# Tag Info

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Note: That depends on what coordinates you use in the resized image. I am assuming that you are using zero-based system (like C, unlike Matlab) and 0 is transformed to 0. Also, I am assuming that you have no skew between coordinates. If you do have a skew, it should be multiplied as well Short answer: Assuming that you are using a coordinate system in ...

9

You have to take images for calibration from different points of view and angles, with as big difference between angles as possible (all three Euler angles should vary), but so that pattern diameter was still fitting to camera field of view. The more views are you using the better calibration will be. That is needed because during the calibration you detect ...

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Andrey mentioned that his solution assumes 0 is transformed to 0. If you are using pixel coordinates this is likely not true when you re-size the image. The only assumption you really need to make is that your image transformation can be represented by a 3x3 matrix (as Andrey demonstrated). To update your camera matrix you can just premultiply it by the ...

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In general case, no, you cannot approximate color of an object from digital image with good accuracy. Human vision system What you see is light reflected from the surface of the object. First the object needs to be illuminated with light. This light has some sort of spectrum (probability density function of photon energy). When the light hits the object, ...

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No, the data in the USB cable is digital. Namely there is either an error in the data or the data is the same at any place along the cable.

5

The physical definition of frequency is the number of oscillations per second and measured in Hertz (Hz) or more general: 1/(duration of one period of a periodic event in seconds). In your case the periodic event is a frame and it occurs 120 times per second so the frequency is 120 Hz. If you want to find out the sampling frequency of the ADC device in the ...

5

Film isn't absolutely "analog", as in continuous. Every individual silver halide molecule, after exposure and development, is either metalized or not; and there are a finite number of these molecules in every frame of film, thus quantizing the exposure measurement. However the density and location of the film grains and silver halide molecules is semi-...

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If you have the extrinsics then it is very easy. Having extrinsics is the same as having "camera pose" and the same as having the homography. Check this post in stackoverflow. You have extrinsics, also called camera pose, which is described as a translation and a rotation: \displaystyle Pose =\begin{bmatrix}R|t \end{bmatrix} = \begin{bmatrix}R_{11} &... 4 Here is a list of 'best practices' for camera calibration which I originally posted here: https://calib.io/blogs/knowledge-base/calibration-best-practices Choose the right size calibration target. Large enough to properly constrain parameters. Preferably it should cover approx. half of the total area when seen fronto-parallel in the camera images. Perform ... 4 I decided to post this answer here because a while back, this came up as the top result in Google and its suggestions helped me. So I decided to share my experience too. Having spent countless hours trying to get the best stereo calibration on a Kinect, I shared my tips and findings in a blog post here. Although it is geared towards stereo calibration and ... 4 If you know the exact response of the camera, you can convert the brightness samples of each pixel to a linear intensity scale and perform the averaging there. That will make your whole problem intensity linear and should solve all your problems. However, I would strongly recommend using a more advanced exposure algorithm. For example you could introduce a ... 3 No, it is not a good idea. Typically, you want to limit the volume of space relative to the camera in which you want to do your measurements, and you want to use a single calibration target appropriate for that volume. In theory, you can calibrate with any number of different patterns. In practice, however, most implementations assume that the pattern is ... 3 The dataset appears to be in the public domain, and has been posted at this site here. The pictures (24 in total) were originally released in 1991 on a CD. The original media with the actual image data could be requested via interlibrary loan; here is a WorldCat link to the repositories holding this item. The CD was originally distributed by Apple ... 3 The most you can do is to capture a color table with reference colors (e.g. GretagMacbeth) and then calibrate the device (e.g. your camera) using software that generates an ICC profile to it. This allows you to represent RGB colors from camera sensor in device-independent model (CIE XYZ). When you have calibrated monitor, you can map colors between devices ... 3 Generally Bayer mask (also known as CFA = color filter array) imagers are the most popular. 3Chip cameras cost considerably more - you are not only paying for 3x as many chips but the alignment in manufacturing is expensive and the range of cameras is much more limited. The largest are HD (1920x1080) although a lot of cameras actually have much smaller ... 3 @Ben - number of views depends the camera and the final accuracy required. With very high quality, low distortion lenses (high-end 35mm SLR) using lots of chessboard images to map the distortions can be unstable - since the distortions are fractions of a pixel. You still need several shots with the board (or camera) rotated since the image centre is ... 2 You have two options, use back projection or projection between two planes (homography). With back projection you take a pseudo inverse of you camera matrixP$and multiply the result with your homogenous presentation of image point: $$P = K\begin{bmatrix}R & -R\textbf{C}\end{bmatrix} \\ \textbf{X}_{reprojected} = P^+\textbf{x}$$ Now you have a ... 2 You can't know the 3d position of the second point. It can be any of the points on the ray from your center of the camera until infinity. You can do the following: Create a predefined 3d space which resembles the real life scene Get more points of images from a different angle, using the intersection of the rays from different angles, you can get an ... 2 If X are known 3D points, then the translation is given simply by: $$T=\frac{\sum_{i=1}^{n}{X'_{i}-RX_{i}}}{n}$$ In case of points being in projected coordinates, then more work have to be done. You can express the 3D points in homogenous coordinates: $$X=(x, y, z, w)$$ where$w$is 1 by default. Then both the translation and rotation can be joined via ... 2 You should photograph a known calibration target with your system, (like this one or the one that I put in the thumbnail). This will allow you discovering the OECF of your system. You can either calculate it by yourself, given the technical data sheet or use third-party software like Imatest, ImageAnalyzer or others. 2 Yes, the center pixel and focal length in pixels will change, as described in the link above. However, if you learn distortion parameters (radial and tangential) then they shouldn't change as resolution changes because they operate on the projective image plane (before multiplying by camera matrix) instead of pixel coordinates (after multiplying by camera ... 2 Here you have two options. 1) Color calibrate your cameras (radiometric calibration) to the same reference. Then try matching. There are many academic works on this topic, one of them being: http://research.microsoft.com/en-us/um/people/yasumat/papers/rankcalib_PAMI12_preprint.pdf 2) Use a vignetting invariant (or color invariant) feature descriptor. For ... 2 Shadow has very specific properties that makes it very clear way of making it distinguishable from the regular object. A lot of work in the area of background subtraction and surveillance has been using this to eliminate the shadows or to avoid them being mistaken as the actual object (or human). As observed by Daniel Grest To distinguish the shadows ... 2 You are correct: to calibrate a camera you need a correspondence between 3D world points and 2D image points. The problem is that the 3D points cannot be co-planar, so people were building 3D calibration rigs, e. g. a box made of checkerboards. One image of a rig like that would be enough to calibrate, but those rigs are hard to build, because you have to ... 2 Try reducing exposure time if your camera allows that. With the short exposure time the conveyor belt should be well-lit. If that is not enough, a common solution is to use a flash which is synchronized to the camera. If a sync signal is not readily available from the camera and you can't trigger the camera with your own sync signal, you probably could find ... 2 What could cause such a shape? Could it be due to the laser source which could not be perfectly single-mode? Or could it be due to the sensor? There is nothing particularly wrong with the "shape", but there are a few things you can do on the sensor and data processing side, to improve the extraction of an accurate profile. Your "biggest" problem is ... 2 This is a long topic to fully explain. I will try to write shortly, so please excuse the brevity. Standard computer vision projection (ignoring distortion like Houdini) follows: $$\mathbf{x} = \lambda \mathbf{K}[\mathbf{R}\mathbf{X} +\mathbf{t} ]$$$\mathbf{R}$is a$3x3$orthogonal matrix,$\mathbf{t}$is a$3x1$translation vector. Camera position$\...

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This is due to the optimization problem being rather high-dimensional (around 11 parameters). With only a single observation of the calibration board, there would be multiple possible combinations of parameters explaining the observed feature point locations (unless a very constrained camera-model is used). Only a sufficient number of sufficiently ...

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If I understand you correctly, what you are asking for is called camera response function (CRF) and in general it is nonlinear and depends on camera device. For fixed device denote its CRF by $f$. Then $f$ maps the set of possible scene irradiances (or illuminances) $\mathcal{I}$ at a given spatial location to the set of possible pixel intensity values \$\...

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The characteristics of the shadow are as follows: It is always dark regardless to the colour of the object or the coloor of the light used to make a shadow It only shows dark outline of the object It is formed in the opposite direction to the source of light The size of the shadow depends upon the distance between the source of light and opaque object The ...

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