Wolfram Virtue Technology Conference:	November 2025

ALΣXANDΣR	cage free thinkin' from the SandBox
12 pages; 2100 words, 18 Slides, approx.20 min. 9/24/25.

FOUR STEPS

for the Parametric Unification of Two Central Force Fields; Gravity and Nuclear.
Joining Classic Big with Quanta Small.

VENUE:
Wolfram Virtue Technology Conference:	November 2025

ALΣXANDΣR	cage free thinkin’ from the SandBox

KEY WORDS:
Discovery Curve, Definition Curve, Hybrid Curve

ABSTRACT

Our domain of being, human SpaceTime experience, is surface acceleration explored by Galileo. Newton, our second SpaceTime, provides a path to quantum existence, our third  SpaceTime. Fall from Newtons outer-space to the surface acceleration curve of Galileo, penetrate  acceleration field nuclear emptiness, and fall into the inversed world of quanta existence.

STEP ONE

Recognize three Parametric Space Time Squares.

S & T1 : Galileo' s Uniform Acceleration Field; incline plane.

S & T2 : Sir Isaac Newton' s Universal Law of Gravity; planetary period motion.

S & T3 : late 19 th and early 20 th Century collective, Quantum Thermodynamic Fields.

STEP2: CURVED SPACE DIVISION ASSEMBLY

BASIC GRAVITY FIELD CURVED SPACE DIVISION ASSEMBLY

INDEPENDENT CURVE⟶1Cos[t], 1Sin[t]
DEPENDENT CURVE⟶{t,((t^2)/(-4(p)))+r}

LATUS RECTUM RESIDENT CURVE

is the average energy diameter of every (Subscript[M, 1])(Subscript[M, 2]) system

unity curves:  one part motion and one part potential.

unity curves only happen at a slope (+/-1) event

({{MARS, KM PARAMETERS, CSDA comparative, CSDA PARAMETERS}, {perihelion, 206620000, (206620000.0/227925000)*2, 1.8131}, {aphelion, 249230000, (249230000.0/227925000)*2, 2.1870}, {average, 227925000, (227925000/227925000)*2, 2}, {ASI(potential), 113962500, (227925000/227925000)*1, 1}, {average v, 24.13, \[Placeholder], 24.13}})

(206620000.0/227925000)*2 ⟶ 1.8131
Establish Mathematica template for energy level:
((t^2)/(-4))+1/.t->1.8131 -> 0.1782
((t^2)/(-4))+1/.t->2.1870 ->(-0.1957)

ParametricPlot[{{Cos[t],Sin[t]},{t,((t^2)/(-4))+1},{1.8131,t},{2.1870,t},{t,0.1782},{t,(-0.1957)}},{t,-π,2π},PlotRange->{{-2,3},{-1,1}}]

CONSTRUCT HIGH ENERGY MOTIVE CURVE OF MARS

CONSTRUCT LOW ENERGY MOTIVE CURVE OF MARS

STEP(3): index solution curves

(13^(1/5))⟶ ((13^(1/5)))⟶(t, ((t^(index(5)))/(-2))+(radicand/2))⟶(t, ((t^5)/(-2))+(13/2))

only three indices are used: ((SC^(1/0)),(SC^(1/1)),(SC^(1/2))  )        (abscissa, ordinate); CrossOverΔ

CURVE(j), rest(OC2): (t, ((t^0)/(-2))+((OC(2))/2))						A2: ((Sqrt[1;Galileo]), ((OC(n))/2))

CURVE(i), registration(SC3): (t, ((t^1)/(-2))+((SC(3))/2))				B2: ((Sqrt[SC(n)]), ((OC(n))/2))

CURVE(f)potential(disc(a2): (t, ((t^2)/(-2))+((SC(3))/2))					C2: ((Sqrt[SC(n)]), ((SC(n))/2)-((Sqrt[SC(n)])/2))

Is this the one and only perfect odd #9? Itself a degree two happening. Straight up exponent and inverse exponent connection. Forever linked in exponent curved space; back and forth. 3&9. The squaring of SpaceCurve 3 brings us to SpaceCurve9. the square root of SpaceCurve 9 brings us to space curve 3.

STEP4: Index Solution Curves; jump average energy curves of orbit using x-over’s

SpaceCurve(8) ∝ ((((Sqrt[micro space curvature])))^(-1)) ⟶(1/((((1/64)))^(1/2))) ∝SquareSpace(8)

Establish crossover triangle to transition from displacement (SpaceCurve(2), to Galileo' surface acceleration curve(SpaceCurve(1). Crossover’s will help us navigate across SC(2) to SC(1).

Thank You for Attending

Bonding Like Period Element Atoms Using Parametric Geometry and Z#

WOLFRAM VIRTUAL TECHNOLOGY CONFERENCE 
OCTOBER 2020
ALEXANDER CEO SAND BOX GEOMETRY LLC

EXPLORING SPACETIME SQUARE2; HEAT of FUSION and VAPORIZATION

at standard atmosphere pressure and 0° centigrade

FINDING NUCLEAR CORNER of PROTIUM Z#1  S&T2	((1/8),(9/8)); (1, 2)

1. Let (p) be initial focal radius (1/4) of neighborhood (J); latent heat nuclear stress meter. Then, shaping curve of nucleus will be focal radius (p).

{(1/4)Cos[t], (1/4)Sin[t]+0}

If we let a radius (p) be aligned (co - incident on) nuclear squaring asymptote, we have a right triangle having hypotenuse (p).

Solve[((((1/4)))^2)==2(((t))^2),t]

{{t->-(1/(4 (Sqrt[2])))},{t->(1/(4 (Sqrt[2])))}}

radius of binding energy curve:		{ (1/(4 (Sqrt[2])))Cos[t], (1/(4 (Sqrt[2])))Sin[t]+0}.

Solve[2(t^2)==((((1/(4(Sqrt[2])))))^2),t]

{{t->-(1/8)},{t->(1/8)}}

abscissa = (1/8) and ordinate = Z # + abscissa: ((1/8),(9/8))