spherical polar geometry change in elevation angle












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how to calculate change in elevation angle if you know coordinates of two point on surface of sphere. let us say assume that a point move on the surface of sphere from [x1 y1 z1 ] = [0.1 0.1 0.9899] to [x2 y2 z2] = [-0.1 -0.1 0.9899].






geometry spherical-coordinates spheres spherical-trigonometry







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    how to calculate change in elevation angle if you know coordinates of two point on surface of sphere. let us say assume that a point move on the surface of sphere from [x1 y1 z1 ] = [0.1 0.1 0.9899] to [x2 y2 z2] = [-0.1 -0.1 0.9899].






    geometry spherical-coordinates spheres spherical-trigonometry







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      $begingroup$


      how to calculate change in elevation angle if you know coordinates of two point on surface of sphere. let us say assume that a point move on the surface of sphere from [x1 y1 z1 ] = [0.1 0.1 0.9899] to [x2 y2 z2] = [-0.1 -0.1 0.9899].






      geometry spherical-coordinates spheres spherical-trigonometry







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      how to calculate change in elevation angle if you know coordinates of two point on surface of sphere. let us say assume that a point move on the surface of sphere from [x1 y1 z1 ] = [0.1 0.1 0.9899] to [x2 y2 z2] = [-0.1 -0.1 0.9899].






      geometry spherical-coordinates spheres spherical-trigonometry




      geometry spherical-coordinates spheres spherical-trigonometry






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      edited Jan 21 at 9:55









      Bernard

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      asked Jan 21 at 9:37









      p singhp singh

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          There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition
          $$bbox{ vec{p} = left [ begin{matrix} x \ y \ z end{matrix} right ] = left [ begin{matrix}
          r costheta cosvarphi \
          r costheta sinvarphi \
          r sintheta end{matrix} right ] tag{1}label{NA1} }$$
          where $varphi$ is the angle on the $xy$ plane, with $0text{°}$ being on the positive $x$ axis, and $90text{°}$ on the positive $y$ axis; and $theta$ being the angle to the $xy$ plane, with $0text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90text{°}$ referring to the hemisphere with negative $z$. In other words, $theta$ is the elevation angle.



          The inverse of $eqref{NA1}$ is
          $$bbox{begin{cases}
          r = sqrt{x^2 + y^2 + z^2} \
          varphi = arctan(y, x) \
          theta = arcsin(z / r) end{cases} tag{2}label{NA2}}$$



          Thus, if you have a vector $(x, y, z)$, its elevation angle is
          $$bbox{ theta = arcsinleft( frac{z}{sqrt{x^2 + y^2 + z^2}} right) tag{3}label{NA3}}$$



          If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is
          $$bbox{ theta_2 - theta_1 = arcsinleft( frac{z_2}{sqrt{x_2^2 + y_2^2 + z_2^2}} right) - arcsinleft( frac{z_1}{sqrt{x_1^2 + y_1^2 + z_1^2}} right) }tag{4}label{NA4}$$
          Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.



          For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)



          Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869text{°}$, or $1.42889$ radians in both cases.






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            $begingroup$

            There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition
            $$bbox{ vec{p} = left [ begin{matrix} x \ y \ z end{matrix} right ] = left [ begin{matrix}
            r costheta cosvarphi \
            r costheta sinvarphi \
            r sintheta end{matrix} right ] tag{1}label{NA1} }$$
            where $varphi$ is the angle on the $xy$ plane, with $0text{°}$ being on the positive $x$ axis, and $90text{°}$ on the positive $y$ axis; and $theta$ being the angle to the $xy$ plane, with $0text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90text{°}$ referring to the hemisphere with negative $z$. In other words, $theta$ is the elevation angle.



            The inverse of $eqref{NA1}$ is
            $$bbox{begin{cases}
            r = sqrt{x^2 + y^2 + z^2} \
            varphi = arctan(y, x) \
            theta = arcsin(z / r) end{cases} tag{2}label{NA2}}$$



            Thus, if you have a vector $(x, y, z)$, its elevation angle is
            $$bbox{ theta = arcsinleft( frac{z}{sqrt{x^2 + y^2 + z^2}} right) tag{3}label{NA3}}$$



            If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is
            $$bbox{ theta_2 - theta_1 = arcsinleft( frac{z_2}{sqrt{x_2^2 + y_2^2 + z_2^2}} right) - arcsinleft( frac{z_1}{sqrt{x_1^2 + y_1^2 + z_1^2}} right) }tag{4}label{NA4}$$
            Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.



            For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)



            Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869text{°}$, or $1.42889$ radians in both cases.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition
              $$bbox{ vec{p} = left [ begin{matrix} x \ y \ z end{matrix} right ] = left [ begin{matrix}
              r costheta cosvarphi \
              r costheta sinvarphi \
              r sintheta end{matrix} right ] tag{1}label{NA1} }$$
              where $varphi$ is the angle on the $xy$ plane, with $0text{°}$ being on the positive $x$ axis, and $90text{°}$ on the positive $y$ axis; and $theta$ being the angle to the $xy$ plane, with $0text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90text{°}$ referring to the hemisphere with negative $z$. In other words, $theta$ is the elevation angle.



              The inverse of $eqref{NA1}$ is
              $$bbox{begin{cases}
              r = sqrt{x^2 + y^2 + z^2} \
              varphi = arctan(y, x) \
              theta = arcsin(z / r) end{cases} tag{2}label{NA2}}$$



              Thus, if you have a vector $(x, y, z)$, its elevation angle is
              $$bbox{ theta = arcsinleft( frac{z}{sqrt{x^2 + y^2 + z^2}} right) tag{3}label{NA3}}$$



              If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is
              $$bbox{ theta_2 - theta_1 = arcsinleft( frac{z_2}{sqrt{x_2^2 + y_2^2 + z_2^2}} right) - arcsinleft( frac{z_1}{sqrt{x_1^2 + y_1^2 + z_1^2}} right) }tag{4}label{NA4}$$
              Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.



              For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)



              Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869text{°}$, or $1.42889$ radians in both cases.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition
                $$bbox{ vec{p} = left [ begin{matrix} x \ y \ z end{matrix} right ] = left [ begin{matrix}
                r costheta cosvarphi \
                r costheta sinvarphi \
                r sintheta end{matrix} right ] tag{1}label{NA1} }$$
                where $varphi$ is the angle on the $xy$ plane, with $0text{°}$ being on the positive $x$ axis, and $90text{°}$ on the positive $y$ axis; and $theta$ being the angle to the $xy$ plane, with $0text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90text{°}$ referring to the hemisphere with negative $z$. In other words, $theta$ is the elevation angle.



                The inverse of $eqref{NA1}$ is
                $$bbox{begin{cases}
                r = sqrt{x^2 + y^2 + z^2} \
                varphi = arctan(y, x) \
                theta = arcsin(z / r) end{cases} tag{2}label{NA2}}$$



                Thus, if you have a vector $(x, y, z)$, its elevation angle is
                $$bbox{ theta = arcsinleft( frac{z}{sqrt{x^2 + y^2 + z^2}} right) tag{3}label{NA3}}$$



                If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is
                $$bbox{ theta_2 - theta_1 = arcsinleft( frac{z_2}{sqrt{x_2^2 + y_2^2 + z_2^2}} right) - arcsinleft( frac{z_1}{sqrt{x_1^2 + y_1^2 + z_1^2}} right) }tag{4}label{NA4}$$
                Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.



                For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)



                Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869text{°}$, or $1.42889$ radians in both cases.






                share|cite|improve this answer









                $endgroup$



                There are several ways to define a spherical coordinate system. The term elevation angle is used with the definition
                $$bbox{ vec{p} = left [ begin{matrix} x \ y \ z end{matrix} right ] = left [ begin{matrix}
                r costheta cosvarphi \
                r costheta sinvarphi \
                r sintheta end{matrix} right ] tag{1}label{NA1} }$$
                where $varphi$ is the angle on the $xy$ plane, with $0text{°}$ being on the positive $x$ axis, and $90text{°}$ on the positive $y$ axis; and $theta$ being the angle to the $xy$ plane, with $0text{°}$ referring to the $xy$ plane ($z = 0$), positive angles up to $90text{°}$ referring to the hemisphere with positive $z$, and negative angles dow to $-90text{°}$ referring to the hemisphere with negative $z$. In other words, $theta$ is the elevation angle.



                The inverse of $eqref{NA1}$ is
                $$bbox{begin{cases}
                r = sqrt{x^2 + y^2 + z^2} \
                varphi = arctan(y, x) \
                theta = arcsin(z / r) end{cases} tag{2}label{NA2}}$$



                Thus, if you have a vector $(x, y, z)$, its elevation angle is
                $$bbox{ theta = arcsinleft( frac{z}{sqrt{x^2 + y^2 + z^2}} right) tag{3}label{NA3}}$$



                If we compare two points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the change in the elevation angle is
                $$bbox{ theta_2 - theta_1 = arcsinleft( frac{z_2}{sqrt{x_2^2 + y_2^2 + z_2^2}} right) - arcsinleft( frac{z_1}{sqrt{x_1^2 + y_1^2 + z_1^2}} right) }tag{4}label{NA4}$$
                Unfortunately, I do not see any way of simplifying that expression, unless we place some restrictions on the coordinates.



                For example, when $z_1 = z_2$, and $x_1^2 + y_1^2 = x_2^2 + y_2^2$, the elevation angle does not change. (Essentially, the points are then on the same latitude.)



                Therefore, in OP's case, the elevation angle does not change between $(0.1, 0.1, 0.9899)$ and $(-0.1, -0.1, 0.9899)$; it is $81.869text{°}$, or $1.42889$ radians in both cases.







                share|cite|improve this answer












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                answered Jan 21 at 16:13









                Nominal AnimalNominal Animal

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