How to calculate distance from the International Space Station given coordinates?
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How would one calculate how far away a point is (latitude/longitude) from the international space station given its latitude/longitude/altitude? The distance would be direct as if drawing a straight line from the two points, even if on the other side of the Earth.
geometry algorithms
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
add a comment |
up vote
2
down vote
favorite
How would one calculate how far away a point is (latitude/longitude) from the international space station given its latitude/longitude/altitude? The distance would be direct as if drawing a straight line from the two points, even if on the other side of the Earth.
geometry algorithms
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
5
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How would one calculate how far away a point is (latitude/longitude) from the international space station given its latitude/longitude/altitude? The distance would be direct as if drawing a straight line from the two points, even if on the other side of the Earth.
geometry algorithms
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
How would one calculate how far away a point is (latitude/longitude) from the international space station given its latitude/longitude/altitude? The distance would be direct as if drawing a straight line from the two points, even if on the other side of the Earth.
geometry algorithms
geometry algorithms
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
edited Aug 31 '15 at 21:01
Rob Arthan
28.8k42866
28.8k42866
asked Aug 31 '15 at 20:40
Orbit
1111
asked Aug 31 '15 at 20:40
Orbit
1111
asked Aug 31 '15 at 20:40
Orbit
1111
1111
5
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03
add a comment |
5
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03
5
5
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03
add a comment |
1 Answer
1
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up vote
2
down vote
If $r$ is the distance from the Earth's centre, $phi$ is the longitude in radians (increasing from zero at Greenwich as one goes eastwards) and $theta$ is the
latitude in radians (increasing from zero at the Equator as one goes northwards), then
the $(x,y,z)$ coordinates are given by:
$$p(r,phi, theta) = r(cos phi cos theta, cos phi sin theta, sin phi)$$
Given two sets of coordinates $(r_k,phi_k, theta_k)$, the distance is given
by $|p(r_1,phi_1, theta_1)-p(r_2,phi_2, theta_2)|$.
In your case, we can take $r_1=R$, $r_2=R+A$, where $R$ is the radius of the Earth (I'm assuming a nice sphere, of course) and $A$ is the altitude of the
station above the surface of the Earth.
answered Aug 31 '15 at 21:07
copper.hat
125k559159
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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up vote
2
down vote
If $r$ is the distance from the Earth's centre, $phi$ is the longitude in radians (increasing from zero at Greenwich as one goes eastwards) and $theta$ is the
latitude in radians (increasing from zero at the Equator as one goes northwards), then
the $(x,y,z)$ coordinates are given by:
$$p(r,phi, theta) = r(cos phi cos theta, cos phi sin theta, sin phi)$$
Given two sets of coordinates $(r_k,phi_k, theta_k)$, the distance is given
by $|p(r_1,phi_1, theta_1)-p(r_2,phi_2, theta_2)|$.
In your case, we can take $r_1=R$, $r_2=R+A$, where $R$ is the radius of the Earth (I'm assuming a nice sphere, of course) and $A$ is the altitude of the
station above the surface of the Earth.
answered Aug 31 '15 at 21:07
copper.hat
125k559159
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
add a comment |
up vote
2
down vote
If $r$ is the distance from the Earth's centre, $phi$ is the longitude in radians (increasing from zero at Greenwich as one goes eastwards) and $theta$ is the
latitude in radians (increasing from zero at the Equator as one goes northwards), then
the $(x,y,z)$ coordinates are given by:
$$p(r,phi, theta) = r(cos phi cos theta, cos phi sin theta, sin phi)$$
Given two sets of coordinates $(r_k,phi_k, theta_k)$, the distance is given
by $|p(r_1,phi_1, theta_1)-p(r_2,phi_2, theta_2)|$.
In your case, we can take $r_1=R$, $r_2=R+A$, where $R$ is the radius of the Earth (I'm assuming a nice sphere, of course) and $A$ is the altitude of the
station above the surface of the Earth.
answered Aug 31 '15 at 21:07
copper.hat
125k559159
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
add a comment |
up vote
2
down vote
up vote
2
down vote
If $r$ is the distance from the Earth's centre, $phi$ is the longitude in radians (increasing from zero at Greenwich as one goes eastwards) and $theta$ is the
latitude in radians (increasing from zero at the Equator as one goes northwards), then
the $(x,y,z)$ coordinates are given by:
$$p(r,phi, theta) = r(cos phi cos theta, cos phi sin theta, sin phi)$$
Given two sets of coordinates $(r_k,phi_k, theta_k)$, the distance is given
by $|p(r_1,phi_1, theta_1)-p(r_2,phi_2, theta_2)|$.
In your case, we can take $r_1=R$, $r_2=R+A$, where $R$ is the radius of the Earth (I'm assuming a nice sphere, of course) and $A$ is the altitude of the
station above the surface of the Earth.
answered Aug 31 '15 at 21:07
copper.hat
125k559159
If $r$ is the distance from the Earth's centre, $phi$ is the longitude in radians (increasing from zero at Greenwich as one goes eastwards) and $theta$ is the
latitude in radians (increasing from zero at the Equator as one goes northwards), then
the $(x,y,z)$ coordinates are given by:
$$p(r,phi, theta) = r(cos phi cos theta, cos phi sin theta, sin phi)$$
Given two sets of coordinates $(r_k,phi_k, theta_k)$, the distance is given
by $|p(r_1,phi_1, theta_1)-p(r_2,phi_2, theta_2)|$.
In your case, we can take $r_1=R$, $r_2=R+A$, where $R$ is the radius of the Earth (I'm assuming a nice sphere, of course) and $A$ is the altitude of the
station above the surface of the Earth.
answered Aug 31 '15 at 21:07
copper.hat
125k559159
edited Aug 31 '15 at 22:43
answered Aug 31 '15 at 21:07
copper.hat
125k559159
answered Aug 31 '15 at 21:07
copper.hat
125k559159
answered Aug 31 '15 at 21:07
copper.hat
125k559159
125k559159
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
add a comment |
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
in your equation, what is the p in p(r,ϕ,θ) as well as k in (rk,ϕk,θk)?
– Orbit
Aug 31 '15 at 22:10
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
$p$ is a function $p:[0,infty) times [0,2 pi] times [-pi,pi] to mathbb{R}^3$. The $k$ is $1$ or $2$ and just selects one of a pair of coordinates.
– copper.hat
Aug 31 '15 at 22:29
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
That just gave me more questions =/ haha. If you have the time, could you provide an example using lat/long/alt with your first answers equations? I think that would help me understand the equation.
– Orbit
Aug 31 '15 at 22:35
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
I'm not sure what you are having issues with. Just substitute the various values in and compute. The radius of the Earth is about 6,400km, the ISS is about 400km above the surface, so $r_1 =6400, r_2 = 6800$. Then substitute the longitudes & latitudes (in radians) to compute the $(x,y,z)$ coordinates of the two points and then compute the Euclidean distance between the two points.
– copper.hat
Aug 31 '15 at 22:42
add a comment |
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5
Maybe convert your location and the location of ISS into rectangular coordinates?
– peterwhy
Aug 31 '15 at 20:55
If you want it to be very precise, you also need to know the radius of Earth at both points. IOW distance from the center of Earth rather than the altitude.
– Jyrki Lahtonen
Aug 31 '15 at 21:03