Mixing
Closest point on line segment of a great circle
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If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.
Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?
spherical-coordinates spherical-geometry geodesic spherical-trigonometry
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
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add a comment |
$begingroup$
If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.
Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?
spherical-coordinates spherical-geometry geodesic spherical-trigonometry
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
$endgroup$
add a comment |
$begingroup$
If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.
Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?
spherical-coordinates spherical-geometry geodesic spherical-trigonometry
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
$endgroup$
If I have a sphere of radius R, and two points $A$ and $B$ on its surface, at $(R, theta_A,phi_A)$ and $(R, theta_B,phi_B)$ respectively in spherical coordinates. Call $AB$ the geodesic from $A$ to $B$, i.e. the segment of the great circle connecting the two points.
Given a third point $P$ at $(R, theta_P,phi_P)$, how can I find out the point $Q in AB$ which is closest to $P$ (in the geodesic sense)?
spherical-coordinates spherical-geometry geodesic spherical-trigonometry
spherical-coordinates spherical-geometry geodesic spherical-trigonometry
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
asked Nov 2 '18 at 11:38
mitchusmitchus
310315
310315
add a comment |
add a comment |
1 Answer
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There are a couple of online references.
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$
Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
$Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$
This can be rephrased by means of compounded vector triple product.
answered Jan 27 at 21:43
rrogersrrogers
418312
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are a couple of online references.
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$
Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
$Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$
This can be rephrased by means of compounded vector triple product.
answered Jan 27 at 21:43
rrogersrrogers
418312
$endgroup$
add a comment |
$begingroup$
There are a couple of online references.
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$
Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
$Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$
This can be rephrased by means of compounded vector triple product.
answered Jan 27 at 21:43
rrogersrrogers
418312
$endgroup$
add a comment |
$begingroup$
There are a couple of online references.
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$
Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
$Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$
This can be rephrased by means of compounded vector triple product.
answered Jan 27 at 21:43
rrogersrrogers
418312
$endgroup$
There are a couple of online references.
Mathoverflow
https://mathoverflow.net/questions/101776/altitudes-of-a-triangle
There is a more extensive theory at:
https://www.researchgate.net/publication/305401152_Projective_configuration_theorems_old_wine_into_new_wineskins
page 29. And, no I haven't read the whole article :) but it looks powerful.
Setting R=1.
The “$times$” product below is the normal cross product normalized to 1.
Here is an intuitive method using the picture below with C taking the place of P.
Consider the great circle passing through $A,B$ and say that the Cartesian coordinates of $A,B,C$ are $left[a_{x},a_{y},a_{z}right],left[b_{x},b_{y},b_{z}right],left[c_{x},c_{y},c_{z}right]$
Call the great circle the “equator”.
Define the “North pole” $N$ as the intercept of $Atimes B$ with the spherical surface; i.e. normalized.
Now every great circle passing through the North pole is a geodesic path intercepting the equator.
Thus the polar great circle through $C$ gives the geodesic altitude of $C$ above the equator at $Q$.
At this point, we could revert to Spherical coordinates and find $Q$.
Alternately we can stay with Cartesian coordinates.
We observe that the vector to Q is orthogonal to the axes of the great circles $A,B$ and $text{N,C }$ giving
$Q=left(Atimes Bright)timesleft(left(Atimes Bright)times Cright)$
This can be rephrased by means of compounded vector triple product.
answered Jan 27 at 21:43
rrogersrrogers
418312
edited Jan 28 at 15:29
answered Jan 27 at 21:43
rrogersrrogers
418312
answered Jan 27 at 21:43
rrogersrrogers
418312
answered Jan 27 at 21:43
rrogersrrogers
418312
418312
add a comment |
add a comment |
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