# Declination as a function of orbital position

Reading this Wikipedia article and following the derivations (https://en.wikipedia.org/wiki/Solar_irradiance#Milankovitch_cycles), I have found an equation I can't see where it comes from. Perhaps it's simple but my positional astronomy is on its basics.

It says that if $$\theta$$ the conventional polar angle describing a planetary orbit ($$\theta = 0$$ at the vernal equinox), $$\epsilon$$ is the obliquity, the declination $$\delta$$ as function of the orbital position is $$\delta = \epsilon \sin\theta$$.

Could you please explain or point to some resource? Thx.

Briefly, the declination varies smoothly from $$-\epsilon$$ to $$\epsilon$$ & back as $$\theta$$ ranges from 0 to $$2\pi$$. The function $$\delta=\epsilon\sin \theta$$ is the simplest way to do that. However, this formula is just a reasonable approximation. Using a spherical right triangle, we get $$\sin\epsilon = \frac{\sin\delta}{\sin\theta}$$ so $$\sin\delta=\sin\epsilon\sin\theta$$
• Let me a comment (I cannot insert a draw) in order to see if I am understanding: $\theta$ is the hypotenuse of that right triangle, measured over the ecliptic, $\delta$ is the cathetus from the Equator to the point where the Sun is and $\epsilon$ is the angle opposite to $\delta$. The fact that we measure $\theta$ along the ecliptic instead of along the Equator is slightly weird to me... – David Feb 21 '19 at 16:51
• @David Pretty much. The vertex of the triangle with the angle $\epsilon$ is where the ecliptic crosses the (celestial) equator. If it were a flat triangle, we'd have $\sin\epsilon=\delta / \theta$. We measure $\theta$ along the ecliptic because the Sun "travels" along the ecliptic. – PM 2Ring Feb 21 '19 at 17:06