*Post by Ray Murphy*----------

*Post by S***@aol.com*Does 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