Hi,

This seems inherently a simple task, but I'm having real problems with it!

I have a plane, let's call it P, with its direction defined by its normal vectorn, and the camera distance, d, both of which I know:

$\displaystyle P: \textbf{n}.\textbf{x} = d $

Upon P, lie a set of coordinates:

$\displaystyle

C = \left(\begin{array}{ccc}

x_1 & & x_n\\

y_1 & ... & y_n\\

z_1 & & z_n\\

\end{array}\right)

$

I need to rotate these coordinates onto the x-y plane to give a top-down view. I can't move the camera - it's fixed.

My initial attempt was to work out the rotation angles from the normal (x-rotation = phi, y-rotation = theta and z-rotation = psi):

$\displaystyle

\phi = cos^{-1}( n_z )

\psi = sin^{-1}( \frac{-n_x}{sin(\phi}})

$

I could then do:

$\displaystyle

C' = R_x(\phi) . R_z(\psi) . \left(\begin{array}{c}x_i\\y_i\\z_i\end{array}\rig ht)

$

(where (xi,yi,zi) is each coordinate in C).

For some reason, this worked sometimes, but other times the angle was completely wrong!

So I tried something else. When the P is rotated to align with the x-y plane we have:

$\displaystyle

R .

\left(\begin{array}{c}

X\\

Y\\

Z

\end{array}\right)

=

\left(\begin{array}{c}

X'\\

Y'\\

0

\end{array}\right)

$

And we know the combined x-y-z rotation matrix:

For ease of reading, hereafter we have:

$\displaystyle

r_3 = [ a, b, c ], where

\begin{array}{l}

a = \sin(\theta)\sin(\phi) \\

b = -\sin( \theta )\cos( \phi ) \\

c = \cos( \theta )

\end{array}

$

Since we don't know what X' and Y' will be, we only use the bottom row of the rotation

matrix. For three coordinates on the original plane, we then have 3 equations with which

to find phi and theta:

$\displaystyle

\begin{array}{l}

a.x_1 + b.y_1 + c.z_1 = 0 \\

a.x_2 + b.y_2 + c.z_2 = 0 \\

a.x_3 + b.y_3 + c.z_3 = 0 \\

\end{array}

$

Which gives rise to the linear system

$\displaystyle

A = \left(

\begin{array}{ccc}

x_1 & y_1 & z_1 \\

x_2 & y_2 & z_2 \\

x_3 & y_3 & z_3 \\

\end{array}

\right)

$

$\displaystyle

x =

\left(

\begin{array}{c}

a \\

b \\

c \\

\end{array}

\right)

$

$\displaystyle

Ax = \textbf{0}

$

Which we solve using Singular Value Decomposition to get a,b and c (take the last column of the "V" matrix). We can then work out phi and theta:

$\displaystyle

\begin{array}{l}

\theta = cos^{-1}( c )\\

\phi = cos^{-1}( \frac{b}{sin(\theta)} )

\end{array}

$

These values were then subbed into the compound rotation matrix given above, and psi was set to 0 giving R. Finally I multiplied each coordinate by R:

$\displaystyle

C' = R(\phi, \theta, \psi) \,. \, C

$

This again, didn't work. Although I'm convinced that in theory it should. Is anyone able to spot any errors I might have made? Or can anyone contribute a different idea? This has taken me the last week of work so any advice is most gratefully received!!!

Thanks!