math - Kinect intrinsic parameters from field of view

Question

Microsoft state that the field of view angles for the Kinect are 43 degrees vertical and 57 horizontal (stated here) . Given these, can we calculate the intrinsic parameters i.e. focal point and centre of projection? I assume centre of projection can be given as (0,0,0)?

Thanks

EDIT: some more information on what I'm trying to do

I have a dataset of images recorded with a Kinect, I am trying to convert pixel positions (x_screen,y_screen and z_world (in mm)) to real world coordinates.

If I know the camera is placed at point (x',y',z') in the real world coordinate system, is it sufficient to find the real world coordinates by doing the following:

x_world = (x_screen - c_x) * z_world / f_x
y_world = (y_screen - c_y) * z_world / f_y

where c_x = x' and c_y = y' and f_x, f_y is the focal length? And also how can I find the focal length given just knowledge of the field of view?

Thanks

Answer 1

If you equate the world origin (0,0,0) with the camera focus (center of projection as you call it) and you assume the camera is pointing along the positive z-axis, then the situation looks like this in the plane x=0:

Here the axes are z (horizontal) and y (vertical). The subscript v is for "viewport" or screen, and w is for world.

If I get your meaning correctly, you know h, the screen height in pixels. Also, zw, yv and xv. You want to know yw and xw. Note this calculation has (0,0) in the center of the viewport. Adjust appropriately for the usual screen coordinate system with (0,0) in the upper left corner. Apply a little trig:

tan(43/2) = (h/2) / f = h / (2f),  so f = h / ( 2 tan(43/2) )

and similar triangles

yw / zw = yv / f            also              xw / zw = xv / f

Solve:

yw = zw * yv / f            and                xw = zw * xv / f

Note this assumes the "focal length" of the camera is equal in the x-direction. It doesn't have to be. For best accuracy in xw, you should recalculate with f = w / 2 tan(57/2) where w is the screen width. This is because f isn't a true focal length. It's just a constant of conversion. If the pixels of the camera are square and optics have no aberrations, these two f calculations will give the same result.

NB: In a deleted (improper) article the OP seemed to say that it isn't zw that's known but the length D of the hypotenuse: origin to (xw,yw,zw). In this case just note zw = D * f / sqrt(xv2 + yv2 + f2) (assuming camera pixels are square; some scaling is necessary if not). They you can proceed as above.