Hi,

I wasn't sure if this should be in geometry or trig, so apologies if I picked the wrong one.

My supervisor hinted that the normal of a plane, as viewed from a camera (using the pinhole model) could be obtained from the yaw rotation ($\displaystyle \psi$) and angle of elevation ($\displaystyle \theta$) as per:

$\displaystyle

\textbf{\underline{n}}' = \left(

\begin{array}{ccc}

\cos^{-1}( \psi )\\

\cos( \theta ) \\

\sin( \theta )

\end{array}

\right) \quad , \quad

\textbf{\underline{n}} = \frac{\textbf{\underline{n}}'}{||\textbf{\underlin e{n}}'||}

$

I've since realised that the $\displaystyle \cos^{-1}( \psi ) $ is wrong for the x component, and believe it to be $\displaystyle cos( 2\,\pi - \psi )$, although I'm not sure and can't prove it as this is only a hunch. Would anyone be able to help me understand and/or correct this?

EDIT:

I've read up on my vector theory and know that the x component should be $\displaystyle \cos( \psi ) $, however this gives me a yaw angle of $\displaystyle 85^\circ$ on an example normal I have, which seems quite unlikely. This could be due to the components being normalised to make the normal vector unit, although I don't know to what extent it would have been normalised. To find my normal, I used a decomposition method from a full camera homography.

Many thanks in advance