extract rotation, scale values from 2d transformation matrix
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]
matrix.rotate(45 / 180 * PI)
matrix.scale(3, 4)
matrix.translate(50, 100)
matrix.rotate(30 / 180 * PI)
matrix.scale(-2, 4)
now my matrix have values [a, b, c, d, tx, ty]
lets forget about the process开发者_如何学Goes above and imagine that we have only the values a, b, c, d, tx, ty
how can i find total rotation and scale values via a, b, c, d, tx, ty
sorry for my english
Thanks your advance
EDIT
I think it should be an answer somewhere...
i just tried in Flash Builder (AS3) like this
var m:Matrix = new Matrix;
m.rotate(.25 * Math.PI);
m.scale(4, 5);
m.translate(100, 50);
m.rotate(.33 * Math.PI);
m.scale(-3, 2.5);
var shape:Shape = new Shape;
shape.transform.matrix = m;
trace(shape.x, shape.y, shape.scaleX, shape.scaleY, shape.rotation);
and the output is:
x = -23.6
y = 278.8
scaleX = 11.627334873920528
scaleY = -13.54222263865791
rotation = 65.56274134518259 (in degrees)
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)
andsy = sqrt(b² + d²)
- The rotation angle is
t = atan(c/d)
ort = atan(-b/a)
as also they should be the same.
Otherwise you don't have a valid rotation matrix.
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.
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!
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;
}
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.
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