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1

$begingroup$

I have a point $P'$ given by known Cartesian coordinates $(x',y',z')$ and an oblate spheroidal surface parameterized by
beginalign
x &= a coshmu cosnu cosphi \
y &= a coshmu cosnu sinphi \
z &= a sinhmu sinnu
endalign
where $a,mu$ are known constants. For an oblate spheroid, $mu$ is similar to a radial coordinate, $nu$ is similar to a latitude, and $phi$ is the same azimuthal angle from spherical coordinates. So for an oblate spheroidal surface, $mu$ is fixed.

When you draw a line from the point $P'$ perpendicular to the oblate spheroid surface ($P'$ is outside the surface), it will intersect the surface at some point $P$ at coordinates $(x,y,z)$. I need to find the location of this point (it will intersect the surface at 2 points, but I want the intersection closest to $P'$ and normal to the surface).

To do this, I write the equation of that line containing both $P'$ and $P$ as:
$$
mathbfx' = mathbfx + t hate_mu
$$
where $mathbfx$ is the point I want to find, $t$ is unknown, and $hate_mu$ is the outward normal vector to the oblate spheroid surface at the point $mathbfx$:
$$
hate_mu = 1 over sqrtsinh^2mu + sin^2nu
beginpmatrix
sinhmu cosnu cosphi \
sinhmu cosnu sinphi \
coshmu sinnu
endpmatrix
$$
So we have 3 equations and 2 unknowns: $t$ and $nu$. As I mentioned before, $a$ and $mu$ are known constants, $mathbfx'$ is given, and the angle $phi$ can be determined from symmetry, since both points $P$ and $P'$ will share the same azimuth angle.

Since $mu$ and $phi$ are known, the goal is to solve for $nu$ since that would immediately yield the coordinates $(x,y,z)$. I don't see an easy way to solve this equation analytically - I could use numerical methods but it seems to me that there should be an analytical solution to this. Does anyone see how?

geometry coordinate-systems surfaces

share|cite|improve this question

edited Mar 15 at 17:51

vibe

asked Mar 15 at 17:12

vibevibe

1748

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After a literature search, it seems that a numerical solution is required: link.springer.com/article/10.1007/s00190-011-0514-7
  $endgroup$
  – vibe
  Mar 15 at 18:35
  

  
  
  
  

add a comment | 

1

$begingroup$

I have a point $P'$ given by known Cartesian coordinates $(x',y',z')$ and an oblate spheroidal surface parameterized by
beginalign
x &= a coshmu cosnu cosphi \
y &= a coshmu cosnu sinphi \
z &= a sinhmu sinnu
endalign
where $a,mu$ are known constants. For an oblate spheroid, $mu$ is similar to a radial coordinate, $nu$ is similar to a latitude, and $phi$ is the same azimuthal angle from spherical coordinates. So for an oblate spheroidal surface, $mu$ is fixed.

When you draw a line from the point $P'$ perpendicular to the oblate spheroid surface ($P'$ is outside the surface), it will intersect the surface at some point $P$ at coordinates $(x,y,z)$. I need to find the location of this point (it will intersect the surface at 2 points, but I want the intersection closest to $P'$ and normal to the surface).

To do this, I write the equation of that line containing both $P'$ and $P$ as:
$$
mathbfx' = mathbfx + t hate_mu
$$
where $mathbfx$ is the point I want to find, $t$ is unknown, and $hate_mu$ is the outward normal vector to the oblate spheroid surface at the point $mathbfx$:
$$
hate_mu = 1 over sqrtsinh^2mu + sin^2nu
beginpmatrix
sinhmu cosnu cosphi \
sinhmu cosnu sinphi \
coshmu sinnu
endpmatrix
$$
So we have 3 equations and 2 unknowns: $t$ and $nu$. As I mentioned before, $a$ and $mu$ are known constants, $mathbfx'$ is given, and the angle $phi$ can be determined from symmetry, since both points $P$ and $P'$ will share the same azimuth angle.

Since $mu$ and $phi$ are known, the goal is to solve for $nu$ since that would immediately yield the coordinates $(x,y,z)$. I don't see an easy way to solve this equation analytically - I could use numerical methods but it seems to me that there should be an analytical solution to this. Does anyone see how?

geometry coordinate-systems surfaces

share|cite|improve this question

edited Mar 15 at 17:51

vibe

asked Mar 15 at 17:12

vibevibe

1748

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After a literature search, it seems that a numerical solution is required: link.springer.com/article/10.1007/s00190-011-0514-7
  $endgroup$
  – vibe
  Mar 15 at 18:35
  

  
  
  
  

add a comment | 

$begingroup$

I have a point $P'$ given by known Cartesian coordinates $(x',y',z')$ and an oblate spheroidal surface parameterized by
beginalign
x &= a coshmu cosnu cosphi \
y &= a coshmu cosnu sinphi \
z &= a sinhmu sinnu
endalign
where $a,mu$ are known constants. For an oblate spheroid, $mu$ is similar to a radial coordinate, $nu$ is similar to a latitude, and $phi$ is the same azimuthal angle from spherical coordinates. So for an oblate spheroidal surface, $mu$ is fixed.

When you draw a line from the point $P'$ perpendicular to the oblate spheroid surface ($P'$ is outside the surface), it will intersect the surface at some point $P$ at coordinates $(x,y,z)$. I need to find the location of this point (it will intersect the surface at 2 points, but I want the intersection closest to $P'$ and normal to the surface).

To do this, I write the equation of that line containing both $P'$ and $P$ as:
$$
mathbfx' = mathbfx + t hate_mu
$$
where $mathbfx$ is the point I want to find, $t$ is unknown, and $hate_mu$ is the outward normal vector to the oblate spheroid surface at the point $mathbfx$:
$$
hate_mu = 1 over sqrtsinh^2mu + sin^2nu
beginpmatrix
sinhmu cosnu cosphi \
sinhmu cosnu sinphi \
coshmu sinnu
endpmatrix
$$
So we have 3 equations and 2 unknowns: $t$ and $nu$. As I mentioned before, $a$ and $mu$ are known constants, $mathbfx'$ is given, and the angle $phi$ can be determined from symmetry, since both points $P$ and $P'$ will share the same azimuth angle.

Since $mu$ and $phi$ are known, the goal is to solve for $nu$ since that would immediately yield the coordinates $(x,y,z)$. I don't see an easy way to solve this equation analytically - I could use numerical methods but it seems to me that there should be an analytical solution to this. Does anyone see how?

geometry coordinate-systems surfaces

share|cite|improve this question

edited Mar 15 at 17:51

vibe

asked Mar 15 at 17:12

vibevibe

1748

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:51

vibe

asked Mar 15 at 17:12

vibevibe

1748

share|cite|improve this question

edited Mar 15 at 17:51

vibe

asked Mar 15 at 17:12

vibevibe

1748

asked Mar 15 at 17:12

vibevibe

1748

asked Mar 15 at 17:12

vibevibe

1748