Skyscript Astrology Forum

Resouces for Bianchini formula:

The Primary directions of Regiomontanus & william Lilly - Rumen Kolev, page 26
Prim?rdirektion - R?diger Plantiko page 114-115
E.C.K?hr - Berechnung der Eregniszeiten page 302
Astrolo.Gallica Book 22 - page 167-172-Jerome Cardan

Petr

I think the Bianchini interpolation may have been based upon the fact that
00? ?sin 90? = 1
60? ?sin 30? = 0.5
90? ?sin 00? = 0
120??sin-30?=-0.5
180??sin-90?= 1
Other aspects are irrelevant in traditional, so no interpolation for these.
The semi-sextile 30? would give? sin 60?=0.866 or the square root of 3/4.
The semi-square 45? would give? sin 45?=0.707 or the square root of 1/2.

The problem is that we are dealing with spherical trigonometry here. I think Curtis' formula would work in plane trigonometry in which the proportions approach would apply with the formulae for plane trigonometry. In a spherical trigonometrical setting however, this formula would give a kind of curve as result, not a great circle on the sphere. Just compare it with the ecliptic as depicted in a starmap based upon the equator (as it usually is). It's a sinusoidal 'wave'. In a sphere this wave appears as a great circle. Had the ecliptic in the starmap been depicted as a straight line upwards from Capricorn to Cancer and a straight line downwards back to Capricorn, then it would appear as curves when viewed in a sphere. Thus for spherical trigonometry other sine formulae are needed. (see first pages in Makransky 1 p.6 on the Napier formulae).

That is the scheme from the famous Argoli's Tabulae Primi Mobilis:
In the example Venus has a Northern latitude of 4?; her opposition has 4? S of latitude because falls in the opposite side, the square has no latitude because it falls on the ecliptic, the sextile 2? N latitude because it falls on the same side, the trine 2?? S latitude because it falls in the opposite side. That's how Argoli explains the matter.

margherita

Eddy wrote:I think the Bianchini interpolation may have been based upon the fact that
00? ?sin 90? = 1
60? ?sin 30? = 0.5
90? ?sin 00? = 0
120??sin-30?=-0.5
180??sin-90?= 1
Other aspects are irrelevant in traditional, so no interpolation for these.
The semi-sextile 30? would give? sin 60?=0.866 or the square root of 3/4.
The semi-square 45? would give? sin 45?=0.707 or the square root of 1/2.

The problem is that we are dealing with spherical trigonometry here. I think Curtis' formula would work in plane trigonometry in which the proportions approach would apply with the formulae for plane trigonometry. In a spherical trigonometrical setting however, this formula would give a kind of curve as result, not a great circle on the sphere. Just compare it with the ecliptic as depicted in a starmap based upon the equator (as it usually is). It's a sinusoidal 'wave'. In a sphere this wave appears as a great circle. Had the ecliptic in the starmap been depicted as a straight line upwards from Capricorn to Cancer and a straight line downwards back to Capricorn, then it would appear as curves when viewed in a sphere. Thus for spherical trigonometry other sine formulae are needed. (see first pages in Makransky 1 p.6 on the Napier formulae).

Ok. So I'm thinking I should write the formula like this:

// Aspect is the number of degrees between 0 and 180
function TDirPoint.GetBianchiniLatitude(EcLatitude, Aspect: Real): Real;
var
Proportion: Real;
begin
Proportion := RadToDeg(Sin(DegToRad(90 - Aspect))) / 180;
Result := EcLatitude * Proportion * 2;
end;

Delphi's Math unit uses radians, so conversion between degrees and radians must be used. I don't like hard coding special cases because you run the risk of breaking code when the component expands into other uses (such as minor aspects) which is why I defined Aspect as a real number rather than an integer or an enumerated type with case statements. Writing it into the TDirPoint class will set up memory for each promissor or significator object so simply assigning the ecliptic latitude with the boolean types for different types of latitude will initialize the applicable aspect arrays.

Proportion * 2 is necessary because (90 - Aspect) is only half the range. You want the dividend to drop out so that multiplication by zero will do its job... sin (0) = 0. I prefer math that keeps sign of the argument in place, rather than finding which quadrant a point is in using the |absolute| values and then using if .. then statements to replace the sign of the argument which if you read the old algorithms and take them straight, you will be tempted to do, but is a lousy way to write software.

It's been a while since I've used radians but I'm not sure if I understand the formula. If I applied it well (using radians) I got ca. +1.91 for the sextile. Bianchini method gives 3.

90-60=30. converted to radians ? 30*pi/180=0.5236
install radians in calculator ? sin 0.5236=0.5
converting this to degrees? 28.6479
divided by 180?28.6479/180=0.15915 is proportion
applied to latitude of 6?
6*0.15915*2=1.90986

Unless I understood the formula wrongly I'm afraid it isn't correct.

Anyhow, I think the use of proportion shouldn't be applied in the spherical trigonometry. Proportions work in plane trigonometry. E.g. a=6millimeter, c(hypotenuse)=57.4 millimeter A(angle opposite to a)=6? then the sine formula for plane trigonometry is 6/57.4=sinA = 0.10453 Inversed sinA=6
When on one third of the distance of 57.4 thus ca. 19.13 multiply this with sinA then you get 2 which is 1/3d of the distance. So here the proportions can be applied, but in spherical you come close to Bianchini's 3.

I think the only way to use is with the conversion formulae of coordinate systems http://en.wikipedia.org/wiki/Ecliptic_c ... rdinates_2
If you apply the second formula, equatorial to ecliptic, to the case then consider the degrees along the tilted plane (as being the 'ecliptic') in case of latitude 6?, this is used as the '?' value in the formula (which is 23.44 or the axial tilt in the equator ecliptic conversion)

You can use it and the answers will be:
60? ? 2.995884850672 (Bianchini method ? 3)
45? ? 4.2387560929649615 (Bianchine method ? 4.242640687119286)
30? ? 5.193770658894835 (Bianchini method ? 5.196152422706632)
Although you don't use the minor aspects I added them so you can get a grasp on the formula. I apllied the Bianchini method in the way I believe the reasoning would be according to my post of 9:06UT today. You can also apply this to 23.44 to compare this with declination tables of the Sun.

If you want to put in into a computer program then the Bianchini method would be sufficient. Considering the minimal differences of the usually small latitudes of the (classical) planets with the spherical trigonometrical results, this wouldn't be troublesome.

margherita wrote:That is the scheme from the famous Argoli's Tabulae Primi Mobilis

I wonder when this was used for the first time. do you know this Margherita?

I assume that after Placidus the approach changed into the more modern view of 'in mundo' with also the mundane aspects in the natal which can be very different from the aspects along the ecliptic.

Eddy wrote: I assume that after Placidus the approach changed into the more modern view of 'in mundo' with also the mundane aspects in the natal which can be very different from the aspects along the ecliptic.

One might hope so. But I don't think so.

- Ed

What is the most common alternative to Bianchini latitude used?

When dealing with latitude of an aspect, is there a variation that has the same latitude throughout (on the small circle usually)?

zoidsoft wrote:What is the most common alternative to Bianchini latitude used?

When dealing with latitude of an aspect, is there a variation that has the same latitude throughout (on the small circle usually)?

The Placidus proportional point method.

Ed F wrote:

zoidsoft wrote:What is the most common alternative to Bianchini latitude used?

When dealing with latitude of an aspect, is there a variation that has the same latitude throughout (on the small circle usually)?

The Placidus proportional point method.

Is this a method I have missed, or do you refer simply to the mundane aspects?

The division by 180 is no longer needed. The Sin(90 - Aspect) is the proportion so no * 2 either. As it stands now:

// Aspect is the number of degrees between 0 and 180
function TDirPoint.GetBianchiniLatitude(EcLatitude, Aspect: Real): Real;
begin
Result := EcLatitude * RadToDeg(Sin(DegToRad(90 - Aspect)));
end;

The Math unit in Delphi uses the Extended data type which is a Real type with high accuracy so lets leave that out for simplicity. Using trig calculator: (given a -4 for latitude and 120 for aspect)

R = EcLatitude * Sin (90 - Aspect)
R = -4 * Sin (90 -120)
R = -4 * Sin (-30)
R = -4 * -0.5
R = 2.0

Another... lets do opposition:

R = EcLatitude * Sin (90 - Aspect)
R = -4 * Sin (90 -180)
R = -4 * Sin (-90)
R = -4 * -1.0
R = 4.0

Another (45 degrees):

R = EcLatitude * Sin (90 - Aspect)
R = -4 * Sin (90 - 45)
R = -4 * Sin (45)
R = -4 * 0.70710678118654752440084436210485
R = -2.8284271247461900976033774484194

This seems to be working right now.

Martin Gansten wrote:

Ed F wrote:

zoidsoft wrote:What is the most common alternative to Bianchini latitude used?

When dealing with latitude of an aspect, is there a variation that has the same latitude throughout (on the small circle usually)?

The Placidus proportional point method.

Is this a method I have missed, or do you refer simply to the mundane aspects?

Mundane aspects mainly. Though there's no reason you couldn't use the declination circle of the zodiacal intercept of the significator. While I understand that it's the historical way of doing it, I still see little sense in using zodiacal positions in a mundane context. They strike me as incidental.

- Ed

Eddy wrote:I think the only way to use is with the conversion formulae of coordinate systems http://en.wikipedia.org/wiki/Ecliptic_c ... rdinates_2
If you apply the second formula, equatorial to ecliptic, to the case then consider the degrees along the tilted plane (as being the 'ecliptic') in case of latitude 6?, this is used as the '?' value in the formula (which is 23.44 or the axial tilt in the equator ecliptic conversion)

There is often more than one way toward achieving mathematical proofs... I plot the points first in ecliptic latitude and longitude (in a Promissor or Significator object like this: Jupiter.EcLongitude[0], Jupiter.EcLatitude[0] (0 = bodily conj in the array 0..7, 1 is sextile, 2 is first square, etc), then do a conversion to right ascension and declination (with arrays Jupiter.Rasc[0], etc). So far the most parsimonious mathematical formulae I've seen for Placidian directions has come from Ed Falis.