Extreme lat and lon for ellipse on sphere

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I have 2 focal points of ellipse defined on a sphere: $F_1 = (q_1, p_1)$ and $F_2 = (q_2, p_2)$ and length of major axis $2a$. $R$ of a sphere is 1.

$q_1, q_2$ are latitudes

$p_1, p_2$ are longitudes

I need to find 4 points:

• point on ellipse with max latitude


• point on ellipse with min latitude


• point on ellipse with max longitude


• point on ellipse with min longitude

I tried to convert points to vectors and use a property that sum of angles
$$sphericalangle F_1OP + sphericalangle F_2OP$$ is constant for each $P$ on ellipse, but it lead me nowhere. Any idea how it can be solved?

geometry

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edited Jul 20 at 16:00

KReiser

7,53511230

asked Jul 20 at 13:59

kosmo16

427

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What is an ellipse on a sphere?
  – H. Gutsche
  Jul 20 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
  – Narlin
  Jul 20 at 17:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
  – Oscar Lanzi
  Jul 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
  – kosmo16
  Jul 24 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
  – kosmo16
  Aug 1 at 13:09
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

I have 2 focal points of ellipse defined on a sphere: $F_1 = (q_1, p_1)$ and $F_2 = (q_2, p_2)$ and length of major axis $2a$. $R$ of a sphere is 1.

$q_1, q_2$ are latitudes

$p_1, p_2$ are longitudes

I need to find 4 points:

• point on ellipse with max latitude


• point on ellipse with min latitude


• point on ellipse with max longitude


• point on ellipse with min longitude

I tried to convert points to vectors and use a property that sum of angles
$$sphericalangle F_1OP + sphericalangle F_2OP$$ is constant for each $P$ on ellipse, but it lead me nowhere. Any idea how it can be solved?

geometry

share|cite|improve this question

edited Jul 20 at 16:00

KReiser

7,53511230

asked Jul 20 at 13:59

kosmo16

427

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What is an ellipse on a sphere?
  – H. Gutsche
  Jul 20 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
  – Narlin
  Jul 20 at 17:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
  – Oscar Lanzi
  Jul 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
  – kosmo16
  Jul 24 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
  – kosmo16
  Aug 1 at 13:09
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

I have 2 focal points of ellipse defined on a sphere: $F_1 = (q_1, p_1)$ and $F_2 = (q_2, p_2)$ and length of major axis $2a$. $R$ of a sphere is 1.

$q_1, q_2$ are latitudes

$p_1, p_2$ are longitudes

I need to find 4 points:

• point on ellipse with max latitude


• point on ellipse with min latitude


• point on ellipse with max longitude


• point on ellipse with min longitude

I tried to convert points to vectors and use a property that sum of angles
$$sphericalangle F_1OP + sphericalangle F_2OP$$ is constant for each $P$ on ellipse, but it lead me nowhere. Any idea how it can be solved?

geometry

share|cite|improve this question

edited Jul 20 at 16:00

KReiser

7,53511230

asked Jul 20 at 13:59

kosmo16

427

I have 2 focal points of ellipse defined on a sphere: $F_1 = (q_1, p_1)$ and $F_2 = (q_2, p_2)$ and length of major axis $2a$. $R$ of a sphere is 1.

$q_1, q_2$ are latitudes

$p_1, p_2$ are longitudes

I need to find 4 points:

• point on ellipse with max latitude


• point on ellipse with min latitude


• point on ellipse with max longitude


• point on ellipse with min longitude

I tried to convert points to vectors and use a property that sum of angles
$$sphericalangle F_1OP + sphericalangle F_2OP$$ is constant for each $P$ on ellipse, but it lead me nowhere. Any idea how it can be solved?

geometry

share|cite|improve this question

edited Jul 20 at 16:00

KReiser

7,53511230

asked Jul 20 at 13:59

kosmo16

427

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 16:00

KReiser

7,53511230

edited Jul 20 at 16:00

KReiser

7,53511230

edited Jul 20 at 16:00

KReiser

7,53511230

7,53511230

asked Jul 20 at 13:59

kosmo16

427

asked Jul 20 at 13:59

kosmo16

427

asked Jul 20 at 13:59

kosmo16

427

427

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What is an ellipse on a sphere?
  – H. Gutsche
  Jul 20 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
  – Narlin
  Jul 20 at 17:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
  – Oscar Lanzi
  Jul 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
  – kosmo16
  Jul 24 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
  – kosmo16
  Aug 1 at 13:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What is an ellipse on a sphere?
  – H. Gutsche
  Jul 20 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
  – Narlin
  Jul 20 at 17:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
  – Oscar Lanzi
  Jul 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
  – kosmo16
  Jul 24 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
  – kosmo16
  Aug 1 at 13:09
  

  
  
  
  

3

3

What is an ellipse on a sphere?
– H. Gutsche
Jul 20 at 16:52

What is an ellipse on a sphere?
– H. Gutsche
Jul 20 at 16:52

1

1

Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
– Narlin
Jul 20 at 17:43

Define an ellipse as the locus of points which is a constant summed distance away from two focal points, whenever the distance is calculated as the shortest path along the surface of the sphere. I believe this turns out to be the intersection of a cylinder and a sphere.
– Narlin
Jul 20 at 17:43

1

1

Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
– Oscar Lanzi
Jul 20 at 21:06

Alas @Narlin, I do not think it is that simple. I believe that the problem is soluble with spherical trig but I will need a bigger block of time to formulate an answer.
– Oscar Lanzi
Jul 20 at 21:06

As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
– kosmo16
Jul 24 at 8:23

As @Narlin mentioned ellipse is the locus of points which is a constant summed distance away from two focal points.
– kosmo16
Jul 24 at 8:23

@OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
– kosmo16
Aug 1 at 13:09

@OscarLanzi Can you give me a clue how to approach this problem, maybe I will be able to find the answer by myself.
– kosmo16
Aug 1 at 13:09

add a comment | 

1 Answer
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0
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A partial answer based on a comment:

Let $A, B$ be two points on the sphere (Earth) with radius taken as unity. Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points. Define $N$ as the North Pole with $p(N)=+pi/2, q(N)$ arbitrary.

Draw $triangle ABN$. Then $|AN|=pi/2-p(A), |BN|=pi/2-p(B), |angle N|=pm (q(A)-q(B))$. Now from the Law of Cosines

$cos(|AB|)=sin(p(A))sin(p(B))+cos(p(A))cos(p(B))cos(q(A)-q(B))$.

(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-pi/2$. Use the identity $sin(pi/2+theta)=sin(pi/2-theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.

Good luck!

share|cite|improve this answer

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

add a comment | 

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1 Answer
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active

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A partial answer based on a comment:

Let $A, B$ be two points on the sphere (Earth) with radius taken as unity. Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points. Define $N$ as the North Pole with $p(N)=+pi/2, q(N)$ arbitrary.

Draw $triangle ABN$. Then $|AN|=pi/2-p(A), |BN|=pi/2-p(B), |angle N|=pm (q(A)-q(B))$. Now from the Law of Cosines

$cos(|AB|)=sin(p(A))sin(p(B))+cos(p(A))cos(p(B))cos(q(A)-q(B))$.

(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-pi/2$. Use the identity $sin(pi/2+theta)=sin(pi/2-theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.

Good luck!

share|cite|improve this answer

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

add a comment | 

up vote
0
down vote

A partial answer based on a comment:

Let $A, B$ be two points on the sphere (Earth) with radius taken as unity. Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points. Define $N$ as the North Pole with $p(N)=+pi/2, q(N)$ arbitrary.

Draw $triangle ABN$. Then $|AN|=pi/2-p(A), |BN|=pi/2-p(B), |angle N|=pm (q(A)-q(B))$. Now from the Law of Cosines

$cos(|AB|)=sin(p(A))sin(p(B))+cos(p(A))cos(p(B))cos(q(A)-q(B))$.

(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-pi/2$. Use the identity $sin(pi/2+theta)=sin(pi/2-theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.

Good luck!

share|cite|improve this answer

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

add a comment | 

up vote
0
down vote

up vote
0
down vote

A partial answer based on a comment:

Let $A, B$ be two points on the sphere (Earth) with radius taken as unity. Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points. Define $N$ as the North Pole with $p(N)=+pi/2, q(N)$ arbitrary.

Draw $triangle ABN$. Then $|AN|=pi/2-p(A), |BN|=pi/2-p(B), |angle N|=pm (q(A)-q(B))$. Now from the Law of Cosines

$cos(|AB|)=sin(p(A))sin(p(B))+cos(p(A))cos(p(B))cos(q(A)-q(B))$.

(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-pi/2$. Use the identity $sin(pi/2+theta)=sin(pi/2-theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.

Good luck!

share|cite|improve this answer

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

A partial answer based on a comment:

Let $A, B$ be two points on the sphere (Earth) with radius taken as unity. Define $p(A), q(A)$ as the latitude and longitude of $A$ on the Earth, analogously for other points. Define $N$ as the North Pole with $p(N)=+pi/2, q(N)$ arbitrary.

Draw $triangle ABN$. Then $|AN|=pi/2-p(A), |BN|=pi/2-p(B), |angle N|=pm (q(A)-q(B))$. Now from the Law of Cosines

$cos(|AB|)=sin(p(A))sin(p(B))+cos(p(A))cos(p(B))cos(q(A)-q(B))$.

(As an exercise you may want to derive the same formula replacing $N$ with $S$ where$p(S)=-pi/2$. Use the identity $sin(pi/2+theta)=sin(pi/2-theta)$.) Use this formula for the distance from any point on the ellipse to the foci, adding up the inverse cosines to equal the major axis.

Good luck!

share|cite|improve this answer

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 13:59

edited Aug 1 at 13:59

edited Aug 1 at 13:59

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

answered Aug 1 at 13:52

Oscar Lanzi

9,96911632

9,96911632

add a comment | 

add a comment | 

 

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