Geek asks and answers

How to rotate and scale homography

I need help,

I get the homography from the server, so I want to normalize this homography in my application coordinate system, when I try to represent the object in coordinates, the server application generates the following 4 points:

obtained [96.629539, 217.31934; 97.289948, 167.21941; 145.69249, 168.28044; 145.69638, 219.84604]

and my application generates the following 4 points:

local [126.0098, 55.600437; 262.39163, 53.98035; 259.41382, 195.34763; 121.48138, 184.95235]

I represent these points in a graph, R (received), P (local)

enter image description here

It seems that the generated square rotates and scales, so I would like to know if there is a way to apply this rotation to the scale for server homography in order to have the same homography as my application homography.

, , , .

, , , findhomography, .

homography = findHomography (srcPoints, dstPoints, match_mask, RANSAC, 10);

!!!

+5
2

, . . - "" , - "" .

enter image description here

OpenCV getAffineTransform , 3 ( ). numpy :

r = array([[97.289948, 167.21941], [96.629539, 217.31934], [145.69638, 219.84604]], np.float32)
l = array([[126.0098, 55.600437], [121.48138, 184.95235], [259.41382, 195.34763]], np.float32)
A = cv2.getAffineTransform(r, l)

:

array([[  2.81385763e+00,  -5.32961421e-02,  -1.38838108e+02],
       [  7.88519054e-02,   2.58291747e+00,  -3.83984986e+02]])

r, , l, , :

# split affine warp into rotation, scale, and/or shear + translation matrix
T = mat(A[:, 2]).T
matrix([[-138.83810801],
        [-383.98498637]])

A = mat(A[:, 0:2])
matrix([[ 2.81385763, -0.05329614],
        [ 0.07885191,  2.58291747]])

# apply warp to r to get l
r = mat(r).T
A*r + T
# gives
matrix([[ 126.00980377,  121.48137665,  259.41381836],
        [  55.60043716,  184.9523468 ,  195.34762573]])
# which equals
l = mat(l).T
matrix([[ 126.00980377,  121.48137665,  259.41381836],
        [  55.60043716,  184.9523468 ,  195.34762573]], dtype=float32)

, , Markus Jarderot, OpenCV getPerspectiveTransform.

, !

+4

Maple, .

> with(LinearAlgebra);

> # The server coordinates
  pa := [[96.629539, 217.31934], [97.289948, 167.21941], [145.69249, 168.28044],
         [145.69638, 219.84604]]:

> # The local coordiantes
  pb := [[126.0098, 55.600437], [262.39163, 53.98035], [259.41382, 195.34763],
         [121.48138, 184.95235]]:

> # The placeholder variables for the transformation (last one is '1', because it
  # is scale-invariant)
  T := [seq]([seq](`if`(i = 3 and j = 3, 1, t[i, j]), j = 1 .. 3), i = 1 .. 3):
  V := convert(map(op, T)[1 .. -2], set):

> # Transformation function (Matrix multiplication + divide with 3rd coordinate)
  trans := (p, T) -> [
      (T[1, 1]*p[1]+T[1, 2]*p[2]+T[1, 3])/(T[3, 1]*p[1]+T[3, 2]*p[2]+T[3, 3]),
      (T[2, 1]*p[1]+T[2, 2]*p[2]+T[2, 3])/(T[3, 1]*p[1]+T[3, 2]*p[2]+T[3, 3])
  ]:

> # Transform pa, and construct the equation system
  pat := map(trans, pa, T):
  eqs := {op}(zip((p1, p2) -> op(zip(`=`, p1, p2)), pat, pb)):

> # Solve for the transform variables
  sol := solve(eqs, V):

> # Populate the transform
  eval(T, sol);

:

[[  .1076044020,   -3.957029830,    1074.517140  ],
 [ 4.795375318,      .3064507355,   -430.7044862 ],
 [ 0.3875626264e-3, 0.3441632491e-2,   1         ]]

, T * <x, y, 1>.

void ServerToLocal(double serverX, double serverY, double *localX, double *localY)
{
    double w;
    w = 0.3875626264e-3 * serverX + 0.3441632491e-2 * serverY + 1.0;
    *localX = (.1076044020 * serverX - 3.957029830 * serverY + 1074.517140) / w;
    *localY = (4.795375318 * serverX + .3064507355 * serverY - 430.7044862) / w;
}

http://alumni.media.mit.edu/~cwren/interpolator/

C, .

+2

More articles:


All Articles