New quantification of angularity

[mikestar13]

This makes some changes in the width of grounds, the angular side of an angle has 10 degree of foreground and the cadent side has only nine degrees of foreground and the width of background is correspondingly narrowed, assuming the boundaries are now 80% and 20%. I could of course play with the numbers a bit but do we in essence want 1 degree shaved off each foreground and background?

[Jim Eshelman]

Yes, I know the boundaries aren't exactly what I've said I want or suggest. Nonetheless, by excitement is that a single equation generates a single curve that hits all the fundamental markers (exactly defines exactly angular, exactly cadent, exactly mid [twice] and is all but a coarse shave from the other points).

If nothing else, it gives me a pretty picture to use in the angularity chapter of the next book, which I'm writing at the moment

I suggest not spending energy on reworking the curve or grounds boundaries unless some of us actually work with this and give meaningful feedback on it. My current expectation is that, since the difference are only at the weak trails of the humps, one can't tell one way or the other in real life practice.

[mikestar13]

I've been working on other things besides the angularity curve, and it took me a month to get around to studying this. I think your equations do fit the cosine model well. Having the 80% mark at 10 and 81 is slightly awkward, but foreground planets that far from the angle aren't all influential anyway, just somewhat more expressive. I could easily implement it in TMSA 1.0, the minor angle curve (derived from the aspect curve) should probably stay as in. The seems reasonable as they are mathematically ecliptic squares to the major angles or in the case of Ea/Wa, right ascension squares to Mc, though different conceptually. If I do this, I can change the foreground specifications for the classes to % strength (and convert existing option files on first use in the program in the software, so users won't have to do anything). The mid-quadrant curve for mundane astrology can remain as is, since as far as I can see, there is no repressiveness curve in ingresses.

[mikestar13]

Also another thought, I could leave the foreground boundaries in degrees, and just accept the percentage number will be slightly different on opposite sides of an angle/cadent cusp. Also, though I've been using them, background marks just aren't as important in practice.

[Jim Eshelman]

My current thoughts: My excitement with this equation and approach is that it gives the whole pattern in a single equation. I'm not "sold" on it to the extent of all the percentages being precisely right - maybe yes, maybe no, but it's so subtle that I'm loathe to abandon something that has worked so well as the flat-number boundaries for angularity.

I'm also not inclined to ask you to do extra work while you already have so much you're trying to finish for just the next stage. The greater elegance of a single equation is intellectually satisfying but MAY not add anything practical to us.

There is also the fact that (our strength % be damned!) individual users will want different angularity orbs (and I want them in different contexts) and our thinking is still in terms of fixed bilateral orb boundaries (though perhaps it shouldn't be).

And, in most cases, it makes no great difference in practice.

My current thought, then, is that ONLY IF YOU ARE MOVED TO TAKE ON THIS CHANGE, use it for calculating precision % curve only, and continue to measure "grounds" based on the hard aspect divisions. The result: Both the % curve and the "grounds bounds" are evident side by side on the chart table - someone can see either of these at a glance and use whichever they want in close calls. Over the long run, this would give us input on which approach works better (even though there is almost no difference in practice).

[mikestar13]

I'll see what inspiration wills as I'm coding other features. The user selection for boundaries will be in degrees with bilateral symmetry as at present, only the percentages will differ slightly, should I decide the change is worth making. In any event, the basic scheme of 100% = exact foreground, 0% = exact background, 50% = exact middleground will be retained.

[cris25]

The image would look like this ?
the red zone is where the planets have more strength and the blue zone is where they have little strength ?

https://imgur.com/ERSNDKa

[Jim Eshelman]

Exactly. (Isn't that the same as what I gave above in the first post?)

[cris25]

Yes , sorry I'm taking my first steps , and I like to do even the drawings by myself xD .

[Mike V]

Here is the awkward math.

Calculate angular and cadent curves as before. In Excel, I have COS(4x) and COS(4(x-60))* (-1). (Mike will have this or his own way of doing it.)

The "faded p scale" - angularity curve (which happens to be in my column P, hence the name) - is the prior result (call it p) multiplied the same number scaled 0 to 1 instead of -1 to +1: p*((p2+1)/2)

The same effect for the repressiveness score (call it q) took a bit of head scratching to get it to work out right, but it turns out to be: q*(1-((q2+1)/2))

This looks clumsy and forced, but it's really a single simple thing: As the score changes from maximum to minimum, it is multiplied by itself to that its contribution fades from 100% to 0% on the same scale (at the same rate).

After getting these two numbers, you just add them. The sums range from +1.25 to -1.25. Divide the score, therefore, by 1.25 to get a "pretty" number.

Here is the value of each degree from the angle (each degree in the mundoscope quadrant) expressed as %.
...

I know this is much lower down the list, but I was getting my hands into all of the mathy stuff at once...

There's something I'm not getting about applying these formulas, because I get different results than your full tabular breakdown.

Here is how I interpreted your explanation; can you let me know what I got wrong?
x = mundoscope quadrant degree 0-90
angular score P = COS(4 * x))
cadency score Q = COS(4(x - 60)) * (-1)

feed those into the below equations... I bet this is what I got wrong, since now I'm not sure that "q2" and "p2" refer to multiplication like I thought they did...

"faded P score" = p * ((2 * p + 1) / 2)
"repressiveness score" = q * (1 - ((2 * q + 1) / 2))

final "pretty" number = (faded P score + repressiveness score) / 1.25

When I do this, I get a cycle that looks like this:
0: 1.20
1: 1.17
2: 1.12
...
29: -0.05
30: 0
31: 0.05
...
59: -1.17
60: -1.20
61: -1.22 (??)
...

Weird stuff. I've tried a couple of different tweaks and I get similar results - there's a curve, but it doesn't cap out at +/- 1.25, and it almost but not quite lines up neatly with 0/30/60/90.

[Jim Eshelman]

feed those into the below equations... I bet this is what I got wrong, since now I'm not sure that "q2" and "p2" refer to multiplication like I thought they did...

Sorry Copying from a spreadsheet where they represent cell numbers.

q2 is the cell that has q - it just means q. Ditto for p2.

Here's how I wrote it in the book:

1. Calculate the initial angularity strength: A = cos 4x

2. To taper (fade) the importance of A, multiply it by the same value (but scaled 0 to 1 instead of -1 to +2). The “faded” angularity score (A2) is: A x ((A+1)/2)

3. Calculate the cadency strength, C, which is the same in principle but shifted 60° along the prime vertical and reversed (made negative): C = -cos(4(x-60)) Obtain C2 by fading C in the same way as A. After adjusting for the flipped curve, the equation becomes: Cx(1-((C+1)/2))

4. Add A2 and C2.

5. Because this produces numbers ranging be-tween +1.25 and -1.25, divide by 1.25 so scores fall between +1.00 (+100%) and -1.00 (-100%). Add 1 and divide by 2 to get a score between 100% and 0%.

[Mike V]

Thank you, I get the same values you do now.

Except...

5. Because this produces numbers ranging be-tween +1.25 and -1.25, divide by 1.25...

Once again, I get numbers between +1.125 and -1.125, not 1.25. I divide by 1.125 and get the same values you do. Is there a 1 hiding in the tenths place in one of your intermediate values?

[Jim Eshelman]

5. Because this produces numbers ranging be-tween +1.25 and -1.25, divide by 1.25...

Once again, I get numbers between +1.125 and -1.125, not 1.25. I divide by 1.125 and get the same values you do. Is there a 1 hiding in the tenths place in one of your intermediate values?

You are correct. Thanks for catching this. I'd copied it wrong several places (I only originally had it embedded in an equation.) I've fixed the chapter where I gave it.

BTW, in looking (the last day or so) at the F/M/B grounds from this equation - I'm actually startled to see it as more descriptive. There are some charts with highly discernible distinctions from all of this. I had expected a subtle shift - I find instead that there are really substantive shifts that both seem more mathematically appropriate to the placement AND are better descriptive of the person. This intrigues me. I'm making it my standard for now and it's causing me to think about some people's charts (including mine) differently than I have. If first impressions persist, I might suggest renaming the current "Cadent" to "TM Classic" and naming the current "Eureka!" as "Cadent" (swapping their placement on the Options row).