View source for Mneme World Generator/Logic Specification (Feb 22 2026)

= Mneme Placement Logic: Formula Reference Guide =

This document isolates and explains the core mathematical formulas used to place planets and determine orbital stability within the Mneme Generator.

== 1. The Anchor Formulas (System Mapping) ==

Before placing any planets, the system's "zones" must be defined. These are entirely dependent on the Star's Luminosity ($L_{Star}$).

=== Formula 1.1: The Snow Line ($D_{snow}$) ===

Purpose: Determines the exact distance (in AU) where water freezes in a vacuum. This is the "seed point" where Gas Giants are most likely to form.

The Formula:
$$D_{snow} = 2.7 \times \sqrt{L_{Star}}$$

Variables:

$L_{Star}$: The Star's Luminosity relative to Sol (Sol = 1.0).

=== Formula 1.2: The Habitable Zone Core ($HZ_{center}$) ===

Purpose: Finds the "Goldilocks" center where Earth-like temperatures are most likely.

The Formula:
$$HZ_{center} = \sqrt{L_{Star}}$$

Edges:

Inner Edge: $0.95 \times HZ_{center}$

Outer Edge: $1.37 \times HZ_{center}$

== 2. The Dominance Formulas (Hill Stability) ==

These formulas determine if two planets can coexist at their specific distances without colliding or ejecting one another.

=== Formula 2.1: The Hill Sphere Ratio ($\mu$) ===

Purpose: Calculates the raw "Gravitational Weight Class" of a planet relative to its star. This ratio remains constant regardless of where the planet is placed.

The Formula:
$$\mu = \sqrt[3]{\frac{m_1 + m_2}{3 \times M_{Star}}}$$

Variables:

$m_1, m_2$: Mass of the two planets interacting (in Earth Masses).

$M_{Star}$: Mass of the central star (in Earth Masses. 1 Solar Mass = 333,000 Earth Masses).

=== Formula 2.2: The Mutual Hill Radius ($R_H$) ===

Purpose: Translates the Hill Sphere Ratio ($\mu$) into an actual physical distance (in AU) based on where the planets are currently sitting.

The Formula:
$$R_H = a_{avg} \times \mu$$
(Where $a_{avg}$ is the average distance of the two planets: $\frac{a_1 + a_2}{2}$)

Concept: A planet's gravitational "bubble" physically widens the further it gets from the star.

=== Formula 2.3: The Required Safety Space ($W_{req}$) ===

Purpose: Determines the absolute minimum gap (in AU) needed between two orbits to prevent chaos over a billion years.

The Formula:
$$W_{req} = K \times R_H$$

The "K" Constant (The Rule of K):

$K = 5$ (The Mneme Standard / "Rule of Five"). This creates a densely packed, but stable system.

$K = 8$ (Sol Standard). Creates a relaxed system like our own Jupiter and Saturn.

== 3. The Action Formulas (The Shove) ==

When a heavy planet forces its way into an occupied zone, the lighter/outer planet must be moved outward to maintain stability.

=== Formula 3.1: The Shove Adjustment ($a_{new\_outer}$) ===

Purpose: Calculates the exact new orbital distance for a planet that has been "pushed" by a heavier inner neighbor.

The Formula:
$$a_{new\_outer} = a_{inner} + W_{req} + Variance$$

Variables:

$a_{inner}$: The position of the heavy planet that claimed the spot.

$W_{req}$: The Required Safety Space calculated in Formula 2.3 (using $K=5$).

$Variance$: A small random addition (e.g., $+0.01$ to $+0.1$ AU) so the pushed planet isn't sitting exactly on the mathematical edge of destruction.

== 4. Multi-Star Stability Formulas (The Rule of Five & Barycenters) ==

When a system has two or more stars, their massive gravities dictate the orbital structure. Stars orbit a shared center of mass called a Barycenter.

=== Formula 4.1: The Barycenter Calculation (Center of Mass) ===

Purpose: Determines the exact point in space two stars orbit around, and how far each star sits from that central point.

The Formulas:

Distance of Star 1 to Barycenter ($r_1$):
$$r_1 = D_{total} \times \frac{M_2}{M_1 + M_2}$$

Distance of Star 2 to Barycenter ($r_2$):
$$r_2 = D_{total} \times \frac{M_1}{M_1 + M_2}$$

Variables:

$D_{total}$: The total distance between the two stars.

$M_1, M_2$: The mass of Star 1 and Star 2.

Note: If $M_1$ is vastly heavier than $M_2$, the Barycenter will be deep inside $M_1$ (like the Earth and the Sun). If they are equal mass, the Barycenter is exactly in the middle.

=== Formula 4.2: The Binary Forbidden Zone ($FZ$) ===

Purpose: Defines the orbital space rendered uninhabitable/unstable by the Companion Star.

The Formulas:

Inner Limit: $FZ_{in} = \frac{D_{total}}{5}$

Outer Limit: $FZ_{out} = D_{total} \times 5$

Rule: No planet can exist between $FZ_{in}$ and $FZ_{out}$ relative to the Barycenter.

=== Formula 4.3: S-Type Placement (Far Companion / Circumstellar) ===

Purpose: Use this formula if you want planets to orbit only one specific star. The Companion must be pushed far enough away to leave the inner system safe.

The Formula:
$$D_{min} = a_{max\_desired} \times 5$$

Outcome: To keep a planet safe orbiting Star A at $4.0$ AU, Star B MUST be placed at $\ge 20.0$ AU.

=== Formula 4.4: P-Type Placement (Close Companion / Circumbinary) ===

Purpose: Use this formula if you want planets to orbit both stars at once (like Tatooine). The stars must be placed extremely close together to act as a single gravitational point.

The Formula:
$$D_{max} = \frac{a_{min\_desired}}{5}$$

Outcome: To keep a planet safe orbiting the Barycenter at $0.8$ AU, Star A and Star B MUST be separated by no more than $0.16$ AU.

=== Formula 4.5: Hierarchical Multi-Star (The Alpha Centauri Model) ===

Purpose: To safely map a stable 3-star (Trinary) system.

The Logic:

Calculate the Inner Binary (e.g., Alpha Centauri A & B). They orbit their Barycenter ($BC_{AB}$) at distance $D_{AB}$.

Treat the Inner Binary as a Single Star Point with mass $M_{AB} = M_A + M_B$.

Place the Tertiary Star (e.g., Proxima Centauri / Star C).

Stability Check: Star C must be placed far outside the Forbidden Zone of the Inner Binary.
$$D_C \ge (D_{AB} \times 5) \times \text{Safety Factor}$$

Real World Example: Alpha Centauri A & B are ~11-35 AU apart. Proxima Centauri is ~13,000 AU away, making it incredibly stable as a hierarchical trinary.

== Example Walkthrough: "The Shove" in Action ==

The Setup: A star ($M_{Star} = 333,000$). We have a Gas Giant ($m_1 = 300$) trying to settle at 3.0 AU ($a_1$).

The Victim: An Ice Giant ($m_2 = 15$) is currently sitting at 3.2 AU ($a_2$).

Calculate Ratio ($\mu$): $\sqrt[3]{(300 + 15) / 999,000} \approx 0.068$

Calculate Average Distance ($a_{avg}$): $(3.0 + 3.2) / 2 = 3.1$ AU.

Calculate Hill Radius ($R_H$): $3.1 \times 0.068 \approx 0.21$ AU.

Calculate Required Space ($W_{req}$): $5 \times 0.21 = \mathbf{1.05 \text{ AU}}$.

The Test: The current gap is only $0.2$ AU. The required gap is $1.05$ AU. Conflict detected.

The Shove: The Ice Giant is pushed out to $3.0 \text{ (Giant Pos)} + 1.05 \text{ (Safety)} = \mathbf{4.05 \text{ AU}}$.

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