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There is a code on Matlab Fileexchange that is relevant to your problem: http://www.mathworks.com/matlabcentral/fileexchange/28155-inscribedrectangle/content/html/Inscribed_Rectangle_demo.html Update I wrote this tutorial article on computing largest inscribed rectangles based on the above link from Atul Ingle. The algorithm first searches for largest ...


8

There are various ways to measure distance and 3D using a single camera. These include: Moving the camera and taking another image - this simulates a 2 camera system. If you are imaging an object of a known size, then you can know the distance with some simple trigonometry. You can project a known pattern from a known projector location to reconstruct 3D. ...


3

I guess this is what Olli implies: Using a lens distortion matrix if necessary, the center of the ball in the picture gives you an elevation and azimuth of your ball relative to the focal point of your camera. The size of the ball image gives you distance. That's a point in 3D space. From there, it's really only a manner of grammar school level math to ...


2

Take a look on Generalized/constrained Procrustes Problems. It should be sufficient to update singular values in $\Sigma$. The $\det(SVD)=-1$ case is discussed in more detail in 1987 paper of Arun, Huang and Blostein: "Least-squares fitting of two 3-D point sets".


2

I guess this is a straightforward non-linear optimization problem (to be solved with Newton variations, such as Trust-Region methods), where you don't even need to compute the Jacobian analytically. It appears to me that the optimization problem is written over $K_i$, and thus is the input to the cost function. To compute the cost, at each call to this ...


2

So if you continue the substitution you get: $R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} + X_t - X_t)^2 + R_B^2)} $ $R_t = \sqrt{((-R_BK_X/\sqrt{(K_R^2 - K_X^2)} )^2 + R_B^2)} $ $R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2)} $ $R_t = \sqrt{(R_B^2K_X^2/(K_R^2 - K_X^2) + R_B^2 (K_R^2 - K_X^2)/(K_R^2 -K_X^2))}$ $R_t = \sqrt{(R_B^2K_R^2/(K_R^2 - K_X^2)...


1

In real world implementations, is the line: for theta = [theta_min to theta_max ] required? . . . Since we apply Canny edge detector to an image, prior to Hough Transform, we can obtain angle of gradient(θ) for a point (x, y) from Sobel operators. This can be noted down against (x, y) and later used in Hough Transform algorithm, eliminating one ...


1

Bear with me, as this is really only tangential¹ to my area of expertise, but: Maybe the question is not how to extend wavelets to work on non-orientable surfaces, but to interpret said surface as something where wavelets natively work well. Enter differential geometry (differential topology? I don't know.): As you've notice: if you only consider a part ...


1

Conceptual partial answer: You can use 3 mics to find position in 3D space, except you can't tell if it's above the plane or below the plane (unless you already know there's nothing above the plane, of course). Since your 4 microphones are in the same plane, the 4th microphone is redundant and doesn't help you know whether the source is above or below the ...


1

I am not completely confident with the full theory of wavelet frames, however I will share some pointers and my beliefs, do not take any of the following for a truth. The property you show is (partly) a characterization of an orthogonal system. In Hernandez & Weiss, A first course on wavelets, 1996, you find Theorem 2.12: If $\psi\in L^2(\mathbb{R})$ ...


1

Assumptions From your question, I assume you already have an algorithm that does produce the transformation you want, but you have to play with the parameters manually. Proposed solution First, convert the images to grayscale (this is mainly for convenience). You might play around a bit with the channels of your images - do you need all channels, or do ...


1

The suggested technique seems to be to gather points of interest, calculate a feature vector for them, and then find matching pairs of points in both images, followed by a calculation of the transform. Here is a sample project: http://www.codeproject.com/Articles/95453/Automatic-Image-Stitching-with-Accord-NET Here is a paper that claims improvement by ...


1

My translation should have described a rotation around the center of the cube, but it was wrong. I changed it to: rotate45 * (translation - center) + center and I get the cube in the middle with the right angle.


1

There are two different things you could be talking a out: delay and reverb. I think what you are looking for is reverb. Two ways to do reverb are multitap reverb and convolution reverb. Multitap is faster but convolution is higher quality. Here are some links with more info and example c++ code: http://blog.demofox.org/2015/03/17/diy-synth-multitap-...


1

A very simple model is a sum of attenuated and delayed replicas of the signal. Such signal can be be generated using FIR filter. The length of the filter is the "echo time", i.e. how long the echoes last. The simplest approach would be to generate sparse impulse response. Build a filter with $n$ non-zero coefficient and fading amplitude (for example the ...


1

Yes it is possible to compute the extrinsics given the intrisics, some points in 3D and their projections in the image. If all your 3D points are in the same plane, then the math for computing the extrinsics is explained in the paper by Zhengyou Zhang, which is the basis for the camera calibration code in OpenCV. If your 3D points are not co-planar, then ...


1

This will somewhat depend on the spectrum and properties of the signals and what your dominant source of noise, distortion, jitter, non-linearietes etc. are. Ideally you can resolve every point in space with 3 non-coplanar microphones. A very simple solution is one with 4 microphones: one in the origin and three one unit step displaced in x, y, and z ...


1

Stereo vision is the best way to get the depth map. However if you want to find coordinates of the end point and not the actual depth, monocular vision is possible. This is done by having knowledge of the environment or by using monocular cues like 'depth by defocus', variation in hue, saturation, obstruction in the image and so on. This might not provide ...


1

Based on the comment: ya i can move camera and scene is fixed You can take one snapshot, move the camera sideways (or orbit the scene) and take another snapshot. This becomes your stereo pair. This won't work if the scene isn't fixed. This will work to a great extent if the camera is moving much faster than the scene. This will also work somewhat (with ...


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