perspective

I am attempting to project a series of 3D points onto the screen using a perspective camera matrix. I do not have world space (or consider it being an identity matrix) and my camera does not have camera space (or consider it an identity matrix), I do have a 4x4 matrix for my object space. I am taking the object matrix and multiplying it by the camera perspective matrix, generated with the following method: Matrix4 createPerspectiveMatrix( Float fov, Float aspect, Float near, Float far ) { Float fov2 = (fov/2) * (Math.PI/180); Float tan = Math.tan(fov2); Float f = 1 / tan; return new Matrix4 (

While dragging the perspective tool bar to a different place in eclipse Juno, my eclipse got hang, and after restarting eclipse, the perspective bar gone. I tried googling but didnt find a solution to display that bar again. Anyone got a solution? There are two things you can try: Reset your current perspective (Window/Reset perspective...). Although your configured set of views will get lost, it might bring back the perspective switcher. If that does not work, then you should reset the entire saved workbench GUI. This can be done by deleting the .metadata/.plugins/org.eclipse.e4.workbench

I got a custom perspecitve with an editor area declared in plugin.xml only. In addtion, I got a custom view, that is opened programatically. I want to stack this view with the editor area. If I set it in the perspective extension as initially visible, everything works fine. <view id="my.viewID" minimized="false" relationship="stack" relative="org.eclipse.ui.editorss" visible="false"> </view> However, if I set visible="false" , and use the following code to open it programmatically, it always appears in the bottom area stacked with the ConsoleView. IViewPart viewPart = PlatformUI.getWorkbench()

I'm trying to skew one single corner of my div background as shown at the green checkmark in the image below: In CSS3 I'm however unable to achieve that, skewing completely skews every corner. I just want to skew the bottom right corner to the left (say 25px) and maintain the perspective (as shown in the image above). background-image: url('http://rtjansen.nl/images/stackoverflow.png'); -webkit-transform: skew(-45deg); Fiddle: http://jsfiddle.net/3eX5j/ My code is: div { width: 300px; height:80px; margin-left:40px; background-image: url('http://rtjansen.nl/images/stackoverflow.png'); -webkit

lets say we have a div with 500x500px size and we rotate it on the x axis via css 45 degrees considering a webkit-perspective value of 1600px. How would you calculate the absolute dimensions of the displayed trapezium? (width, max-height, angles) I did only figure out a formula that calculates the width but without considering the perspective, so the value differ some pixels (JavaScript): var absoluteWidth = Math.cos(45 * (Math.PI / 180)) * 500); EDIT: Here is the spec about the -webkit-perspective function: perspective(<number>) specifies a perspective projection matrix. This matrix maps a

For a project I am trying to create a perspective distortion of an image to match a DVD case front template. So I want to automate this using ImageMagick (CLI) but I have a hard time understanding the mathematical aspects of this transformation. convert \ -verbose mw2.png \ -alpha set \ -virtual-pixel transparent \ -distort Perspective-Projection '0,0 0,0 0,0 0,0' \ box.png This code is en empty set of coordinates, I have read the documentation thoroughly but I can't seem to understand what parameter represents what point. The documentation gives me variables and names where I have no clue

I'm making a canvas-based game engine and am wondering if anyone has any good info on how to achieve an overhead view perspective. What I'm looking for is somewhere halfway between the traditional birds eye view and the old SNES mode7 view. Just a slight angle to give the illusion of 3D. I'm trying to figure out what is going to be the best way to deal with the perspective skewing. I'm not doing rotations so 3D matrix stuff would be going overboard, but I need to be able to deal with rendering the map layers at a consistent angle and it'd be nice if the angle was adjustable. I also need to

I'm working with a homography calculated by OpenCV. I currently use this homography to transform points using the function below. This function performs the task I require however I have no clue how it actually works. Can anyone explain, line by line exactly, the logic/theory behind the last 3 lines of code, I understand that this transforms the point x,y but I'm unclear as to why this works: Why are Z , px and py calculated in this way, what do the elements in h correspond to? Your comments are greatly appreciated :) double h[9]; homography = cvMat(3, 3, CV_64F, h); CvMat ps1 = cvMat(MAX