Calculate distance (disparity) OpenCV - opencv

-- Update 2 --
The following article is really useful (although it is using Python instead of C++) if you are using a single camera to calculate the distance: Find distance from camera to object/marker using Python and OpenCV
Best link is Stereo Webcam Depth Detection. The implementation of this open source project is really clear.
Below is the original question.
For my project I am using two camera's (stereo vision) to track objects and to calculate the distance. I calibrated them with the sample code of OpenCV and generated a disparity map.
I already implemented a method to track objects based on color (this generates a threshold image).
My question: How can I calculate the distance to the tracked colored objects using the disparity map/ matrix?
Below you can find a code snippet that gets the x,y and z coordinates of each pixel. The question: Is Point.z in cm, pixels, mm?
Can I get the distance to the tracked object with this code?
Thank you in advance!
cvReprojectImageTo3D(disparity, Image3D, _Q);
vector<CvPoint3D32f> PointArray;
CvPoint3D32f Point;
for (int y = 0; y < Image3D->rows; y++) {
float *data = (float *)(Image3D->data.ptr + y * Image3D->step);
for (int x = 0; x < Image3D->cols * 3; x = x + 3)
{
Point.x = data[x];
Point.y = data[x+1];
Point.z = data[x+2];
PointArray.push_back(Point);
//Depth > 10
if(Point.z > 10)
{
printf("%f %f %f", Point.x, Point.y, Point.z);
}
}
}
cvReleaseMat(&Image3D);
--Update 1--
For example I generated this thresholded image (of the left camera). I almost have the same of the right camera.
Besides the above threshold image, the application generates a disparity map. How can I get the Z-coordinates of the pixels of the hand in the disparity map?
I actually want to get all the Z-coordinates of the pixels of the hand to calculate the average Z-value (distance) (using the disparity map).

See this links: OpenCV: How-to calculate distance between camera and object using image?, Finding distance from camera to object of known size, http://answers.opencv.org/question/5188/measure-distance-from-detected-object-using-opencv/
If it won't solve you problem, write more details - why it isn't working, etc.

The math for converting disparity (in pixels or image width percentage) to actual distance is pretty well documented (and not very difficult) but I'll document it here as well.
Below is an example given a disparity image (in pixels) and an input image width of 2K (2048 pixels across) image:
Convergence Distance is determined by the rotation between camera lenses. In this example it will be 5 meters. Convergence distance of 5 (meters) means that the disparity of objects 5 meters away is 0.
CD = 5 (meters)
Inverse of convergence distance is: 1 / CD
IZ = 1/5 = 0.2M
Size of camera's sensor in meters
SS = 0.035 (meters) //35mm camera sensor
The width of a pixel on the sensor in meters
PW = SS/image resolution = 0.035 / 2048(image width) = 0.00001708984
The focal length of your cameras in meters
FL = 0.07 //70mm lens
InterAxial distance: The distance from the center of left lens to the center of right lens
IA = 0.0025 //2.5mm
The combination of the physical parameters of your camera rig
A = FL * IA / PW
Camera Adjusted disparity: (For left view only, right view would use positive [disparity value])
AD = 2 * (-[disparity value] / A)
From here you can compute actual distance using the following equation:
realDistance = 1 / (IZ – AD)
This equation only works for "toe-in" camera systems, parallel camera rigs will use a slightly different equation to avoid infinity values, but I'll leave it at this for now. If you need the parallel stuff just let me know.

if len(puntos) == 2:
x1, y1, w1, h1 = puntos[0]
x2, y2, w2, h2 = puntos[1]
if x1 < x2:
distancia_pixeles = abs(x2 - (x1+w1))
distancia_cm = (distancia_pixeles*29.7)/720
cv2.putText(imagen_A4, "{:.2f} cm".format(distancia_cm), (x1+w1+distancia_pixeles//2, y1-30), 2, 0.8, (0,0,255), 1,
cv2.LINE_AA)
cv2.line(imagen_A4,(x1+w1,y1-20),(x2, y1-20),(0, 0, 255),2)
cv2.line(imagen_A4,(x1+w1,y1-30),(x1+w1, y1-10),(0, 0, 255),2)
cv2.line(imagen_A4,(x2,y1-30),(x2, y1-10),(0, 0, 255),2)
else:
distancia_pixeles = abs(x1 - (x2+w2))
distancia_cm = (distancia_pixeles*29.7)/720
cv2.putText(imagen_A4, "{:.2f} cm".format(distancia_cm), (x2+w2+distancia_pixeles//2, y2-30), 2, 0.8, (0,0,255), 1,
cv2.LINE_AA)
cv2.line(imagen_A4,(x2+w2,y2-20),(x1, y2-20),(0, 0, 255),2)
cv2.line(imagen_A4,(x2+w2,y2-30),(x2+w2, y2-10),(0, 0, 255),2)
cv2.line(imagen_A4,(x1,y2-30),(x1, y2-10),(0, 0, 255),2)
cv2.imshow('imagen_A4',imagen_A4)
cv2.imshow('frame',frame)
k = cv2.waitKey(1) & 0xFF
if k == 27:
break
cap.release()
cv2.destroyAllWindows()
I think this is a good way to measure the distance between two objects

Related

How Convexity Defect is calculated in OpenCV?

What is the algorithm used in OpenCV function convexityDefects() to calculate the convexity defects of a contour?
Please, describe and illustrate the high-level operation of the algorithm, along with its inputs and outputs.
Based on the documentation, the input are two lists of coordinates:
contour defining the original contour (red on the image below)
convexhull defining the convex hull corresponding to that contour (blue on the image below)
The algorithm works in the following manner:
If the contour or the hull contain 3 or less points, then the contour is always convex, and no more processing is needed. The algorithm assures that both the contour and the hull are accessed in the same orientation.
N.B.: In further explanation I assume they are in the same orientation, and ignore the details regarding representation of the floating point depth as an integer.
Then for each pair of adjacent hull points (H[i], H[i+1]), defining one edge of the convex hull, calculate the distance from the edge for each point on the contour C[n] that lies between H[i] and H[i+1] (excluding C[n] == H[i+1]). If the distance is greater than zero, then a defect is present. When a defect is present, record i, i+1, the maximum distance and the index (n) of the contour point where the maximum located.
Distance is calculated in the following manner:
dx0 = H[i+1].x - H[i].x
dy0 = H[i+1].y - H[i].y
if (dx0 is 0) and (dy0 is 0) then
scale = 0
else
scale = 1 / sqrt(dx0 * dx0 + dy0 * dy0)
dx = C[n].x - H[i].x
dy = C[n].y - H[i].y
distance = abs(-dy0 * dx + dx0 * dy) * scale
It may be easier to visualize in terms of vectors:
C: defect vector from H[i] to C[n]
H: hull edge vector from H[i] to H[i+1]
H_rot: hull edge vector H rotated 90 degrees
U_rot: unit vector in direction of H_rot
H components are [dx0, dy0], so rotating 90 degrees gives [-dy0, dx0].
scale is used to find U_rot from H_rot, but because divisions are more computationally expensive than multiplications, the inverse is used as an optimization. It's also pre-calculated before the loop over C[n] to avoid recomputing each iteration.
|H| = sqrt(dx0 * dx0 + dy0 * dy0)
U_rot = H_rot / |H| = H_rot * scale
Then, a dot product between C and U_rot gives the perpendicular distance from the defect point to the hull edge, and abs() is used to get a positive magnitude in any orientation.
distance = abs(U_rot.C) = abs(-dy0 * dx + dx0 * dy) * scale
In the scenario depicted on the above image, in first iteration, the edge is defined by H[0] and H[1]. The contour points tho examine for this edge are C[0], C[1], and C[2] (since C[3] == H[1]).
There are defects at C[1] and C[2]. The defect at C[1] is the deepest, so the algorithm will record (0, 1, 1, 50).
The next edge is defined by H[1] and H[2], and corresponding contour point C[3]. No defect is present, so nothing is recorded.
The next edge is defined by H[2] and H[3], and corresponding contour point C[4]. No defect is present, so nothing is recorded.
Since C[5] == H[3], the last contour point can be ignored -- there can't be a defect there.

bad distance results using stereo camera

I'm trying to measure distance in real time from stereo pair to a person detected in the scene. First i calibrated both cameras separately with a 9x6 checkerboard (square size of 59 mm) and i obtained a rms error between 0.15 and 0.19 for both cameras. Using the obtained parameters i calibrated the stereo pair and the rms error was 0.36. Later, I rectified, undistort and remap the stereo pair giving me this result:
rectified and undistorted stereo
Done that, I computed stereo correspondence using stereoSGBM. That's how i did:
Mat imgDisp= Mat(frame1.cols, frame1.rows,CV_16S);
cvtColor(frame1, frame1, CV_BGR2GRAY);
cvtColor(frame2, frame2, CV_BGR2GRAY);
//parameters for stereoSGBM
stereo.SADWindowSize = 3;
stereo.numberOfDisparities = 144;
stereo.preFilterCap = 63;
stereo.minDisparity = -39;
stereo.uniquenessRatio = 10;
stereo.speckleWindowSize = 100;
stereo.speckleRange = 32;
stereo.disp12MaxDiff = 1;
stereo.fullDP = false;
stereo.P1 = 216;
stereo.P2 = 864;
double minVal; double maxVal;
minMaxLoc(imgDisp, &minVal, &maxVal);
return imgDisp;
I attached the result from stereoSGBM here: disparity map.
For detect a person in the scene I used hog + svm (the default people dectector) and tracked that person with optical flow (cvCalcOpticalFlowPyrLK()). Using the disparity map obtained in the stereo correspondence process i obtained the disparity for each corner tracked from one person as follow:
int x= cornersA[k].x;
int y= cornersA[k].y;
short pixVal= mapaDisp.at<short>(y,x);
float dispFeatures= pixVal/ 16.0f;
with the disparity for each corner tracked for one person in the scene I computed the maxim disparity and computed the depth in that pixel using the formula ((focal*baseline)/disp):
float Disp =maxDisp_v[p];
cout<< "max disp"<< Disp<<endl;
float d = ((double)(879.85* 64.32)/(double)(Disp))/10; //distance in cms.
** for focal length I calculated the average between fx and fy obtained in the cameras matrix [3x3] parameters:
CM1: [9.0472706037497187e+02 0. 3.7829164759284492e+02
0. 8.4576999835299739e+02 1.8649783393160138e+02
0. 0. 1.]
CM2: [9.1390904648169953e+02 0. 3.5700689147467887e+02 0.
8.5514555697053311e+02 2.1723345133656409e+02 0. 0. 1.]
so fx camera1: 904.7; fy camera1: 845.7; fx camera2: 913.9; fy camera2: 855.1
** The result of T[0,0] matrix matched with the baseline that I measure manuallly so I assumed that's correct baseline.
**due to the square size of checkerboard is in mm i assumed that baseline must be in the same unit, that's why I'm put 64.32 mm in baseline.
The result of distance is aprox. 55 cms but the real distance is 300 cms. I have checked many times but the measured distance is still incorrect: distanceResult
Help me please!, I have no idea what i'm doing wrong.
*** I'm using opencv 2.4.9 in osx system.
I think you are making a mistake with units:
focal length is provided in pixels,
baseline is provided in cm
disparity is provided in pixels.
Right?
According to formula you have pix*cm/pix = cm. But you devide it by 10 and get dm. So you have the distance around 55dm which is twice bigger then 300. Which is not a bad case for you approach.
You cannot use the simple parallel-cameras triangulation formula on rectified images, because you need to undo the rectification homographies.
Use cv2.reprojectImageTo3D

Estimating distance from camera to ground plane point

How can I calculate distance from camera to a point on a ground plane from an image?
I have the intrinsic parameters of the camera and the position (height, pitch).
Is there any OpenCV function that can estimate that distance?
You can use undistortPoints to compute the rays backprojecting the pixels, but that API is rather hard to use for your purpose. It may be easier to do the calculation "by hand" in your code. Doing it at least once will also help you understand what exactly that API is doing.
Express your "position (height, pitch)" of the camera as a rotation matrix R and a translation vector t, representing the coordinate transform from the origin of the ground plane to the camera. That is, given a point in ground plane coordinates Pg = [Xg, Yg, Zg], its coordinates in camera frame are given by
Pc = R * Pg + t
The camera center is Cc = [0, 0, 0] in camera coordinates. In ground coordinates it is then:
Cg = inv(R) * (-t) = -R' * t
where inv(R) is the inverse of R, R' is its transpose, and the last equality is due to R being an orthogonal matrix.
Let's assume, for simplicity, that the the ground plane is Zg = 0.
Let K be the matrix of intrinsic parameters. Given a pixel q = [u, v], write it in homogeneous image coordinates Q = [u, v, 1]. Its location in camera coordinates is
Qc = Ki * Q
where Ki = inv(K) is the inverse of the intrinsic parameters matrix. The same point in world coordinates is then
Qg = R' * Qc + Cg
All the points Pg = [Xg, Yg, Zg] that belong to the ray from the camera center through that pixel, expressed in ground coordinates, are then on the line
Pg = Cg + lambda * (Qg - Cg)
for lambda going from 0 to positive infinity. This last formula represents three equations in ground XYZ coordinates, and you want to find the values of X, Y, Z and lambda where the ray intersects the ground plane. But that means Zg=0, so you have only 3 unknowns. Solve them (you recover lambda from the 3rd equation, then substitute in the first two), and you get Xg and Yg of the solution to your problem.

Field of view of a GoPro camera

I have calibrated my GoPro Hero 4 Black using Camera calibration toolbox for Matlab and calculated its fields of view and focal length using OpenCV's calibrationMatrixValues(). These, however, differ from GoPro's specifications. Istead of 118.2/69.5 FOVs I get 95.4/63.4 and focal length 2.8mm instead of 17.2mm. Obviously something is wrong.
I suppose the calibration itself is correct since image undistortion seems to be working well.
Can anyone please give me a hint where I made a mistake? I am posting my code below.
Thanks.
Code
cameraMatrix = new Mat(3, 3, 6);
for (int i = 0; i < cameraMatrix.height(); i ++)
for (int j = 0; j < cameraMatrix.width(); j ++) {
cameraMatrix.put(i, j, 0);
}
cameraMatrix.put(0, 0, 582.18394);
cameraMatrix.put(0, 2, 663.50655);
cameraMatrix.put(1, 1, 582.52915);
cameraMatrix.put(1, 2, 378.74541);
cameraMatrix.put(2, 2, 1.);
org.opencv.core.Size size = new org.opencv.core.Size(1280, 720);
//output parameters
double [] fovx = new double[1];
double [] fovy = new double[1];
double [] focLen = new double[1];
double [] aspectRatio = new double[1];
Point ppov = new Point(0, 0);
org.opencv.calib3d.Calib3d.calibrationMatrixValues(cameraMatrix, size,
6.17, 4.55, fovx, fovy, focLen, ppov, aspectRatio);
System.out.println("FoVx: " + fovx[0]);
System.out.println("FoVy: " + fovy[0]);
System.out.println("Focal length: " + focLen[0]);
System.out.println("Principal point of view; x: " + ppov.x + ", y: " + ppov.y);
System.out.println("Aspect ratio: " + aspectRatio[0]);
Results
FoVx: 95.41677635378488
FoVy: 63.43170132212425
Focal length: 2.8063085232812504
Principal point of view; x: 3.198308916796875, y: 2.3934605770833333
Aspect ratio: 1.0005929569269807
GoPro specifications
https://gopro.com/help/articles/Question_Answer/HERO4-Field-of-View-FOV-Information
Edit
Matlab calibration results
Focal Length: fc = [ 582.18394 582.52915 ] ± [ 0.77471 0.78080 ]
Principal point: cc = [ 663.50655 378.74541 ] ± [ 1.40781 1.13965 ]
Skew: alpha_c = [ -0.00028 ] ± [ 0.00056 ] => angle of pixel axes = 90.01599 ± 0.03208 degrees
Distortion: kc = [ -0.25722 0.09022 -0.00060 0.00009 -0.01662 ] ± [ 0.00228 0.00276 0.00020 0.00018 0.00098 ]
Pixel error: err = [ 0.30001 0.28188 ]
One of the images used for calibration
And the undistorted image
You have entered 6.17mm and 4.55mm for the sensor size in OpenCV, which corresponds to an aspect ratio 1.36 whereas as your resolution (1270x720) is 1.76 (approximately 16x9 format).
Did you crop your image before MATLAB calibration?
The pixel size seems to be 1.55µm from this Gopro page (this is by the way astonishingly small!). If pixels are squared, and they should be on this type of commercial cameras, that means your inputs are not coherent. Computed sensor size should be :
[Sensor width, Sensor height] = [1280, 720]*1.55*10^-3 = [1.97, 1.12]
mm
Even if considering the maximal video resolution which is 3840 x 2160, we obtain [5.95, 3.35]mm, still different from your input.
Please see this explanation about equivalent focal length to understand why the actual focal length of the camera is not 17.2 but 17.2*5.95/36 ~ 2.8mm. In that case, compute FOV using the formulas here for instance. You will indeed find values of 93.5°/61.7° (close to your outputs but still not what is written in the specifications because there probably some optical distortion due to the wide angles).
What I do not understand though, is how the focal distance returned can be right whereas sensor size entered is wrong. Could you give more info and/or send an image?
Edits after question updates
On that cameras, with a working resolution of 1280x720, the image is downsampled but not cropped so what I said above about sensor dimensions does not apply. The sensor size to consider is indeed the one used (6.17x4.55) as explained in your first comment.
The FOV is constrained by the calibration matrix inputs (fx, fy, cx, cy) given in pixels and the resolution. You can check it by typing:
2*DEGRES(ATAN(1280/(2*582.18394))) (=95.416776...°)
This FOV value is smaller than what is expected, but by the look of the undistorted image, your MATLAB distortion model is right and the calibration is correct. The barrel distortion due to the wide angle seems well corrected by the the rewarp you applied.
However, MATLAB toolbox uses a pinhole model, which is linear and cannot account for intrinsic parameters such as lens distortion. I assume this from the page :
https://fr.mathworks.com/help/vision/ug/camera-calibration.html
Hence, my best guess is that unless you find a model which fits more accurately the Gopro camera (maybe a wide-angle lens model), MATLAB calibration will return an intrinsic camera matrix corresponding to the "linear" undistorted image and the FOV will indeed be smaller (in the case of barrel distortion). You will have to apply distortion coefficients associated to the calibration to retrieve the actual FOV value.
We can see in the corrected image that side parts of the FOV get rejected out of bounds. If you had warped the image entirely, you would find that some undistorted pixels coordinates exceed [-1280/2;+1280/2] (horizontally, and idem vertically). Then, replacing opencv.core.Size(1280, 720) by the most extreme ranges obtained, you would hopefully retrieve Gopro website values.
In conclusion, I think you can rely on the focal distance value that you obtained if you make measurements in the center of your image, otherwise there is too much distortion and it doesn't apply.

How to calculate distance between two rectangles? (Context: a game in Lua.)

Given two rectangles with x, y, width, height in pixels and a rotation value in degrees -- how do I calculate the closest distance of their outlines toward each other?
Background: In a game written in Lua I'm randomly generating maps, but want to ensure certain rectangles aren't too close to each other -- this is needed because maps become unsolvable if the rectangles get into certain close-distance position, as a ball needs to pass between them. Speed isn't a huge issue as I don't have many rectangles and the map is just generated once per level. Previous links I found on StackOverflow are this and this
Many thanks in advance!
Not in Lua, a Python code based on M Katz's suggestion:
def rect_distance((x1, y1, x1b, y1b), (x2, y2, x2b, y2b)):
left = x2b < x1
right = x1b < x2
bottom = y2b < y1
top = y1b < y2
if top and left:
return dist((x1, y1b), (x2b, y2))
elif left and bottom:
return dist((x1, y1), (x2b, y2b))
elif bottom and right:
return dist((x1b, y1), (x2, y2b))
elif right and top:
return dist((x1b, y1b), (x2, y2))
elif left:
return x1 - x2b
elif right:
return x2 - x1b
elif bottom:
return y1 - y2b
elif top:
return y2 - y1b
else: # rectangles intersect
return 0.
where
dist is the euclidean distance between points
rect. 1 is formed by points (x1, y1) and (x1b, y1b)
rect. 2 is formed by points (x2, y2) and (x2b, y2b)
Edit: As OK points out, this solution assumes all the rectangles are upright. To make it work for rotated rectangles as the OP asks you'd also have to compute the distance from the corners of each rectangle to the closest side of the other rectangle. But you can avoid doing that computation in most cases if the point is above or below both end points of the line segment, and to the left or right of both line segments (in telephone positions 1, 3, 7, or 9 with respect to the line segment).
Agnius's answer relies on a DistanceBetweenLineSegments() function. Here is a case analysis that does not:
(1) Check if the rects intersect. If so, the distance between them is 0.
(2) If not, think of r2 as the center of a telephone key pad, #5.
(3) r1 may be fully in one of the extreme quadrants (#1, #3, #7, or #9). If so
the distance is the distance from one rect corner to another (e.g., if r1 is
in quadrant #1, the distance is the distance from the lower-right corner of
r1 to the upper-left corner of r2).
(4) Otherwise r1 is to the left, right, above, or below r2 and the distance is
the distance between the relevant sides (e.g., if r1 is above, the distance
is the distance between r1's low y and r2's high y).
Actually there is a fast mathematical solution.
Length(Max((0, 0), Abs(Center - otherCenter) - (Extent + otherExtent)))
Where Center = ((Maximum - Minimum) / 2) + Minimum and Extent = (Maximum - Minimum) / 2.
Basically the code above zero's axis which are overlapping and therefore the distance is always correct.
It's preferable to keep the rectangle in this format as it's preferable in many situations ( a.e. rotations are much easier ).
Pseudo-code:
distance_between_rectangles = some_scary_big_number;
For each edge1 in Rectangle1:
For each edge2 in Rectangle2:
distance = calculate shortest distance between edge1 and edge2
if (distance < distance_between_rectangles)
distance_between_rectangles = distance
There are many algorithms to solve this and Agnius algorithm works fine. However I prefer the below since it seems more intuitive (you can do it on a piece of paper) and they don't rely on finding the smallest distance between lines but rather the distance between a point and a line.
The hard part is implementing the mathematical functions to find the distance between a line and a point, and to find if a point is facing a line. You can solve all this with simple trigonometry though. I have below the methodologies to do this.
For polygons (triangles, rectangles, hexagons, etc.) in arbitrary angles
If polygons overlap, return 0
Draw a line between the centres of the two polygons.
Choose the intersecting edge from each polygon. (Here we reduce the problem)
Find the smallest distance from these two edges. (You could just loop through each 4 points and look for the smallest distance to the edge of the other shape).
These algorithms work as long as any two edges of the shape don't create angles more than 180 degrees. The reason is that if something is above 180 degrees then it means that the some corners are inflated inside, like in a star.
Smallest distance between an edge and a point
If point is not facing the face, then return the smallest of the two distances between the point and the edge cornerns.
Draw a triangle from the three points (edge's points plus the solo point).
We can easily get the distances between the three drawn lines with Pythagorean Theorem.
Get the area of the triangle with Heron's formula.
Calculate the height now with Area = 12⋅base⋅height with base being the edge's length.
Check to see if a point faces an edge
As before you make a triangle from an edge and a point. Now using the Cosine law you can find all the angles with just knowing the edge distances. As long as each angle from the edge to the point is below 90 degrees, the point is facing the edge.
I have an implementation in Python for all this here if you are interested.
This question depends on what kind of distance. Do you want, distance of centers, distance of edges or distance of closest corners?
I assume you mean the last one. If the X and Y values indicate the center of the rectangle then you can find each the corners by applying this trick
//Pseudo code
Vector2 BottomLeftCorner = new Vector2(width / 2, heigth / 2);
BottomLeftCorner = BottomLeftCorner * Matrix.CreateRotation(MathHelper.ToRadians(degrees));
//If LUA has no built in Vector/Matrix calculus search for "rotate Vector" on the web.
//this helps: http://www.kirupa.com/forum/archive/index.php/t-12181.html
BottomLeftCorner += new Vector2(X, Y); //add the origin so that we have to world position.
Do this for all corners of all rectangles, then just loop over all corners and calculate the distance (just abs(v1 - v2)).
I hope this helps you
I just wrote the code for that in n-dimensions. I couldn't find a general solution easily.
// considering a rectangle object that contains two points (min and max)
double distance(const rectangle& a, const rectangle& b) const {
// whatever type you are using for points
point_type closest_point;
for (size_t i = 0; i < b.dimensions(); ++i) {
closest_point[i] = b.min[i] > a.min[i] ? a.max[i] : a.min[i];
}
// use usual euclidian distance here
return distance(a, closest_point);
}
For calculating the distance between a rectangle and a point you can:
double distance(const rectangle& a, const point_type& p) const {
double dist = 0.0;
for (size_t i = 0; i < dimensions(); ++i) {
double di = std::max(std::max(a.min[i] - p[i], p[i] - a.max[i]), 0.0);
dist += di * di;
}
return sqrt(dist);
}
If you want to rotate one of the rectangles, you need to rotate the coordinate system.
If you want to rotate both rectangles, you can rotate the coordinate system for rectangle a. Then we have to change this line:
closest_point[i] = b.min[i] > a.min[i] ? a.max[i] : a.min[i];
because this considers there is only one candidate as the closest vertex in b. You have to change it to check the distance to all vertexes in b. It's always one of the vertexes.
See: https://i.stack.imgur.com/EKJmr.png
My approach to solving the problem:
Combine the two rectangles into one large rectangle
Subtract from the large rectangle the first rectangle and the second
rectangle
What is left after the subtraction is a rectangle between the two
rectangles, the diagonal of this rectangle is the distance between
the two rectangles.
Here is an example in C#
public static double GetRectDistance(this System.Drawing.Rectangle rect1, System.Drawing.Rectangle rect2)
{
if (rect1.IntersectsWith(rect2))
{
return 0;
}
var rectUnion = System.Drawing.Rectangle.Union(rect1, rect2);
rectUnion.Width -= rect1.Width + rect2.Width;
rectUnion.Width = Math.Max(0, rectUnion.Width);
rectUnion.Height -= rect1.Height + rect2.Height;
rectUnion.Height = Math.Max(0, rectUnion.Height);
return rectUnion.Diagonal();
}
public static double Diagonal(this System.Drawing.Rectangle rect)
{
return Math.Sqrt(rect.Height * rect.Height + rect.Width * rect.Width);
}
Please check this for Java, it has the constraint all rectangles are parallel, it returns 0 for all intersecting rectangles:
public static double findClosest(Rectangle rec1, Rectangle rec2) {
double x1, x2, y1, y2;
double w, h;
if (rec1.x > rec2.x) {
x1 = rec2.x; w = rec2.width; x2 = rec1.x;
} else {
x1 = rec1.x; w = rec1.width; x2 = rec2.x;
}
if (rec1.y > rec2.y) {
y1 = rec2.y; h = rec2.height; y2 = rec1.y;
} else {
y1 = rec1.y; h = rec1.height; y2 = rec2.y;
}
double a = Math.max(0, x2 - x1 - w);
double b = Math.max(0, y2 - y1 - h);
return Math.sqrt(a*a+b*b);
}
Another solution, which calculates a number of points on the rectangle and choses the pair with the smallest distance.
Pros: works for all polygons.
Cons: a little bit less accurate and slower.
import numpy as np
import math
POINTS_PER_LINE = 100
# get points on polygon outer lines
# format of polygons: ((x1, y1), (x2, y2), ...)
def get_points_on_polygon(poly, points_per_line=POINTS_PER_LINE):
all_res = []
for i in range(len(poly)):
a = poly[i]
if i == 0:
b = poly[-1]
else:
b = poly[i-1]
res = list(np.linspace(a, b, points_per_line))
all_res += res
return all_res
# compute minimum distance between two polygons
# format of polygons: ((x1, y1), (x2, y2), ...)
def min_poly_distance(poly1, poly2, points_per_line=POINTS_PER_LINE):
poly1_points = get_points_on_polygon(poly1, points_per_line=points_per_line)
poly2_points = get_points_on_polygon(poly2, points_per_line=points_per_line)
distance = min([math.sqrt((a[0] - b[0])**2 + (a[1] - b[1])**2) for a in poly1_points for b in poly2_points])
# slower
# distance = min([np.linalg.norm(a - b) for a in poly1_points for b in poly2_points])
return distance

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