计算旋转的2D图像的平移值和旋转角度 [英] Calculating translation value and rotation angle of a rotated 2D image
问题描述
我有两个图像，其中一个是原始图像，第二个是变形图像。
我必须找出多少度转换图像被旋转使用3x3变换矩阵。此外，我需要找到从原点翻译多远。
两个图像都是灰度图并保存在矩阵变量中。它们的大小相同 [350 500] 。
我找到了一些演讲笔记，例如问题。你正在做的是在一个图像中给定坐标，在另一个图像中给出相应的坐标，你正在尝试恢复用于将点从一个图像扭曲到另一个图像的组合的平移和旋转矩阵。 / p>
您可以通过将两个矩阵相乘在一起，将旋转和平移合并为一个矩阵。乘法只是将两个操作合成在一起。你会得到：
H = [cos（theta）-sin（theta）tx]
[sin theta）cos（theta）ty]
[0 0 1]
基本上，你想要找到的是以下关系：
xi_after = H * xi_before
H 是将坐标从一个图像映射到另一个图像所需的组合旋转和平移矩阵。 H 也是一个3 x 3矩阵，并且知道右下方的条目（第3行，第3列）为1，它使事情变得更容易。此外，假设你的点在增强坐标系中，我们基本上想从第一个图像（x_i，y_i）到其他（x_i'，y_i'）：
p_i * x_i'] [h11 h12 h13] [x_i]
[p_i * y_i'] = [h21 h22 h23] * [y_i]
[p_i] [h31 h32 1] [1]
p_i 的比例是考虑单应性缩放和消失点。让我们执行该方程的矩阵向量乘法。我们可以忽略第三个元素，因为它对我们没有用（现在）：
p_i * x_i'= h11 * x_i + h12 * y_i + h13
p_i * y_i'= h21 * x_i + h22 * y_i + h23
$ b b
现在让我们来看看第三个元素。我们知道 p_i = h31 * x_i + h32 * y_i + 1 。因此，将 p_i 代入每个方程式，并重新排列以解决 x_i'和 y_i'，我们得到：
x_i'= h11 * x_i + h12 * y_i + h13-h31 * x_i * x_i'-h32 * y_i * x_i'
y_i'= h21 * x_i + h22 * y_i + h23-h31 * x_i * y_i'-h32 * y_i * y_i'
您现在拥有的是每个唯一点对的两个方程式。我们现在可以做的是建立一个过度确定的方程组。取每对，并从中构建两个方程。然后将其转换为矩阵形式，即：
Ah = b
A 将是使用来自第一图像的坐标从每组方程构建的系数矩阵， b 将是第二个图像的每一对点， h 将是您要解决的参数。最终，你终于解决了以矩阵形式重新表达的线性方程组：
您可以解决可以执行的向量 h 通过最小二乘法。在MATLAB中，您可以通过以下方式实现：
h = A \ b;
一个sidenote：如果图像之间的移动只是一个旋转和平移， h32在我们求解参数后都为零。
注意：这种方法只有在你使用的时候才能工作。至少 具有唯一点对。因为有8个参数要解决，并且每个点有两个方程，所以 A 必须至少具有8的等级，以使系统一致（如果你想要在循环中引入一些线性代数术语）。 如果你想要一些MATLAB代码，让我们假设你的积分存储在 sourcePoints 和 targetPoints 。 sourcePoints 来自第一张图片， targetPoints 则用于第二张图片。显然，两个图像之间应该有相同的点数。假设 sourcePoints 和 targetPoints 都存储为 M x 2 matrix。第一列包含 x 坐标，第二列包含 y 坐标。
numPoints = size（sourcePoints，1）;
％//将数据转换为double以确保
sourcePoints = double（sourcePoints）;
targetPoints = double（targetPoints）;
％//提取相关数据
xSource = sourcePoints（：，1）;
ySource = sourcePoints（：，2）;
xTarget = targetPoints（：，1）;
yTarget = targetPoints（：，2）;
％//创建帮助向量
vec0 = zeros（numPoints，1）;
vec1 = ones（numPoints，1）;
xSourcexTarget = -xSource。* xTarget;
ySourcexTarget = -ySource。* xTarget;
xSourceyTarget = -xSource。* yTarget;
ySourceyTarget = -ySource。* yTarget;
％//构建矩阵
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
％//构建RHS向量
b = [xTarget; yTarget];
％//通过最小二乘法求解单应性
h = A \ b;
％//重构为3 x 3矩阵（可选）
％//必须转置为重构形式
％//在列主格式中
h 9）= 1; ％//加入，h33是1，我们改造之前
hmatrix = reshape（h，3，3）';
完成后，您有一个组合的旋转和平移矩阵。如果您想要 x 和 y 翻译，只需选择第3栏，<$ c $中的第1和第2行c> hmatrix 。但是，我们也可以使用 h 本身的向量，因此h13将是元素3，h23将是元素数6.如果你想要旋转角度，简单地对行1,2和列1,2进行相应的反三角函数。对于 h 向量，这将是元素1,2,4和5。有点不一致，取决于你选择哪些元素，因为这是由最小二乘法解决。获得良好的总体角度的一种方式可能是找到所有4个元素的角度，然后做某种平均。
参考资料
之前我通过Leow学到了单应性Wee Kheng的计算机视觉课程。我告诉你的是根据他的幻灯片： http：// www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf 。看看幻灯片30-32如果你想知道我从哪里拉这种材料。但是，MATLAB代码我自己写道：）
I have two images which one of them is the Original image and the second one is Transformed image.
I have to find out how many degrees Transformed image was rotated using 3x3 transformation matrix. Plus, I need to find how far translated from origin.
Both images are grayscaled and held in matrix variables. Their sizes are same [350 500].
I have found a few lecture notes like this.
Lecture notes say that I should use the following matrix formula for rotation:
For translation matrix the formula is given:
Everything is good. But there are two problems:
- I could not imagine how to implement the formulas using MATLAB.
- The formulas are shaped to find
x',y'values but I already have gotx,x',y,y'values. I need to find rotation angle (theta) andtxandty. - I want to know the equivailence of
x,x',y,y'in the the matrix.
I have got the following code:
rotationMatrix = [ cos(theta) sin(theta) 0 ; ...
-sin(theta) cos(theta) 0 ; ...
0 0 1];
translationMatrix = [ 1 0 tx; ...
0 1 ty; ...
0 0 1];
But as you can see, tx, ty, theta variables are not defined before used. How can I calculate theta, tx and ty?
PS: It is forbidden to use Image Processing Toolbox functions.
This is essentially a homography recovery problem. What you are doing is given co-ordinates in one image and the corresponding co-ordinates in the other image, you are trying to recover the combined translation and rotation matrix that was used to warp the points from the one image to the other.
You can essentially combine the rotation and translation into a single matrix by multiplying the two matrices together. Multiplying is simply compositing the two operations together. You would this get:
H = [cos(theta) -sin(theta) tx]
[sin(theta) cos(theta) ty]
[ 0 0 1]
The idea behind this is to find the parameters by minimizing the error through least squares between each pair of points.
Basically, what you want to find is the following relationship:
xi_after = H*xi_before
H is the combined rotation and translation matrix required to map the co-ordinates from the one image to the other. H is also a 3 x 3 matrix, and knowing that the lower right entry (row 3, column 3) is 1, it makes things easier. Also, assuming that your points are in the augmented co-ordinate system, we essentially want to find this relationship for each pair of co-ordinates from the first image (x_i, y_i) to the other (x_i', y_i'):
[p_i*x_i'] [h11 h12 h13] [x_i]
[p_i*y_i'] = [h21 h22 h23] * [y_i]
[ p_i ] [h31 h32 1 ] [ 1 ]
The scale of p_i is to account for homography scaling and vanishing points. Let's perform a matrix-vector multiplication of this equation. We can ignore the 3rd element as it isn't useful to us (for now):
p_i*x_i' = h11*x_i + h12*y_i + h13
p_i*y_i' = h21*x_i + h22*y_i + h23
Now let's take a look at the 3rd element. We know that p_i = h31*x_i + h32*y_i + 1. As such, substituting p_i into each of the equations, and rearranging to solve for x_i' and y_i', we thus get:
x_i' = h11*x_i + h12*y_i + h13 - h31*x_i*x_i' - h32*y_i*x_i'
y_i' = h21*x_i + h22*y_i + h23 - h31*x_i*y_i' - h32*y_i*y_i'
What you have here now are two equations for each unique pair of points. What we can do now is build an over-determined system of equations. Take each pair and build two equations out of them. You will then put it into matrix form, i.e.:
Ah = b
A would be a matrix of coefficients that were built from each set of equations using the co-ordinates from the first image, b would be each pair of points for the second image and h would be the parameters you are solving for. Ultimately, you are finally solving this linear system of equations reformulated in matrix form:
You would solve for the vector h which can be performed through least squares. In MATLAB, you can do this via:
h = A \ b;
A sidenote for you: If the movement between images is truly just a rotation and translation, then h31 and h32 will both be zero after we solve for the parameters. However, I always like to be thorough and so I will solve for h31 and h32 anyway.
NB: This method will only work if you have at least 4 unique pairs of points. Because there are 8 parameters to solve for, and there are 2 equations per point, A must have at least a rank of 8 in order for the system to be consistent (if you want to throw in some linear algebra terminology in the loop). You will not be able to solve this problem if you have less than 4 points.
If you want some MATLAB code, let's assume that your points are stored in sourcePoints and targetPoints. sourcePoints are from the first image and targetPoints are for the second image. Obviously, there should be the same number of points between both images. It is assumed that both sourcePoints and targetPoints are stored as M x 2 matrices. The first columns contain your x co-ordinates while the second columns contain your y co-ordinates.
numPoints = size(sourcePoints, 1);
%// Cast data to double to be sure
sourcePoints = double(sourcePoints);
targetPoints = double(targetPoints);
%//Extract relevant data
xSource = sourcePoints(:,1);
ySource = sourcePoints(:,2);
xTarget = targetPoints(:,1);
yTarget = targetPoints(:,2);
%//Create helper vectors
vec0 = zeros(numPoints, 1);
vec1 = ones(numPoints, 1);
xSourcexTarget = -xSource.*xTarget;
ySourcexTarget = -ySource.*xTarget;
xSourceyTarget = -xSource.*yTarget;
ySourceyTarget = -ySource.*yTarget;
%//Build matrix
A = [xSource ySource vec1 vec0 vec0 vec0 xSourcexTarget ySourcexTarget; ...
vec0 vec0 vec0 xSource ySource vec1 xSourceyTarget ySourceyTarget];
%//Build RHS vector
b = [xTarget; yTarget];
%//Solve homography by least squares
h = A \ b;
%// Reshape to a 3 x 3 matrix (optional)
%// Must transpose as reshape is performed
%// in column major format
h(9) = 1; %// Add in that h33 is 1 before we reshape
hmatrix = reshape(h, 3, 3)';
Once you are finished, you have a combined rotation and translation matrix. If you want the x and y translations, simply pick off column 3, rows 1 and 2 in hmatrix. However, we can also work with the vector of h itself, and so h13 would be element 3, and h23 would be element number 6. If you want the angle of rotation, simply take the appropriate inverse trigonometric function to rows 1, 2 and columns 1, 2. For the h vector, this would be elements 1, 2, 4 and 5. There will be a bit of inconsistency depending on which elements you choose as this was solved by least squares. One way to get a good overall angle would perhaps be to find the angles of all 4 elements then do some sort of average. Either way, this is a good starting point.
References
I learned about homography a while ago through Leow Wee Kheng's Computer Vision course. What I have told you is based on his slides: http://www.comp.nus.edu.sg/~cs4243/lecture/camera.pdf. Take a look at slides 30-32 if you want to know where I pulled this material from. However, the MATLAB code I wrote myself :)
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