Formulas and Calculations

v1.3March 2026Author: Marco Gipp

This document presents the mathematical formulas of the Law of Equalization and demonstrates their application through concrete examples.

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1. The Intrinsic Energy Formula

Basic Formula

$$E=\rho \cdot V\cdot S\cdot k$$

Variable	Meaning	Unit
$E$	Intrinsic energy	Joules
$\rho$	Density of matter	kg/m³
$V$	Volume of matter	m³
$S$	Stability factor (binding energy)	dimensionless, 0–1
$k$	Material constant (material-dependent)	material-dependent

Alternative Formulation (via mass)

Since $\rho =m/V$:

$$E=\frac{m\cdot S\cdot k}{V}$$

What the Formula Describes

Intrinsic energy is NOT equal to mass. The formula accounts for: density (storage capacity per volume), volume (spatial extension), stability (molecular binding energy), material constant (specific material properties).

Why this formula is more precise than $E=m{c}^2$: Energy is not defined exclusively through mass. The structure of matter plays a role ($S$ factor). No speed of light required ($c$ drops out). Material-dependent — different elements = different energy.

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2. Sample Calculations

Iron Block

Parameter	Value
Density ($\rho$)	7,874 kg/m³
Volume ($V$)	0.01 m³
Stability factor ($S$)	0.9
Constant ($k$)	1.5
Intrinsic energy	0.106 J

$$E=7,874\cdot 0,01\cdot 0,9\cdot 1,5=0,106\text{ J}$$

Copper Block

Parameter	Value
Density ($\rho$)	8,960 kg/m³
Volume ($V$)	0.01 m³
Stability factor ($S$)	0.85
Constant ($k$)	1.4
Intrinsic energy	0.106 J

$$E=8,96\cdot 0,01\cdot 0,85\cdot 1,4=0,106\text{ J}$$

Observation: Despite different materials, objects can have the same intrinsic energy when the parameters balance each other out.

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3. Top 20 Elements by Intrinsic Energy

(based on $\rho$, $V=0,01{\text{ m}}^3$, $S$, $k$)

Rank	Element	Density (kg/m³)	$S$	$k$	Intrinsic Energy (J)
1	Osmium	22,610	0.95	1.5	322.19
2	Iridium	22,560	0.94	1.5	318.10
3	Tungsten	19,250	1.00	1.5	288.75
4	Platinum	21,450	0.93	1.4	279.28
5	Rhenium	21,020	0.91	1.4	267.79
6	Gold	19,320	0.90	1.4	243.43
7	Uranium	18,900	0.85	1.3	208.85
8	Tantalum	16,650	0.89	1.2	177.82
9	Rhodium	12,410	0.92	1.4	159.84
10	Mercury	13,534	0.80	1.3	140.75
11	Molybdenum	10,280	0.89	1.4	128.09
12	Thorium	11,724	0.88	1.2	123.81
13	Silver	10,490	0.86	1.3	117.28
14	Lead	11,340	0.78	1.1	97.30
15	Cobalt	8,900	0.87	1.2	92.92
16	Nickel	8,908	0.85	1.2	90.86
17	Copper	8,960	0.84	1.2	90.32
18	Iron	7,874	0.88	1.3	90.08
19	Chromium	7,190	0.81	1.1	64.06
20	Zinc	7,140	0.75	1.1	58.91

Important: Highest intrinsic energy $\ne$ highest mass. The stability factor ($S$) plays a decisive role. Tungsten has $S=1,0$ (highest stability), hence rank 3 despite lower density than platinum.

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4. Planetary Positions Formula

Basic Formula

$$\frac{r_i}{r_j}={\left(\frac{E_i}{E_j}\right)}^{1/3}$$

Variable	Meaning
$r_i,r_j$	Orbital radii of two planets (in m)
$E_i,E_j$	Intrinsic energies of the planets (in J)

Interpretation: The orbital radius of a planet relates to the cube root of its intrinsic energy relative to a reference planet.

No gravitation required! Only energy ratios.

Why the Cube Root?

Since we live in three-dimensional space, the system pressure of the sun (System 2) distributes volumetrically. The cube root corrects the dimensions from energy (volume/mass) to distance (radius). This is pure 3D geometry: $V\propto r^3$, therefore $r\propto E^{1/3}$.

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5. Test: Earth vs. Mars

Masses (proxy for intrinsic energy): Earth: $5,97\times 10^{24}$ kg, Mars: $6,42\times 10^{23}$ kg

$$\begin{array}{rl}\frac{E_{\text{Earth}}}{E_{\text{Mars}}}& =\frac{5,97\times 10^{24}}{6,42\times 10^{23}}\approx 9,30\\ (9,30{)}^{1/3}& \approx 2,10\text{(theory)}\\ \frac{r_{\text{Mars}}}{r_{\text{Earth}}}& =\frac{2,279\times 10^{11}}{1,496\times 10^{11}}\approx 1,52\text{(observation)}\end{array}$$

Deviation: ~28%

Interpretation: Intrinsic energy $\ne$ mass. Mars has cooled (lower active energy); Earth has a hot iron core (higher active energy). System pressure of the sun is not homogeneous. Reference-point interpolation is still pending. Nevertheless: correct order of magnitude without a gravitational constant.

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6. Test: Earth vs. Jupiter

Without Correction

$$\begin{array}{rl}\frac{E_{\text{Jupiter}}}{E_{\text{Earth}}}& =\frac{1,898\times 10^{27}}{5,97\times 10^{24}}\approx 318\\ (318{)}^{1/3}& \approx 6,84\text{(theory)}\\ \frac{r_{\text{Jupiter}}}{r_{\text{Earth}}}& =\frac{7,785\times 10^{11}}{1,496\times 10^{11}}\approx 5,20\text{(observation)}\end{array}$$

Deviation: ~31%

With Correction (Jupiter: 70% Hydrogen & Helium)

Jupiter consists predominantly of gases with low binding energy (low $S$). The conventional mass therefore overestimates its effective intrinsic energy. At an effectiveness factor of 75%:

$$318\times 0,75=238,5\Rightarrow (238,5{)}^{1/3}\approx 6,2$$

Deviation (corrected): ~19%

With reference-point interpolation (in development) and more precise intrinsic energy values (instead of mass as proxy), this deviation would decrease further.

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7. Reference-Point Interpolation (in Development)

Every system has a reference point A (boundary, e.g., heliopause) and reference point B (center, e.g., sun).

$$r=r_A+\alpha (E,\text{System})\cdot (r_B-r_A)$$

Where $\alpha$ depends on: intrinsic energy of the object, position in the system, system pressure gradient.

Status: Formula in development; foundational principle established.

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8. General Equalization Formula (in Development)

$$\frac{dE}{dt}=k\cdot A\cdot \frac{E_1-E_2}{d}$$

Variable	Meaning
$dE/dt$	Energy flow per time
$k$	Medium constant
$A$	Contact surface area
$E_1-E_2$	Energy differential
$d$	Distance between systems

Status: Conceptual; refinement to follow.

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9. "Energy Dominates Energy" — Practical Applications of the Formula

Water Jet Cutting: Why Water Cuts Steel

A water jet under extreme pressure (~4,000 bar) cuts effortlessly through steel. Classical physics explains this through "kinetic energy" and "material removal" — separate formulas for separate aspects.

In the Law of Equalization: The same basic formula. The water is massively overloaded by the pressure — its demand energy per contact surface exceeds the intrinsic energy of the steel. "Energy always dominates energy": the higher-energy system (water jet) dominates the lower-energy system (steel at the contact point).

$$E_{\text{jet}}>E_{\text{steel}}\Rightarrow \text{steel loses (Variant 3: destruction)}$$

The same formula also explains why hydrogen can diffuse through steel containers: its extremely high intrinsic energy relative to its minimal material mass "dominates" the binding energy of the metal lattice.

Diamond vs. Glass: Why Diamond Scratches

Diamond ($S\approx 1,0$, $k\approx 1,5$) has the highest intrinsic energy per volume of all natural materials. Glass ($S\approx 0,5$) has significantly less. On contact, the diamond dominates — Variant 3 (destruction) occurs in the glass, not in the diamond.

Ball Against Wall vs. Ball Against Glass

Ball against wall: The wall has higher intrinsic energy → ball bounces back (Variant 2: return).

Ball against glass: The ball (in motion = overloaded) has higher intrinsic energy than the stationary glass → glass shatters (Variant 3: destruction).

The principle is always the same: Compare the intrinsic energies of both systems. The higher-energy one wins. No separate formula for "hardness," "momentum," or "elasticity" needed — everything reduces to $E_1$ vs. $E_2$.

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10. Comparison with Existing Formulas

$E=\rho \cdot V\cdot S\cdot k$ vs. $E=m{c}^2$

Aspect	Einstein ($E=m{c}^2$)	Law of Equalization ($E=\rho VSk$)
Energy from	Mass $\times$ $c^2$	Material properties
Constant	Speed of light	Material constant
Material-dependent	No	Yes
Structure-dependent	No	Yes ($S$ factor)
Speed of light	Required	Not required

Planetary Positions vs. Newton $F=G\frac{m_1{m}_2}{r^2}$

Aspect	Newton	Law of Equalization
Mechanism	Attractive force	Pressure equalization
Action at a distance	Yes (mysterious)	No (direct contact)
Constant	$G$ (universal)	No mysterious constant
Explanation	Describes effect	Explains cause

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11. Open Questions

1. Refinement of the reference-point interpolation
2. Derivation of $S$ and $k$ from atomic physics (electron configuration, nuclear binding energy)
3. System pressure function: how does pressure vary with distance from center?
4. Experimental validation: which tests are possible with current technology?
5. Application of the planetary formula to all 8 planets and exoplanet systems

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