I have an star image like this
and a list of the center of the detected stars which looks like this
Name;RA(deg);Dec(deg);PixelX;PixelY;RPmag;Gmag;BPmag;Parallax;Flags
;313.311502;44.387289;2516;1458;4.850;4.728;4.681;5.8088;000800f0
;313.327406;45.181692;236;2256;4.517;5.190;5.715;3.6700;000800f0
;314.580976;44.472083;3131;4138;4.660;5.304;5.779;14.4950;00800f00
;314.471585;44.788107;2143;4206;4.358;5.673;7.615;1.9323;000800f0
; ...
I want to detect the contour/size of the star at the x,y coordinates of each entry and then write a white dot in the size of the star to a new image.
Has anyone an idea how to use the contour detection of opencv/cv2 on given coordinates?
Related
When you make a line profile of all x-values or all y-values the extraction from each pixel is clear. But when you take a line profile along a diagonal, how does DM choose which pixels to use in the one dimensional readout?
Not really a scripting question, but I'm rather certain that it uses bi-linear interpolation between the grid-points along the drawn line. (And if perpendicular integration is enabled, it does so in an integral.) It's the same interpolation you would get for a "rotate" image.
In fact, you can think of it as a rotate-image (bi-linearly interpolated) with a 'cut-out' afterwards, potentially summed/projected onto the new X-axis.
Here is an example
Assume we have a 5 x 4 image, which gives the grid as shown below.
I'm drawing top-left corners to indicate the coordinates system pixel convention used in DigitalMicrgraph, where
(x/y)=(0/0) is the top-left corner of the image
Now extract a LineProfile from (1/1) to (4/3). I have highlighted the pixels for those coordinates.
Note, that a Line drawn from the corners seems to be shifted by half-a-pixel from what feels 'natural', but that is the consequence of the top-left-corner convention. I think, this is why a LineProfile-Marker is shown shifted compared to f.e. LineAnnotations.
In general, this top-left corner convention makes schematics with 'pixels' seem counter-intuitive. It is easier to think of the image simply as grid with values in points at the given coordinates than as square pixels.
Now the maths.
The exact profile has a length of:
As we can only have profiles with integer channels, we actually extract a LineProfile of length = 4, i.e we round up.
The angle of the profile is given by the arc-tangent of dX and dY.
So to extract the profile, we 'rotate' the grid by that angle - done by bilinear interpolation - and then extract the profile as grid of size 4 x 1:
This means the 'values' in the profile are from the four points:
Which are each bi-linearly interpolated values from four closest points of the original image:
In case the LineProfile is averaged over a certain width W, you do the same thing but:
extract a 2D grid of size L x W centered symmetrically over the line.i.e. the grid is shifted by (W-1)/2 perpendicular to the profile direction.
sum the values along W
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'm currently implementing a mesh extrusion algorithm for plane shapes, let's assume for a rectangle.
When I extrude this rectangle I create four new sides (resulting in 8 new triangles) and a new bottom for the 3d shape.
This works fine when I duplicate all vertices so that my final cube has 24 of them. But I'd like to avoid these extra vertices now so that I have only 8 vertices. Unfortunately, in this case I do not know how to calculate the UV coordinates and I keep getting wrong results as shown in the image below.
The correct result would look like this (with duplicated faces):
My first question is: Is it possible to generate a good uv map with just 8 vertices (and hence 8 uv coordinates) for a cube?
Second is: How? :)
Thanks for your help.
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 to implement a depth image base rendering. Given a 2D image and a depth map, the algorithm will generate a virtual view - what the scene would look like if a camera was placed in a different position. I wrote this function, V is the matrix with the pixel of 2d view, D the pixels from depth map and camera shift a parameter.
Z=1.1-D./255; is a normalization. I try to follow this instruction:
For each pixel in the depth map, compute the disparity that results from the depth, For each pixel in the source 2D image, find a new location for it in the virtual view: old location + disparity of that specific pixel.
The function doesnt work very well, what's wrong?
function[virtualView]=renderViews(V,D,camerashift)
Z=1.1-D./255;
[M,N]=size(Z);
for m=1:M
for n=1:N
d=camerashift/Z(m,n);
shift=round(abs(d));
V2(m,n)=V(m+shift,n);
end
end
imshow(V2)