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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
edited Jul 20 at 16:00
KReiser
7,53511230
asked Jul 20 at 13:59
kosmo16
427
add a comment |Â
up vote
3
down vote
favorite
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
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 |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
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
edited Jul 20 at 16:00
KReiser
7,53511230
asked Jul 20 at 13:59
kosmo16
427
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 |Â
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 |Â
1 Answer
1
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oldest
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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!
answered Aug 1 at 13:52
Oscar Lanzi
9,96911632
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
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!
answered Aug 1 at 13:52
Oscar Lanzi
9,96911632
add a comment |Â
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!
answered Aug 1 at 13:52
Oscar Lanzi
9,96911632
add a comment |Â
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!
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!
answered Aug 1 at 13:52
Oscar Lanzi
9,96911632
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 |Â
add a comment |Â
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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