extract rotation, scale values from 2d transformation matrix

Question

how can i extract rotation, scale and translation values from 2d transformation matrix? i mean a have a 2d transformation

matrix = [1, 0, 0, 1, 0, 0]

matri         


        

Answer 1

Not all values of a,b,c,d,tx,ty will yield a valid rotation sequence. I assume the above values are part of a 3x3 homogeneous rotation matrix in 2D

    | a  b  tx |
A = | c  d  ty |
    | 0  0  1  |

which transforms the coordinates [x, y, 1] into:

[x', y', 1] = A * |x|
                  |y|
                  |z|

• Thus set the traslation into [dx, dy]=[tx, ty]
• The scale is sx = sqrt(a² + c²) and sy = sqrt(b² + d²)
• The rotation angle is t = atan(c/d) or t = atan(-b/a) as also they should be the same.

Otherwise you don't have a valid rotation matrix.

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The above transformation is expanded to:

x' = tx + sx (x Cos θ - y Sin θ)
y' = ty + sy (x Sin θ + y Cos θ)

when the order is rotation, followed by scale and then translation.

Answer 2

I ran into this problem today and found the easiest solution to transform a point using the matrix. This way, you can extract the translation first, then rotation and scaling.

This only works if x and y are always scaled the same (uniform scaling).

Given your matrix m which has undergone a series of transforms,

var translate:Point;
var rotate:Number;
var scale:Number;

// extract translation
var p:Point = new Point();
translate = m.transformPoint(p);
m.translate( -translate.x, -translate.y);

// extract (uniform) scale
p.x = 1.0;
p.y = 0.0;
p = m.transformPoint(p);
scale = p.length;

// and rotation
rotate = Math.atan2(p.y, p.x);

There you go!

Answer 3

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

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;
}

Answer 4

If in scaling you'd scaled by the same amount in x and in y, then the determinant of the matrix, i.e. ad-bc, which tells you the area multiplier would tell you the linear change of scale too - it would be the square root of the determinant. atan( b/a ) or better atan2( b,a ) would tell you the total angle you have rotated through.

However, as your scaling isn't uniform, there is usually not going to be a way to condense your series of rotations and scaling to a single rotation followed by a single non-uniform scaling in x and y.