19 -122.0829062425667,37.42151569778871,19 -122.0828502269665,37.42176282576465,19 -122.0829435788635,37.42176776969635,19 -122.083217411188,37.42179248552686,19 -122.0835970430103,37.4217480074456,19 -122.0839455556771,37.42169364237603,19 -122.0840077894637,37.42176283815853,19 -122.084113587521,37.42174801104392,19 -122.0840762473784,37.42171341292375,19 -122.0841447047739,37.42167881534569,19 -122.084144704223,37.42181720660197,19 -122.0842503333074,37.4218170700446,19 -122.0844371128284,37.42177253003091,19 </coordinates> </LinearRing> </outerBoundaryIs> </Polygon> </Placemark> </Folder> <Folder> <name>LineString</name> <open>0</open> <Placemark> <LineString> <tessellate>1</tessellate> <coordinates> -112.0814237830345,36.10677870477137,0 -112.0870267752693,36.0905099328766,0 </coordinates> </LineString> </Placemark> </Folder> <Folder> <name>GroundOverlay</name> <open>0</open> <GroundOverlay> <name>Large-scale overlay on terrain</name> <description>Overlay shows Mount Etna erupting on July 13th, 2001.</description> <Icon> <href>http://code.google.com/apis/kml/documentation/etna.jpg</href> </Icon> <LatLonBox> <north>37.91904192681665</north> <south>37.46543388598137</south> <east>15.35832653742206</east> <west>14.60128369746704</west> </LatLonBox> </GroundOverlay> </Folder> <Folder> <name>ScreenOverlays</name> <open>0</open> <ScreenOverlay> <name>screenoverlay_dynamic_top</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/dynamic_screenoverlay.jpg</href> </Icon> <overlayXY x="0" y="1" xunits="fraction" yunits="fraction"/> <screenXY x="0" y="1" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="1" y="0.2" xunits="fraction" yunits="fraction"/> </ScreenOverlay> <ScreenOverlay> <name>screenoverlay_dynamic_right</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/dynamic_right.jpg</href> </Icon> <overlayXY x="1" y="1" xunits="fraction" yunits="fraction"/> <screenXY x="1" y="1" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="0" y="1" xunits="fraction" yunits="fraction"/> </ScreenOverlay> <ScreenOverlay> <name>Simple crosshairs</name> <visibility>0</visibility> <description>This screen overlay uses fractional positioning to put the image in the exact center of the screen</description> <Icon> <href>http://code.google.com/apis/kml/documentation/crosshairs.png</href> </Icon> <overlayXY x="0.5" y="0.5" xunits="fraction" yunits="fraction"/> <screenXY x="0.5" y="0.5" xunits="fraction" yunits="fraction"/> <rotationXY x="0.5" y="0.5" xunits="fraction" yunits="fraction"/> <size x="0" y="0" xunits="pixels" yunits="pixels"/> </ScreenOverlay> <ScreenOverlay> <name>screenoverlay_absolute_topright</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/top_right.jpg</href> </Icon> <overlayXY x="1" y="1" xunits="fraction" yunits="fraction"/> <screenXY x="1" y="1" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="0" y="0" xunits="fraction" yunits="fraction"/> </ScreenOverlay> <ScreenOverlay> <name>screenoverlay_absolute_topleft</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/top_left.jpg</href> </Icon> <overlayXY x="0" y="1" xunits="fraction" yunits="fraction"/> <screenXY x="0" y="1" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="0" y="0" xunits="fraction" yunits="fraction"/> </ScreenOverlay> <ScreenOverlay> <name>screenoverlay_absolute_bottomright</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/bottom_right.jpg</href> </Icon> <overlayXY x="1" y="-1" xunits="fraction" yunits="fraction"/> <screenXY x="1" y="0" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="0" y="0" xunits="fraction" yunits="fraction"/> </ScreenOverlay> <ScreenOverlay> <name>screenoverlay_absolute_bottomleft</name> <visibility>0</visibility> <Icon> <href>http://code.google.com/apis/kml/documentation/bottom_left.jpg</href> </Icon> <overlayXY x="0" y="-1" xunits="fraction" yunits="fraction"/> <screenXY x="0" y="0" xunits="fraction" yunits="fraction"/> <rotationXY x="0" y="0" xunits="fraction" yunits="fraction"/> <size x="0" y="0" xunits="fraction" yunits="fraction"/> </ScreenOverlay> </Folder> </Document> </kml> and my code is : function initialize() { if (GBrowserIsCompatible()) { var map = new GMap2(document.getElementById("map_canvas")); var center=new GLatLng(39.9493, 116.3975); map.setCenter(center, 13); var geoXml = new GGeoXml("SamplesInMaps.kml"); <!--Place KML on Map --> map.addOverlay(geoXml); } } but ,i don't successful ,, do you know how to do this.. thanks

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  • libgdx ActorGestureListener.pan() parameters not moving actor in smooth line

    - by Roar Skullestad
    I override the pan method in ActorGestureListener to implement dragging actors in libgdx (scene2d). When I move individual pieces on a board they move smoothly, but when moving the whole board, the x and y coordinates that is sent to pan is "jumping", and in an increasingly amount the longer it is dragged. These are an example of the deltaY coordinates sent to pan when dragging smoothly downwards: 1.1156368 -0.13125038 -1.0500145 0.98439217 -1.0500202 0.91877174 -0.984396 0.9187679 -0.98439026 0.9187641 -0.13125038 This is how I move the camera: public void pan (InputEvent event, float x, float y, float deltaX, float deltaY) { cam.translate(-deltaX, -deltaY); I have been using both the delta values sent to pan and the real position values, but similar results. And since it is the coordinates that are wrong, it doesn't matter whether I move the board itself or the camera. What could the cause be for this and what is the solution? When I move camera only half the delta-values, it moves smoothly but only at half the speed of the mouse pointer: cam.translate(-deltaX / 2, -deltaY / 2); It seems like the moving of camera or board affects the mouse input coordinates. How can I drag at "mouse speed" and still get smooth movements? (This question was also posted on stackoverflow: http://stackoverflow.com/questions/20693020/libgdx-actorgesturelistener-pan-parameters-not-moving-actor-in-smooth-line)

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  • How to implement physical effect, perspective effect on Android

    - by asedra_le
    I'm researching about 2D game for Android to implement an Android Game Project. My project looks nearly like PaperToss. Instance of throwing a page, my game will throw a coin. Suppose that I have a coin put in three-dimensional that have coordinates at A(x,y,z). I throw that point ahead, after 1/100 second, that coin move from A(x,y,z) to A'(x',y',z'). By this way, I have two problems need to solve. Determine the formulas can be used to compute the coordinates of the coin at time t. This problem is under-researching. I have no idea to solve this problem. Mapping three-dimensional points to a two-dimensional and use those new coordinates (a two-dimensional coordinates) to draw our coin on screen. I have found two solutions for this problem: Orthographic projection & Perspective projection However, my old friend said that OpenGL supports to solve problems like my problems. Any body have experiences about my problems? Help me please :) Thank for reading my question.

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  • how to make HLSL effect just for lighning without texture mapping?

    - by naprox
    I'm new to XNA, i created an effect and just want to use lightning but in default effect that XNA create we should do texture mapping or the model appears 'RED', because of this lines of code in the effect file: float4 PixelShaderFunction(VertexShaderOutput input) : COLOR0 { float4 output = float4(1,0,0,1); return output; } and if i want to see my model (appear like when i use basiceffect) must do texture mapping by UV coordinates. but my model does not have UV coordinates assigned or its UV coordinates is not exported. and if i do texture mapping i got error. (i do texture mapping by this line of code in vertexshaderfunction and other necessary codes) output.UV= input.UV i have many of this models and want to work with them.(my models are in .FBX format) when i use Bassiceffect i have no problem and model appears correctly. how can i use "just" lightnings in my custom effects? and don't do texture mapping (because i have no UV coordinates in my models) and my model be look like when i use BasicEffect? if you need my complete code Here it is: http://www.mediafire.com/?4jexhd4ulm2icm2 here is inside of my Model Using BasicEffect http://i.imgur.com/ygP2h.jpg?1 and this is my code for drawing with or without BasicEffect inside of my draw() method: Matrix baseWorld = Matrix.CreateScale(Scale) * Matrix.CreateFromYawPitchRoll(Rotation.Y, Rotation.X, Rotation.Z) * Matrix.CreateTranslation(Position); foreach(ModelMesh mesh in Model.Meshes) { Matrix localWorld = ModelTransforms[mesh.ParentBone.Index] * baseWorld; foreach(ModelMeshPart part in mesh.MeshParts) { Effect effect = part.Effect; if (effect is BasicEffect) { ((BasicEffect)effect).World = localWorld; ((BasicEffect)effect).View = View; ((BasicEffect)effect).Projection = Projection; ((BasicEffect)effect).EnableDefaultLighting(); } else { setEffectParameter(effect, "World", localWorld); setEffectParameter(effect, "View", View); setEffectParameter(effect, "Projection", Projection); setEffectParameter(effect, "CameraPosition", CameraPosition); } } mesh.Draw(); } setEffectParameter is another method that sets effect parameter if i use my custom effect.

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  • What's a good way to check that a player has clicked on an object in a 3D game?

    - by imja
    I'm programming a 3D game (using C++ and OpenGL), and I have a few 3D objects in the scene, we can say they are boxes for this example. I want to let the player click on those boxes to select them (ie. they might change color) with the typical restriction like if more than one box is located where the user clicked, only the one closest to the camera would get selected. What would be the best way to do this? The fact that these objects go through several transforms before getting to window coordinates is what makes this a bit tricky. One approach I thought about was that if the player clicks on the screen, I could normalize the x,y coordinates of mouse click and then transform the bounding box coordinates of the objects into clip-space so that I could compare then to the normalized mouse coordinates. I guess I could then do some sort of ray-box collision test to see if any objects lie as the path of the mouse click. I'm afraid I might be over complicating it. Any better methods out there?

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  • Maze not generating properly. Out of bounds exception. need quick fix

    - by Dan Joseph Porcioncula
    My maze generator seems to have a problem. I am trying to generate something like the maze from http://mazeworks.com/mazegen/mazetut/index.htm . My program displays this http://a1.sphotos.ak.fbcdn.net/hphotos-ak-snc7/s320x320/374060_426350204045347_100000111130260_1880768_1572427285_n.jpg and the error Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1 at Grid.genRand(Grid.java:73) at Grid.main(Grid.java:35) How do I fix my generator program? import java.awt.*; import java.awt.Color; import java.awt.Component; import java.awt.Graphics; import javax.swing.*; import java.util.ArrayList; public class Grid extends Canvas { Cell[][] maze; int size; int pathSize; double width, height; ArrayList<int[]> coordinates = new ArrayList<int[]>(); public Grid(int size, int h, int w) { this.size = size; maze = new Cell[size][size]; for(int i = 0; i<size; i++){ for(int a =0; a<size; a++){ maze[i][a] = new Cell(); } } setPreferredSize(new Dimension(h, w)); } public static void main(String[] args) { JFrame y = new JFrame(); y.setLayout(new BorderLayout()); Grid f = new Grid(25, 400, 400); y.add(f, BorderLayout.CENTER); y.setSize(450, 450); y.setVisible(true); y.setDefaultCloseOperation(y.EXIT_ON_CLOSE); f.genRand(); f.repaint(); } public void push(int[] xy) { coordinates.add(xy); int i = coordinates.size(); coordinates.ensureCapacity(i++); } public int[] pop() { int[] x = coordinates.get((coordinates.size())-1); coordinates.remove((coordinates.size())-1); return x; } public int[] top() { return coordinates.get((coordinates.size())-1); } public void genRand(){ // create a CellStack (LIFO) to hold a list of cell locations [x] // set TotalCells = number of cells in grid int TotalCells = size*size; // choose a cell at random and call it CurrentCell int m = randomInt(size); int n = randomInt(size); Cell curCel = maze[m][n]; // set VisitedCells = 1 int visCel = 1,d=0; int[] q; int h,o = 0,p = 0; // while VisitedCells < TotalCells while( visCel < TotalCells){ // find all neighbors of CurrentCell with all walls intact if(maze[m-1][n].countWalls() == 4){d++;} if(maze[m+1][n].countWalls() == 4){d++;} if(maze[m][n-1].countWalls() == 4){d++;} if(maze[m][n+1].countWalls() == 4){d++;} // if one or more found if(d!=0){ Point[] ls = new Point[4]; ls[0] = new Point(m-1,n); ls[1] = new Point(m+1,n); ls[2] = new Point(m,n-1); ls[3] = new Point(m,n+1); // knock down the wall between it and CurrentCell h = randomInt(3); switch(h){ case 0: o = (int)(ls[0].getX()); p = (int)(ls[0].getY()); curCel.destroyWall(2); maze[o][p].destroyWall(1); break; case 1: o = (int)(ls[1].getX()); p = (int)(ls[1].getY()); curCel.destroyWall(1); maze[o][p].destroyWall(2); break; case 2: o = (int)(ls[2].getX()); p = (int)(ls[2].getY()); curCel.destroyWall(3); maze[o][p].destroyWall(0); break; case 3: o = (int)(ls[3].getX()); p = (int)(ls[3].getY()); curCel.destroyWall(0); maze[o][p].destroyWall(3); break; } // push CurrentCell location on the CellStack push(new int[] {m,n}); // make the new cell CurrentCell m = o; n = p; curCel = maze[m][n]; // add 1 to VisitedCells visCel++; } // else else{ // pop the most recent cell entry off the CellStack q = pop(); m = q[0]; n = q[1]; curCel = maze[m][n]; // make it CurrentCell // endIf } // endWhile } } public int randomInt(int s) { return (int)(s* Math.random());} public void paint(Graphics g) { int k, j; width = getSize().width; height = getSize().height; double htOfRow = height / (size); double wdOfRow = width / (size); //checks verticals - destroys east border of cell for (k = 0; k < size; k++) { for (j = 0; j < size; j++) { if(maze[k][j].checkWall(2)){ g.drawLine((int) (k * wdOfRow), (int) (j * htOfRow), (int) (k * wdOfRow), (int) ((j+1) * htOfRow)); }} } //checks horizontal - destroys north border of cell for (k = 0; k < size; k++) { for (j = 0; j < size; j++) { if(maze[k][j].checkWall(3)){ g.drawLine((int) (k * wdOfRow), (int) (j * htOfRow), (int) ((k+1) * wdOfRow), (int) (j * htOfRow)); }} } } } class Cell { private final static int NORTH = 0; private final static int EAST = 1; private final static int WEST = 2; private final static int SOUTH = 3; private final static int NO = 4; private final static int START = 1; private final static int END = 2; boolean[] wall = new boolean[4]; boolean[] border = new boolean[4]; boolean[] backtrack = new boolean[4]; boolean[] solution = new boolean[4]; private boolean isVisited = false; private int Key = 0; public Cell(){ for(int i=0;i<4;i++){wall[i] = true;} } public int countWalls(){ int i, k =0; for(i=0; i<4; i++) { if (wall[i] == true) {k++;} } return k;} public boolean checkWall(int x){ switch(x){ case 0: return wall[0]; case 1: return wall[1]; case 2: return wall[2]; case 3: return wall[3]; } return true; } public void destroyWall(int x){ switch(x){ case 0: wall[0] = false; break; case 1: wall[1] = false; break; case 2: wall[2] = false; break; case 3: wall[3] = false; break; } } public void setStart(int i){Key = i;} public int getKey(){return Key;} public boolean checkVisit(){return isVisited;} public void visitCell(){isVisited = true;} }

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  • Vertex data split into separate buffers or one one structure?

    - by kiba2
    Is it better to have all vertex data in one structure like this: class MyVertex { int x,y,z; int u,v; int normalx, normaly, normalz; } Or to have each component (location, normal, texture coordinates) in separate arrays/buffers? To me it always seemed logical to keep the data grouped together in one structure because they'd always be the same for each instance of a shared vertex and that seems to be true for things like character models (ex: the normal should be an average of adjacent normals for smooth lighting). One instance where this doesn't seem to work is other kinds of meshes like say a cube where the texture coordinates for each may be the same but that causes them to be different where the vertices are shared. Does everybody normally keep them separate? Won't this make them less space efficient if there needs to be an instance of texture coordinates and normals for each triangle vertex (They won't be indexed)? Can OpenGL even handle this mixing of indexed (for location) vs non-indexed buffers in the same VBO?

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  • How do I calculate the motion of 2 massive bodies in space?

    - by 1224
    I'm writing code simulating the 2-dimensional motion of two massive bodies with gravitational fields. The bodies' masses are known and I have a gravitational force equation. I know from that force I can get a differential equation for coordinates. I know that I once I solve this equation I will get the coordinates. I will need to make up some initial position and some initial velocity. I'd like to end up with a numeric solver for the ordinal differential equation for coordinates to get the formulas that I can write in code. Could someone break down how from laws and initial conditions we get to the formulas that calculate x and y at time t?

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