Ecliptic Longitude and Right Ascension of the Sun

[Helios]

Here's a question. Where do the ecliptic longitude and the right ascension ( of the Sun )reach their maximum difference? What is the declination for these events? What calendar dates do you suppose this happens? I know the answer. I've never heard anyone make this observation except me. These events don't even have a name?! We can even name them here and now for the first time!

 

[giann_tee]

What are you saying? Ecliptic longitude is measured on one great circle, and right ascension on another one. The sun travels the ecliptic in non-uniform way strictly speaking due to eccentricity of the orbit. Let's say its about a degree a day. Then, sun's coordinate point on the equator also travels like a shadowy projection. If obliquity was really big, the speed on the equator of that point (defined by intersection of great circle containing north celestial pole P and the sun) travels a little quicker starting from gamma point, then slower, then quicker and there I give up. What is the difference between the two?

 

[Helios]

OK, here's a hint. The relationship between the two is

tan \alpha = tan \lambda cos \epsilon

Right Ascension ( alpha )
Ecliptic Longitude ( lambda )
Obilquity Angle ( epsilon ) = 23.439°

Find where \lambda - \alpha reaches the maximum ( 4 times per year ).

 

[giann_tee]

I wrote all equations for triangle declination-ascension-longitude but I don't know what to do next.
sin(alpha)=cos(e)sin(lambda)
Right? Lambda is changing at constant rate (pretend), alpha is whatever.

 

[Helios]

That's not it.
write
f = \lambda - \alpha = \lambda - arctan( tan\lambda cos\epsilon)

then find df/d\lambda
set this to zero to find maximum
then values for \lambda
and \alpha will follow.

 

[giann_tee]

That's not it.
write
f = \lambda - \alpha = \lambda - arctan( tan\lambda cos\epsilon)

then find df/d\lambda
set this to zero to find maximum
then values for \lambda
and \alpha will follow.

I haven't reverse engineered your equation, normally I'd trust it but it seems more complicated than necessary. I used program "Derive" to solve the first differential as an equation but all solutions seem to be complex! It appears absurdly complicated.

Clearly, equator and ecliptic are two great circles on same sphere at an angle. Great circle ecliptic_poles-sun is rotating and so is celestial_poles-sun circle which is tracing sun on equator and recording declination. All 3 coordinates are one-on-one and one on date of tropical year (short term). Hence let's see the tables with numbers...

 

[giann_tee]

...and I get

for l=0 to 360
RA=acos[cos(l) / asin( sin(e)sin(l) )]

 

[giann_tee]

May 31st 2009, angular difference worth almost 2.5 degrees. How about that?

 

[Helios]

Well, the month is right, but you're still cold

 

[giann_tee]

I protest! First its irrelevant thing to know. Second I didn't design these awful arc sine functions.

 

[Helios]

Irrelevant? It's not irrelevant. Know that there's an "Equation of Time" that corrects sundial time to local mean time. The Sun runs fast or slow compared the mean time. Should we correct for obliquity, we must know
\lambda- \alpha
The fact that it has extrema should be interesting to any calculus student. Oh well.

So there's another feature of these same points!
Firstly, Ecliptic Slope (sigma) is related to declination by
cos\epsilon = cos d cos\sigma

Find where the declination equals the ecliptic slope or the negative of the ecliptic slope.
Then use
sin d = sin \lambda sin \epsilon

This approach is too easy, a dead give-away to the problem.