Elisabeth Queen Consort Spain
02 Avr 1545 JUL CAL
friday JUL
Fontainebleau | lat 48° 24' 10" | N 2°42' E
France
---------------------------------
natal 23h 30' 0"
lmt 23h 19' 12"
tu 23h 19' 12"
tsn 12h 56' 40"
---------------------------------
timezone meantime
Equation of time 0h 0' 20"
ΔT 0h 2' 32"
---------------------------------
source : https://www.astro.com/astro-databank/Elisabeth,_Queen_Consort_of_Spain_(1545)
French-Spanish royalty, the eldest daughter of Henry II of France and Catherine de' Medici. Elisabeth married Philip II of Spain ("Philip the Catholic"), son of Charles V, Holy Roman Emperor, and Isabella of Portugal in 1559. Elisabeth's first pregnancy in 1564 ended with a miscarriage of twin girls. She later gave birth to Infanta Isabella Clara Eugenia of Spain on 12 August 1566, and then to Isabella's younger sister Catherine Michelle of Spain on 10 October 1567. Elisabeth had another miscarriage on 3 October 1568, and died the same day, along with her newborn infant daughter.
theme markers
conj MO -JU in I but unfortunate because MO D and JU F.
conj SA - ASC with SA r unfortunate
opp ARGOL (β Persei) - ASC unfortunate
conj VE - MA in VI unfortunate (VE and MA P)
Hyleg : MO
Alchocoden : JU
angular house but taking into account the dignities, only lesser years : mundane natal -> 27.6 Y.
If we take SU for hyleg, the alchocoden is MA : mundane years -> 25 Y
mundane chart
| MUND | |||||||
| RAY ORB ±5 | SU | MO | ME | VE | MA | JU | SA |
| SU | |||||||
| MO | |||||||
| ME | 60 | ||||||
| VE | 60 | 90 | |||||
| MA | 60 | 90 | 0 | ||||
| JU | 0 | 60 | |||||
| SA |
In the case of Elisabeth Queen Consort Spain we have the table below which allows us to estimate the breakdown of aspects between the different planets and the alchocoden.
| MUND | SU | MO | ME | VE | MA | JU | SA | rgo |
| MA | 61,73 | 142,18 | 86,31 | 3,20 | 0,00 | 149,69 | 167,42 | |
| P | 61,73 | 86,31 | 3,20 | 0,00 | ||||
| 60 | 0 | 90 | 0 | 0 | 0 | 0 | ||
| 9,50 | 0,00 | -1,11 | 0,67 | 0,00 | 0,00 | 0,00 | 9,06 | |
| 60 | 0 | 90 | 0 | 0 | 0 | 0 | ||
| 0,50 | -1,00 | 0,17 | 0,08 | -0,08 | 0,08 | 0,08 | ||
| E | D | P | P | P | F | T |
So, we can add the lesser Y of MA (cadent = 19 Y) and the Σ 9.06 = 24.06 Y
Primary directions
| Lat | Dec | AR | MD | SA | HA | |
| SU | - | 8,79 N | 20,84 | -7,65 N | 79,97 N | 87,62 W |
| □MA | 0 S | -3 S | 353,07 | -20,11 N | 93,39 N | 113,5 E |
MD = meridian distance (from MC if SA f SU is diurnal or IC if Sa f is nocturnal) – SA = semi-arc (if f is diurnal, SA f SU is D and all MD’s and SA’s are D, otherwise N – HA = horizontal distance (from the nearest horizon W or E for f SU and m □MA)
DIRECTION : □MA conj SU
---------------------------------
We must take into account an important element: the ascensional difference (DA); it can be observed on the graph in a dotted line (measured between the horizon and the axis of the pole). This is the difference between Right Ascension (AR) and Oblique Ascension (OA). This is the difference between Right Ascension (AR) and Oblique Ascension (OA). DA is always calculated in absolute value |DA| and it is added or subtracted from 90° (SA = 90° corresponds to a point on the equator cut by the horizon; depending on whether a star approaches or moves away from the line of horizon, SA is > 90° or < 90°, i,e, (+) depending on whether it is diurnal and northern ; or nocturnal and southern ; (-) depending on whether it is diurnal and southern ; or nocturnal and northern.
sin(DA) = -tan(φ)tan(δ)
φ = latitude 48,4 N
δ SU = 8,79 +
DA-SU = 10,03°
δ □MA =-3 -
DA-□MA =3,39°
We will first use the Placidus system of mundane directions. The simplest system is that of Choisnard-Fomalhaut. First you need to retrieve the data from the SA (semi-arc) and the DM (meridian distance) of the nocturnal point because the altitude of SU is -32,41°. important note: the SA and DM of the two points are always counted diurnal if the first point (here SU) is above the horizon even if the second is below. They are counted nightly if the first point (SU) is below the horizon regardless of the position of the second point.
For DMs, they are counted in AR from the diurnal meridian if the fixed point SU is diurnal, and from the nocturnal meridian if it is nocturnal.
nocturnal meridian MC = 13,18°
AR SU = 20,83°
AR □MA = 353,07°
SA N (d+) SU = 79,97°
DM N SU = -7,65°
For the significator □MA altitude (h) =-41,43°. so :
= 93,39°
DM N □MA = -20,11°
Then we compute Saf/DMf (so : SA f [ 79,97°] / DM f [ -7,65°])
Sa f / DM f =10,45
and the angle x = SAm x DM f/SA f, so : SA m [ 93,39°] x DM f [ -7,65°]/SA f [ 79,97°]
x = 351,07°
We find the direction by DMm - x, so : DM m [ -20,11°] ± x [351,07]
We must now have regard to the double ± sign of the last expression; in the case where f (SU) and m (□MA) are on either side of the meridian, the direction arc is obtained by taking the sum (instead of the difference) of the two quantities DMm and x. this is the case here so sign = (-)
the computation of the arc requires, depending on the case, a reduction of 360° (so arc modulo 360°)
---------------------------------
arc D =29,04°
---------------------------------
in the technical sense, It is a direct direction but in the astrological sense, it is a true converse direction since it is an aspect considered as a promissor which goes towards the significator. ; so the m point is an aspect (here □MA) and the f point is a planet or an axis, (here SU)
We can now compute the converse direction : point f is directed towards point m, i.e. the star is directed towards the aspect. This is where the problem of the orientation of the primum mobile arises because it is not concevable to rotate the local sphere in both directions… It does not seem convenient to postulate that the arc of direction is counted in the order of the signs of the zodiac (when it is direct, i.e. when one directs a promissor towards a significator): indeed, the ecliptic has nothing to do with a direction since this one depends only on the diurnal movement ( primum mobile). It is therefore otherwise that we must pass judgment on this.
That time, we compute Sa m / DM m (so : SA m [ 93,39°] / DM f [ -20,11°])
Sa m / DM m =0,58
and the angle x = SA f x DM m/SA m, so : SA f [ 79,97°] x DM m [ -20,11°] / SA m [ 93,39°]
x = -17,22°
We find the direction by DM f - x, so : DM f [ -7,65°] ± x [-17,22°]
We must now have regard to the double ± sign of the last expression; in the case where m (□MA) and f (SU) are on either side of the meridian, the direction arc is obtained by taking the sum (instead of the difference) of the two quantities DM f and x. this is the case here : so, signe = (-)
---------------------------------
arc C =24,87°
---------------------------------
Now we can study the same direction with the Regiomontanus system. To obtain the arc of direction between two signifying points (planets in body, aspect versus planet, planet versus axis) one must find AO (oblique ascent) of f and of m, calculated under the pole of f.
The formulas to use can be found either in the Dictionnaire astrologique of Henri Joseph Gouchon (Dervy Livres, 1937) pp. 266-267, or in his Horoscope annuel simplifié (Dervy, 1973) p.181. Other formulas can be found in Les moyens de pronostic en astrologie, Max Duval (editions traditionnelles, 1986) and Domification et transits (Editions traditionnelles, 1985). We can also cite by André Boudineau : Les bases scientifiques de l’astrologie (Chacornac, 1937) These are references in French but there are many other references in English or German of a less obvious but equally valid use.
First, compute the ascensional difference under f (SU) : cot DAP f = (cot de f x cot lat) / in DM f ± cot DM f, i,e, : cot (DAP f) = (Cot dec f[8,79°] x Cot Lat [48,4°]) /sin DM f [7,65°] ± cot DM f [7,65°]
DAPf = 1,13°
We find the pole of f (SU) by formula : tan(pole f) = sin (DAP f) x cot (dec f) i,e, tan(pole f) = tan f [178,39°] x cot f [8,79°]
pole SU regio =10,27°
(1) We need now the DAP of m (□MA) under the pole of f, sin (DAP m) = tan (pole f) x tan (DEC m), i,e, : (SU) : sin (DAPm/f) = tan [10,27°] x tan [-3°]
DAP m/f = -0,54°
then we find for the points located in the eastern part of the chart : AO f = AR f± DAP f ; sign (+) if Dec f boreal or sign (–) if Dec f Austral ; so : AO f SU = 22,44° and AO m = AR m ± DAP m ; idem for sign ; so AO m□MA = 353,62°
---------------------------------
arc D Regio = -29,91°
---------------------------------
We are now going to compute the converse Regiomontanus direction corresponding to the arc f SU / p □MA
First, compute the ascensional difference under m (□MA) : cot DAPm = (cot dec m x cot lat)/sin DM m ± cot DM m, i,e, : Cot(DAP m) = (Cot decm[-3°] x Cot Lat [48,4°]) / Sin DM f [159,89°] ± Cot DM m [339,89°]
DAP m = 1,23°
We find the pole of m (□MA) by formula : Tan(pole m) = Sin (DAP m) x Cot (dec m) i,e, Tan(pole m) = Sin m [1,1°] x Cot m [-3°]
pole □MA regio =-20,17°
We need now the DAP of f (SU) under the pole of m, Sin (DAP f) = Tan (pole m) x Tan (DEC f), i,e, : (□MA) : Sin (DAP f/m) = Tan[20,15°] x Tan [8,79°]
DAP f/m = 3,25°
then we find for the points located in the eastern part of the chart : AO m = AR m ± DAP m ; sign (+) if Dec m boreal or sign (–) if Dec m Austral ; so : AO m □MA = 354° and AO f = AR f ± DAP f ; idem for sign ; so AO f SU = 24,09°
---------------------------------
arc C Regio = 23,41°
---------------------------------
H.J. Gouchon [l’Horoscope Annuel simplifié, Dervy, 1973, p, 181-182 and Dictionnaire astrologique, p, 277, 1937-1942, Gouchon ed., 1975, Dervy, but be careful because in DAP's equation, the double sign ± was mistakenly replaced by the sign (-) ] advises to avoid errors, to always place the star A1 (for us f, i.e. SU) in the eastern houses; in fact it is enough to change the registration number of the house based on the transformation (IV-V-VI) -> (X-XI-XII) and (VII-VIII-IX) -> (I-II-III ) to adapt the double sign ± in the calculation of DAP f or DAP m; moreover, this sign must be reversed if |DM| > 90°.
For the Regiomontanus directions, there is another mode of computing, mentioned by Gouchon (Dictionnaire astrologique, op. cit., p. 276) and especially Martin Gansten (Primary directions, pp. 155-157, the Wessex Astrologer, 2009)
This method consists at computing first 3 auxiliary angles before the pole. It then joins the other method. Contrary to what Gouchon says, I find it easier than the previous one because we avoid the double sign ± in the determination of DAP f.
So, initially, we have A => Tan f = tan dec f [8,79°] / cos DM f [7,65°]
A = 8,87°
Then : B = Lat [48,4°] + A [8,87°]
B = 57,27°
And, Tang C = Cot DM f [7,65°] x Cos B [57,27°] / Cos A [8,87°]
C = -76,21°
Then, we have Sin pole f = Cos C [-76,21°] x Sin LG [48,4°]
---------------------------------
So, pole SU regio = 10,27°
---------------------------------
Now go back to (1)
For m □MA; we have : A => Tan m = tan dec m [-3°] / cos DM m [339,89°]
A = -3,19°
Then : B = Lat [48,4°] + A [-3,19°]
B = 45,21°
And, Tang C = Cot DM m [-20,11°] x Cos B [45,21°] / Cos A [-3,19°]
C = -62,57°
Then, we have Sin pole m = Cos C [-62,57°] x Sin LG [48,4°]
---------------------------------
So, pole □MA regio = 20,15°
---------------------------------
Now go back to (1)
ARMILLARY SPHERAE
This armillary sphere presents us with a true stereographic projection of the DIRECTION : □MA conj SU
We see the superior meridian upper the pole 48,4° N LAT , the inferior meridian, and the other great circles : equator, ecliptic λ, latitude circle β, azimuth circle A and horary circle H
the zenith with colat 41,6° and the prime vertical
the horizon with ecliptic inclination of 41,74° and the ecliptic pole at 48,26°
the line Nord-Sud, as a circle, is the equinoctial colure ; the meridian circle can be considered as the solsticial colure (i,e, the equinoctial colure is a meridian passing through the equinoctial points ; and the solsticial colure is a meridian passing through the solsticial points). The colures therefore divide the apparent annual path of the Sun into four parts which determine the seasons,,,
Ascensional difference (DA) for f SU is = sin DA = -tan(lat [48,4]) x tan(dec f [8,79])
---------------------------------
so DA f SU = 10,03°
---------------------------------
Ascensional difference (DA) for m □MA is = sin DA = -tan(lat [48,4]) x tan(dec f [-3])
---------------------------------
so DA m □MA = 3,39°
---------------------------------
You see also an almucantar circle for the mundane primary directions : actually the altitude of m □MA = -29,68° ; , This altitude corresponds to that of point f SU (alt f = -30,67), assumed to have remained fixed during the displacement of the diurnal movement.
note that if the m point is a counter parallel, it is retrograde (and it is not a zodiacal aspect because one use declination to compute mundane parallel),
We can see too two or three parallels of declination ; for point m □MA with dashed line (between equator and equinoctial colure) to design the m DA (see above) ; for point f SU (idem) and for a star (Algol i,e, β Persei or another if present in the sky path of the natal chart here : α Cma i,e, sirius),
Then we find the index for rising, transit and setting the two points f and m,
