Post by Ray Murphy----------
Post by S***@aol.comDoes anyone happen to have a copy of The New Waites Compendium Of
Natal Astrology, which has "universal tables of houses" that
include Colin Evans' "natural graduation" method?
[....]
RM: As a matter of interest, I've never heard of the book, or of
universal tables of houses, so it's probably pretty rare.
Ray
The term "universal tables of houses" refers not to a specific house
system, but to a set of tables which include many different house
systems-- in other words, Placidus, Regiomontanus, Campanus, Natural
Graduation, and so on.
I do have the book in a box somewhere, but can't find it right now.
When I finally locate it, I'll post the results of my investigation.
In the meantime, I received an email from Martin that quoted a brief
passage from another book, which described the Natural Graduation well
enough to confirm my tentative understanding of it, and which
essentially confirmed another brief description that I had found on the
internet. The results of this method-- which I call "Natural Graduation
A" (see below)-- agree with the house cusps given by the StarLogin
freeware program, but I was trying to see if they agree with the values
in the house tables which Colin Evans gave in "The New Waite's
Compendium."
The Janus commercial software program also includes the Natural
Graduation house system, so if anyone here happens to have Janus, or
access to it, I'd love to get some confirmation from you, so I can
determine whether the results from Janus agree with the results from
StarLogin.
In trying to "reverse engineer" the Natural Graduation system, I have
come up with at least five possible variations of it, which I am
calling A, B, C, D, and E. Variation A is the one that agrees with
StarLogin, and which conforms to every description of the system that
I've yet seen. Variation B is like A, but is based on whole houses
instead of half houses (see below). Variations A and B are based on a
geometric progression, or multiplication (see below). Variation C is
like variation A, but uses an arithmetic progression, or addition.
Variation D is like variation C (arithmetic), but is based on whole
houses instead of half houses. And variation E is similar to variation
D (arithmetic, and using whole houses), but the progression is not
constant. Variation E gives the house cusps referred to as
"Neo-Porphyry" by the Astrolog freeware program, which I had always
assumed was just another name for the Natural Graduation system, but it
appears that I may have been mistaken about that. If so, then I don't
know who devised the Neo-Porphyry system, or if it might have been a
faulty implementation of the Natural Graduation system.
Anyway, the basic concept behind Colin Evans' Natural Graduation system
is that the Porphyry system is generally correct, but that the division
of the quadrants should produce houses which gradually and smoothly
increase in size, starting from the center of the smaller quadrants,
and ending at the center of the larger quadrants. If we bisect the
quadrants, we can label their midpoints as M1, M2, M3, and M4, where
M1=ASC/IC, M2=IC/DSC, M3=DSC/MC, and M4=MC/ASC (i.e., M1 is the
midpoint of Quadrant I, M2 is the midpoint of Quadrant II, M3 is the
midpoint of Quadrant III, and M4 is the midpoint of Quadrant IV). It
should be noted that these four points, or midpoints, will be exactly
square or opposite each other. And if we determine the cusps of the
houses in the Eastern Hemisphere (or Quadrant I and Quadrant IV), then
the house cusps of the Western Hemisphere will be their opposites. So
we only need to worry about the MC, ASC, M1 (or ASC/IC), and M4 (or
MC/ASC), with M1 being exactly square M4.
The idea is that, if we know the ASC and the MC, then we can easily
calculate M1 (ASC/IC) and M4 (MC/ASC), and use the distances between
the ASC and M1, as well as the distance between the ASC and M4, to
determine the sizes of houses 11, 12, 1, and 2. Then we can use these
house sizes, along with M1 or M4, to determine the cusps of all twelve
houses.
Variation A -- This is consistent with the descriptions I have seen for
the Natural Graduation system, so it is presumably the "correct"
variation. The region between M1 and M4 is divided into six half houses
(such that the entire circle is divided into 24 half houses), with a
constant ratio or multiplying factor existing between each subsequent
half house, from the middle of the smaller quadrant to the middle of
the larger quadrant.
In other words, let's suppose that Quadrant I is smaller than Quadrant
IV. Then the progression will start at M1 (which is the midpoint of the
2nd House), and continue to M4 (which is the midpoint of the 11th
House). If we divide the 90 degrees between M1 and M4 into six half
houses, then we can label these half houses as A (first half of 2nd
House), B (second half of 1st House), C (first half of 1st House), D
(second half of 12th House), E (first half of 12th House), and F
(second half of 11th House). In that case, A will be the smallest half
house, and F will be the largest half house, with each half house being
a constant multiple of the previous half house, such that
B/A=C/B=D/C=E/D=F/E=X, where X is the constant multiple. Thus, B=A*X,
C=B*X, D=C*X, E=D*X, and F=E*X. We can simplify this to B=A*X, C=A*X^2,
D=A*X^3, E=A*X^4, and F=A*X^5.
Since M1 - M4 = 90 degrees, and the arc between M1 and M4 is divided
into the half houses A, B, C, D, E, and F, it follows that
A + B + C + D + E + F = 90 degrees, or
A + A*X + A*X^2 + A*X^3 + A*X^4 + A*X^5 = 90 degrees, or
A * (1 + X + X^2 + X^3 + X^4 + X^5) = 90 degrees, or
A = 90 / (1 + X + X^2 + X^3 + X^4 + X^5).
Also, M1 - ASC is half the width of Quadrant I (which for now we are
assuming is the smaller quadrant), so if we define M=(M1-ASC)/2, we
also get
A + B + C = M, or
A + A*X + A*X^2 = M, or
A * (1 + X + X^2) = M.
Note that, if we know the ASC and the MC, then we also know M, but we
want to find A and X. I suppose we can solve this somehow, but I am
doing it by a recursive routine. That is, I pick a value for X, divide
90 by (1+X+X^2+X^3+X^4+X^5) to find A, and then multiply A times
(1+X+X^2) to see how close it is to M. Then I pick another value for X,
solve for A, see how close this comes to M, and keep picking new values
of X until I can find the value which satisfies the two equations given
above.
For example, suppose that ASC = 0 Aries, and MC = 0 Sagittarius. Then
M1 = 0 Taurus, and M4 = 0 Aquarius. Also, M = 30 degrees. It happens
that X must be greater than or equal to 1, so we can start by
considering X=1. Then we get A=90/(1+X+X^2+X^3+X^4+X^5)=90/6=15. This
then gives us A*(1+X+X^2)=15*3=45, which is greater than 30 (or M). If
we keep trying new values of X, and increase or decrease X until we can
match M, we end up with a value between 1.2599210 and 1.2599211 for X.
(My spreadsheet program doesn't go beyond 7 decimal places, so I am
stopping there.) I'll use 1.2599211 for X. Then, if we plug that value
into A=90/(1+X+X^2+X^3+X^4+X^5), we get a value of 7.7976305 for A.
This means that
A = 7.7976305 degrees,
B = 9.8243992 degrees,
C = 12.3779679 degrees,
D = 15.5952629 degrees,
E = 19.6488008 degrees, and
F = 24.7559387 degrees.
For the sizes of the houses, we get
H2 = A + A = 15.5952610 degrees,
H1 = B + C = 22.2023671 degrees,
H12 = D + E = 35.2440637 degrees, and
H11 = F + F = 49.5118774 degrees.
Note that the sizes of the other houses are H3=H1, H4=H12, H5=H11,
H6=H12, H7=H1, H8=H2, H9=H1, and H10=H12.
If we start at the MC, and add the sizes of each house, we get
MC = 240 = 0:00 Sag
Cusp 11 = MC+H10 = 240+35.2440637 = 275.2440637 = 5:15 Cap
Cusp 12 = C11+H11 = 275.2440637+49.5118774 = 324.7559411 = 24:45 Aqu
ASC = C12+H12 = 324.7559411+35.2440637 = 360.0000048 = 0:00 Ari
Cusp 2 = ASC+H1 = 0.0000048+22.2023671 = 22.2023719 = 22:12 Ari
Cusp 3 = C2+H2 = 22.2023719+15.5952610 = 37.7976329 = 7:48 Tau.
Variation B -- This is the same as variation A, except we use whole
houses instead of half houses. Thus, if we divide Quadrant IV and
Quadrant I into six houses, the sizes of the houses (beginning at the
IC and moving backward to the MC) can be given as B, A, B, C, D, and C,
where B is the size of houses 3 and 1, A is the size of house 2, C is
the size of houses 12 and 10, and D is the size of house 11, with A
being the smallest, D being the largest, and B=A*X, C=B*X, and D=C*X.
If we call M the size of the smaller quadrant (which we are supposing
is Quadrant I), then we have
B + A + B + C + D + C = 180 degrees, or
A + 2*A*X + 2*A*X^2 + A*X^3 = 180, or
A * (1 + 2*X + 2*X^2 + X^3) = 180, or
A = 180 / (1 + 2*X + 2*X^2 + X^3).
We also have
B + A + B = M, or
A + 2*A*X = M.
I'm not going to discuss this variation further for now, but it could
be solved recursively as in variation A.
Variation C -- This is like variation A, but it uses an arithmetic
progression instead of a geometric progression. Thus, if we label the
half houses and other points as we did in variation A, we get
B = A + X,
C = B + X,
D = C + X,
E = D + X, and
F = E + X.
Thus,
A + B + C + D + E + F = 90 degrees, or
A + A+X + A+X+X + A+X+X+X + A+X+X+X+X + A+X+X+X+X+X = 90, or
6*A + 15*X = 90, or
6*A = 90 - 15*X, or
A = (90 - 15*X) / 6, or
A = 15 - 5*X / 2.
Also,
A + B + C = M, or
A + A+X + A+X+X = M, or
3*A + 3*X = M, or
3*(15-5*X/2) + 3*X = M, or
45 - 15*X/2 + 3*X = M, or
90 - 15*X + 6*X = 2*M, or
90 - 9*X = 2*M, or
90 - 2*M = 9*X, or
9*X = 90 - 2*M, or
X = (90 - 2*M) / 9, or
X = 10 - 2*M / 9.
This variation is much simpler to solve, because if we know ASC and MC,
then we know M, and we can easily find X and A, then get B, C, D, E,
and F. I won't give an example right now.
Variation D -- This is like variation C, except using whole houses
rather than half houses. Thus,
B + A + B + C + D + C = 180 degrees, or
A + A+X + A+X + A+X+X + A+X+X + A+X+X+X = 180, or
6*A + 9*X = 180, or
6*A = 180 - 9*X, or
A = (180 - 9*X) / 6, or
A = 30 - 3*X / 2.
Also,
B + A + B = M, or
A + A+X + A+X = M, or
3*A + 2*X = M, or
3*(30-3*X/2) + 2*X = M, or
90 - 9*X/2 + 2*X = M, or
180 - 9*X + 4*X = 2*M, or
180 - 5*X = 2*M, or
180 - 2*M = 5*X, or
5*X = 180 - 2*M, or
X = (180 - 2*M) / 5, or
X = 36 - 2*M / 5.
Note that in this case, M is equal to the whole smaller quadrant,
rather than half of the smaller quadrant.
Variation E -- This is similar to variation D, and also uses whole
houses as in variation D, but we have
B = A + X,
C = B + 2*X, and
D = C + X.
In this case, the amount of the increase (or X) is not constant, but is
instead doubled between B and C. Thus,
B + A + B + C + D + C = 180 degrees, or
A + A+X + A+X + A+X+X+X + A+X+X+X + A+X+X+X+X = 180, or
6*A + 12*X = 180, or
6*A = 180 - 12*X, or
A = (180 - 12*X) / 6, or
A = 30 - 2*X.
Also,
B + A + B = M, or
A + A+X + A+X = M, or
3*A + 2*X = M, or
3*(30-2*X) + 2*X = M, or
90 - 6*X + 2*X = M, or
90 - 4*X = M, or
90 - M = 4*X, or
4*X = 90 - M, or
X = (90 - M) / 4, or
X = 22.5 - M / 4.
Remember, in this case (as in variation D), M is the whole smaller
quadrant, rather than half of the smaller quadrant. This variation
produces the house cusps which the Astrolog freeware program calls the
Neo-Porphyry system.
The interesting thing about the arithmetic-progression variations (or
variations C, D, and E) is that they produce the same type of pattern
as found in the Porphyry system, where the degrees and minutes of
houses 2, 6, 8, and 12 are the same, and the degrees and minutes of
houses 3, 5, 9, and 11 are the same. On the other hand, the
geometric-progression variations (or variations A and B) do not fit
that pattern. The reason I mention this is because, the way I remember
it (and my memory is admittedly hazy), the house tables which Colin
Evans gave in "The New Waite's Compendium" fit the Porphyry pattern,
which would indicate that he actually used an arithmetic progression
rather than a geometric progression, even though his description of the
method clearly suggests a geometric progression. That's why I want to
check with the tables of houses in "The New Waite's Compendium."
Until I manage to locate my copy of that book, I would appreciate it if
someone with the Janus program could erect a chart using the Natural
Graduation system, and post the house cusps as given by the Janus
program, so I can compare them with the results of the five variations
described above. Please use the following chart data:
January 1, 1950
6:00 p.m. GMT
0W00, 50N00
Michael Rideout