# How to tell which of the ecliptic cross horizon angles is to the east?

Jean Meeus' Astronomical Algorithms, 2nd ed. p. 99 gives a formula for finding the two longitudes where the ecliptic crosses the horizon:

$$\tan\lambda = \frac{-\cos\theta}{\sin\epsilon \tan\phi + \cos\epsilon\sin\theta}$$

where "$$\epsilon$$ is the obliquity of the ecliptic, $$\phi$$ is the latitude of the observer, and $$\theta$$ is the local sidereal time, returns $$\lambda$$ the longitudes of the two points of the ecliptic which are (180 degrees apart) on the horizon".

The software I am using then applies $$\arctan$$ to get $$\lambda$$ and adds $$\pi$$ if $$\lambda < 0$$. Then it returns $$\lambda$$ and $$\lambda + \pi$$.

I need to know which of the two angles returned points to the easterly direction. Is there any way to tell which is which? I have all of the information needed to perform the calculation (UT, local time, sidereal time, latitude, longitude), but I can't see how they are related.

Thanks much!

• $\lambda$ must be the ecliptic longitude, but what are the other values? – JohnHoltz May 18 '20 at 12:27
• @JohnHoltz Good question, I've updated the text to say include what each value means. – Shawn Lauzon May 19 '20 at 2:29

These steps can be used to find which of the two horizons is to the east:

The local sidereal time equals the right ascension ($$\alpha$$) of the meridian. From equation (13.3) that transforms the ecliptical into equatorial coordinates, the ecliptic longitude of the meridian ($$\lambda_m$$) can be calculated. Since the ecliptic latitude is zero ($$\beta=0$$) on the ecliptic,

$$\frac{\tan(\alpha)}{\cos(\epsilon)} = \tan(\lambda_m)$$

Note that $$\lambda_m$$ must be in the same quadrant as $$\alpha$$. If this is not the case, add 180 degrees. Of the two ecliptic longitudes calculated to intersect the horizon, the one that is approximately 90 degrees greater than the adjusted $$\lambda_m$$ is the one that is in the east.

• Thank you! I'll try this out tomorrow and let you know if it works. – Shawn Lauzon May 19 '20 at 14:17
• Unfortunately this isn't giving me the result I expect for my test case. I'm using $\theta=2.571rad$ and $\epsilon=0.409rad$. If I put those results in my original formula I get the two angles $.861rad$ and $4.003rad$ of which I know the latter is the correct. If I use your formula, substituting $\theta$ for $\alpha$, I get $-.610rad$ to which I can add $\pi$ and arrive at $2.531rad$ which is not either of the expected values. Do you have any advice? – Shawn Lauzon May 20 '20 at 3:00
• Thanks for your comment @JohnHoltz, and that works for the original problem, but not for another one. With a sidereal time of 3.575 rad, the two longitudes calculated are 1.449 (Q1) and 4.591 rad (Q3) (the second one is east). $\lambda_m$ is calculated to be 0.467 rad (Q1). Adding $\pi/2$ = 2.0381 rad (Q2) which does not point to the correct answer. I imagine I'm doing something wrong, but can't tell what. – Shawn Lauzon May 20 '20 at 8:35