Equation for ASC & MC

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Equation for ASC & MC

March 11, 2023, 11:17 a.m. Mark Rusborn 2 min. to read

In the previous article, we introduced the concept of sidereal time. The sidereal day begins at 00:00, when 0° Aries culminates, and 0° Cancer rises. Therefore, by knowing the current sidereal hour, one can predict the exact degree of the ecliptic that rises and culminates.

Let's derive formulas for ASC and MC. All we know is local coordinates (geographic latitude) and local sidereal time (expressed in RAMC terms).

ASC Equation

Let's look at the figure below:

Fig. 1 - ASC.

We have two right spherical triangles:

  • First with the sides $OA + AD$, $ASC$ and angle $\epsilon$ between them
  • Second with the sides $D$, $AD$ and angle $\epsilon$

Here we use the following notation:

Let's use our handy equations for spherical triangles.

From (3) we have

$$ \sin(AD + OA) = \frac{\tan D}{\tan\epsilon}\tag{1.a} $$ $$ \sin AD = \frac{\tan D}{\tan(90° - \phi)}\tag{1.b} $$

From (6) we have

$$ \tan ASC = \frac{\tan(AD + OA)}{\cos\epsilon}\tag{1.c} $$

Let's divide $(1.a)$ by $(1.b)$ and expand the sine of two angles by ($2$)

$$ \tan AD = \frac{\sin OA \tan\epsilon\tan\phi}{1-\cos OA\tan\epsilon\tan\phi}\tag{1.d} $$

We will rewrite $\tan(AD + OA)$ in (1.c) in form:

$$ \frac{\tan AD + \tan OA }{1 - \tan AD \tan OA } $$

By substituting $(1.d)$ in the last equation, we have the formula for the ASC:

$$ \tan ASC = \frac{\sin OA}{\cos\epsilon\cos OA - \tan\phi\sin\epsilon} $$

As we discussed earlier, the $OA_{ASC} = RAMC + 90°$. It means that:

$$ \tan ASC = \frac{-\cos RAMC} {\cos\epsilon\sin RAMC + \tan\phi\sin\epsilon}\tag{2} $$

MC Equation

Let's look at the figure below:

Fig. 2 - MC.

We have the right triangle with sides $360° - RAMC$, $360°-MC$, and angle $\epsilon$ between them.

From the ($9$) it follows that:

$$ \tan(360° - MC) = \cos\epsilon \tan(360° - RAMC) $$

It gives us the equation for the MC:

$$ \tan MC = \frac{\tan RAMC}{\cos\epsilon}\tag{2} $$

Bottom Line

We have derived equations for ASC and MC for a given sidereal time $t = RAMC / 15$ and a given geographical latitude $\phi$

$$\begin{align} &\tan ASC = \frac{-\cos RAMC} {\cos\epsilon\sin RAMC + \tan\phi\sin\epsilon}\tag{1} \\\ &\tan MC = \frac{\tan RAMC}{\cos\epsilon}\tag{2} \end{align}$$

Mark Rusborn

Mark Rusborn

I am a former Soviet physicist, now a professional astrologer. I have been lucky to get familiar with eminent scientists, including the Nobel laureate in physics. It helped me a lot in the proper structuring of thinking.

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