从2D图像用单校准相机获得世界坐标 [英] Getting world coordinates from 2D image using single calibrated camera

问题描述

我想从一个单一的图像校准我的相机没有一个棋盘。为此，我用下面的应用程序（ http://w3.impa.br/ 〜臧/ qtcalib / index.html的），它使用蔡琴的方法来校准相机。

I am trying to calibrate my camera from a single image without a chessboard. To this end, I used the following application (http://w3.impa.br/~zang/qtcalib/index.html) which uses Tsai's method to calibrate the camera.

应用程序将返回intristic，旋转和平移矩阵。

The application returns intristic, rotation and translation matrix.

在这之后，我想计算世界从2D图像坐标。因此，与二维坐标（x，y）和矩阵（intristic，旋转，平移），我想获得的三维坐标（X，Y，Z）。喂我的程序

After that, I would like to calculate world coordinates from the 2d image. Thus, feeding my program with 2d coordinates (x,y) and the matrices (intristic, rotation, translation) I would like to get the 3D coordinates (X,Y,Z).

我跟一些相关的线程的指令（<一href="http://stackoverflow.com/questions/7836134/get-3d-coord-from-2d-image-pixel-if-we-know-extrinsic-and-intrinsic-parameters">get从2D图像像素3D坐标，如果我们知道外在和内在的参数的），但结果并不如我所期待的。另外，我不知道哪里是我的原点（0,0,0）。

I followed the instruction of some relative threads (get 3d coord from 2d image pixel if we know extrinsic and intrinsic parameters) but the results were not as I was expecting. Plus I have no idea where is my origin (0,0,0).

我是什么做错了吗？

推荐答案

该射影变换是

X = K * T * X

x = K * T * X

其中

K：内在
T：外在的（你的旋转和平移）。变换X转换成照相机坐标系。
X：在图像
像素坐标 X：三维坐标

K : intrinsic
T : extrinsic (your rotation and translation). Transforms X into the camera coordinate system.
x : pixel coordinate in image
X : 3D coordinate

投射二维像素三维比得上在三维空间中的线，因而有三维的无限多点可能。为了得到一个普通的三维坐标，你必须选择一个深度值。

Projecting a 2D pixel to 3D is comparable to a line in 3D space, thus there are an infinite number of 3D points possible. To get an ordinary 3D coordinate, you have to choose a depth value.

您可以决定你所得到的三维坐标原点：

You can decide your origin of the resulting 3D coordinate:

• 摄像机坐标系，[0; 0; 0]为相机中心：
  X'= INV（K）* X

• Camera coordinate system, [0;0;0] is at the camera center:
  X' = inv(K) * x

您棋盘或仔格的坐标系，[0; 0; 0]通常是在一个角落：
X'= INV（T）* INV（K）* X

Your chessboard or Tsai-grid coordinate system, [0;0;0] is usually at one of the corners:
X' = inv(T) * inv(K) * x

然而，如已经提到的，X'为更像一条线。你需要它的规模得到适当的三维坐标。

However, as already mentioned, X' is more like a line. You need to scale it to get a proper 3D coordinate.

它有时很难知道什么是仿射变换做。如果它不这样做你想要什么，咋工程的办法是与它的逆试试吧：）

It's sometimes difficult to know what a affine transformation is doing. In case it doesn't do what you want, a quick engineering approach could be to try it with its inverse :)

如果不帮你，你可以通过你的矩阵。

If that does not help you, you can pass your matrices.

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