CODEWITHSUNDEEP
actionscript-3flashmath
tzl
tzl

Reputation: 1590

get skew or rotation value from affine transformation matrix

I'm trying to get the skew values out of a transformation matrix in a flash movie clip. The transformation matrix is represented by 

a b tx
c d ty
0 0 1

I have no information on what kind of transformation is performed and which comes first. I do know that in flash, you may only rotate OR skew a movie clip (correct me if I am wrong). I can get scale values from scaleX and scaleY properties of the movie clip. I believe translation does not quite matter i can just equate tx and ty to zero.

so my question has 2 parts. How do I determine if a skew or a rotation had been applied, and how do I get the respective values?

Upvotes: 5

Views: 9939

Answers (4)

Simon Epskamp
Simon Epskamp

Reputation: 9976

The term for this is matrix decomposition. Here is a solution that includes skew as described by Frédéric Wang.

Works when transforms are applied in this order: skew, scale, rotate, translate.

function decompose_2d_matrix(mat) {
  var a = mat[0];
  var b = mat[1];
  var c = mat[2];
  var d = mat[3];
  var e = mat[4];
  var f = mat[5];

  var delta = a * d - b * c;

  let result = {
    translation: [e, f],
    rotation: 0,
    scale: [0, 0],
    skew: [0, 0],
  };

  // Apply the QR-like decomposition.
  if (a != 0 || b != 0) {
    var r = Math.sqrt(a * a + b * b);
    result.rotation = b > 0 ? Math.acos(a / r) : -Math.acos(a / r);
    result.scale = [r, delta / r];
    result.skew = [Math.atan((a * c + b * d) / (r * r)), 0];
  } else if (c != 0 || d != 0) {
    var s = Math.sqrt(c * c + d * d);
    result.rotation =
      Math.PI / 2 - (d > 0 ? Math.acos(-c / s) : -Math.acos(c / s));
    result.scale = [delta / s, s];
    result.skew = [0, Math.atan((a * c + b * d) / (s * s))];
  } else {
    // a = b = c = d = 0
  }

  return result;
}

Upvotes: 5

Clément Barthes
Clément Barthes

Reputation: 41

You need to do a polar decomposition. This Wikipedia article explains how it works: http://en.wikipedia.org/wiki/Polar_decomposition Here is the code I wrote for my own program using the OpenCV library.

const double PI = 3.141592653;
    cv::Mat rotationOutput = cv::Mat::zeros(warp00.size(),CV_64F);
    cv::Mat_<double>::iterator rotIter = rotationOutput.begin<double>();
    cv::Mat_<double>::iterator warp00Iter = warp00.begin<double>();
    cv::Mat_<double>::iterator warp01Iter = warp01.begin<double>();
    cv::Mat_<double>::iterator warp10Iter = warp10.begin<double>();
    cv::Mat_<double>::iterator warp11Iter = warp11.begin<double>();

    for(; warp00Iter != warp00.end<double>(); ++warp00Iter, ++warp01Iter, ++warp10Iter,
        ++warp11Iter, ++rotIter){
        cv::Matx22d fMatrix(*warp00Iter,*warp01Iter, *warp10Iter, *warp11Iter);
        cv::Matx22d cMatrix;
        cv::Matx22d cMatSqrt(0.,0.,0.,0.);
        cv::mulTransposed(fMatrix, cMatrix, true);
        cv::Matx21d eigenVals;
        cv::Matx22d eigenVecs;
        if((cMatrix(0,0) !=0.) && (cMatrix(1,1) !=0.)){
            if(cv::eigen(cMatrix,true,eigenVals,eigenVecs)){
                cMatSqrt = eigenVecs.t()*
                        cv::Matx22d(sqrt(eigenVals(0,0)),0.,0.,sqrt(eigenVals(1,0)))*eigenVecs;
            }
        }
        cv::Matx22d rMat = fMatrix*cMatSqrt.inv();
        *rotIter = atan(rMat(1,0)/rMat(0,0));
        
    }

warp00, warp01, warp10 and warp11 contains the first 4 params of the affine transform (translation params warp02 and warp12 are not needed). IN your case it would be a,b,c,d. You'll notice in the wikipedia article that you need to compute the square root of a matrix. The only way to do so is by computing the eigen values, then compute their square roots and rotate the diagonal matrix back to the original coordinate system. It's complicated, but it is the only way to compute the rotations when you have an affine transform. In my case, I only cared about the rotations, so my code won't give you the skew.

Upvotes: 2

beaker
beaker

Reputation: 16801

The 2D rotation matrix is

cos(theta) -sin(theta)
sin(theta)  cos(theta)

so if you have no scaling or shear applied,

   a = d
and
   c = -b
and the angle of rotation is
   theta = asin(c) = acos(a)

If you've got scaling applied and can recover the scaling factors sx and sy, just divide the first row by sx and the second by sy in your original transformation matrix and then recover the rotation angle as above.

If you've got a shear (skew) applied anywhere in there, I'm with the previous commenters, it might not be possible except in very limited cases (such as shear in only one known direction at a time and in a known order).

Upvotes: 3

Vesper
Vesper

Reputation: 18747

First, you can do both skew and rotate, but you have to select the order first. A skew matrix is explained here, to add a skew matrix to a transformation you create a new matrix and do yourTransformMatrix.concat(skewMatrix);

I can't currently say if you can retrieve values for transformation in terms of "rotation angle", "skew_X angle", "skew_Y angle", "translation_X","translation_Y", this in general is a nonlinear equation system which might not have a solution for a specific matrix.

Upvotes: 0

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