Timeline for What is the angle between the North Celestial Pole and the Galactic Equator?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2022 at 13:24 | comment | added | John Alexiou | I edited the answer to use $\delta$ as the unknown, instead of $\theta$. | |
| Jan 28, 2022 at 13:23 | history | edited | John Alexiou | CC BY-SA 4.0 |
replaced theta with delta.
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| Jan 28, 2022 at 13:16 | comment | added | John Alexiou | @Astrofan - transform the equation to $$ \sin(δ)\cos(23.43928°)-\sin(192.8583°)\cos(δ)\sin(23.439289°) =\cos( 60.18894°) $$ and get the $A$, $B$ and $C$ coefficients by matching and using $\delta = \theta$ in notation. | |
| Jan 28, 2022 at 4:33 | comment | added | user52025 | None of the above responses answer the original question, at least not in a way that I can understand. If I ask Wolfram to solve arccos(sin(δ)cos(23.43928°)sin(192.8583°)cos(δ)sin(23.439289°))-60.18894°=0 the result is: δ = 0.473325 Radians converting Radian to Degrees, I get δ = 0.473325*(180/π) = 27.12837° Using Stellarium, I get δ = 27° 7' 41.7" = 27.12825° So δ = 27.128 is close enough for me. In the answers given above, I don't know what A, B or C mean, so this is as close as I'll ever get to teasing δ out of arccos(sin(δ)cos(23.43928°)sin(192.8583°)cos(δ)sin(23.439289°))=60.18894°. | |
| Nov 30, 2021 at 23:46 | comment | added | John Alexiou | No, as I stated above $\varphi = {\rm atan}(A/B)$. | |
| Nov 30, 2021 at 22:30 | comment | added | user52025 | φ = cos-1(A^2 + B^2 - C^2)/( 4 A B)? | |
| Nov 11, 2021 at 0:35 | comment | added | user52025 | Thanks for the answer, and the work you put into this, but I'm wondering what happened to φ in these calculations. I'm just a layman, with an undergrad's understanding of trig, and little to none of spherical trig. But I think I may have found another hint in an old post from the Stack Exchange: physics.stackexchange.com/questions/88663/… by user6972: δ=sin−1(sin(β)∗cos(ϵ)+cos(β)∗sin(ϵ)∗sin(λ)) However, I still can't figure out what the values are for β, ϵ, and λ. If I did, I could "punch them into a spreadsheet" and find φ ! | |
| Nov 11, 2021 at 0:12 | comment | added | John Alexiou | @user52025 in (3) $\theta = \varphi + \psi$ and $\varphi = {\rm atan}(A/B)$ | |
| Nov 10, 2021 at 13:24 | comment | added | John Alexiou | @DavidHammen, great suggestion. I have added this to the answer. | |
| Nov 10, 2021 at 13:23 | history | edited | John Alexiou | CC BY-SA 4.0 |
added 205 characters in body
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| Nov 10, 2021 at 9:18 | comment | added | David Hammen | My answer was your essentially equations (4) to (8). There's no need to write that up. There is one interesting case you didn't cover: What if $B+C=0$? | |
| Nov 10, 2021 at 9:05 | comment | added | John Alexiou | @DavidHammen - You should submit your answer anyway, as there are always things to be learned from the nuances of different approaches. | |
| Nov 10, 2021 at 9:02 | comment | added | David Hammen | You beat me to it. My partially written answer goes down the drain! Personally I like the tangent half angle expression (your equation 8) because that form makes the quadrant restrictions go away. The 360° (or $2\pi$) issue doesn't go away, but who cares? | |
| Nov 10, 2021 at 8:44 | history | edited | John Alexiou | CC BY-SA 4.0 |
added 218 characters in body
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| Nov 10, 2021 at 8:35 | history | answered | John Alexiou | CC BY-SA 4.0 |