Planar object Homography calculation and Pose Estimation

Question

I'm trying to implement my own Homography and Pose estimation in OpenCV. Suppose I have a square image as a model which I want to locate in input frame from camera. My question is about how to prepare model data to calculate Homography.

I did the following procedure:

1-I extracted 3 corresponding set in both images.

 Input features:       Model Features:
p1_Input(278,278)---> p1_model(137,273)
p2_Input(317,235)---> p2_model(176,230)
p3_Input(217,204)---> p3_model(76,199)  //all in pixel

2- Solving P3P problem:

a) normalizing input points using camera Intrinsic parameters:

u.x=(p1_Input.x - cx) / fx 
u.y=(p1_Input.y - cy) / fy  
u.z=1

v.x=(p2_Input.x - cx) / fx 
v.y=(p2_Input.y - cy) / fy  
v.z=1

w.x=(p3_Input.x - cx) / fx 
w.y=(p3_Input.y - cy) / fy  
w.z=1

b) normalizing the lenght in order to have a unit vector:

d = sqrt(u.x*u.x + u.y*u.y + 1);
u.x=u.x/d;
u.y=u.y/d;
u.z=u.z/d;  // and same for v and w

c)Finding distance between u and camera focus.(same for v and w) by solving p3p and storing in a,b and c

d)computing 3D coordinates:

A_Input.x=a*u.x;
A_Input.y=a*u.y;
A_Input.z=a*u.z;  // same for B_input and C_Input

like:
A_Input:(-0.0899342 ,0.0570672 ,0.976046)
B_Input:(-0.0197703 ,-0.0194311 ,0.955101) 
C_Input:(-0.197233 ,-0.0746457 ,0.967379)

3-Computing Homography

My question is arising here. How should I prepare and modify p1_model,p2_model and p3_model to be prepared for Homography computing?

Obviously A_Input is 3D vector with normalized date while p1_model is 2D vector in pixel.

After solving the problem the rest would be as follows:

a)finding centeroid point for both set.

b)finding H using this formula dot prodoct

4- finding rigid transform using H and SVD

more detail is available here and here

you need 4 points to uniquely define a perspective transformation, are you making an affine approximation? — Hammer
@Hammer No I used 3points to determine rigid transform rather perspective transform — batista cori
I am still a bit confused, do you want a 2d rigid transform or a 3d one? You can't get a 3d rigid transform unless you have 3d coordinates but it looks like you want to use the rigid transform to find your 3d coordinates (step c). It seems a bit like pulling yourself up by your bootstraps. — Hammer
I'm asking how to prepare 3D coordinate for calculation 6DOF rigid transform. As you can see I assume z=1 for the first set — batista cori

Hammer Hammer · Accepted Answer · 2012-10-22T16:47:48

As far as coordinate systems go, there is no specific "format" that you need to put your points in to get good results from a P3P (or PnP) algorithm, just be consistent. The solution the algorithm is looking for given a correspondence between p1 and p2 is

p2 = [resulting_3x4_transform]*p1;

for each correspondence. It may be advisable for numerical reasons to not have any coordinate values be very large or very small but other than that, feel free to put your points in any coordinate system you like.

Planar object Homography calculation and Pose Estimation

1 Answers