相机单应性 [英] Camera homography

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本文介绍了相机单应性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在学习相机矩阵的东西。我已经知道我可以通过在物体空间的平面中使用四个点来获得相机的单应性(3 * 3矩阵)。我想知道我们是否可以在平面上没有四个点的情况下获得同形像?如果是,我如何获得矩阵?我应该看什么公式?

I am learning camera matrix stuff. I already known that I can get the homography of the camera (3*3 matrix) by using four points in a plane in object space. I want to know if we can get the homagraphy with four points not in a plane? If yes, how can I get the matrix? What formulas should I look at?

我也将单应性与另一个概念相混淆:如果我想将点从一个坐标转换为另一个坐标,则只需要知道三个点即可。系统。那么,为什么在计算单应性时需要四个点呢?

I also confused homography with another concept: I only need to know three points if I want to convert from points from one coordinate to another coordinate system. So why we need four points in computing homography?

推荐答案

Homography映射点
1.在平面上指向另一个点plane
2.在纯粹的相机旋转或缩放过程中,以3D投影的点(不必在同一平面上)。

Homography maps points 1. On plane to points at another plane 2. Projections of points in 3D (no obligatory lying on the same plane) during a pure camera rotation or zoom.

可以轻松地验证后者如果您查看在传感器平面旋转时连接点的光线:绿色是两个传感器位置,黑色是3d对象

The latter can be easily verified if you look at the rays that connect points while sensor plane rotates: green are two sensor positions and black is a 3d object

由于单应性是在投影之间而不是在3D中的对象之间,因此您不必在意这些投影所代表的含义。我同意,但这可能会造成混淆。例如,您可以将相机指向3D场景(不是平坦的!),然后旋转相机,场景的两个结果图片将通过单应性进行关联。顺便说一下,这是图像全景的基础。

Since Homography is between projections and not between objects in 3D you don’t care what these projections represent. But this can be confusing, I agree. For example you can point your camera at 3D scene (that is not flat!), then rotate your camera and the two resulting pictures of the scene will be related by homography. This is, by the way, a foundation for image panoramas.

您提到的三点对应关系可能会被归类为一种称为仿射的变换(在发生透视效果时会在大变焦时发生)消失)或在3D空间中进行刚性旋转和平移。两者都需要3点对应,但是前者只需要2D点,而后者则需要3D点。后一种情况具有6DOF(旋转3,平移3),而每个对应关系提供2DOF,因此6/2 = 3对应关系。同形异义词有8个自由度,因此应该有8/2 = 4个对应关系。

Three point correspondences you mentioned may be reladte to a transformation called Affine (happens during large zooms when a perspective effects disappears) or to the finding a rigid rotation and translation in 3D space. Both require 3 point correspondences but the former needs only 2D points while the latter needs 3D points. The latter case has 6DOF ( 3 for rotation and 3 for translation) while each correspondence provides 2DOF, hence 6/2=3 correspondences. Homography has 8 DOF so there should be 8/2=4 correspondences;

下面是一个小图,解释了当原始正方形向前倾斜时仿射变换和同形异义变换之间的区别。在仿射情况下,远侧与近侧的长度相同的透视效果可以忽略。在同形照相术的情况下,另一端较短。

Below is a little diagram that explains the difference between affine and homographs transformation when the original square tilts forward. In affine case the perspective effect is negligible that is far side has the same length as a near one. In the case of Homography the far side is shorter.

这篇关于相机单应性的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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