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What are the vertical and horizontal focal lengths?

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Problem Detail: 

In this paper,

the author mentioned "the vertical and horizontal focal lengths" but didn't have any clear definition.

My understanding about "focal length" is the distance between the lens and the image plane.

Does anyone have the idea about these focal lengths?

Thank you.

Asked By : willSapgreen
Answered By : masterxilo

Note that they assume a pinhole camera model. The term 'focal length' means something different here than it does with a lens camera.

All you ever really need to know about the pinhole camera model can be expressed as: $$x_\mbox{imageplane} := x_\mbox{world}/z_\mbox{world}$$ $$y_\mbox{imageplane} := y_\mbox{world}/z_\mbox{world}$$ This describes how points in the scene project to an image plane 1 unit in front of the camera origin (center of projection & center of world coordinate system), with the optical axis along the z direction.

All the other values usually discussed in connection with pinhole cameras 'just' describe which rectangular part of the image plane is mapped to the rectangle of pixels that is your image. Or, conversely, how pixel coordinates can be transformed back into points on the image plane (and from there, can be transformed to directions in the scene).

Of course this rectangle-on-a-2d-plane can be defined by a 2d point $b$ and two 2d vectors $v_1, v_2$ such that:

$$\begin{bmatrix}x_\mbox{image}\\y_\mbox{image}\end{bmatrix} := \begin{bmatrix}\vec{v_1}&\vec{v_2}\end{bmatrix} \begin{bmatrix}x_\mbox{imageplane}\\y_\mbox{imageplane}\end{bmatrix} + b$$ If we abbreviate this as $(x_\mbox{image},y_\mbox{image}) = f(x_\mbox{imageplane},y_\mbox{imageplane})$ the part of the imageplane that your image depicts is the set $f^{-1}([0,\mbox{width}]\times[0,\mbox{height}])$ which, as I said, is a rectangle1.

The 'other values', for example:

  • optical center: $c_x, c_y$ or $c_u, c_v$
  • horizontal and vertical 'focal length': $f_x, f_y$ or $f_u, f_v$
  • aspect ratio
  • field of view angle $\alpha$
  • shear (you don't usually want that)

are just used to describe special cases of the above formula.

In your case, $b = (c_u, c_v)$ and $v_1 = (f_u,0), v_2 = (0,f_v)$ such that

$$\begin{bmatrix}x_\mbox{image}\\y_\mbox{image}\end{bmatrix} := \begin{bmatrix}f_u x_\mbox{imageplane} + c_u\\f_v y_\mbox{imageplane} + c_v\end{bmatrix} $$

1 Note that it is not the rectangle with corners $b, b+v_1,b+v_2,b+v_1+v_2$. To find it you'd have to compute the inverse of the function $f$.

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