<p>Consider the projected angular velocity of the sun along a parallel of declination. I was told that it is <span class="math-container">$$\frac{\cos\varepsilon}{\cos\delta_{\odot}} \frac{d\ell_{\odot}}{dt} \tag{1}$$</span></p>
<p>where we let <span class="math-container">$\ell$</span> be the ecliptic longitude of the sun and <span class="math-container">$\varepsilon$</span> be the obliquity of the ecliptic.</p>
<p>The explanation given is to verify that when: <span class="math-container">$\delta_{\odot} = 0$</span> (Equinox), we have, <span class="math-container">$\cos\varepsilon \frac{d\ell_{\odot}}{dt}$</span>; <span class="math-container">$\delta_{\odot} = \varepsilon$</span> (Solstice) gives <span class="math-container">$\frac{d\ell_{\odot}}{dt}$</span>.</p>
<p>This seems more like a hint to me, that the function takes on a particular form and values at end points. <em><strong>Hence, how can we complete the derivation, using the hints given above?</strong></em></p>
<p>I am aware that the alternative derivation is to differentiate <span class="math-container">$\tan \alpha_{\odot} = \cos \varepsilon \tan{\ell_{\odot}}$</span> (After some substitutions, multiply by <span class="math-container">$\cos \delta_{\odot}$</span>). <em><strong>However, I prefer a more intuitive, geometric argument.</strong></em></p>
<hr />
<p>For more context:
Projecting onto the equator,
<span class="math-container">$$\frac{d \alpha_{\odot}}{dt} = \frac{\cos\varepsilon}{\cos^2 \delta_{\odot}} \frac{d\ell_{\odot}}{dt}$$</span></p>
<p>Thanks in advance!</p>

CC BY-SA 4.0
2022-09-22T13:42:08.720Z
To me, it's not clear what you're asking, you seem to have everything needed.
Greg Miller
CC BY-SA 4.0
2022-09-22T23:57:31.870Z
@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 $\frac{d\ell_{\odot}}{dt}$.
Cheng
<coordinate>
Derivation: Formula for projection of sun's motion onto parallel of declination
Cheng
# Derivation: Formula for projection of sun's motion onto parallel of declination

(Asked by: username_0 on 2022-09-22)

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.***

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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!

**Tags:** coordinate
> **Comments:**
> **username_1**: To me, it's not clear what you're asking, you seem to have everything needed.  
> **username_0**: @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 $\frac{d\ell_{\odot}}{dt}$.  

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## 0 answers

