A bit of background
I am writing a simple ray tracer in C++. I have most of the core complete but don't understand how to retrieve the world coordinate of a pixel on the image plane. I need this location so that I can cast the ray into the world.
Currently I have a Camera with a position(aka my perspective reference point), a direction (vector) which is not normalized. The directions length signifies the center of the image plane and which way the camera is facing.
There are other values associated with the camera but they should not be relevant.
My image coordinates will range from -1 to 1 and the perspective(focal length), will change based on the distance of the direction associated with the camera.
What I need help with
I need to go from pixel coordinates (say [0, 256] in an image 256 pixels on each side) to my world coordinates.
I will also want to program this so that no matter where the camera is placed and where it is directed, that I can find the pixel in the world coordinates. (Currently the camera will almost always be centered at the origin and will look down the negative z axis. I would like to program this with the future changes in mind.) It is also important to know if this code should be pushed down into my threaded code as well. Otherwise it will be calculated by the main thread and then the ray will be used in the threaded code.
(source: in.tum.de)
I did not make this image and it is only there to give an idea of what I need.
Please leave comments if you need any additional info. Otherwise I would like a simple theory/code example of what to do.
Basically you have to do the inverse process of V * MVP which transforms the point to unit cube dimensions. Look at the following urls for programming help
http://nehe.gamedev.net/article/using_gluunproject/16013/ https://sites.google.com/site/vamsikrishnav/gluunproject
Related
I made an object tracker that calculates the position of an object recorded in a live camera feed using stereoscopic cameras. The math was simple, once you know the camera distance and orientation. However, now I thought it would be nice to allow me to quickly extract all these parameters, so when I change my setup or cameras I will be able to quickly calibrate it again.
To calculate the object position I made some simplifications/assumptions, which made the math easier: the cameras are in the same YZ plane, so there is only a distance in x between them. Their tilt is also just in the XY plane.
To reverse the triangulation I thought a test pattern (square) of 4 points of which I know the distances to each other would suffice. Ideally I would like to get the cameras' positions (distances to test pattern and each other), their rotation in X (and maybe Y and Z if applicable/possible), as well as their view angle (to translate pixel position to real world distances - that should be a camera constant, but in case I change cameras, it is quite a bit to define accurately)
I started with the same trigonometric calculations, but always miss parameters. I am wondering if there is an existing solution or a solid approach. If I need to add parameter (like distances, they are easy enough to measure), it's no problem (my calculations didn't give me any simple equations with that possibility though).
I also read about Homography in opencv, but it seems it applies to 2D space only, or not?
Any help is appreciated!
I'm trying to infer an object's direction of movement using dense optical flow in OpenCV. I'm using calcOpticalFlowFarneback() to get flow coordinates and cartToPolar() to acquire vector angles which would indicate direction.
To interpret the results I need to know the reference point for measuring the angle. I have found this blog post indicating that the range of angles is 360°. That tells me that the angle measurement would go along the lines of the unit circle. I couldn't make out much more than that.
The documentation for cartToPolar() doesn't cover this and my attempts at testing it have failed.
It seems that the angle produced by cartToPolar() is in reference to the unit circle rotated clockwise by 90° centered on the image coordinate starting point in the top left corner. It would look like this.
I came to this conclusion by using the dense optical flow example provided by OpenCV. I replaced the line hsv[...,0] = ang*180/np.pi/2 with hsv[...,0] = ang*180/np.pi to get correct angle conversion from radians. Then I tested a video with people moving from top right to bottom left and vice versa. I sampled the dominant color with GIMP and got RGB values which I converted to HSV values. Hue value corresponds to the angle in degrees.
People moving from top right to bottom left produced an angle of about 300° and people moving the other way round produced an angle of about 120°. This hinted at the way the unit circle is positioned.
Looking at the code, fastAtan32f is used to compute the angles. and that seems to be a atan2 implementation.
First, this Calculating camera ray direction to 3d world pixel helped me a bit in understanding what the virtual camera setup is like. I don't understand how the vectors work in this setup, and I thought normalized device coordinates had to be used which led me to this page http://www.scratchapixel.com/lessons/3d-basic-lessons/lesson-6-rays-cameras-and-images/building-primary-rays-and-rendering-an-image/. What I am trying to do is build a ray tracer, and as the question states, find out the pixels position in order to shoot out a ray. What I really, really really would like, is an actually example showing a virtual camera setup, screen resolution and how to calculate a pixels position, then transform to world space coordinates. Experts!, Thank you for your help! :D
Multiply a matrix by the coordinates. What matrix? There are lots of choices. For example XNA uses a projection matrix, view matrix and world matrix. Applying all of them transforms pixel coordinates into world coordinates or vice versa. Breaking it down this way helps to understand the different transformations going on so you can more easily construct the matrices.
Isn't this webpage providing you already with 4 pages of explanation on how these rays are built? It seems like you haven't made the effort to read the content of the link you are referring to. I would suggest you read it first, try to understand it, maybe look at the source code they provide and come back with a real question regarding what you potentially don't understand.
It's all there, and I am not going to re-write what these people seem to have put a lot of energy already to explain! (nor should anybody else really ...).
I would like to calculate the distance between my camera and a recognized "object".
The recognized "object" is a black rectangle sticker on a white board for example. I know the values of the rectangle (x,y).
Is there a method that I can use to calculate the distance with the values of my original rectangle, and the values of the picture of the rectangle I took with the camera?
I searched the forum for answeres, but none of the were specified to calculate the distance with these attributes.
I am working on a robot called Nao from Aldebaran Robotics, I am planing to use OpenCV to recognize the black rectangle.
If you could compute the angle taken up by the image of the target, then the distance to the target should be proportional to cot (i.e. 1/tan) of that angle. You should find that the number of pixels in the image corresponded roughly to the angles, but I doubt it is completely linear, especially up close.
The behaviour of your camera lens is likely to affect this measurement, so it will depend on your exact setup.
Why not measure the size of the target at several distances, and plot a scatter graph? You could then fit a curve to the data to get a size->distance function for your particular system. If your camera is close to an "ideal" camera, then you should find this graph looks like cot, and you should be able to find your values of a and b to match dist = a * cot (b * width).
If you try this experiment, why not post the answers here, for others to benefit from?
[Edit: a note about 'ideal' cameras]
For a camera image to look 'realistic' to us, the image should approximate projection onto a plane held infront of the eye (because camera images are viewed by us by holding a planar image in front of our eyes). Imagine holding a sheet of tracing paper up in front of your eye, and sketching the objects silhouette on that paper. The second diagram on this page shows sort of what I mean. You might describe a camera which achieves this as an "ideal" camera.
Of course, in real life, cameras don't work via tracing paper, but with lenses. Very complicated lenses. Have a look at the lens diagram on this page. For various reasons which you could spend a lifetime studying, it is very tricky to create a lens which works exactly like the tracing paper example would work under all conditions. Start with this wiki page and read on if you want to know more.
So you are unlikely to be able to compute an exact relationship between pixel length and distance: you should measure it and fit a curve.
It is a big topic. If you want to proceed from a single image, take a look at this old paper by A. Criminisi. For an in-depth view, read his Ph.D. thesis. Then start playing with the OpenCV routines in the "projective geometry" sectiop.
I have been working on Image/Object Recognition as well. I just released a python programmed android app (ported to android) that recognizes objects, people, cars, books, logos, trees, flowers... anything:) It also shows it's thought process as it "thinks" :)
I've put it out as a test for 99 cents on google play.
Here's the link if you're interested, there's also a video of it in action:
https://play.google.com/store/apps/details?id=com.davecote.androideyes
Enjoy!
:)
I have given an assignment of to project a object in 3D space into a 2D plane using simple graphics in C. The question is that a cube is placed in fixed 3D space and there is camera which is placed in a position whose co-ordinates are x,y,z and the camera is looking at the origin i.e. 0,0,0. Now we have to project the cube vertex into the camera plane.
I am proceeding with the following steps
Step 1: I find the equation of the plane aX+bY+cZ+d=0 which is perpendicular to the line drawn from the camera position to the origin.
Step 2: I find the projection of each vertex of the cube to the plane which is obtained in the above step.
Now I want to map those vertex position which i got by projection in step 2 in the plane aX+bY+cZ+d=0 into my screen plane.
thanks,
I don't think that by letting the z co-ordinate equals zero will lead me to the actual mapping. So any help to figure out this.
You can do that in two simple steps:
Translate the cube's coordinates to the camera's system (using
rotation), such that the camera's own coordinates in that system are x=y=z=0 and the cube's translated z's are > 0.
Project the translated cube's coordinates onto a 2d plain by dividing its x's and y's by their respective z's (you may need to apply a constant scaling factor here for the coordinates to be reasonable for the screen, e.g. not too small and within +/-half the screen's height in pixels). This will create the perspective effect. You can now draw pixels using these divided x's and y's on the screen assuming x=y=0 is the center of it.
This is pretty much how it is done in 3d games. If you use cube vertex coordinates, then you get projections of its sides onto the screen. You may then solid-fill the resultant 2d shapes or texture-map them. But for that you'll have to first figure out which sides are not obscured by others (unless, of course, you use a technique called z-buffering). You don't need that for a simple wire-frame demo, though, just draw straight lines between the projected vertices.