Closest point on line segment of a great circle












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If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.



Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?






spherical-coordinates spherical-geometry geodesic spherical-trigonometry







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    $begingroup$


    If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.



    Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?






    spherical-coordinates spherical-geometry geodesic spherical-trigonometry







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.



      Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?






      spherical-coordinates spherical-geometry geodesic spherical-trigonometry







      share|cite|improve this question









      $endgroup$




      If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.



      Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?






      spherical-coordinates spherical-geometry geodesic spherical-trigonometry




      spherical-coordinates spherical-geometry geodesic spherical-trigonometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 2 '18 at 11:38









      mitchusmitchus

      310315




      310315






















          1 Answer
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          $begingroup$

          There are a couple of online references.



          Mathoverflow
          https://mathoverflow.net/questions/101776/altitudes-of-a-triangle

          There is a more extensive theory at:
          https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins

          page 29. And, no I haven't read the whole article :) but it looks powerful.

          Setting R=1.

          The “$times$” product below is the normal cross product normalized to 1.

          Here is an intuitive method using the picture below with C taking the place of P.

          Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$

          Call the great circle the “equator”.

          Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.

          Now every great circle passing through the North pole is a geodesic path intercepting the equator.

          Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.

          At this point, we could revert to Spherical coordinates and find $Q$.

          Alternately we can stay with Cartesian coordinates.

          We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
          $Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$



          This can be rephrased by means of compounded vector triple product.



          enter image description here






          share|cite|improve this answer











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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            There are a couple of online references.



            Mathoverflow
            https://mathoverflow.net/questions/101776/altitudes-of-a-triangle

            There is a more extensive theory at:
            https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins

            page 29. And, no I haven't read the whole article :) but it looks powerful.

            Setting R=1.

            The “$times$” product below is the normal cross product normalized to 1.

            Here is an intuitive method using the picture below with C taking the place of P.

            Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$

            Call the great circle the “equator”.

            Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.

            Now every great circle passing through the North pole is a geodesic path intercepting the equator.

            Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.

            At this point, we could revert to Spherical coordinates and find $Q$.

            Alternately we can stay with Cartesian coordinates.

            We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
            $Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$



            This can be rephrased by means of compounded vector triple product.



            enter image description here






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              There are a couple of online references.



              Mathoverflow
              https://mathoverflow.net/questions/101776/altitudes-of-a-triangle

              There is a more extensive theory at:
              https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins

              page 29. And, no I haven't read the whole article :) but it looks powerful.

              Setting R=1.

              The “$times$” product below is the normal cross product normalized to 1.

              Here is an intuitive method using the picture below with C taking the place of P.

              Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$

              Call the great circle the “equator”.

              Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.

              Now every great circle passing through the North pole is a geodesic path intercepting the equator.

              Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.

              At this point, we could revert to Spherical coordinates and find $Q$.

              Alternately we can stay with Cartesian coordinates.

              We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
              $Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$



              This can be rephrased by means of compounded vector triple product.



              enter image description here






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                There are a couple of online references.



                Mathoverflow
                https://mathoverflow.net/questions/101776/altitudes-of-a-triangle

                There is a more extensive theory at:
                https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins

                page 29. And, no I haven't read the whole article :) but it looks powerful.

                Setting R=1.

                The “$times$” product below is the normal cross product normalized to 1.

                Here is an intuitive method using the picture below with C taking the place of P.

                Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$

                Call the great circle the “equator”.

                Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.

                Now every great circle passing through the North pole is a geodesic path intercepting the equator.

                Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.

                At this point, we could revert to Spherical coordinates and find $Q$.

                Alternately we can stay with Cartesian coordinates.

                We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
                $Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$



                This can be rephrased by means of compounded vector triple product.



                enter image description here






                share|cite|improve this answer











                $endgroup$



                There are a couple of online references.



                Mathoverflow
                https://mathoverflow.net/questions/101776/altitudes-of-a-triangle

                There is a more extensive theory at:
                https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins

                page 29. And, no I haven't read the whole article :) but it looks powerful.

                Setting R=1.

                The “$times$” product below is the normal cross product normalized to 1.

                Here is an intuitive method using the picture below with C taking the place of P.

                Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$

                Call the great circle the “equator”.

                Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.

                Now every great circle passing through the North pole is a geodesic path intercepting the equator.

                Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.

                At this point, we could revert to Spherical coordinates and find $Q$.

                Alternately we can stay with Cartesian coordinates.

                We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
                $Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$



                This can be rephrased by means of compounded vector triple product.



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 28 at 15:29

























                answered Jan 27 at 21:43









                rrogersrrogers

                418312




                418312






























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