# The Geomantic Emblems and their Rulerships

March 17, 2013 2 Comments

Last time I brought up the geomantic emblems (previously called geomantic superfigures, 256 16-line “figures” that each contain all 16 geomantic figures within themselves), I described a few bits about the elemental representation and force within each figure. In the process, I described a method where each geomantic emblem can be elementally analyzed and given an “elemental essential” rulership, by taking the “pure elemental” lines, and also how to split up the emblems into four figures to give them an entire geomantic chart as background. However, I also mentioned that all 256 emblems could be reduced to a set of 16 by rotating them around; in other words, there are 16 sets of 16 topologically equivalent geomantic emblems. 16 is a significant number in geomancy, as my astute readers may have noticed, and I brought up how tempting and tantalizing it would be to assign a set of rulerships that correspond these 16 sets of geomantic emblems to the 16 figures of geomancy. I didn’t have the method done just then, but I’ve finally come up with a way to link the two sets of symbols. The correspondences are, using the list from last time:

**Laetitia:**1000010011010111**Carcer:**1000010011110101**Fortuna Minor:**1000010100110111**Puer:**1000010100111101**Acquisitio:**1000010110011110**Populus:**1000010110100111**Coniunctio:**1000010111100110**Albus:**1000010111101001**Tristitia:**1000011001011110**Rubeus:**1000011010010111**Amissio:**1000011010111100**Puella:**1000011011110010**Fortuna Maior:**1000011110010110**Caput Draconis:**1000011110100101**Cauda Draconis:**1000011110101100**Via:**1000011110110010

How did I go about finding these correspondences? A lot of math, hand-wringing, and sangria, that’s for sure. If, dear reader, you’re interested in finding out how I corresponded the figures to the emblems, please continue after the break, but I’m going to warn you. This post is long and at times tedious, and is full of binary mathematics and lots of 1s and 0s. This post is only for the hardcore geomancy geeks like me out there, and it helps to have a solid footing in computer science, basic/low-level programming exercises, and binary/discrete mathematics. Even I’m kinda shocked by how lengthy and pointlessly in-depth this post is, if that’s any indication of what you’re in for. If you want to stop reading now, I forgive you and completely understand. If you want to find out why I allocated the above emblems and their rotated variants to the figures like I did above, read on. Either way, expect another post in the near future on how to use these emblems, their geomantic rulership, and elemental analyses in magic and divination!

To be clear about what I’m doing, I’m trying to figure out a way to assign the sixteen cycles of emblems (or emblematic cycles) to the sixteen geomantic figures. I went over a way that relies on the lines of a particular emblem before to find one kind of ruling figure for a set of sixteen emblems, but that left me feeling unsatisfied, since 16 sets of 16 emblems each effectively share the same structure, just rotated a little differently. Different rotations of the same emblematic cycle would change elemental ruler due to the line positions of the bits, but the underlying structure would remain the same. What I wanted to do was to find a way to assign the emblematic cycles themselves to the geomantic figures regardless of “starting point” or orientation. This requires looking at the structure itself in such a way that any analysis or interpretation would yield the same results independent of “starting point”; the elemental method of interpretation last time (taking the “fire of fire” line, etc.) is based on “starting point”, so any similar method here would have to be thrown out. The best way to go about doing this is to find four binary methods of grouping the emblematic cycles that are independent of each other and equally distribute the sixteen emblematic cycles among themselves (i.e. eight per group). Once this is done, each would be assigned a different element (fire, air, water, earth), and the choice would be translated into a 1 or 0. This would yield a four-bit binary number, or a geomantic figure.

First, let’s start with the same listing of emblematic cycles from last time. They’re all conveniently aligned to all start with 100001…, so let’s start from there.

- 1000010011010111
- 1000010011110101
- 1000010100110111
- 1000010100111101
- 1000010110011110
- 1000010110100111
- 1000010111100110
- 1000010111101001
- 1000011001011110
- 1000011010010111
- 1000011010111100
- 1000011011110010
- 1000011110010110
- 1000011110100101
- 1000011110101100
- 1000011110110010

Keep in mind that, because emblems can be cycled, you could all start these so that they start with the part of the emblem that goes 1111 and still have valid emblems. It’s probably better to picture them as rings or bands instead of strings to emphasize their cyclic nature. We want a common starting point for all the figures, just to keep them aligned, but this is only a matter of convenience.

The first thing to notice is that, because these emblems contain all sixteen geomantic figures, there’s going to be repeated values of some bits. Populus, for instance, appears where you take those four consecutive 0s, and Via when you take the four consecutive 1s. Carcer occurs where you take the two consecutive 0s with a 1 on each end, and so forth. However, if you ignore the repeated values of each bit and “collapse” them into a single bit, you end up with patterns that look like the following:

- 1000010011010111 →
**10**1**01**010 - 1000010011110101 →
**10**1**01**010 - 1000010100110111 →
**10**101**01**0 - 1000010100111101 →
**10**101**01**0 - 1000010110011110 → 1
**0**10**101**0 - 1000010110100111 →
**10**10**1**01**0** - 1000010111100110 → 1
**0**10**101**0 - 1000010111101001 →
**10**10**1**01**0** - 1000011001011110 → 1
**010**10**1**0 - 1000011010010111 →
**101**01**0**10 - 1000011010111100 → 1
**01**010**10** - 1000011011110010 → 1
**01**0**10**10 - 1000011110010110 → 1
**010**10**1**0 - 1000011110100101 →
**101**01**0**10 - 1000011110101100 → 1
**01**010**10** - 1000011110110010 → 1
**01**0**10**10

We find that, by collapsing the repeated bits into a single bit (the ones that were collapsed are presented in boldface above), we get alternating 8-bit sequences of 1s and 0s. This also applies to the 1 at the end of an emblematic cycle and the 1 at the beginning, with the final 1 being “folded into” the first if this case happens (emblematic cycles 1, 2, 3, 4, 6, 8, 10, 14). The alternating sequence of bits makes sense, because we can’t have multiple Populi or Viae consecutively without throwing off the bit count or the alternation. If we transform these sequences such that a “collapsed” (bold) bit becomes a 0 (indicating duplication) and a “standalone” (unbolded) bit becomes a 1 (indicating duplication), we get the following orders:

- 00100111
- 00100111
- 00111001
- 00111001
- 10110001
- 00110110
- 10110001
- 00110110
- 10001101
- 00011011
- 10011100
- 10010011
- 10001101
- 00011011
- 10011100
- 10010011

Looking at that list (we’ll call them the change sequence of the collapsed cycles), we find that there are only eight sequences that each appear twice, allowing for a two-fold grouping of the cycles. Using Roman numerals to mark these sequences, we have the following:

**00100111:**emblematic cycles 1 and 2**00111001:**emblematic cycles 3 and 4**10110001:**emblematic cycles 5 and 7**00110110:**emblematic cycles 6 and 8**10001101:**emblematic cycles 9 and 13**00011011:**emblematic cycles 10 and 14**10011100:**emblematic cycles 11 and 15**10010011:**emblematic cycles 12 and 16

Knowing that the 256 emblems can all come from different rotations of the 16 emblematic cycles, we can similarly group the above eight change sequences into two based on whether they’re rotated or not. In this case, we find the two sequence cycles 00100111 and 11011000, which we’ll term the 111 and 000 sequence cycles, respectively. By circularly rotating these cycles around, we get different change sequences, which are derived from the pattern of repeated and standalone bits in the emblematic cycles. The sixteen emblematic cycles and eight sequence cycles can then be divided up into these two groups:

**111 Sequence Cycle Group**- emblematic cycles 1, 2, 3, 4, 11, 12, 15, 16
- sequence cycles I, II, VII, VIII

**000 Sequence Cycle Group**- emblematic cycles 5, 6, 7, 8, 9, 10, 13, 14
- sequence cycles III, IV, V, VI

Let’s investigate the sequence cycles within the two groups above. By performing single-bit binary addition (0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0), also known as the exclusive or (returns 1 if both bits are different, returns 0 if both bits are the same), we can obtain a new binary sequence. This is how one “adds” figures in geomancy, so let’s try it with the sequence cycles. It turns out that, because there are four sequence cycles within the 111 and 000 groups, there are twelve possible ways of adding them together (six within the 111 group and six within the 000 group). Adding two sequence cycles yields an 8-bit number, as follows:

**111 Sequence Cycle Group**- I + II = 00100111 + 00111001 = 00011110
- I + VII = 00100111 + 10011100 = 10111011
- I + VIII = 00100111 + 10010011 = 10110100
- II + VII = 00111001 + 10011100 = 10100101
- II + VIII = 00111001 + 10010011 = 10101010
- VII + VIII = 10011100 + 10010011 = 00001111

**000 Sequence Cycle Group**- III + IV = 10110001 + 00110110 = 10000111
- III + V = 10110001 + 10001101 = 00111100
- III + VI = 10110001 + 00011011 = 10101010
- IV + V = 00110110 + 10001101 = 10111011
- IV + VI = 00110110 + 00011011 = 00101101
- V + VI = 10001101 + 00011011 =10010110

If we break down the 8-bit results into two 4-bit results, transforming them into geomantic figures (where the leftmost bit represents the fire line, a 0 represents a passive element, and a 1 represents an active element), and then add the geomantic figures, we get the following:

**111 Sequence Cycle Group**- 00011110 → 0001 + 1110 = 1111 → Tristitia + Cauda Draconis = Via
- 10111011 → 1011 + 1011 = 0000 → Puella + Puella = Populus
- 10110100 → 1011 + 0100 = 1111 → Puella + Rubeus = Via
- 10100101 → 1010 + 0101 = 1111 → Amissio + Acquisitio = Via
- 10101010 → 1010 + 1010 = 0000 → Amissio + Amissio = Populus
- 00001111 → 0000 + 1111 = 1111 → Populus + Via = Via

**000 Sequence Cycle Group**- 10000111 → 1000 + 0111 = 1111 → Laetitia + Caput Draconis = Via
- 00111100 → 0011 + 1100 = 1111 → Fortuna Maior + Fortuna Minor = Via
- 10101010 → 1010 + 1010 = 0000 → Amissio + Amissio = Populus
- 10111011 → 1011 + 1011 = 0000 → Puella + Puella = Populus
- 00101101 → 0010 + 1101 = 1111 → Albus + Puer = Via
- 10010110 → 1001 + 0110 = 1111 → Carcer + Coniunctio = Via

Notice that there are four combinations of sequences that yield Populus, while the other eight yield Via. These groups that together yield Populus (sequence cycles I and VII, II and VIII, III and VI, and IV and V) are important, and help us distinguish between individual emblematic/sequence cycles. Even if we had a different set of rotated sequence cycles to add, the combinations of their respective 4-bit patterns would still yield Populus and Via in the same orders. Using letters to distinguish between these groups:

- sequence cycles I and VII; emblematic cycles 1, 2, 11, 15
- sequence cycles II and VIII; emblematic cycles 3, 4, 12, 16
- sequence cycles III and VI; emblematic cycles 5, 7, 10, 14
- sequence cycles IV and V; emblematic cycles 6, 8, 9, 13

Let’s try the addition trick within the individual sequence cycle groups. Consider group A, which contains emblematic cycles 1, 2, 11, and 15. We know that emblematic cycles 1 and 2 share the same sequence cycle, as do 11 and 15. So, adding the sequence cycles for emblematic cycles within the same group will yield the boring result of 00000000. However, adding the sequence cycles for emblematic cycles not in the same cycle but within the same group (e.g. emblematic cycles 1 and 11, 2 and 16, 5 and 10) yields other results we can look at:

- I + VII = 00100111 + 10011100 = 10111011
- II + VIII = 00111001 + 10010011 = 10101010
- III + VI = 10110001 + 00011011 = 10101010
- IV + V = 00110110 + 10001101 = 10111011

Notice that the addition of sequence cycles in groups A and D have two 0s in their results (i.e. they overlap two times), while the addition for groups B and C have four 0s (i.e. they overlap four times). We’ll call these the 2-0 and 4-0 addition groups, based on how the sequence cycles between them interact with each other or how many times the sequences overlap between each other. So, we have this information as well:

**2-0 Addition Group**- emblematic cycles 1, 2, 11, 15, 6, 8, 9, 13
- sequence cycles I, IV, V, VII
- sequence cycle groups A, D

**4-0 Addition Group**- emblematic cycles 5, 6, 7, 8, 9, 10, 11, 14
- sequence cycles II, III, VI, VIII
- sequence cycle groups B, C

Now, consider that we have the 111 sequence cycle groups (groups A and B), and the 000 sequence cycle groups (groups C and D), alond with the 2-0 addition group (groups A and D) with the 4-0 addition group (groups B and C). We now have the beginnings of a decision tree for our sequence cycle groups:

- 111 sequence cycle, 2-0 addition group
- 111 sequence cycle, 4-0 addition group
- 000 sequence cycle, 4-0 addition group
- 000 sequence cycle, 2-0 addition group

So now we have a four-way grouping of the emblematic cycles, each with four emblematic cycles each. If this is starting to sound like the beginnings of an either/or chain that would yield a 4-bit number (i.e. a geomantic figure) for each emblematic cycle, you’re doing well. If you’re still confused at this point about what the hell I’m doing, take a break, get some coffee, and review some of the steps from before. We’re halfway done.

Let’s back up a bit now and go back to the collapsed sequences from above, the ones with bolded bits to indicate collapsed rows in the emblematic cycles:

- 1000010011010111 →
**10**1**01**010 - 1000010011110101 →
**10**1**01**010 - 1000010100110111 →
**10**101**01**0 - 1000010100111101 →
**10**101**01**0 - 1000010110011110 → 1
**0**10**101**0 - 1000010110100111 →
**10**10**1**01**0** - 1000010111100110 → 1
**0**10**101**0 - 1000010111101001 →
**10**10**1**01**0** - 1000011001011110 → 1
**010**10**1**0 - 1000011010010111 →
**101**01**0**10 - 1000011010111100 → 1
**01**010**10** - 1000011011110010 → 1
**01**0**10**10 - 1000011110010110 → 1
**010**10**1**0 - 1000011110100101 →
**101**01**0**10 - 1000011110101100 → 1
**01**010**10** - 1000011110110010 → 1
**01**0**10**10

In each geomantic emblematic cycle, there will always be a mixture of standalone bits and collapsed bits, four of each to be precise. Half the bits are 1s and the other half are 0s. Of the collapsed bits, two will always be 0s (indicating where Populus or Carcer lie in the emblem), and two will always be 1s (indicating where Via or Coniunctio lie in the emblem). Further, two of the collapsed bits will be derived from four bits (indicating either Populus or Via), and two will be derived from two bits (indicating either Carcer or Coniunctio). Let’s focus on the collapsed bits, this time counting how many bits they were collapsed from in the original emblematic cycle, starting from the leftmost collapsed bit:

- 1000010011010111 →
**10**1**01**010 → 4, 4, 2, 2 - 1000010011110101 →
**10**1**01**010 → 4, 2, 4, 2 - 1000010100110111 →
**10**101**01**0 → 4, 2, 2, 4 - 1000010100111101 →
**10**101**01**0 → 4, 2, 4, 2 - 1000010110011110 → 1
**0**10**101**0 → 4, 2, 2, 4 - 1000010110100111 →
**10**10**1**01**0**→ 4, 2, 2, 4 - 1000010111100110 → 1
**0**10**101**0 → 4, 4, 2, 2 - 1000010111101001 →
**10**10**1**01**0**→ 4, 4, 2, 2 - 1000011001011110 → 1
**010**10**1**0 → 4, 2, 2, 4 - 1000011010010111 →
**101**01**0**10 → 4, 2, 2, 4 - 1000011010111100 → 1
**01**010**10**→ 4, 2, 4, 2 - 1000011011110010 → 1
**01**0**10**10 → 4, 2, 4, 2 - 1000011110010110 → 1
**010**10**1**0 → 4, 4, 2, 2 - 1000011110100101 →
**101**01**0**10 → 4, 4, 2, 2 - 1000011110101100 → 1
**01**010**10**→ 4, 4, 2, 2 - 1000011110110010 → 1
**01**0**10**10 → 4, 4, 2, 2

All the above counts start with 4 due to the alignment of the emblematic cycles, all starting with 10000…, but this is okay. In fact, for this bit of analysis, we’re going to ignore that first count of 4 anyway, since the below trick will work regardless of alignment with the same caveat as above. This means we have a group of three numbers, two of which are 2 and the third is 4, referring to a sequence of bits that translate to 00, 11, and 1111. These groups can combine in exactly four ways: 00111111, 11001111, 11110011, and 11111100. The emblematic cycles that have these combinations are:

**00111111:**emblematic cycles 1, 2, 3, 4**11001111:**emblematic cycles 5, 6, 9, 10**11110011:**emblematic cycles 7, 8, 13, 14**11111100:**emblematic cycles 11, 12, 15, 16

We can group the above into two categories: the ones that start with 1111 and the ones that end with 1111. We’ll call these the V1 and V2 categories, standing for “Via first” and “Via second” (if we look at these eight-bit numbers as two geomantic figures). So, we now have

**V1 Collapsed Bit Group:**emblematic cycles 7, 8, 11, 12, 13, 14, 15, 16**V2 Collapsed Bit Group:**emblematic cycles 1, 2, 3, 4, 5, 6, 9, 10

Because of the way the V1 and V2 collapsed bit groups contain the emblematic cycles, they’re independent of both the 111 and 000 sequence cycle groups and the addition groups. Now we have three methods of grouping, which yields eight groups of two emblematic cycles:

**000, 2-0, V1:**emblematic cycles 8, 13**000, 2-0, V2:**emblematic cycles 6, 9**000, 4-0, V1:**emblematic cycles 7, 14**000, 4-0, V2:**emblematic cycles 5, 10**111, 2-0, V1:**emblematic cycles 11, 12**111, 2-0, V2:**emblematic cycles 1, 2**111, 4-0, V1:**emblematic cycles 15, 16**111, 4-0, V2:**emblematic cycles 3, 4

This is awesome, but keep in mind that we’ve been working so far off of the 8-bit sequence cycles that were reduced from the complete 16-bit emblematic cycle. 8-bit numbers can only be uniquely identified based on three binary choices or classifications, while we need four choices. So, at this point, we need to step back and take a look at the emblematic cycles themselves. It’d be nice if we could point to some “crucial” or “seed” point in the emblematic cycles that would decide this for us, but when we deal with cycles, we’re effectively dealing with circles. We need to find some way to distinguish between the pairs given above in a regular manner that makes sense across all groupings. Mathematically, this makes sense, but it also paves the way for the element for this last choice: earth. If we consider the other three elements (fire, air, and water) to be immaterial without an earthy basis, then those three elements provide an “ideal”, “astral”, or “imaginary” form that can lay out a blueprint. When it comes time for that blueprint or idea to be manifested, it can be done in one of two ways, which we’ll call masculine or feminine. This last choice is strictly how an idea becomes real, how a vague form of change can give birth to a concrete form that changes. It’s like developing an algebraic formula from integral calculus.

While we’re talking about elements, let’s also assign elements to the three choices we have so far:

**000/111 Sequence Grouping: Fire.**This was the first pattern we picked up on, based on how the emblems manifest based on their changes. These sequences help provide the manner of expansion within the figures, acting as a kind of blueprint that can still be easily molded or reshaped into other blueprints by rotating them around. This is closest to the element of fire, which provides the idea and force for things grow, similar to the the original idea or design behind something. We’ll assign 000 sequence groups the bit 0 for being more passive and at rest at once, and 111 sequence groups the bit 1, for being more active and in flux at once.**2-0/4-0 Addition Grouping: Air.**This was the second pattern we picked up on, by crossing the boundaries laid by the sequence groups and finding how different emblematic cycles connected with each other. The ones that formed a collection through adding into Populus were marked as being within the same addition group bonded more cohesively than others, and air is the element associated with things sticking to or working with each other. We’ll assign cycles in the 4-0 addition group the bit 1 for having more overlap and being more interactive, and the cycles with 2-0 as their addition group the bit 0 for having less interaction.**V1/V2 Collapsed Bit Grouping: Water.**This was the third pattern we picked up on, through seeing how the figures expanded in their own manner. Although the sequence groups specify what to expand, the collapsed bit groups indicate by how much, providing the growth or famine in the positions to be expanded. Just as water takes the shape of its container while retaining the same volume, so too will these values of bit expansion/contraction fill up the spaces given to them before. Cycles in the V1 group will be given the bit 1 for being more fluid and active at first, and cycles in the V2 group will be given the bit 0 for being more passive and at rest at first.

For reasons above, we know this fourth and last decision must be for the element of earth. But what might that choice be? For that, we need to inspect the pairs of the emblematic cycles themselves. Here they are, based on the groupings given above:

**000, 2-0, V1**- emblematic cycle 8: 1000010111101001
- emblematic cycle 13: 1000011110010110

**000, 2-0, V2**- emblematic cycle 6: 1000010110100111
- emblematic cycle 9: 1000011001011110

**000, 4-0, V1**- emblematic cycle 7: 1000010111100110
- emblematic cycle 14: 1000011110100101

**000, 4-0, V2**- emblematic cycle 5: 1000010110011110
- emblematic cycle 10: 1000011010010111

**111, 2-0, V1**- emblematic cycle 11: 1000011010111100
- emblematic cycle 12: 1000011011110010

**111, 2-0, V2**- emblematic cycle 1: 1000010011010111
- emblematic cycle 2: 1000010011110101

**111, 4-0, V1**- emblematic cycle 15: 1000011110101100
- emblematic cycle 16: 1000011110110010

**111, 4-0, V2**- emblematic cycle 3: 1000010100110111
- emblematic cycle 4: 1000010100111101

Although we need to decide based on the structure of the emblematic cycles, it’s not readily apparent what that choice might be. All of the emblematic cycles must contain an equal number of 0s and 1s, and all combinations of “components” (100001, 1001, 101, 101, 11, and 1111) must appear and overlap in equal amounts that sometimes cross group boundaries. All the pairs of emblematic cycles above must change equally to balance each other out in the end. Finding any mathematical pattern in this instance seemed futile, especially since this appears to be dependent on the structure itself. In the end, the rule I’m going by is, starting after the figure Tristitia as it appears in the cycle, whichever cycle in the above groups reaches five active lines first is the “active” earth line. We’ll call these cycles “speedy”, while the ones that fall behind “slow”. This is another way of saying “pick the cycle that has 1111 appear first, or if both cycles have 1111 appearing at the same time, pick the one that has 11 appearing closest after that”. This is the only time we rely on the order of the bits within the emblematic cycles themselves which uses an anchor point, but this is the only way to accomplish this method. So, we have the following bit speed groups:

**Speedy:**emblematic cycles 2, 4, 5, 9, 12, 13, 14, 16**Slow:**emblematic cycles 1, 3, 6, 7, 8, 10, 11, 15

And that’s our fourth and final decision! Putting them all together, we have the following table:

Emblematic Cycle | Sequence Group | Addition Group | Collapsed Bit Group | Bit Speed | Binary Sequence | Geomantic Figure |
---|---|---|---|---|---|---|

1 | 111 | 2-0 | V2 | Slow | 1000 | Laetitia |

2 | 111 | 2-0 | V2 | Speedy | 1001 | Carcer |

3 | 111 | 4-0 | V2 | Slow | 1100 | Fortuna Minor |

4 | 111 | 4-0 | V2 | Speedy | 1101 | Puer |

5 | 000 | 4-0 | V2 | Speedy | 0101 | Acquisitio |

6 | 000 | 2-0 | V2 | Slow | 0000 | Populus |

7 | 000 | 4-0 | V1 | Slow | 0110 | Coniunctio |

8 | 000 | 2-0 | V1 | Slow | 0010 | Albus |

9 | 000 | 2-0 | V2 | Speedy | 0001 | Tristitia |

10 | 000 | 4-0 | V2 | Slow | 0100 | Rubeus |

11 | 111 | 2-0 | V1 | Slow | 1010 | Amissio |

12 | 111 | 2-0 | V1 | Speedy | 1011 | Puella |

13 | 000 | 2-0 | V1 | Speedy | 0011 | Fortuna Maior |

14 | 000 | 4-0 | V1 | Speedy | 0111 | Caput Draconis |

15 | 111 | 4-0 | V1 | Slow | 1110 | Cauda Draconis |

16 | 111 | 4-0 | V1 | Speedy | 1111 | Via |

And, presented in a simpler list form:

**Laetitia:**1000010011010111**Carcer:**1000010011110101**Fortuna Minor:**1000010100110111**Puer:**1000010100111101**Acquisitio:**1000010110011110**Populus:**1000010110100111**Coniunctio:**1000010111100110**Albus:**1000010111101001**Tristitia:**1000011001011110**Rubeus:**1000011010010111**Amissio:**1000011010111100**Puella:**1000011011110010**Fortuna Maior:**1000011110010110**Caput Draconis:**1000011110100101**Cauda Draconis:**1000011110101100**Via:**1000011110110010

Is this system of correspondences perfect? Probably not. For one, this is entirely new research without anything to back it up. I’ve never seen this written about or spoken about in any tradition of geomancy, and I have no reason to believe that it was ever done or devised before. For two, I developed these methods of deciding which emblems get which element active or passive almost arbitrarily, based on how I came across them and how it felt. This was largely a mathematical exercise, but there are some spiritual components to this that directed what to look for and how to combine them (when I wasn’t just throwing math at the emblems to see what’d stick). For three, there are a few things that bug me about the emblematic cycles. For instance, every geomantic figure has an inverse (Populus and Via, Puer and Albus, etc.). It might be expected, then, that the geomantic emblematic cycles would correspond to this same method, such that the emblematic cycle of Populus would have Via as its inverse (take all the points in the emblematic cycle, invert them, and rotate it accordingly to sync it up with the rest of the cycles). However, the inverse emblematic cycle of Populus is actually Caput Draconis. The inverses of the emblematic cycles are, in order as above:

- 16
- 4
- 15
- 2
- 8
- 14
- 9
- 5
- 7
- 13
- 12
- 11
- 10
- 6
- 3
- 1

It might be that there’s another method of organizing or allotting the emblematic cycles to the figures so that the inverses of the figures would agree with the inverses of the emblematic cycles. However, the inverse function is the only one that might possibly apply to both geomantic figures and their respective emblematic cycles. Reversion, for instance, won’t work; the reversion of Populus is also Populus, but no such emblematic cycle could possibly revert to itself; the same goes with conversion, as well. Maybe it doesn’t make sense to apply these functions to the cycles directly, instead interpreting them elementally (as in the previous post).

And, dear reader, since you’ve made it this far (I hope), I really thank you for sticking around. This was a puzzle that took me a good few days of constant thinking to solve, and though I’m wary of how strong the system is, it’s at least a good start. I hope you at least got the gist of what I was trying to do; if you have any questions about the hows or whys of the above post (the more specific the better), please let me know down in the comments. We’ll pick up next time with what these emblems can be used for in geomantic divination and magic, and how to make really awesome designs incorporating different depictions of the emblems.

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