How to calculate ray












0












$begingroup$


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?



enter image description here










share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is it that you want to know?
    $endgroup$
    – amd
    Jan 16 at 0:48










  • $begingroup$
    @amd I update question - now it is more clear?
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:08










  • $begingroup$
    This is described in almost all tutorials and books on the subject.
    $endgroup$
    – lightxbulb
    Jan 16 at 6:45










  • $begingroup$
    @lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:55










  • $begingroup$
    Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
    $endgroup$
    – lightxbulb
    Jan 16 at 7:29
















0












$begingroup$


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?



enter image description here










share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is it that you want to know?
    $endgroup$
    – amd
    Jan 16 at 0:48










  • $begingroup$
    @amd I update question - now it is more clear?
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:08










  • $begingroup$
    This is described in almost all tutorials and books on the subject.
    $endgroup$
    – lightxbulb
    Jan 16 at 6:45










  • $begingroup$
    @lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:55










  • $begingroup$
    Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
    $endgroup$
    – lightxbulb
    Jan 16 at 7:29














0












0








0





$begingroup$


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?



enter image description here










share|cite|improve this question











$endgroup$




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?



enter image description here







analytic-geometry






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 12:36







Kamil Kiełczewski

















asked Jan 15 at 23:04









Kamil KiełczewskiKamil Kiełczewski

1267




1267












  • $begingroup$
    What exactly is it that you want to know?
    $endgroup$
    – amd
    Jan 16 at 0:48










  • $begingroup$
    @amd I update question - now it is more clear?
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:08










  • $begingroup$
    This is described in almost all tutorials and books on the subject.
    $endgroup$
    – lightxbulb
    Jan 16 at 6:45










  • $begingroup$
    @lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:55










  • $begingroup$
    Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
    $endgroup$
    – lightxbulb
    Jan 16 at 7:29


















  • $begingroup$
    What exactly is it that you want to know?
    $endgroup$
    – amd
    Jan 16 at 0:48










  • $begingroup$
    @amd I update question - now it is more clear?
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:08










  • $begingroup$
    This is described in almost all tutorials and books on the subject.
    $endgroup$
    – lightxbulb
    Jan 16 at 6:45










  • $begingroup$
    @lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
    $endgroup$
    – Kamil Kiełczewski
    Jan 16 at 6:55










  • $begingroup$
    Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
    $endgroup$
    – lightxbulb
    Jan 16 at 7:29
















$begingroup$
What exactly is it that you want to know?
$endgroup$
– amd
Jan 16 at 0:48




$begingroup$
What exactly is it that you want to know?
$endgroup$
– amd
Jan 16 at 0:48












$begingroup$
@amd I update question - now it is more clear?
$endgroup$
– Kamil Kiełczewski
Jan 16 at 6:08




$begingroup$
@amd I update question - now it is more clear?
$endgroup$
– Kamil Kiełczewski
Jan 16 at 6:08












$begingroup$
This is described in almost all tutorials and books on the subject.
$endgroup$
– lightxbulb
Jan 16 at 6:45




$begingroup$
This is described in almost all tutorials and books on the subject.
$endgroup$
– lightxbulb
Jan 16 at 6:45












$begingroup$
@lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
$endgroup$
– Kamil Kiełczewski
Jan 16 at 6:55




$begingroup$
@lightxbulb In all sources that I know - there are similar description to above, but no explicite formulas
$endgroup$
– Kamil Kiełczewski
Jan 16 at 6:55












$begingroup$
Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
$endgroup$
– lightxbulb
Jan 16 at 7:29




$begingroup$
Scratchapixel, pbrt-book, ray tracing in one weekend are all free access, and all of those explain how this is done.
$endgroup$
– lightxbulb
Jan 16 at 7:29










1 Answer
1






active

oldest

votes


















0












$begingroup$

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)



$$t = T-E, qquad b = wtimes t $$



$$
t_n = frac{t}{||t||}, qquad
b_n = frac{b}{||b||}, qquad
v_n = t_ntimes b_n \
$$



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.






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    1 Answer
    1






    active

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    votes






    active

    oldest

    votes









    0












    $begingroup$

    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)



    $$t = T-E, qquad b = wtimes t $$



    $$
    t_n = frac{t}{||t||}, qquad
    b_n = frac{b}{||b||}, qquad
    v_n = t_ntimes b_n \
    $$



    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.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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)



      $$t = T-E, qquad b = wtimes t $$



      $$
      t_n = frac{t}{||t||}, qquad
      b_n = frac{b}{||b||}, qquad
      v_n = t_ntimes b_n \
      $$



      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.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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)



        $$t = T-E, qquad b = wtimes t $$



        $$
        t_n = frac{t}{||t||}, qquad
        b_n = frac{b}{||b||}, qquad
        v_n = t_ntimes b_n \
        $$



        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.






        share|cite|improve this answer











        $endgroup$



        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)



        $$t = T-E, qquad b = wtimes t $$



        $$
        t_n = frac{t}{||t||}, qquad
        b_n = frac{b}{||b||}, qquad
        v_n = t_ntimes b_n \
        $$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 24 at 10:35

























        answered Jan 21 at 21:37









        Kamil KiełczewskiKamil Kiełczewski

        1267




        1267






























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