There are many possible camera positions + orientations where the bounding box would fit inside the view frustum. But any procedure would select one specific camera position and orientation. If you would consider bounding spheres, one solution could be to first change orientation to look at bounding sphere center then move back sufficiently (negative look … Read more
Relevant steps: Go to “Help” -> “Install New Software…”. In the new dialog, in the “Work with:” pull-down list, select “Indigo – http://download.eclipse.org/releases/indigo”. (Note: Replace Indigo and the download list with download locations for your eclipse version) Select the following packages to install: “Web, XML, Java EE and OSGI Enterprise Development”. “Programming langauges -> Eclipse … Read more
This is my drag distort proposal (share you knowledge, Q&A-style). Live example is in http://jsfiddle.net/xjHUk/278/ and the main code is this: (only output window: http://jsfiddle.net/xjHUk/279/embedded/result/) function transferPoint (xI, yI, source, destination) { var ADDING = 0.001; // to avoid dividing by zero var xA = source[0].x; var yA = source[0].y; var xC = source[2].x; var … Read more
The purpose of the 4 parameters is to define a view frustum, like this: where nothing outside of the frustum should be is visible on screen ( To accomplish this, the parameters are used to calculate a 4×4 matrix, which is then used to transform each vertex into the so-called clip space. There, testing if … Read more
So after a few days of trying to solve it, I came up with a solution (Ignore the blue dots on the second image): As promised, here’s a complete copy of the code: – (void)confirmedImage { cv::Mat originalRot = [self cvMatFromUIImage:_sourceImage]; cv::Mat original; cv::transpose(originalRot, original); originalRot.release(); cv::flip(original, original, 1); CGFloat scaleFactor = [_sourceImageView contentScale]; CGPoint … Read more
To switch between perspectives. Windows: Ctrl+F8 Mac: cmd+F8
To apply a perspective transformation you first have to know four points in a plane A that will be mapped to four points in a plane B. With those points, you can derive the homographic transform. By doing this, you obtain your 8 coefficients and the transformation can take place. The site http://xenia.media.mit.edu/~cwren/interpolator/ (mirror: WebArchive), … Read more
So, first problem is corner order. They must be in the same order in both vectors. So, if in the first vector your order is:(top-left, bottom-left, bottom-right, top-right) , they MUST be in the same order in the other vector. Second, to have the resulting image contain only the object of interest, you must set … Read more
Preferences -> Run/Debug -> Perspectives -> Open the associated perspective when application suspends