$r = ycos(theta) - xsin(theta) $ derivation for Hough Transform












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$begingroup$


I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.



Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.



enter image description here



Many thanks in advance










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$endgroup$

















    0












    $begingroup$


    I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.



    Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.



    enter image description here



    Many thanks in advance










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.



      Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.



      enter image description here



      Many thanks in advance










      share|cite|improve this question











      $endgroup$




      I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.



      Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.



      enter image description here



      Many thanks in advance







      polar-coordinates parametric






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      edited Jan 6 at 17:20







      zcahfg2

















      asked Jan 6 at 17:09









      zcahfg2zcahfg2

      382212




      382212






















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          $begingroup$

          Starting with $y=mx+c$,
          m is the slope of the line =$tan(theta)$



          We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
          So that
          $$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
          $$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
          $$ c=r/cos(theta))$$



          Replacing $m$ and $c$ in $y=mx+c$, you get
          $$y=xtan(theta)+r/cos(theta)$$



          The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).






          share|cite|improve this answer









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            $begingroup$

            Starting with $y=mx+c$,
            m is the slope of the line =$tan(theta)$



            We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
            So that
            $$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
            $$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
            $$ c=r/cos(theta))$$



            Replacing $m$ and $c$ in $y=mx+c$, you get
            $$y=xtan(theta)+r/cos(theta)$$



            The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Starting with $y=mx+c$,
              m is the slope of the line =$tan(theta)$



              We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
              So that
              $$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
              $$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
              $$ c=r/cos(theta))$$



              Replacing $m$ and $c$ in $y=mx+c$, you get
              $$y=xtan(theta)+r/cos(theta)$$



              The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Starting with $y=mx+c$,
                m is the slope of the line =$tan(theta)$



                We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
                So that
                $$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
                $$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
                $$ c=r/cos(theta))$$



                Replacing $m$ and $c$ in $y=mx+c$, you get
                $$y=xtan(theta)+r/cos(theta)$$



                The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).






                share|cite|improve this answer









                $endgroup$



                Starting with $y=mx+c$,
                m is the slope of the line =$tan(theta)$



                We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
                So that
                $$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
                $$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
                $$ c=r/cos(theta))$$



                Replacing $m$ and $c$ in $y=mx+c$, you get
                $$y=xtan(theta)+r/cos(theta)$$



                The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 18:46









                Al-CAl-C

                386




                386






























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