In ray-tracing technique critical point is to calculate rays which came out from eye $E$ to target $T$ through pixel $P_{ij}$ on viewport. The "viewport" is represented as rectangle divided to square pixels - this rectangle is perpendicular to line which go through points $E$, $C$ (viewport center) and $T$. The ray (red line on image) is represented by point $E$ and unit vector $r_{ij}$ (not shown in picture but it lay on red line) - below is picture which show "geometry" - but what are the formulas to calculate $r_{ij}$?
The given input values are:
- eye position $E$,
- target position $T$,
- field of view $\theta$ (angle, for human eye $\approx 90^\circ$),
- number of square pixels $k$ (horizontal direction) and $m$ (vertical direction).
- we also know vertical $w$ vector usually equal to $w=[wx,wy,wz]=[0,1,0]$ (not shown on picture) which indicate where is up and where is down
The orthogonal vectors $v$ and $b$ (and $t$) on picture are determined by $w$ and $t=T-E$ and maybe will useful in $r_{ij}$ calculations. The $d$ and pixel size is arbitrary and don't change the result because of fixed $\theta$.
Question: How to calculate unit vector $r_{ij}$ knowing input values described above?
IDEA: lets find position of center of each pixel $P_{ij}$ which allows us to easily find ray which starts at $E$ and go thought that pixel. To do it we find first $P_{1m}$ and find others by move on vievports plane.
ASSUMPTION: Below we introduce formulas which includes distance $d$ between eye and viewport however this value will be reduced during ray $r_{ij}$ normalization (so you might as well accept that $d=1$ and remove it from calculations).
PRECALCULATIONS: First we calculate normalized vectors $v_n, b_n$ from picutre (which are parallel to viewport plane and give as direction for shifting)
$$t = T-E, \qquad b = w\times t $$
$$ t_n = \frac{t}{||t||}, \qquad b_n = \frac{b}{||b||}, \qquad v_n = t_n\times b_n \\ $$
notice: $C=E+t_nd$, then we calculate viewport size divided by 2 and including aspect ratio $\frac{m}{k}$
$$g_x=\frac{h_x}{2} =d \tan \frac{\theta}{2}, \qquad g_y =\frac{h_y}{2} = g_x \frac{m}{k}$$
and then we calculate shifting vectors $q_x,q_y$ on viewport $x,y$ direction and viewport left upper pixel
$$ q_x = \frac{2g_x}{k-1}b_n, \qquad q_y = \frac{2g_y}{m-1}v_n, \qquad p_{1m} = t_n d - g_xb_n - g_yv_n$$
CALCULATIONS: notice that $P_{ij} = E + p_{ij}$ and ray $R_{ij} = P_{ij} -E = p_{ij}$ so normalized ray $r_{ij}$ is
$$ p_{ij} = p_{1m} + q_x(i-1) + q_y(j-1)$$ $$ r_{ij} = \frac{p_{ij}}{||p_{ij}||} $$
TEST: above formulas wast tested here (works in browser)
SUMMARY: The above form is convenient to use it in shaders where in shader kernel we perform only final calculation based on prcarculated $q_x,q_y$ and $p_{1m}$. Wiki here.