perspective

问题 I know that transforming a square into a trapezoid is a linear transformation, and can be done using the projective matrix, but I'm having a little trouble figuring out how to construct the matrix. Using the projective matrix to translate, scale, rotates, and shear is straightforward. Is there a simple projective matrix which will transform a square to a trapezoid? 回答1: a,b,c,d are the four corners of your 2D square. a,b,c,d are expressed in homogeneous coordinate and so they are 3x1 matrices

问题 What's the meaning of m34 of the structure CATransform3D, I only know it can change the perspective, but what's the meaning when the value is -0.001 and 0.001? 回答1: You can find the full details here. Note that Apple uses the reversed multiplication order for projection (relative to the given link) so all matrix multiplications are reversed and all matrices are transposed. A short description of the meaning: m34 = 1/z distance to projection plane (the 1/e z term in the reference link) + for

问题 I am trying to do a perspective correction of a tilted rectangle ( a credit card), which is tilted in all the 4 directions. I could find its four corners and the respective angles of its tilt but I cannot find the exact location of the coordinates, where it has to be projected. I am using cv2.getPerspectiveTransform to do the transformation. I have the aspect ratio of the actual card (the non tilted one), I want such coordinates such that the original aspect ratio is preserved. I have tried

问题 I'm looking for an algorithm to fit a bounding box inside a viewport (in my case a DirectX scene). I know about algorithms for centering a bounding sphere in a orthographic camera but would need the same for a bounding box and a perspective camera. I can not just change the FOV because this app has FOV as a user editable variable, so it must move the camera. I have most of the data: I have the up-vector for the camera I have the center point of the bounding box I have the look-at vector

问题 PIL's Image.transform has a perspective-mode which requires an 8-tuple of data but I can't figure out how to convert let's say a right tilt of 30 degrees to that tuple. Can anyone explain it? 回答1: 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:/

问题 I'm trying to do a perspective transformation of a set of points in order to achieve a deskewing effect: http://nuigroup.com/?ACT=28&fid=27&aid=1892_H6eNAaign4Mrnn30Au8d I'm using the image below for tests, and the green rectangle display the area of interest. I was wondering if it's possible to achieve the effect I'm hoping for using a simple combination of cv::getPerspectiveTransform and cv::warpPerspective . I'm sharing the source code I've written so far, but it doesn't work. This is the

问题 I have a DIV element with CSS3 perspective on it. The DIV contains 2 child DIVs, and one of these has a translation on the z-axis. This should result in one DIV in front of the other, and so the one in behind should be blocked from view. However, the visibility of these DIVs seem to be controlled by the order in which they appear in the HTML, instead of being handled automatically by CSS3 perspective. In the code below, even though #div3 is behind #div2 on the z-axis, it shows up in front

问题 How can I apply a perspective transformation on an image using only the PHP GD library? I don't want to use a function someone else made I want to UNDERSTAND what's going on 回答1: I honestly don't know how to describe mathematically a perspective distortion. You could try searching the literature for that (e.g. Google Scholar). See also in the OpenGL documentation, glFrustum. EDIT : Interestingly, starting with version 8, Mathematica has a ImagePerspectiveTransformation. In the relevant part,

问题 I've got an UIView and I want to add perspective to it. Basically I got 4 other views that the user will drag in order to create the perspective. I'd like to indicate 4 points where the UIView's corners will be. I investigated it and I figured out I have to use CALayer to do that, but I'm still no clearer on how to create this kind of transform on an iPhone. Any help, pointers or example code snippets would be really appreciated! Thanks! 回答1: Start off by taking a look at the pertinent

问题 I'm trying to implement a custom perspective switcher toolbar to replace eclipse's built-in one. I couldn't get the toolbar to display, and it was shown to me that due to a bug with the dynamic element in a menu contribution, I have to use a control element instead, as described in the workaround to the dynamic bug. I have a toolbar displaying following that approach, but I cannot figure out how to update it dynamically. The workaround instruction is to call ContributionItem#fill(CoolBar, int