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Intersection spheroid-polar plane in parametric or spherical coordinates
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The Earth is assumed to be a WGS84 ellipsoid (oblate spheroid) $E$.
$E: (x^2+y^2)/R_eq^2+z^2/R_pol^2=1$
With $R_eq > R_pol$, a point $M$ is outside $E$, and $P$ is its polar plane, that intersects $E$, forming an ellipse $S$.
How to get the coordinates of this $S$ in parametric form or in spherical form?
My goal is to iterate over the parameters of $S$ to get the coordinates of all the points forming $S$.
I tried many methods, adviced from others' help, but they all complicated
spherical-coordinates parametrization intersection-theory
asked Jul 16 at 11:49
Khaled
436
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up vote
1
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The Earth is assumed to be a WGS84 ellipsoid (oblate spheroid) $E$.
$E: (x^2+y^2)/R_eq^2+z^2/R_pol^2=1$
With $R_eq > R_pol$, a point $M$ is outside $E$, and $P$ is its polar plane, that intersects $E$, forming an ellipse $S$.
How to get the coordinates of this $S$ in parametric form or in spherical form?
My goal is to iterate over the parameters of $S$ to get the coordinates of all the points forming $S$.
I tried many methods, adviced from others' help, but they all complicated
spherical-coordinates parametrization intersection-theory
asked Jul 16 at 11:49
Khaled
436
One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The Earth is assumed to be a WGS84 ellipsoid (oblate spheroid) $E$.
$E: (x^2+y^2)/R_eq^2+z^2/R_pol^2=1$
With $R_eq > R_pol$, a point $M$ is outside $E$, and $P$ is its polar plane, that intersects $E$, forming an ellipse $S$.
How to get the coordinates of this $S$ in parametric form or in spherical form?
My goal is to iterate over the parameters of $S$ to get the coordinates of all the points forming $S$.
I tried many methods, adviced from others' help, but they all complicated
spherical-coordinates parametrization intersection-theory
asked Jul 16 at 11:49
Khaled
436
The Earth is assumed to be a WGS84 ellipsoid (oblate spheroid) $E$.
$E: (x^2+y^2)/R_eq^2+z^2/R_pol^2=1$
With $R_eq > R_pol$, a point $M$ is outside $E$, and $P$ is its polar plane, that intersects $E$, forming an ellipse $S$.
How to get the coordinates of this $S$ in parametric form or in spherical form?
My goal is to iterate over the parameters of $S$ to get the coordinates of all the points forming $S$.
I tried many methods, adviced from others' help, but they all complicated
spherical-coordinates parametrization intersection-theory
asked Jul 16 at 11:49
Khaled
436
asked Jul 16 at 11:49
Khaled
436
asked Jul 16 at 11:49
Khaled
436
asked Jul 16 at 11:49
Khaled
436
436
One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04
add a comment |Â
One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04
One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04
add a comment |Â
1 Answer
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For a spherical Earth defined by $x^2+y^2+x^2=R^2$ and a point $M$ at $(D,0,0)$ the polar plane is defined by $x=d=fracR^2D$, which intersects the sphere in a circle with radius $rho=Rsqrt1-(frac RD)^2$ that can be parametrized by $(x,y,z)=(d,rho cos t,rho sin t)$ with $0le t lt 2pi$.
With point $M$ rotated upwards to $(Dcosphi,0,Dsinphi)$ the parametrization for the similarly rotated circle becomes $(x,y,z)=(dcosphi-rhosinphisin t,rhocos t,dsin phi+rhocosphisin t)$
A final rotation around the $z$-axis relocates point $M$ to $(Dcoslambdacosphi,Dsinlambdacosphi,Dsinphi)$ and the circle to points $(dcoslambdacosphi-rhocoslambdasinphisin t-rhosinlambdacos t,dsinlambdacosphi-rhosinlambdasinphisin t+rhocoslambdacos t,dsin phi+rhocosphisin t)$.
So for the ellipsoid case one could first transform $M=(x,y,z)$ to $(x,y,fracR_eqR_polz)$, find the circle parametrisation for a sphere with radius $R_eq$ and multiply its $z$-component with $fracR_polR_eq$.
answered Jul 17 at 0:16
random
23815
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For a spherical Earth defined by $x^2+y^2+x^2=R^2$ and a point $M$ at $(D,0,0)$ the polar plane is defined by $x=d=fracR^2D$, which intersects the sphere in a circle with radius $rho=Rsqrt1-(frac RD)^2$ that can be parametrized by $(x,y,z)=(d,rho cos t,rho sin t)$ with $0le t lt 2pi$.
With point $M$ rotated upwards to $(Dcosphi,0,Dsinphi)$ the parametrization for the similarly rotated circle becomes $(x,y,z)=(dcosphi-rhosinphisin t,rhocos t,dsin phi+rhocosphisin t)$
A final rotation around the $z$-axis relocates point $M$ to $(Dcoslambdacosphi,Dsinlambdacosphi,Dsinphi)$ and the circle to points $(dcoslambdacosphi-rhocoslambdasinphisin t-rhosinlambdacos t,dsinlambdacosphi-rhosinlambdasinphisin t+rhocoslambdacos t,dsin phi+rhocosphisin t)$.
So for the ellipsoid case one could first transform $M=(x,y,z)$ to $(x,y,fracR_eqR_polz)$, find the circle parametrisation for a sphere with radius $R_eq$ and multiply its $z$-component with $fracR_polR_eq$.
answered Jul 17 at 0:16
random
23815
add a comment |Â
up vote
0
down vote
For a spherical Earth defined by $x^2+y^2+x^2=R^2$ and a point $M$ at $(D,0,0)$ the polar plane is defined by $x=d=fracR^2D$, which intersects the sphere in a circle with radius $rho=Rsqrt1-(frac RD)^2$ that can be parametrized by $(x,y,z)=(d,rho cos t,rho sin t)$ with $0le t lt 2pi$.
With point $M$ rotated upwards to $(Dcosphi,0,Dsinphi)$ the parametrization for the similarly rotated circle becomes $(x,y,z)=(dcosphi-rhosinphisin t,rhocos t,dsin phi+rhocosphisin t)$
A final rotation around the $z$-axis relocates point $M$ to $(Dcoslambdacosphi,Dsinlambdacosphi,Dsinphi)$ and the circle to points $(dcoslambdacosphi-rhocoslambdasinphisin t-rhosinlambdacos t,dsinlambdacosphi-rhosinlambdasinphisin t+rhocoslambdacos t,dsin phi+rhocosphisin t)$.
So for the ellipsoid case one could first transform $M=(x,y,z)$ to $(x,y,fracR_eqR_polz)$, find the circle parametrisation for a sphere with radius $R_eq$ and multiply its $z$-component with $fracR_polR_eq$.
answered Jul 17 at 0:16
random
23815
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For a spherical Earth defined by $x^2+y^2+x^2=R^2$ and a point $M$ at $(D,0,0)$ the polar plane is defined by $x=d=fracR^2D$, which intersects the sphere in a circle with radius $rho=Rsqrt1-(frac RD)^2$ that can be parametrized by $(x,y,z)=(d,rho cos t,rho sin t)$ with $0le t lt 2pi$.
With point $M$ rotated upwards to $(Dcosphi,0,Dsinphi)$ the parametrization for the similarly rotated circle becomes $(x,y,z)=(dcosphi-rhosinphisin t,rhocos t,dsin phi+rhocosphisin t)$
A final rotation around the $z$-axis relocates point $M$ to $(Dcoslambdacosphi,Dsinlambdacosphi,Dsinphi)$ and the circle to points $(dcoslambdacosphi-rhocoslambdasinphisin t-rhosinlambdacos t,dsinlambdacosphi-rhosinlambdasinphisin t+rhocoslambdacos t,dsin phi+rhocosphisin t)$.
So for the ellipsoid case one could first transform $M=(x,y,z)$ to $(x,y,fracR_eqR_polz)$, find the circle parametrisation for a sphere with radius $R_eq$ and multiply its $z$-component with $fracR_polR_eq$.
answered Jul 17 at 0:16
random
23815
For a spherical Earth defined by $x^2+y^2+x^2=R^2$ and a point $M$ at $(D,0,0)$ the polar plane is defined by $x=d=fracR^2D$, which intersects the sphere in a circle with radius $rho=Rsqrt1-(frac RD)^2$ that can be parametrized by $(x,y,z)=(d,rho cos t,rho sin t)$ with $0le t lt 2pi$.
With point $M$ rotated upwards to $(Dcosphi,0,Dsinphi)$ the parametrization for the similarly rotated circle becomes $(x,y,z)=(dcosphi-rhosinphisin t,rhocos t,dsin phi+rhocosphisin t)$
A final rotation around the $z$-axis relocates point $M$ to $(Dcoslambdacosphi,Dsinlambdacosphi,Dsinphi)$ and the circle to points $(dcoslambdacosphi-rhocoslambdasinphisin t-rhosinlambdacos t,dsinlambdacosphi-rhosinlambdasinphisin t+rhocoslambdacos t,dsin phi+rhocosphisin t)$.
So for the ellipsoid case one could first transform $M=(x,y,z)$ to $(x,y,fracR_eqR_polz)$, find the circle parametrisation for a sphere with radius $R_eq$ and multiply its $z$-component with $fracR_polR_eq$.
answered Jul 17 at 0:16
random
23815
answered Jul 17 at 0:16
random
23815
answered Jul 17 at 0:16
random
23815
answered Jul 17 at 0:16
random
23815
23815
add a comment |Â
add a comment |Â
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One could first analyze the problem for a spherical Earth and note that reducing the $z$-coordinate of everything by a factor $frac R_polR_eq$ is a solution for the problem in the question.
– random
Jul 16 at 12:49
I began with intersecting a sphere of radius $a$ with $P: ux+vy+wz=d$, But I got a quadratic equation of a projection of $S$ onto $xOy Plane$
– Khaled
Jul 16 at 14:04