look at the picture above, see a black circle.
black circle Coordinates is lat(126.897453), lon(37.530028)
if the red rectangle is square(vertical and horizontal are 20m), I want to know the blue circle of coordinates
please let me know calculation formula.
In advance, thanks for your answer!
have a good time :)
This task is solved by first transforming the spherical lat/long coordinates to a cartesian x,y coordinates with unit meters.
Then you calculate the location with very basic addition. (x = x-20, y=y-20/2)
Then you transform the location back to lat/long coordinates.
Related
I have 2 input images of a plane where the (static) camera is at an unknown angle. I managed to extract edges and points of interests using opencv. But I'm stuck calculating real angles from the images.
From image #1 I need to calculate the camera angle relative to the plane. I know 3 points on the plane that form a equilateral triangle (angles of 60 degree). The center point of the triangle is also the centerpoint of the plane. However the plane center point on the image is covered by another object.
From image #2 I need to calculate the real angle of an object (Point C) on the plane to one of the 3 points and the plane center point (= line A to B).
How can I calculate the real angle β as if the camera had no angle towards the plane?
Update:
I was looking for a solution for my problem at https://docs.opencv.org/3.4/d9/d0c/group__calib3d.html
There is a number of functions but I couldn't figure out how to apply them to my specific problem.
There is a function to calculate Homography using two images with keypoints but I do not have images of the scene from different camera angles.
Then there is cv::findHomography which Finds a perspective transformation between two planes. I know 4 source points but what are my 4 destination points?
Another one I was looking at is cv::solvePnP and cv::solvePnPRansac but again I only know 4 source points on the plane. I don't know about their 3D correspondence point.
What am I missing?
#Micka: Thanks for your input. I have 4 points for processing the image (the 3 static base points + the object at point C). I can assume these points are all located on the plane at z=0. However I do not have coordinates for a second plane neither the (x,y) of the corresponding 3D points.
Your description does not explicitly say it, but if you can assume that segment AB bisects the base of the triangle, then you have 4 point correspondences between the plane and its image, so you can use cv::findHomography.
I am writing a program. I have, say, a grid of dots on a piece of paper. I fix one end and bend the paper toward the screen, giving me a trapezoidal shape from the camera's point of view. I have the (x,y) camera coordinate of each dot. Is there a simple way I can change these (x,y) to real life (x,y) which should give me a rectangle? I have the camera/real (x,y) of the original flat sheet of paper pre-bend if that helps.
I have looked at 3D Camera coordinates to world coordinates (change of basis?) and Transforming screen coordinates from security camera to real world coordinates.
Look up "homography". The transformation from a plane in 3D space to its image as captured by an ideal pinhole camera is a homography. It can be represented as a 3x3 matrix H that transforms the 3D coordinates X of points in the world to their corresponding homogeneous image coordinates x:
x = H * X
where X is a 3x1 vector of the world point coordinates, and x = [u, v, w]^T is the image point in homogeneous coordinates.
Given a minimum of 4 matches between world and image points (e.g. the corners of a rectangle) you can estimate the parameters of the matrix H. For details, look up "DLT algorithm". In OpenCV the routine to use is findHomography.
I have a database(PostgreSQL) with coordinates of lightning bolts.
I want coordinates inside a circumference sector(SQL Query). I have the radius, angle(Z) and point A of circumference, and coordinates of extremal points([lat1,lng1], [lat2, lng2]).
Thanks a lot!
How do i calculate point D for triangle 2?
I have the the following coordinates for triangle 1:
a(0,0) b(0,78) c(18,39)
point D is located at (0,39) in triangle 1.
now I change the perspective on my triangle by for example moving coordinate b and c.
the new triangle formed is called triangle 2 with coordinates:
a(0,0) b(11,72) c(37,42)
AS YOU CAN SEE POINT D IS NOT IN THE MIDDLE OF LINE a<-->b BECAUSE OF THE CHANGE IN PERSPECTIVE/SKEW.
How do i calculate point d? I have the coordinates abc of triangle 1 & 2.
Preferably answer in programcode rather than using math signs, since i am not a hero at reading math :)
You need to convert point D to barycentric coordinates using the original triangle coordinates, then convert it back to cartesian coordinates using the modified triangle coordinates.
This looks like a good introduction to triangular barycentric coordinates: http://blogs.msdn.com/b/rezanour/archive/2011/08/07/barycentric-coordinates-and-point-in-triangle-tests.aspx
Also, explicit formulae for converting a point in a triangle to barycentric coordinates are given at the end of the Converting to Barycentric Coordinates section of the Wikipedia article “Barycentric coordinate system”.
I guess there are more ways of calculating a coordinate from one perspective to another.
more on the triangle way is written by culebrón here: Transforming captured co-ordinates into screen co-ordinates
At the same link there is another way by using SVD and calculate an H-matrix which can be used to translate any coordinate from one perspective to another. I am going to use this way because i could solve this way in matlab. Next step in objective-c! i had some trouble calculating the same in objective-c. more on that here: calculate the V from A = USVt in objective-C with SVD from LAPACK in xcode
I would like to know how to solve the triangle way too! i could not figure out what a1 and a2 were in culebron's post: https://stackoverflow.com/a/1690300/1568532 neither the width and height made much sense to me.
Also i would like to know how to calculate the EYE's point of view on a triangle or quadrangle based on 3 or 4 coordinates. if you know the original size of the object.
any ideas on this?
when i search for eye or camera's point of view. there is load of result about photography.
what do i need to use in order to calculate this? maybe some example anyone?
Working with iPhone and Objective C.
I am working on a game and I need to correctly reflect a ball off a circle object. I am trying to do it as a line and circle intersection. I have my ball position outside the circle and I have the new ball position that would be inside the circle at the next draw update. I know the intersect point of the line (ball path) and the circle. Now I want to rotate the ending point of the ball path about the intersection point to get the correct angle of reflection off the tangent.
The following are known:
ball current x,y
ball end x,y
ball radius
circle center x,y
circle radius
intersection point of ball path and circle x and y
I know I need to find the angle of incidence between the tangent line and the incoming ball path which will also equal my angle of reflection. I think once I know those two angles I can subtract them from 180 to get my rotation angle then rotate my end point about the angle of intersection by that amount. I just don't know how.
First, you should note that the center of the ball doesn't have to be inside of the circle to indicate that there's a reflection or bounce. As long as the distance between ball center and circle is less than the radius of the ball, there will be a bounce.
If the radius of the circle is R and the radius of the ball is r, things are simplified if you convert to the case where the circle has radius R+r and the ball has radius 0. For the purposes of collision detection and reflection/bouncing, this is equivalent.
If you have the point of intersection between the (enlarged) circle and the ball's path, you can easily compute the normal N to the circle at that point (it is the unit vector in the direction from the center of the circle to the collision point).
For an incoming vector V the reflected vector is V-2(N⋅V) N, where (N⋅V) is the dot product. For this problem, the incoming vector V is the vector from the intersection point to the point inside the circle.
As for the reflection formula given above, it is relatively easy to derive using vector math, but you can also Google search terms like "calculate reflection vector". The signs in the formula will vary with the assumed directions of V and N. Mathworld has a derivation although, as noted, the signs are different.
I only know the solution to the geometry part.
Let:
r1 => Radius of ball
r2 => Radius of circle
You can calculate the distance between the two circles by using Pythagoras theorem.
If the distance is less than the r1+r2 then do the collision.
For the collision,I would refer you Here. It's in python but I think it should give you an idea of what to do. Hopefully, even implement it in Objective C. The Tutorial By PeterCollingRidge.