Transforming captured co-ordinates into screen co-ordinates

Question

I think this is probably a simple maths question but I have no idea what\'s going on right now.

I\'m capturing the positions of \"markers\" on a webcam and I have a

Answer 1

More of how to do this in objective-c in xcode, related to jacobs post, you can find here: calculate the V from A = USVt in objective-C with SVD from LAPACK in xcode

Answer 2

Due to perspective effects linear or even bilinear transformations may not be accurate enough. Look at correct perspective mapping and more from google on this phrase, may be this is what you need...

Answer 3

Since your input area isn't a rectangle of the same aspect-ratio as the screen, you'll have to apply some sort of transformation to do the mapping.

What I would do is take the proportions of where the inner point is with respect to the outer sides and map that to the same proportions of the screen.

To do this, calculate the amount of the free space above, below, to the left, and to the right of the inner point and use the ratio to find out where in the screen the point should be.

alt text http://img230.imageshack.us/img230/5301/mapkg.png

Once you have the measurements, place the inner point at:

x = left / (left + right)
y = above / (above + below)

This way, no matter how skewed the webcam frame is, you can still map to the full regular rectangle on the screen.

Answer 4

In order to compute the warping, you need to compute a homography between the four corners of your input rectangle and the screen.

Since your webcam polygon seems to have an arbitrary shape, a full perspective homography can be used to convert it to a rectangle. It's not that complicated, and you can solve it with a mathematical function (should be easily available) known as Singular Value Decomposition or SVD.

Background information:

For planar transformations like this, you can easily describe them with a homography, which is a 3x3 matrix H such that if any point on or in your webcam polygon, say x1 were multiplied by H, i.e. H*x1, we would get a point on the screen (rectangular), i.e. x2.

Now, note that these points are represented by their homogeneous coordinates which is nothing but adding a third coordinate (the reason for which is beyond the scope of this post). So, suppose your coordinates for X1 were, (100,100), then the homogeneous representation would be a column vector x1 = [100;100;1] (where ; represents a new row).

Ok, so now we have 8 homogeneous vectors representing 4 points on the webcam polygon and the 4 corners of your screen - this is all we need to compute a homography.

Computing the homography:

A little math: I'm not going to get into the math, but briefly this is how we solve it:

We know that 3x3 matrix H,

H = 

h11 h12 h13
h21 h22 h23
h31 h32 h33

where hij represents the element in H at the ith row and the jth column

can be used to get the new screen coordinates by x2 = H*x1. Also, the result will be something like x2 = [12;23;0.1] so to get it in the screen coordinates, we normalize it by the third element or X2 = (120,230) which is (12/0.1,23/0.1).

So this means each point in your webcam polygon (WP) can be multiplied by H (and then normalized) to get your screen coordinates (SC), i.e.

SC1 = H*WP1
SC2 = H*WP2
SC3 = H*WP3
SC4 = H*WP4
where SCi refers to the ith point in screen coordinates and 
      WPi means the same for the webcam polygon

Computing H: (the quick and painless explanation)

Pseudocode:

for n = 1 to 4
{
    // WP_n refers to the 4th point in the webcam polygon 
    X = WP_n;

    // SC_n refers to the nth point in the screen coordinates
    // corresponding to the nth point in the webcam polygon

    // For example, WP_1 and SC_1 is the top-left point for the webcam
    // polygon and the screen coordinates respectively.

    x = SC_n(1); y = SC_n(2);

    // A is the matrix which we'll solve to get H
    // A(i,:) is the ith row of A

    // Here we're stacking 2 rows per point correspondence on A
    // X(i) is the ith element of the vector X (the webcam polygon coordinates, e.g. (120,230)
    A(2*n-1,:) = [0 0 0 -X(1) -X(2) -1 y*X(1) y*X(2) y];    
    A(2*n,:)   = [X(1) X(2) 1 0 0 0 -x*X(1) -x*X(2) -x];
}

Once you have A, just compute svd(A) which will give decompose it into U,S,V^T (such that A = USV^T). The vector corresponding to the smallest singular value is H (once you reshape it into a 3x3 matrix).

With H, you can retrieve the "warped" coordinates of your widget marker location by multiplying it with H and normalizing.

Example:

In your particular example if we assume that your screen size is 800x600,

WP =

    98   119   583   569
    86   416    80   409
     1     1     1     1

SC =

     0   799     0   799
     0     0   599   599
     1     1     1     1

where each column corresponds to corresponding points.

Then we get:

H = 
   -0.0155   -1.2525  109.2306
   -0.6854    0.0436   63.4222
    0.0000    0.0001   -0.5692

Again, I'm not going into the math, but if we normalize H by h33, i.e. divide each element in H by -0.5692 in the example above,

H =
    0.0272    2.2004 -191.9061
    1.2042   -0.0766 -111.4258
   -0.0000   -0.0002    1.0000

This gives us a lot of insight into the transformation.

• [-191.9061;-111.4258] defines the translation of your points (in pixels)
• [0.0272 2.2004;1.2042 -0.0766] defines the affine transformation (which is essentially scaling and rotation).
• The last 1.0000 is so because we scaled H by it and
• [-0.0000 -0.0002] denotes the projective transformation of your webcam polygon.

Also, you can check if H is accurate my multiplying SC = H*WP and normalizing each column with its last element:

SC = H*WP    

    0.0000 -413.6395         0 -411.8448
   -0.0000    0.0000 -332.7016 -308.7547
   -0.5580   -0.5177   -0.5554   -0.5155

Dividing each column, by it's last element (e.g. in column 2, -413.6395/-0.5177 and 0/-0.5177):

SC
   -0.0000  799.0000         0  799.0000
    0.0000   -0.0000  599.0000  599.0000
    1.0000    1.0000    1.0000    1.0000

Which is the desired result.

Widget Coordinates:

Now, your widget coordinates can be transformed as well H*[452;318;1], which (after normalizing is (561.4161,440.9433).

So, this is what it would look like after warping:

As you can see, the green + represents the widget point after warping.

Notes:

1. There are some nice pictures in this article explaining homographies.
2. You can play with transformation matrices here

MATLAB Code:

WP =[
    98   119   583   569
    86   416    80   409
     1     1     1     1
     ];

SC =[
     0   799     0   799
     0     0   599   599
     1     1     1     1
     ];    

A = zeros(8,9);  

for i = 1 : 4     
    X = WP(:,i);    
    x = SC(1,i); y = SC(2,i);        
    A(2*i-1,:) = [0 0 0 -X(1) -X(2) -1 y*X(1) y*X(2) y];        
    A(2*i,:)   = [X(1) X(2) 1 0 0 0 -x*X(1) -x*X(2) -x];
end

[U S V] = svd(A);

H = transpose(reshape(V(:,end),[3 3]));
H = H/H(3,3);

A

       0           0           0         -98         -86          -1           0           0           0
      98          86           1           0           0           0           0           0           0
       0           0           0        -119        -416          -1           0           0           0
     119         416           1           0           0           0      -95081     -332384        -799
       0           0           0        -583         -80          -1      349217       47920         599
     583          80           1           0           0           0           0           0           0
       0           0           0        -569        -409          -1      340831      244991         599
     569         409           1           0           0           0     -454631     -326791        -799

Answer 5

Try the following: split the original rectangle and this figure with 2 diagonals. Their crossing is (k, l). You have 4 distorted triangles (ab-cd-kl, cd-ef-kl, ef-gh-kl, gh-ab-kl) and the point xy is in one of them.

(4 triangles are better than 2, since the distortion doesn't depend on the diagonal chosen)

You need to find in which triangle point XY is. To do that you need only 2 checks:

1. Check if it's in ab-cd-ef. If true, go on with ab-cd-ef, (in your case it's not, so we proceed with cd-ef-gh).
2. We don't check cd-ef-gh, but already check a half of it: cd-gh-kl. The point is there. (Otherwise it would have been ef-gh-kl)

Here's an excellent algorythm to check if a point is in a polygon, using only it's points.

Now you need only to map the point to the original triangle cd-gh-kl. The point xy is a linear combination of the 3 points:

x = c * a1 + g * a2 + k * (1 - a1 - a2)
y = d * a1 + h * a2 + l * (1 - a1 - a2)
a1 + a2 <= 1

2 variables (a1, a2) with 2 equations. I guess you can derive the solution formulae on your own.

Then you just make a linear combinations of a1&a2 with the corresponding points' co-ordinates in the original rectangle. In this case with W (width) and H (height) it's

X = width * a1 + width * a2 + width / 2 * (1 - a1 - a2)
Y = 0 * a1 + height * a2 + height / 2 * (1 - a1 - a2)

Answer 6

The "Kabcsh Algorithm" does exactly this: it creates a rotation matrix between two spaces given N matched pairs of positions.

http://en.wikipedia.org/wiki/Kabsch_algorithm