Thanks. Let $\Theta$ be the x-y-z coordenates in the plane of the horizon, with the origin $O$ at my position, and z axis pointing to the zenith. $\Pi$ the x'-y'-z' coordinates on the plane of the Sun's "orbit" on the celestial sphere (the ecliptic I think it is called), with the origin at $O$ and intersecting $\Theta$ along the x axis, so angle between $\Theta$ and $\Pi$ is $\alpha$. Assuming the "Sun" rotates in $\Pi$ with angular velocity $\omega$, the position in $\Pi$ reference frame is $s(t)=(cos \omega t, sin \omega t, 0)$. The position in the $Theta$ reference frame is therefor $r(t)=(cos \omega t, cos \alpha sin \omega t, 0)$. Now I just differentiate $r$ to get $\Omega$.
*
Last edited by ENRIQUESTEFANINI; Jun 3rd 2017 at 04:45 PM.
* |