Basic Ray casting vector equality problem












0












$begingroup$


I am working through the basic math of the ray casting algorithm detailed here



The problem I am facing is as follows:



There exists this vector $textbf{r}=(r_x,r_y)$ whose magnitude is not equal to 1 aka not a unit vector



There exists this other vector $textbf{s} = (1,s_y)$ that is in the same direction of $textbf{r}$, but has a different magnitude



The author in the link above manages to calculate the magnitude of $textbf{s}$ in terms of $r_x$ and $r_y$ namely $midtextbf{s}mid = sqrt frac{(r_y)^2 + (r_x)^2}{(r_x)^2}$



He claims to have derived it geometrically (no calculation shown)



I see that since both $textbf{r}$ and $textbf{s}$ point in the same direction, I could equate their unit vectors as such
$(frac{1}{sqrt{(s_y)^2 + 1}},_) = (frac{r_x}{sqrt{(r_x)^2 + (r_y)^2)}},_)$ to obtain $s_y = frac{r_y}{r_x}$



and hence



$midtextbf{s}mid = sqrt{1 + s_y^2} = sqrt{1 + frac{r_y^2}{r_x^2}}$



The reason I'm asking on here is because the equality of the unit vectors to obtain the end resulthere seems forced and to a very large degree contrived. Am I wrong here? If so, how and why? Is the method that I have employed non kosher?






vectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:13










  • $begingroup$
    @JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
    $endgroup$
    – functor-soup
    Dec 10 '18 at 23:16










  • $begingroup$
    No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:17
















0












$begingroup$


I am working through the basic math of the ray casting algorithm detailed here



The problem I am facing is as follows:



There exists this vector $textbf{r}=(r_x,r_y)$ whose magnitude is not equal to 1 aka not a unit vector



There exists this other vector $textbf{s} = (1,s_y)$ that is in the same direction of $textbf{r}$, but has a different magnitude



The author in the link above manages to calculate the magnitude of $textbf{s}$ in terms of $r_x$ and $r_y$ namely $midtextbf{s}mid = sqrt frac{(r_y)^2 + (r_x)^2}{(r_x)^2}$



He claims to have derived it geometrically (no calculation shown)



I see that since both $textbf{r}$ and $textbf{s}$ point in the same direction, I could equate their unit vectors as such
$(frac{1}{sqrt{(s_y)^2 + 1}},_) = (frac{r_x}{sqrt{(r_x)^2 + (r_y)^2)}},_)$ to obtain $s_y = frac{r_y}{r_x}$



and hence



$midtextbf{s}mid = sqrt{1 + s_y^2} = sqrt{1 + frac{r_y^2}{r_x^2}}$



The reason I'm asking on here is because the equality of the unit vectors to obtain the end resulthere seems forced and to a very large degree contrived. Am I wrong here? If so, how and why? Is the method that I have employed non kosher?






vectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:13










  • $begingroup$
    @JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
    $endgroup$
    – functor-soup
    Dec 10 '18 at 23:16










  • $begingroup$
    No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:17














0












0








0





$begingroup$


I am working through the basic math of the ray casting algorithm detailed here



The problem I am facing is as follows:



There exists this vector $textbf{r}=(r_x,r_y)$ whose magnitude is not equal to 1 aka not a unit vector



There exists this other vector $textbf{s} = (1,s_y)$ that is in the same direction of $textbf{r}$, but has a different magnitude



The author in the link above manages to calculate the magnitude of $textbf{s}$ in terms of $r_x$ and $r_y$ namely $midtextbf{s}mid = sqrt frac{(r_y)^2 + (r_x)^2}{(r_x)^2}$



He claims to have derived it geometrically (no calculation shown)



I see that since both $textbf{r}$ and $textbf{s}$ point in the same direction, I could equate their unit vectors as such
$(frac{1}{sqrt{(s_y)^2 + 1}},_) = (frac{r_x}{sqrt{(r_x)^2 + (r_y)^2)}},_)$ to obtain $s_y = frac{r_y}{r_x}$



and hence



$midtextbf{s}mid = sqrt{1 + s_y^2} = sqrt{1 + frac{r_y^2}{r_x^2}}$



The reason I'm asking on here is because the equality of the unit vectors to obtain the end resulthere seems forced and to a very large degree contrived. Am I wrong here? If so, how and why? Is the method that I have employed non kosher?






vectors







share|cite|improve this question









$endgroup$




I am working through the basic math of the ray casting algorithm detailed here



The problem I am facing is as follows:



There exists this vector $textbf{r}=(r_x,r_y)$ whose magnitude is not equal to 1 aka not a unit vector



There exists this other vector $textbf{s} = (1,s_y)$ that is in the same direction of $textbf{r}$, but has a different magnitude



The author in the link above manages to calculate the magnitude of $textbf{s}$ in terms of $r_x$ and $r_y$ namely $midtextbf{s}mid = sqrt frac{(r_y)^2 + (r_x)^2}{(r_x)^2}$



He claims to have derived it geometrically (no calculation shown)



I see that since both $textbf{r}$ and $textbf{s}$ point in the same direction, I could equate their unit vectors as such
$(frac{1}{sqrt{(s_y)^2 + 1}},_) = (frac{r_x}{sqrt{(r_x)^2 + (r_y)^2)}},_)$ to obtain $s_y = frac{r_y}{r_x}$



and hence



$midtextbf{s}mid = sqrt{1 + s_y^2} = sqrt{1 + frac{r_y^2}{r_x^2}}$



The reason I'm asking on here is because the equality of the unit vectors to obtain the end resulthere seems forced and to a very large degree contrived. Am I wrong here? If so, how and why? Is the method that I have employed non kosher?






vectors




vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 10 '18 at 22:58









functor-soupfunctor-soup

989




989












  • $begingroup$
    Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:13










  • $begingroup$
    @JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
    $endgroup$
    – functor-soup
    Dec 10 '18 at 23:16










  • $begingroup$
    No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:17


















  • $begingroup$
    Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:13










  • $begingroup$
    @JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
    $endgroup$
    – functor-soup
    Dec 10 '18 at 23:16










  • $begingroup$
    No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
    $endgroup$
    – John Douma
    Dec 10 '18 at 23:17
















$begingroup$
Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
$endgroup$
– John Douma
Dec 10 '18 at 23:13




$begingroup$
Your way is correct and you get the same answer. Exploiting the fact that two vectors point in the same direction doesn't seem contrived to me. It seems like a good way to solve the problem.
$endgroup$
– John Douma
Dec 10 '18 at 23:13












$begingroup$
@JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
$endgroup$
– functor-soup
Dec 10 '18 at 23:16




$begingroup$
@JohnDouma I see. What about the part of equating their unit vectors? Do you still think that doesn't feel forced?
$endgroup$
– functor-soup
Dec 10 '18 at 23:16












$begingroup$
No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
$endgroup$
– John Douma
Dec 10 '18 at 23:17




$begingroup$
No. It shows that you understand that two vectors that point in the same direction will have the same unit vectors.
$endgroup$
– John Douma
Dec 10 '18 at 23:17










1 Answer
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$begingroup$

The direction of two vectors is the same if the ratio of the components is the same: $$frac{r_y}{r_x}=frac{s_y}{1}$$
That's because you can write $vec s = cvec r$ or $(s_x,s_y)=c(r_x,r_y)$.
We can write this as a system of equations for $x$ and $y$ components. Take $c$ from one equation and replace it into the second.



Now all you need to calculate the magnitude is $$|vec s|=sqrt{s_x^2+s_y^2}=sqrt{1+frac{r_y^2}{r_x^2}}$$






share|cite|improve this answer









$endgroup$













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    $begingroup$

    The direction of two vectors is the same if the ratio of the components is the same: $$frac{r_y}{r_x}=frac{s_y}{1}$$
    That's because you can write $vec s = cvec r$ or $(s_x,s_y)=c(r_x,r_y)$.
    We can write this as a system of equations for $x$ and $y$ components. Take $c$ from one equation and replace it into the second.



    Now all you need to calculate the magnitude is $$|vec s|=sqrt{s_x^2+s_y^2}=sqrt{1+frac{r_y^2}{r_x^2}}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The direction of two vectors is the same if the ratio of the components is the same: $$frac{r_y}{r_x}=frac{s_y}{1}$$
      That's because you can write $vec s = cvec r$ or $(s_x,s_y)=c(r_x,r_y)$.
      We can write this as a system of equations for $x$ and $y$ components. Take $c$ from one equation and replace it into the second.



      Now all you need to calculate the magnitude is $$|vec s|=sqrt{s_x^2+s_y^2}=sqrt{1+frac{r_y^2}{r_x^2}}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The direction of two vectors is the same if the ratio of the components is the same: $$frac{r_y}{r_x}=frac{s_y}{1}$$
        That's because you can write $vec s = cvec r$ or $(s_x,s_y)=c(r_x,r_y)$.
        We can write this as a system of equations for $x$ and $y$ components. Take $c$ from one equation and replace it into the second.



        Now all you need to calculate the magnitude is $$|vec s|=sqrt{s_x^2+s_y^2}=sqrt{1+frac{r_y^2}{r_x^2}}$$






        share|cite|improve this answer









        $endgroup$



        The direction of two vectors is the same if the ratio of the components is the same: $$frac{r_y}{r_x}=frac{s_y}{1}$$
        That's because you can write $vec s = cvec r$ or $(s_x,s_y)=c(r_x,r_y)$.
        We can write this as a system of equations for $x$ and $y$ components. Take $c$ from one equation and replace it into the second.



        Now all you need to calculate the magnitude is $$|vec s|=sqrt{s_x^2+s_y^2}=sqrt{1+frac{r_y^2}{r_x^2}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 23:08









        AndreiAndrei

        11.8k21026




        11.8k21026






























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