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, $$\frac{d \alpha_{\odot}}{dt} = \frac{\cos\varepsilon}{\cos^2 \delta_{\odot}} \frac{d\ell_{\odot}}{dt}$$

Thanks in advance!

  • $\begingroup$ To me, it's not clear what you're asking, you seem to have everything needed. $\endgroup$ Sep 22, 2022 at 13:42
  • $\begingroup$ @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 $\frac{d\ell_{\odot}}{dt}$. $\endgroup$
    – Cheng
    Sep 22, 2022 at 23:57


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