We already mentioned in the article about primary directions that house cusps represent intersections of lines dividing the celestial sphere into 12 parts with the zodiac circle.
This article will derive equations for calculating house cusps in the Placidus system. We will use the mundane position formula in the Placidus system we derived earlier.
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House Cusps Equations
Let's, for simplicity, consider the equation of the 12th house. By definition, the line of the 12th house is the set of points that have passed 1/3 of their ascending path. In other words, for each point, the ratio of the upper meridian distance to the diurnal semi-arc is 1/3. In other words
$$ R = \frac{UMD}{DSA} = 1/3 $$
From equation (3) of the previous article on mundane positions, we will get the equation of the S-curve for the 12th house:
This equation expresses the right ascension $RA$ of any point on the S-curve as a function of declination $D$.
On the other hand, the ecliptic plane is a set of points, described by eq. (1) of the ecliptic plane.
Since the 12th house cusp is an intersection of the S-curve with the ecliptic plane, it has coordinates $(RA, D)$, which are described by the following set of equations:
The set of equations (2) and similar equations for other houses don't have obvious analytical solutions (roots), so they can be solved numerically.
Numerical Calculation (the Old Way)
This method is an iteration of a convergent sequence. For example, we calculate the cusp of the 12th house.
- At the zero iteration, we set aside a third of the diurnal semi-arc $DSA_0$ of the equator from MC. The resulting equatorial coordinate (right ascension) is denoted by $R_0$. Further, from the ecliptic equation we find the declination $D_0$, which corresponds to right ascension we just found.
- Next, we calculate diurnal semi-arc $DSA_1$ corresponding to the found declination.
- After that, we set aside a third of $DSA_1$ from MC and denote the resulting right ascension as $R_1$.
- After that, we repeat the cycle for $R_1$
Convergent right ascension sequence
We get a convergent right ascension sequence $[R_0, R_1, R_2, ... R_n]$ where $\Delta R_n \rightarrow 0$.
If we express the length of the diurnal semi-arc as a function of right ascension $DSA(RA)$, then we get the sequence
and so on
How to Find the Length of the Diurnal Semi-Arc?
From the equation (3) of diurnal semi-arc and equation (1) of ascension difference we get
where $\phi$ is the geographic latitude of the observer.
Taking the cosine of the right and left sides of the equation and substituting equation (2) of the ecliptic, we get
Here $\epsilon$ is the inclination of the ecliptic.
So our algorithm looks like this:
and so on, until $\Delta = RA_{n+1} - RA_n$ reaches the desired order of magnitude.
What is the Problem With This Approach?
First, in this form, the asymptotic solution finds exactly one cusp of the house. However, as shown in the article about the Placidus house system, we have several intersections of the dividing house curve with the ecliptic, that is, several cusps of the 12th or 11th house under certain conditions.
Secondly, this algorithm does not apply to extreme latitudes. It may happen that the ecliptic point with coordinates $RA_n$ never rises or sets. We cannot calculate $DSA(RA)$ in this case. Even if we try, we will get a mathematical expression for the arccosine of a number greater than one, leading to a computational error.
Knowing this feature of the Placidus house system at extreme latitudes, one should apply any other numerical method that finds several roots of equation (2) of this article without going beyond the declinations at which the ascent of the ecliptic points is still possible.
