Mixing
Circumpolar Equation Derivation
$begingroup$
Edit: Another algebraic solution from Astronomy.stackexchange
I am trying to understand how the circumpolar star equation works. Although it is astronomy, this part is pretty much just geometry. The equation is this
$delta + l geq frac{pi}{2}$
The setup is for an observer at latitude $l$, stars whose declination $delta$ fit this equation are circumpolar. What I cannot get past is how to get from the star's lowest altitude $a$ to there. *Declination is measured from the equator and altitude is measured from the horizon for those who don't know." I know that the angle of the cone made by the rotating star is within the $frac{pi}{2}$ limitation. I just can't express it in an algebraic way.
This shows what is meant by circumpolar stars.
This shows the sort of rotation undergone by the celestial sphere due to the observer's latitude. NCP and SCP are North and South Celestial Poles.
In the second picture, it is easy to prove that the altitude $a$ is equal to longitude $l$ and that $c = frac{pi}{2} - l$. What I need to know is how to go about proving that a star of declination $delta$ will always have an altitude $a geq 0$.
Edit: Included pictures in line instead of as a link.
geometry mathematical-astronomy
edited Jan 25 at 21:26
Aretino
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
$endgroup$
add a comment |
$begingroup$
Edit: Another algebraic solution from Astronomy.stackexchange
I am trying to understand how the circumpolar star equation works. Although it is astronomy, this part is pretty much just geometry. The equation is this
$delta + l geq frac{pi}{2}$
The setup is for an observer at latitude $l$, stars whose declination $delta$ fit this equation are circumpolar. What I cannot get past is how to get from the star's lowest altitude $a$ to there. *Declination is measured from the equator and altitude is measured from the horizon for those who don't know." I know that the angle of the cone made by the rotating star is within the $frac{pi}{2}$ limitation. I just can't express it in an algebraic way.
This shows what is meant by circumpolar stars.
This shows the sort of rotation undergone by the celestial sphere due to the observer's latitude. NCP and SCP are North and South Celestial Poles.
In the second picture, it is easy to prove that the altitude $a$ is equal to longitude $l$ and that $c = frac{pi}{2} - l$. What I need to know is how to go about proving that a star of declination $delta$ will always have an altitude $a geq 0$.
Edit: Included pictures in line instead of as a link.
geometry mathematical-astronomy
edited Jan 25 at 21:26
Aretino
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
$endgroup$
$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23
add a comment |
$begingroup$
Edit: Another algebraic solution from Astronomy.stackexchange
I am trying to understand how the circumpolar star equation works. Although it is astronomy, this part is pretty much just geometry. The equation is this
$delta + l geq frac{pi}{2}$
The setup is for an observer at latitude $l$, stars whose declination $delta$ fit this equation are circumpolar. What I cannot get past is how to get from the star's lowest altitude $a$ to there. *Declination is measured from the equator and altitude is measured from the horizon for those who don't know." I know that the angle of the cone made by the rotating star is within the $frac{pi}{2}$ limitation. I just can't express it in an algebraic way.
This shows what is meant by circumpolar stars.
This shows the sort of rotation undergone by the celestial sphere due to the observer's latitude. NCP and SCP are North and South Celestial Poles.
In the second picture, it is easy to prove that the altitude $a$ is equal to longitude $l$ and that $c = frac{pi}{2} - l$. What I need to know is how to go about proving that a star of declination $delta$ will always have an altitude $a geq 0$.
Edit: Included pictures in line instead of as a link.
geometry mathematical-astronomy
edited Jan 25 at 21:26
Aretino
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
$endgroup$
Edit: Another algebraic solution from Astronomy.stackexchange
I am trying to understand how the circumpolar star equation works. Although it is astronomy, this part is pretty much just geometry. The equation is this
$delta + l geq frac{pi}{2}$
The setup is for an observer at latitude $l$, stars whose declination $delta$ fit this equation are circumpolar. What I cannot get past is how to get from the star's lowest altitude $a$ to there. *Declination is measured from the equator and altitude is measured from the horizon for those who don't know." I know that the angle of the cone made by the rotating star is within the $frac{pi}{2}$ limitation. I just can't express it in an algebraic way.
This shows what is meant by circumpolar stars.
This shows the sort of rotation undergone by the celestial sphere due to the observer's latitude. NCP and SCP are North and South Celestial Poles.
In the second picture, it is easy to prove that the altitude $a$ is equal to longitude $l$ and that $c = frac{pi}{2} - l$. What I need to know is how to go about proving that a star of declination $delta$ will always have an altitude $a geq 0$.
Edit: Included pictures in line instead of as a link.
geometry mathematical-astronomy
geometry mathematical-astronomy
edited Jan 25 at 21:26
Aretino
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
edited Jan 25 at 21:26
Aretino
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
edited Jan 25 at 21:26
Aretino
25.3k21445
edited Jan 25 at 21:26
Aretino
25.3k21445
edited Jan 25 at 21:26
Aretino
25.3k21445
25.3k21445
asked Jan 25 at 9:17
abyssmuabyssmu
425
asked Jan 25 at 9:17
abyssmuabyssmu
425
asked Jan 25 at 9:17
abyssmuabyssmu
425
425
$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23
add a comment |
$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23
$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $pi/2-L$.
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
$endgroup$
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
add a comment |
$begingroup$
For a star to be circumpolar, the star must, at the very least, come above the horizon.
$0 leq a leq frac{pi}{2}$
In the celestial sphere coordinate system, $c$ can be found to be
$c = frac{pi}{2} - delta$
In the horizon coordinate system, $c$ can be found to be
$c = l - a_{min}$
Therefore
$l - a_{min} = frac{pi}{2} - delta$
$\$
$l + delta = frac{pi}{2} + a_{min}$
$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.
$l + delta = frac{pi}{2}$
$delta$ must be in the cone made by the star's rotation about its declination angle. Therefore
$frac{pi}{2} - l leq delta$
answered Jan 25 at 20:33
abyssmuabyssmu
425
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $pi/2-L$.
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
$endgroup$
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
add a comment |
$begingroup$
See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $pi/2-L$.
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
$endgroup$
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
add a comment |
$begingroup$
See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $pi/2-L$.
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
$endgroup$
See picture below: a star is circumpolar if it lies within the grey cone of half-width $L$, the same as the latitude. Hence declination of such a star must be greater than $pi/2-L$.
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
answered Jan 25 at 17:09
AretinoAretino
25.3k21445
25.3k21445
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
add a comment |
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
$begingroup$
This is a very elegant solution. I like it. I did manage to find a solution a little while ago, albeit not as clean as this, but it is more so algebraic instead of geometric. I will post the solution as an answer as well. Thank you!
$endgroup$
– abyssmu
Jan 25 at 20:28
add a comment |
$begingroup$
For a star to be circumpolar, the star must, at the very least, come above the horizon.
$0 leq a leq frac{pi}{2}$
In the celestial sphere coordinate system, $c$ can be found to be
$c = frac{pi}{2} - delta$
In the horizon coordinate system, $c$ can be found to be
$c = l - a_{min}$
Therefore
$l - a_{min} = frac{pi}{2} - delta$
$\$
$l + delta = frac{pi}{2} + a_{min}$
$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.
$l + delta = frac{pi}{2}$
$delta$ must be in the cone made by the star's rotation about its declination angle. Therefore
$frac{pi}{2} - l leq delta$
answered Jan 25 at 20:33
abyssmuabyssmu
425
$endgroup$
add a comment |
$begingroup$
For a star to be circumpolar, the star must, at the very least, come above the horizon.
$0 leq a leq frac{pi}{2}$
In the celestial sphere coordinate system, $c$ can be found to be
$c = frac{pi}{2} - delta$
In the horizon coordinate system, $c$ can be found to be
$c = l - a_{min}$
Therefore
$l - a_{min} = frac{pi}{2} - delta$
$\$
$l + delta = frac{pi}{2} + a_{min}$
$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.
$l + delta = frac{pi}{2}$
$delta$ must be in the cone made by the star's rotation about its declination angle. Therefore
$frac{pi}{2} - l leq delta$
answered Jan 25 at 20:33
abyssmuabyssmu
425
$endgroup$
add a comment |
$begingroup$
For a star to be circumpolar, the star must, at the very least, come above the horizon.
$0 leq a leq frac{pi}{2}$
In the celestial sphere coordinate system, $c$ can be found to be
$c = frac{pi}{2} - delta$
In the horizon coordinate system, $c$ can be found to be
$c = l - a_{min}$
Therefore
$l - a_{min} = frac{pi}{2} - delta$
$\$
$l + delta = frac{pi}{2} + a_{min}$
$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.
$l + delta = frac{pi}{2}$
$delta$ must be in the cone made by the star's rotation about its declination angle. Therefore
$frac{pi}{2} - l leq delta$
answered Jan 25 at 20:33
abyssmuabyssmu
425
$endgroup$
For a star to be circumpolar, the star must, at the very least, come above the horizon.
$0 leq a leq frac{pi}{2}$
In the celestial sphere coordinate system, $c$ can be found to be
$c = frac{pi}{2} - delta$
In the horizon coordinate system, $c$ can be found to be
$c = l - a_{min}$
Therefore
$l - a_{min} = frac{pi}{2} - delta$
$\$
$l + delta = frac{pi}{2} + a_{min}$
$a_{min}$ can be set to 0 because it is the lower bound for circumpolarity.
$l + delta = frac{pi}{2}$
$delta$ must be in the cone made by the star's rotation about its declination angle. Therefore
$frac{pi}{2} - l leq delta$
answered Jan 25 at 20:33
abyssmuabyssmu
425
answered Jan 25 at 20:33
abyssmuabyssmu
425
answered Jan 25 at 20:33
abyssmuabyssmu
425
answered Jan 25 at 20:33
abyssmuabyssmu
425
425
add a comment |
add a comment |
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$begingroup$
Declination is measured from the equator going up to 90 at the north pole and -90 at the south. It's the exact same thing as latitude, but for the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 14:11
$begingroup$
In the image for horizon coordinates, what are NCP and SCP?
$endgroup$
– Lee David Chung Lin
Jan 25 at 17:04
$begingroup$
They are the North and South Celestial Poles. They are projections of the N/S poles of Earth onto the celestial sphere.
$endgroup$
– abyssmu
Jan 25 at 20:23