i have a leaflet feature rectangle and i need to pragmatically draw another squre rectangle . see the image dotted rectangle is the one i need to calculate
possible solutions
get center of rectangle create new 4 point from center draw a rectangle
i there any other better solutions for this
Distance from square vertices to rectangle center is
d = rect_width / 2 + rect_height / 2
So their coordinates are
rect_center_x + d; rect_center_y
rect_center_x; rect_center_y - d
rect_center_x - d; rect_center_y
rect_center_x; rect_center_y + d
Related
I am trying to draw a UIBezierPath in swift where i can get it to go from the center of a circle to another circles center but what i really want is for the line to stop at the edge of the end circle but at the point where the line would have intersected the circles edge if the line was continuating to the center of the circle and not stopping at the edge.
I need somehow to calculate the intersection point between the line and the circle. how would i be able to do that when the circles are UIImageViews with width and height of 30 (so radius 15) and i now the centers coordinates of the two circles.
I have found a solution.
From the the set of x and y (center of circles) from now called (x, y) and (x2, y2) a vector can be made that is (x2 - x, y2 - y) where x2 - x i will call a and y2 - y i will call b then the distance of the vector can be calculates as sqrt(a^2 + b^2) where the result of i will call dist.
From this a Unit vector can be made with is (a/dist, b/dist) now all i need to do to get my new coordinates for the arrow is (x,y) as starting coordinates and (x2 + ra/dist, y2 + rb/dist) as the end coordinates where r is the radius of the circle at (x2, y2).
This question already has answers here:
How to straighten a tilted square shape in an image?
(2 answers)
Closed 5 years ago.
I have a cropped image of a card:
The card is a rectangle with rounded corners, is brightly colored, and sits on a relatively dark background.
It is, therefore, easy to differentiate between pixels belonging to the card and pixels belonging to the background.
I want to use MATLAB to rotate the card so its sides are vertical and horizontal (and not diagonal) and create an image of nothing but the straightened card.
I need this to work for any reasonable card angle (say +45 to -45 degrees of initial card rotation).
What would be the best way of doing this?
Thanks!
You can do this by finding the lines made by the edges of the card. The angle of rotation is then the angle between one of the lines and the horizontal (or vertical).
In MATLAB, you can use the Hough line detector to find lines in a binary image.
0. Read the input image
I downloaded your image and renamed it card.png.
A = imread('card.png');
We don't need color information, so convert to grayscale.
I = rgb2gray(A);
1. Detect edges in the image
A simple way is to use the Canny edge detector. Adjust the threshold to reject noise and weak edges.
BW = edge(I, 'canny', 0.5);
Display the detected edges.
figure
imshow(BW)
title('Canny edges')
2. Use the Hough line detector
First, you need to use the Hough transform on the black and white image, with the hough function. Adjust the resolution so that you detect all lines you need later.
[H,T,R] = hough(BW, 'RhoResolution', 2);
Second, find the strongest lines in the image by finding peaks in the Hough transform with houghpeaks.
P = houghpeaks(H, 100); % detect a maximum of 100 lines
Third, detect lines with houghlines.
lines = houghlines(BW, T, R, P);
Display the detected lines to make sure you find at least one along the edge of the card. The white border around the black background in your image makes detecting the right edges a bit more difficult.
figure
imshow(A)
hold on
for k = 1:length(lines)
xy = [lines(k).point1; lines(k).point2];
plot(xy(:,1), xy(:,2), 'LineWidth', 2, 'Color', 'red');
end
title('Detected lines')
3. Calculate the angle of rotation
lines(3) is the left vertical edge of the card. lines(3).point2 is the end of the line that is at the bottom. We want this point to stay where it is, but we want to vector along the line to be aligned with the vector v = [0 -1]'. (The origin is the top-left corner of the image, x is horizontal to the right and y is vertical down.)
lines(3)
ans =
struct with fields:
point1: [179 50]
point2: [86 455]
theta: 13
rho: 184
Simply calculate the angle between the vector u = lines(3).point1 - lines(3).point2 and the vertical vector v.
u = lines(3).point1 - lines(3).point2; % vector along the vertical left edge.
v = [0 -1]; % vector along the vertical, oriented up.
theta = acos( u*v' / (norm(u) * norm(v)) );
The angle is in radians.
4. Rotate
The imrotate function lets you rotate an image by specifying an angle in degrees. You could also use imwarp with a rotation transform.
B = imrotate(A, theta * 180 / pi);
Display the rotated image.
figure
imshow(B)
title('Rotated image')
Then you would have to crop it.
I have background subtracted images as input. The idea is to reduce search areas for person detection by using a smaller search area for the HOG algorithm. The output required is a bounding box around the person and the pixel positions of the box corners.
This is the input image:
This is the required output:
This is what I have tried so far:
x=imread('frame 0080.png');
y=im2bw(x);
s=regionprops(y);
imshow(y);
hold on
for i=1:numel(s)
rectangle('Position',s(i).BoundingBox,'edgecolor','y')
end
This was the output I got:
It looks like you have tried what I have suggested. However, you want the bounding box that encapsulates the entire object. This can easily be done by using the BoundingBox property, then calculating each of the four corners of each rectangle. You can then calculate the minimum spanning bounding box that encapsulates all of the rectangles which ultimately encapsulates the entire object.
I do notice that there is a thin white strip at the bottom of your image, and that will mess up the bounding box calculations. As such, I'm going to cut the last 10 rows of the image before we proceed calculating the minimum spanning bounding box. To calculate the minimum spanning bounding box, all you have to do is take all of the corners for all of the rectangles, then calculate the minimum and maximum x co-ordinates and minimum and maximum y co-ordinates. These will correspond to the top left of the minimum spanning bounding box and the bottom right of the minimum spanning bounding box.
When taking a look at the BoundingBox property using regionprops, each bounding box outputs a 4 element vector:
[x y w h]
x,y denote the top left co-ordinate of your bounding box. x would be the column and y would be the row of the top left corner. w,h denote the width and the height of the bounding box. We would use this and compute top left, top right, bottom left and bottom right of every rectangle that is detected. Once you complete this, stack all of these rectangle co-ordinates into a single 2D matrix, then calculate the minimum and maximum x and y co-ordinates. To calculate the rectangle, simply use the minimum x and y co-ordinates as the top left corner, then calculate the width and height by subtracting the maximum and minimum x and y co-ordinates respectively.
Without further ado, here's the code. Note that I want to extract all of the bounding box co-ordinates in a N x 4 matrix where N denotes the number of bounding boxes that are detected. You would have to use reshape to do this correctly:
% //Read in the image from StackOverflow
x=imread('http://i.stack.imgur.com/mRWId.png');
% //Threshold and remove last 10 rows
y=im2bw(x);
y = y(1:end-10,:);
% //Calculate all bounding boxes
s=regionprops(y, 'BoundingBox');
%// Obtain all of the bounding box co-ordinates
bboxCoords = reshape([s.BoundingBox], 4, []).';
% // Calculate top left corner
topLeftCoords = bboxCoords(:,1:2);
% // Calculate top right corner
topRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) topLeftCoords(:,2)];
% // Calculate bottom left corner
bottomLeftCoords = [topLeftCoords(:,1) topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculate bottom right corner
bottomRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) ...
topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculating the minimum and maximum X and Y values
finalCoords = [topLeftCoords; topRightCoords; bottomLeftCoords; bottomRightCoords];
minX = min(finalCoords(:,1));
maxX = max(finalCoords(:,1));
minY = min(finalCoords(:,2));
maxY = max(finalCoords(:,2));
% Draw the rectangle on the screen
width = (maxX - minX + 1);
height = (maxY - minY + 1);
rect = [minX minY width height];
% // Show the image
imshow(y);
hold on;
rectangle('Position', rect, 'EdgeColor', 'yellow');
This is the image I get:
I have background subtracted images as input. The idea is to reduce search areas for person detection by using a smaller search area for the HOG algorithm. The output required is a bounding box around the person and the pixel positions of the box corners.
This is the input image:
This is the required output:
This is what I have tried so far:
x=imread('frame 0080.png');
y=im2bw(x);
s=regionprops(y);
imshow(y);
hold on
for i=1:numel(s)
rectangle('Position',s(i).BoundingBox,'edgecolor','y')
end
This was the output I got:
It looks like you have tried what I have suggested. However, you want the bounding box that encapsulates the entire object. This can easily be done by using the BoundingBox property, then calculating each of the four corners of each rectangle. You can then calculate the minimum spanning bounding box that encapsulates all of the rectangles which ultimately encapsulates the entire object.
I do notice that there is a thin white strip at the bottom of your image, and that will mess up the bounding box calculations. As such, I'm going to cut the last 10 rows of the image before we proceed calculating the minimum spanning bounding box. To calculate the minimum spanning bounding box, all you have to do is take all of the corners for all of the rectangles, then calculate the minimum and maximum x co-ordinates and minimum and maximum y co-ordinates. These will correspond to the top left of the minimum spanning bounding box and the bottom right of the minimum spanning bounding box.
When taking a look at the BoundingBox property using regionprops, each bounding box outputs a 4 element vector:
[x y w h]
x,y denote the top left co-ordinate of your bounding box. x would be the column and y would be the row of the top left corner. w,h denote the width and the height of the bounding box. We would use this and compute top left, top right, bottom left and bottom right of every rectangle that is detected. Once you complete this, stack all of these rectangle co-ordinates into a single 2D matrix, then calculate the minimum and maximum x and y co-ordinates. To calculate the rectangle, simply use the minimum x and y co-ordinates as the top left corner, then calculate the width and height by subtracting the maximum and minimum x and y co-ordinates respectively.
Without further ado, here's the code. Note that I want to extract all of the bounding box co-ordinates in a N x 4 matrix where N denotes the number of bounding boxes that are detected. You would have to use reshape to do this correctly:
% //Read in the image from StackOverflow
x=imread('http://i.stack.imgur.com/mRWId.png');
% //Threshold and remove last 10 rows
y=im2bw(x);
y = y(1:end-10,:);
% //Calculate all bounding boxes
s=regionprops(y, 'BoundingBox');
%// Obtain all of the bounding box co-ordinates
bboxCoords = reshape([s.BoundingBox], 4, []).';
% // Calculate top left corner
topLeftCoords = bboxCoords(:,1:2);
% // Calculate top right corner
topRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) topLeftCoords(:,2)];
% // Calculate bottom left corner
bottomLeftCoords = [topLeftCoords(:,1) topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculate bottom right corner
bottomRightCoords = [topLeftCoords(:,1) + bboxCoords(:,3) ...
topLeftCoords(:,2) + bboxCoords(:,4)];
% // Calculating the minimum and maximum X and Y values
finalCoords = [topLeftCoords; topRightCoords; bottomLeftCoords; bottomRightCoords];
minX = min(finalCoords(:,1));
maxX = max(finalCoords(:,1));
minY = min(finalCoords(:,2));
maxY = max(finalCoords(:,2));
% Draw the rectangle on the screen
width = (maxX - minX + 1);
height = (maxY - minY + 1);
rect = [minX minY width height];
% // Show the image
imshow(y);
hold on;
rectangle('Position', rect, 'EdgeColor', 'yellow');
This is the image I get:
iPhone SDK and Objective-C
Goal:
I'm trying to calculate the 'x' and 'y' coordinates of 2 circles. I have the inner circle dimensions and want to calculate what the 'x' and 'y' coordinates of the larger outer circle circumference would be to match the same width (distance) along the edge of the larger circle as it does with the inner circle.
In the end, I just need to figure out what the edge x/y points would be for the large circles edge. So that it matches the same as the inner smaller circle. If the width is 10 high on the inner circle, I need to know the x/y points to make it 10 high to the larger circle. To make a rectangle that will extend. Perpendicular lines.
Example:
I'm using the following to calculate the first 2 sets of x/y for the arc on the inner circle to plot points:
- (CGPoint)coordinatePoints:(CGFloat)radius angleDegrees:(CGFloat)degrees xAxis:(CGFloat)x yAxis:(CGFloat)y {
CGFloat pointX = (CGFloat) ((radius * cos((degrees * M_PI) / 180.0f)) + x);
CGFloat pointY = (CGFloat) ((radius * sin((degrees * M_PI) / 180.0f)) + y);
CGPoint points = CGPointMake(pointX, pointY);
return points;
}
I call it for the first 2 positions on the inner circle. I need to figure out how to make it have the distance on the outer circle as well.
CGPoint innerPoints1 = [self coordinatePoints:innerRadius angleDegrees:startingPoint xAxis:x yAxis:y];
CGPoint innerPoints2 = [self coordinatePoints:innerRadius angleDegrees:endingPoint xAxis:x yAxis:y];
If the inner circle radius is 200, and the outer circle radius is 500, I want it to still be the same thickness from the inner circle to the larger outer circle when I plot the points.
// I have these calculated.
CGContextMoveToPoint(context, innerPoints1.x, innerPoints1.y);
CGContextAddLineToPoint(context, innerPoints2.x, innerPoints2.y);
// I need to find the solution for making innerPoints3 and innerPoints4 correctly.
CGContextAddLineToPoint(context, innerPoints3.x, innerPoints3.y);
CGContextAddLineToPoint(context, innerPoints4.x, innerPoints4.y);
I have the coordinates for the inner circle lines for spaced out x/y points. I need to find the proper way to get the same width plotted for the larger circle locations. Circle sizes will always change. Lengths of the lines will be dynamic. As I'm trying to create a polygon, I need to find 2 coordinates on the larger circles, for each segment.
Any help with this would be greatly appreciated.
Information graphics: a comprehensive illustrated reference
Page 74: In the section "Circular Column Graph", my end goal is to be able to produce the same result as displayed in the 3 images.
If the spokes are not too thick, then the arc length is a good approximation of the spoke width:
So first you construct your 2 inner points, with 2 angles (a1 and a2) centered around a main spoke angle (a).
Then you calculate the distance D between these points (or you approximate it by R1*(a2-a1))
Then you take the points on the outer circle with angle values centered around the same main spoke angle: a-0.5*D/R2 and a+0.5*D/R2. These points will be D apart (measured on the arc)