# Derivation: Formula for projection of sun's motion onto parallel of declination

Consider the projected angular velocity of the sun along a parallel of declination. I was told that it is $$\frac{\cos\varepsilon}{\cos\delta_{\odot}} \frac{d\ell_{\odot}}{dt} \tag{1}$$

where we let $$\ell$$ be the ecliptic longitude of the sun and $$\varepsilon$$ be the obliquity of the ecliptic.

The explanation given is to verify that when: $$\delta_{\odot} = 0$$ (Equinox), we have, $$\cos\varepsilon \frac{d\ell_{\odot}}{dt}$$; $$\delta_{\odot} = \varepsilon$$ (Solstice) gives $$\frac{d\ell_{\odot}}{dt}$$.

This seems more like a hint to me, that the function takes on a particular form and values at end points. Hence, how can we complete the derivation, using the hints given above?

I am aware that the alternative derivation is to differentiate $$\tan \alpha_{\odot} = \cos \varepsilon \tan{\ell_{\odot}}$$ (After some substitutions, multiply by $$\cos \delta_{\odot}$$). However, I prefer a more intuitive, geometric argument.

For more context: Projecting onto the equator, $$\frac{d \alpha_{\odot}}{dt} = \frac{\cos\varepsilon}{\cos^2 \delta_{\odot}} \frac{d\ell_{\odot}}{dt}$$

• @Greg Miller I believe the derivation is far from complete... Maybe I'm too innocent to see it... Hence I was wondering whether anyone else knows how to complete it, using the hints given i.e. when: $\delta_{\odot} = 0$ (Equinox), we have, $\cos\varepsilon \frac{d\ell_{\odot}}{dt}$; $\delta_{\odot} = \varepsilon$ (Solstice) gives $\frac{d\ell_{\odot}}{dt}$. Sep 22, 2022 at 23:57