Get angle from x,y of a linear base of an object - c++

Question

How to align/parallelise the green arrow (pic 2) with the linear base.
After applying the basic PCA analysis of OpenCV I was quite happy with the result but I would like to know how to manipulate the center point location and angle to match with the base.
In the pictures provided you can see the green line is off by a few degrees while I would like it to be either on or just parallel to the "base".
Source image:
What I currently get:
#include <iostream>
#include <opencv2/opencv.hpp>
using namespace std;
using namespace cv;
// Function declarations
void drawAxis(Mat&, Point, Point, Scalar, const float);
double getOrientation(const vector<Point> &, Mat&);
void drawAxis(Mat& img, Point p, Point q, Scalar colour, const float scale = 0.2)
{
double angle;
double hypotenuse;
angle = atan2( (double) p.y - q.y, (double) p.x - q.x ); // angle in radians
hypotenuse = sqrt( (double) (p.y - q.y) * (p.y - q.y) + (p.x - q.x) * (p.x - q.x));
double degrees = angle * 180 / CV_PI; // convert radians to degrees (0-180 range)
cout << "Degrees: " << abs(degrees - 180) << endl; // angle in 0-360 degrees range
// Here we lengthen the arrow by a factor of scale
q.x = (int) (p.x - scale * hypotenuse * cos(angle));
q.y = (int) (p.y - scale * hypotenuse * sin(angle));
line(img, p, q, colour, 1, CV_AA);
// create the arrow hooks
p.x = (int) (q.x + 9 * cos(angle + CV_PI / 4));
p.y = (int) (q.y + 9 * sin(angle + CV_PI / 4));
line(img, p, q, colour, 1, CV_AA);
p.x = (int) (q.x + 9 * cos(angle - CV_PI / 4));
p.y = (int) (q.y + 9 * sin(angle - CV_PI / 4));
line(img, p, q, colour, 1, CV_AA);
}
double getOrientation(const vector<Point> &pts, Mat &img)
{
//Construct a buffer used by the pca analysis
int sz = static_cast<int>(pts.size());
Mat data_pts = Mat(sz, 2, CV_64FC1);
for (int i = 0; i < data_pts.rows; ++i)
{
data_pts.at<double>(i, 0) = pts[i].x;
data_pts.at<double>(i, 1) = pts[i].y;
}
//Perform PCA analysis
PCA pca_analysis(data_pts, Mat(), CV_PCA_DATA_AS_ROW);
//Store the center of the object
Point cntr = Point(static_cast<int>(pca_analysis.mean.at<double>(0, 0)),
static_cast<int>(pca_analysis.mean.at<double>(0, 1)));
//Store the eigenvalues and eigenvectors
vector<Point2d> eigen_vecs(2);
vector<double> eigen_val(2);
for (int i = 0; i < 2; ++i)
{
eigen_vecs[i] = Point2d(pca_analysis.eigenvectors.at<double>(i, 0),
pca_analysis.eigenvectors.at<double>(i, 1));
eigen_val[i] = pca_analysis.eigenvalues.at<double>(0, i);
}
// Draw the principal components
circle(img, cntr, 3, Scalar(255, 0, 255), 2);
Point p1 = cntr + 0.02 * Point(static_cast<int>(eigen_vecs[0].x * eigen_val[0]), static_cast<int>(eigen_vecs[0].y * eigen_val[0]));
Point p2 = cntr - 0.02 * Point(static_cast<int>(eigen_vecs[1].x * eigen_val[1]), static_cast<int>(eigen_vecs[1].y * eigen_val[1]));
drawAxis(img, cntr, p1, Scalar(0, 255, 0), 1);
drawAxis(img, cntr, p2, Scalar(255, 255, 0), 5);
double angle = atan2(eigen_vecs[0].y, eigen_vecs[0].x); // orientation in radians
return angle;
}
int main(int, char** argv)
{
// Load image
Mat src = imread("/path/image.jpg");
// Check if image is loaded successfully
if(!src.data || src.empty())
{
cout << "Problem loading image!!!" << endl;
return EXIT_FAILURE;
}
imshow("src", src);
// Convert image to grayscale
Mat gray;
cvtColor(src, gray, COLOR_BGR2GRAY);
// Convert image to binary
Mat bw;
threshold(gray, bw, 50, 255, CV_THRESH_BINARY | CV_THRESH_OTSU);
// Find all the contours in the thresholded image
vector<Vec4i> hierarchy;
vector<vector<Point> > contours;
findContours(bw, contours, hierarchy, CV_RETR_LIST, CV_CHAIN_APPROX_NONE);
for (size_t i = 0; i < contours.size(); ++i)
{
// Calculate the area of each contour
double area = contourArea(contours[i]);
// Ignore contours that are too small or too large
//if (area < 1e2 || 1e5 < area) continue;
if (area > 1e6) continue;
cout << "Area: " << area << endl;
// Draw each contour only for visualisation purposes
drawContours(src, contours, static_cast<int>(i), Scalar(0, 0, 255), 2, 8, hierarchy, 0);
// Find the orientation of each shape
getOrientation(contours[i], src);
}
imshow("output", src);
waitKey(0);
return 0;
}

Answer 1

I don't know nothing about opencv, but this should do.
To project a vector on a plane you have to:
Greenvector - scalarproduct(plane.Normalvector,scalarproduct(plane.Normalvector,Greenvector)/plane.Normalvector.value^2)
Assuming vectors are as common used in math, describing a direction.
As I am at it, thank you, upon checking your question I've found an error within my own code.
edit:/
Are you sure the greenvector ain't on the plane, did you check it or are you assuming it from the picture?