Question about object_localization (single view geometry)

[where2go947]

Thanks for your awesome work. It does help!
But I got confused about a line in code. Could you give me some help? Thanks in advance!

according to the slides (02 single view geometry, 24th):
$Z = \frac{\bar{z}}{\bar{y}} L$
$X = \frac{\bar{x}}{\bar{y}} L$

However, in the code below, $X = \frac{c[0]}{c[2]} * L$ (line 62), which $c[2]$ is actually $\bar{z}$ instead of $\bar{y}$.
So in my opinion, the code is supposed to be: $X = \frac{c[0]}{c[1]} * L$

3dv_tutorial/examples/object_localization.py

Lines 54 to 64 in f295338

	while True:
	img_copy = img.copy()
	if mouse_state['xy_e'][0] > 0 and mouse_state['xy_e'][1] > 0:
	# Calculate object location and height
	c = R.T @ [mouse_state['xy_s'][0] - cx, mouse_state['xy_s'][1] - cy, f]
	h = R.T @ [mouse_state['xy_e'][0] - cx, mouse_state['xy_e'][1] - cy, f]
	if c[1] < 1e-6: # Skip the degenerate case (beyond the horizon)
	continue
	X = c[0] / c[2] * L # Object location X [m]
	Z = c[2] / c[1] * L # Object location Y [m]
	H = (c[1] / c[2] - h[1] / h[2]) * Z # Object height [m]

[where2go947]

By the way, I think equation in slides above $[\bar{x}, \bar{y}, \bar{z}]^T = R^T* [x-c_x, y-c_y, f]^T$ is wrong, which $R^T$ should be $R$. ($R$ is the camera orientation)
The implementation in the CPP file is right: Rc is the camera orientation.

3dv_tutorial/examples/object_localization.cpp

Line 85 in f295338

cv::Point3d c = R.t() * cv::Point3d(drag.xy_s.x - cx, drag.xy_s.y - cy, f);

Actually, R.t()=Rc, which is the camera orientation:

3dv_tutorial/examples/object_localization.cpp

Lines 60 to 63 in f295338

	cv::Matx33d Rc = Rz(cam_ori.z) * Ry(cam_ori.y) * Rx(cam_ori.x);
	cv::Point3d tc = cv::Point3d(0, -L, 0);
	cv::Matx33d R = Rc.t();
	cv::Point3d t = -Rc.t() * tc;

From my perspective, what $R * [x-c_x, y-c_y, f]^T$ does is to represent the point $[x-c_x, y-c_y, f]^T$ under an unrotated coordinate system (its orientation is aligned with the world coordinate system, but the origin keeps unchanged), so that the similar triangles equation $\frac{\bar{x}}{X} = \frac{\bar{y}}{L} = \frac{\bar{z}}{Z}$ would be right.