Determining if a point lies within an ellipse, including the edge
Date : March 29 2020, 07:55 AM
may help you . You have to decide what kind of tolerance your method will use. While your example uses points that are expressible in floating point, there are many points along the border of the ellipse which will not be, and so deciding whether a point is "on the border" isn't clearcut. If you don't much care, then I would suggest making the ellipse slightly "bigger" than you actually want and using the builtin contains() method. If you want to write your own method, it's as simple as taking the formula for an ellipse, plugging in the X and Y values of the point you wish to test, and observing the result: bool isInsideOfOrOnBorderOfEllipse = ((x*x)/(a*a) + (y*y)/(b*b)) <= 1;

how to determine the rotation angle of ellipse and fill area between points in opencv, c++
Tag : cpp , By : hlpimfalling
Date : March 29 2020, 07:55 AM
Any of those help After short check  it seems that OpenCV function calculates middle angle as ma = (angle1 + angle2) / 2 and draws arc through this point. Both (45,45) and (45,45) give the same 90degrees arc through zero, both (315,45) and (45,315) give the same 270degrees arc.

Points of Intersection of an Ellipse and a line after rotating ellipse by angle theta
Tag : python , By : Novi Indrayani
Date : March 29 2020, 07:55 AM
it fixes the issue If you have tested and readytouse solution for axisaligned ellipse, it is much simpler to transform points defining line into ellipse system, find intersection points, then make reverse transformation. For origincentered ellipse and rotation angle theta x1' = x1 * Cos(theta) + y1 * Sin(theta)
y1' =  x1 * Sin(theta) + y1 * Cos(theta)
x1' = (x1  cx) * Cos(theta) + (y1  cy) * Sin(theta)
y1' =  (x1  cx) * Sin(theta) + (y1  cy) * Cos(theta)

Given the midpoints of ellipse, its orientation, minor and major axis length, get n points on the ellipse
Date : March 29 2020, 07:55 AM
this one helps. You can use OpenCV to draw an ellipse. cv2.ellipse(img,(256,256),(100,50),0,0,360,255,1)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
contours, _ = cv2.findContours(gray,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
epsilon = 0.01*cv2.arcLength(contours[0],True)
green_dots = cv2.approxPolyDP(contours[0],epsilon,True)

Can I determine, using only a subset of polygon points, where the interior of the polygon lies?
Date : March 29 2020, 07:55 AM
should help you out The usual approach would be to project a line horizontally left from your given point and find the nearest edge that that line intersects. The direction of that edge tells you whether the point is inside or outside the polygon (with a path direction imposed as per Paul's suggestion). If no edge is encountered, you are, of course, outside. To achieve this, you need each cell of your binary tree to reference every edge that passes through the cell, even if its vertices are outside the cell. You may sometimes need to traverse through multiple cells to find the nearest intersecting edge.

