Camera Calibration | Camera Calibration

00:07:46
https://www.youtube.com/watch?v=GUbWsXU1mac

الملخص

TLDRLe modèle camera linéaire est étayé par une matrice de projection qui est calibrée à l'aide d'un cube ayant des dimensions et motifs connus. Dès qu'un repère est placé sur un coin du cube, les coordonnées 3D sont établies. Des correspondances sont ensuite faites entre les coordonnées 3D et les points d'image 2D via la matrice de projection. Le problème d'optimisation qui en résulte doit être minimisé sous une contrainte de normalisation. La solution finale pour la matrice de projection se trouve en utilisant le vecteur propre associé à la plus petite valeur propre d'une matrice construite à partir des équations correspondantes.

الوجبات الجاهزة

  • 📷 La calibration de caméra utilise une matrice de projection.
  • 🔲 Un cube est utilisé comme référence pour la géométrie connue.
  • 🌐 Les correspondent entre points 3D et 2D sont essentielles.
  • ✏️ Une solution d'optimisation sous contrainte est appliquée.
  • ⚙️ La matrice de projection peut être normalisée à une échelle arbitraire.
  • 🔍 Les équations sont formées en utilisant des matrices.
  • 🧮 La recherche de la matrice de projection se résume à un problème de valeurs propres.
  • 📏 Le vecteur de projection est essentiel pour le calibrage.
  • 🔗 L'approche de moindres carrés contraints est appliquée ici.
  • 🎯 La normalisation est cruciale pour obtenir une solution viable.

الجدول الزمني

  • 00:00:00 - 00:07:46

    Nous avons maintenant un modèle de caméra linéaire décrit par une matrice de projection. Pour calibrer la caméra, nous utilisons un objet de géométrie connue, comme un cube dont nous connaissons les dimensions et la position de chaque point. En plaçant un repère de coordonnées du monde à un des coins du cube, nous prenons une image du cube et établissons des correspondances entre les coordonnées 3D des points sur le cube et leurs coordonnées 2D sur l'image. Ces correspondances peuvent être obtenues manuellement ou automatiquement. Cela nous permet de créer un ensemble d'équations qui relient les points 3D à leurs coordonnées image, et nous cherchons à estimer la matrice de projection qui reste inconnue. La matrice de projection ne peut être déterminée qu'à un facteur d'échelle près, ce qui signifie que si le monde et la caméra sont mis à l'échelle, l'image résultante reste inchangée. Nous allons chercher à minimiser une fonction de perte qui peut être reformulée en un problème d'autovalues, où l'autovecteur correspondant à la plus petite autovaleur donnera la matrice de projection souhaitée.

الخريطة الذهنية

فيديو أسئلة وأجوبة

  • Quel est l'objectif de la calibration de caméra?

    L'objectif est d'estimer la matrice de projection.

  • Pourquoi utiliser un cube pour la calibration?

    Un cube a des dimensions et des motifs visuels connus, ce qui permet de faciliter la correspondance entre les points 3D et 2D.

  • Quelle est la contrainte appliquée lors de la minimisation?

    La contrainte est que le carré de la norme du vecteur de projection soit égal à 1.

  • Comment se manifeste la matrice de projection?

    Elle se manifeste dans des équations reliant les coordonnées d’image 2D et les coordonnées du monde 3D.

  • Qu'est-ce que le 'problème des moindres carrés contraints'?

    C'est un problème d'optimisation où l'on cherche à minimiser une fonction de coût tout en respectant des contraintes imposées.

  • Quel est le lien entre la matrice de projection et les coordonnées homogènes?

    La matrice de projection est suffisante pour transformer les coordonnées homogènes d'un point 3D en coordonnées 2D dans l'image.

  • Comment trouver le vecteur de projection optimal?

    En résolvant le problème des valeurs propres associé à une matrice formée à partir des équations de correspondance.

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الترجمات
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التمرير التلقائي:
  • 00:00:03
    Now we have a linear camera model, which is given by the projection matrix.
  • 00:00:08
    And so we are ready to calibrate the camera, which essentially means estimated the projection matrix.
  • 00:00:15
    We do this by using an object of known geometry, such as this one shown right here.
  • 00:00:21
    This is a cube. We happen to know everything about this cube in terms of its dimensions.
  • 00:00:27
    It also has a visual pattern on it.
  • 00:00:32
    And we know the location of every point in this on this cube.
  • 00:00:38
    Every visual feature in 3-D. So what we do is we essentially place our world coordinate frame.
  • 00:00:46
    And one of the corners of this cube, what that means is that once we had done would say 3D reconstruction,
  • 00:00:53
    if that's our application, then all the points are going to be measured with respect to this coordinate frame right here.
  • 00:01:00
    So we have a world coordinate frame right here. And now we take a single image of this cube, which looks like this.
  • 00:01:08
    So if you consider a single point here, at this point, we know its world coordinates, X,
  • 00:01:15
    W, Y, W, Z, W, let's say that happens to be zero three and four inches in the world
  • 00:01:22
    coordinate frame. And we know it's image, location U v, which is two dimensional, of course U v.
  • 00:01:30
    And that let's say is happens to be fifty six and one hundred and fifteen pixels.
  • 00:01:35
    So we have this correspondence now. Now we can do the same for all other visual features here.
  • 00:01:43
    And we can establish the correspondence which is shown in color right here.
  • 00:01:48
    This can be either done manually by clicking on points in this image or one can do it by using an algorithm.
  • 00:01:55
    Since this pattern is fairly simple, one can imagine writing an algorithm for automatically establishing the correspondence.
  • 00:02:02
    But irrespective of what you end up with is a set of corresponding points 3D to 2D.
  • 00:02:10
    If you take one of these points, you get through our camera model, linear camera model.
  • 00:02:17
    You get a mapping from the point in 3D the world coordinate frame to the image coordinates right here, 2D using our projection matrix.
  • 00:02:29
    This is what we're interested in finding. So you and X W are known, but P is unknown.
  • 00:02:36
    That's what we are trying to estimate.
  • 00:02:39
    Now, if you take this expression and you essentially write it out, then you get these two equations, one for you and one for V.
  • 00:02:47
    The subscript eye corresponds to the point we are looking at so many.
  • 00:02:52
    You go from one through end in terms of corresponding points.
  • 00:02:58
    So for each one of these points, we get an equation, a pair of equations that looks like this.
  • 00:03:04
    So we can rewrite all of those equations in matrix form as follows.
  • 00:03:09
    But you have a single matrix where everything is known to us.
  • 00:03:14
    We call that Matrix A and you write in your projection matrix in terms of a vector here with 12 elements.
  • 00:03:24
    And that's what's unknown. So we have this simple form, which is a B equal to zero.
  • 00:03:31
    Now, before we saw four P, let's take a look at an important property of P, which has to do with its scale.
  • 00:03:38
    So we know that if you express an image point in homogeneous coordinates as such and if you simply scale it with some factor K not equal to zero.
  • 00:03:49
    These two are equivalent to each other, which means they end up giving you the same image coordinates U.
  • 00:03:55
    V. So that means that X.
  • 00:04:00
    W. Times P, which is equal to this is equivalent to X.
  • 00:04:05
    W. Times B multiplied by some scale. Factor K.
  • 00:04:10
    That means that the projection matrices P and K times P produce exactly the same homogeneous pixel coordinates.
  • 00:04:19
    In other words, the projection matrix is only defined up to a scale factor.
  • 00:04:26
    So what does this really mean? This means that if you have some world and a camera and you take an image and now
  • 00:04:34
    let's say that you double the size of the world in terms of the dimensions of objects,
  • 00:04:39
    but also the distances between them.
  • 00:04:42
    And you double the size of the camera, which means the effect of focal length, the pics of the size and everything.
  • 00:04:48
    And you take another image. These two images are going to be identical to one another.
  • 00:04:54
    That been scaling the projection matrix implies simultaneously scaling the world and the camera,
  • 00:05:02
    which does not change the image, which goes to say that we can set the scale of the projection matrix arbitrarily.
  • 00:05:10
    You can choose any scale you wish for the key that we were just talking about.
  • 00:05:14
    So there's multiple ways to do this. One is we can just choose one of the elements of P the twelve elements and simply said that equal to one.
  • 00:05:24
    That would be one way to do it. Or we could just say.
  • 00:05:29
    That the square of the magnitude of the vector B is equal to one.
  • 00:05:32
    And this is what we choose because it's convenient for us in the way we want to formulate the problem.
  • 00:05:38
    So we say P squared is equal to one. So what do we have now? We want to ap equal to zero and P squared equal to one.
  • 00:05:48
    And so we can say that we want to find the P, the vector P that minimizes AP squared such that P squared is equal to one.
  • 00:05:58
    And you can see here that this is exactly the problem is solved in the case of finding the Hamar graphy matrix during IMiDs stitching.
  • 00:06:08
    It's exactly the same problem. It's called the constrained least squares problem.
  • 00:06:12
    So we use exactly the same approach. We can express AP squared as P transpose a transpose AP and P squared s b transpose B.
  • 00:06:24
    So we can now define a lost function, which is B, transpose a transpose a P minus lambda.
  • 00:06:34
    Multiply it times P. Transpose B minus one. That's your loss function.
  • 00:06:39
    And we want to minimize this loss function. In other words, we want to find the P that minimizes N.
  • 00:06:46
    And how do we do that. We find the derivative of that respect to P and set that equal to zero.
  • 00:06:52
    And you get this simple expression right here.
  • 00:06:55
    And it turns out then that finding the P that minimizes L is equivalent to solving the well-known eigenvalue problem.
  • 00:07:05
    That's what you're seeing right here.
  • 00:07:07
    In other words, the P that we're looking for is the eigenvector corresponding to the smallest eigenvalue of a transpose.
  • 00:07:19
    We simply take a transpose and find its eigenvalues and eigenvectors and then eigenvector that corresponds to the smallest eigenvalue of a transpose.
  • 00:07:29
    A is the projection vector. B, that V that minimizes the loss function.
  • 00:07:36
    So once we have that, we can rearrange the elements of P to get our projection matrix B.
الوسوم
  • calibration
  • camera
  • projection matrix
  • 3D reconstruction
  • correspondence
  • linear camera model
  • least squares
  • eigenvalue problem
  • known geometry
  • cube