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The Universal Vector Space and Static Equilibrium Techniques

Paul TE Cusack*

Independent Researcher, BSc E, DULE, 1641 Sandy Point Rd, Saint John, NB, Canada E2K 5E8, Canada

*Corresponding Author:
Cusack PTE
Independent Researcher, BSc E, DULE
1641 Sandy Point Rd, Saint John
NB, Canada E2K 5E8, Canada
Tel: (506) 214-3313
E-mail: [email protected]

Received date: April 30, 2017; Accepted date: June 21, 2017; Published date: June 30, 2017

Citation: Cusack PTE (2017) The Universal Vector Space and Static Equilibrium Techniques. Fluid Mech Open Acc 4: 165. doi: 10.4172/2476-2296.1000165

Copyright: © 2017 Cusack PTE. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

In this paper, the author uses well know Structural Engineering theorems to solve for physical constants for the universe. The universe is in static equilibrium. Thus, these rules of statics apply to the universe. We show how the speed of light is derived to be c=2.997. In keeping with the theme of Astro-Theology, the physical universe are inseparable.

Keywords

Vector space; Force polygon; Static equilibrium; Speed of light

Introduction

Yet another way to analyze the universe is through a vector space, or a force polygon wihth 8 joints, 2 reactions, and 18 memebers. This system will be considered below (Figure 1).

fluid-mechanics-universal-vector-space

Figure 1: The universal vector space.

The Universe can be modelled as a vector space, or alternatively named a force polygon in structural mechanics. Since there are 10 unknowns in AT math, there are 8 joints in this vector space plus two reactions acting at joint 3, Energy and time. There are 18 members connecting every nod to every other node. A member from joint 1 to joint 3 and joint 3 to joint 7 lines up perfectly, which indicates a statically determinate system. These two members will be analyzed below as a structural simply supported beam, what I term, the “universal beam.” The load on these beams is q, and is the resultant of the Energy-time vectors. All the members of this vector space make up the physical constants of the universe (Figure 2).

fluid-mechanics-simply-supported-beams

Figure 2: Two simply supported beams.

The definition of the numeral one is where the fraction meets the multiple. Every other number is either a multiple or a fraction of one. One, is therefore the tipping point. In algebraic terms, the fraction and the multiple meet at he golden mean. Therefor, t^2-t-1=0 is where Energy and time meet -at joint 3 shown in Figure 2.

LOAD:

q=f(E,t)

q=1+sin 60, refer to Figure 3.

fluid-mechanics-universal-load

Figure 3: The universal load, q.

q=t^2-t-1=E

1+sin 60 = t^2-t-1=0

Roots,

t=2.2652, 1.2652=rho

t=rho when E=q

ENERGY: refer to Figure 4.

fluid-mechanics-energy-time-graph

Figure 4: Energy time graph.

E=Mc^2

dE/dt=dM/dt c^2

1=1/t=(2)(2)(3)

1/t=12

t=1/12

Cross Product

s=|E||t| sin t

4/3=E x (1/12)(sin 60 deg.

E=1.0695~1.07=Mass H+

Add an electron:

1.07+0.511=1.581~sin 1

Energy & Graivitational Acceleration

E=W x t

=F x dx t

1-sin 1=(1+sin 1)(d)(1)

[1-sin 1]/[1+sin 1]=1585/1.866=117.7 =M=d

117.7/12=9.8083=g (Figure 5).

fluid-mechanics-universal-beam

Figure 5: The universal beam.

Static Equilibrium

Joints 1,3,7

2n-1

[(dM/dt)n-M]/c=jcritical/c=Ec

When E=0

[(y’)y-y]/c=0

(y’)y-y=0

(dE/dt)E-E=0

Clairnaut

d²E/dt²-E=0

So,

(dE/dt)E=d²E/dt²

(dy/dx)y=d²x²/dt²

y[Ln (dy/dx)=Ln (d²y²/dt²)

y(Ln 1)=y“

y(0)=Ln y”

0=eLn y“

0=y”

a=0

F=Ma

F=0

ΣFx=0 Static Equilibrium

ΣFx=0

q cos 60° -F cos 60°=1.866 (1/2) - (2.667 (1/2)

=0.4005

=t

40% of a cycle =1 rad.

1/250=1/T=1/[1/=t (Figure 6).

fluid-mechanics-cantilever-stresses

Figure 6: Cantilever stresses.

f=My/I

0.0809=M (Pi)/(1/Y)

M=52

Element = Tellurian at fulcrim

ρ=126.7

10.5/0.0809=11.72~1118 Last element in the periodic table of the elements (Figure 7).

fluid-mechanics-areas-under-curve

Figure 7: Areas under curve.

A1=1/2 bh=1/2 (1/81)(1.236)=7.639

A2=1/2(129.5)(10.5)=679.22

Σ[A1+A2]=6.7922-7.639=0.8468

A3=∫(-0.618-0) t²-t-1=2t³/3-³2/2-t=E²/2

E=0

A4=∫(0-1.618)=1.618

Σ[A3+A4]=1.618

Σ[A1+A2]/{A3+A4]=

0.8468/1.618=2.998~c (Figure 8).

fluid-mechanics-chain-link-fence

Figure 8: The chain link fence

d²y/dt²=M/EI

dθ/dt=M/EI

1 rad=M/(0.4233)(1/Y)

Y=202

M=1/1201

1/12.01 ×360=2.9975=c

δ=∫ M/EI x dx

=Σ[A1+A2]

δ/t=δ/1.618=Σ [A1+A2]/Σ[A3 +A4]=d/t=v=c

Divine Mercy Icon

E=Mc²

MαE/c²

dM/dtαdE/dt ×1/[2c]

2α1(1/2(3))

12 α1

12 Apostles

1/12=t =Judas

And

F=t=d

Fd=1+F=1+sin 60°

Fd-F-1=0

(Sin θ)(sin θ)- F-t=0

1/s=Fd

1=Fds

t=E=Fds

t=E=Fs²

=F(4/3)²

=F(16/9)=F √π

s²=√π

s4=π=I=moment of inertia=t

s4=Ω = linear operator

Now, for he Great Abomination of Revelation:

y=δ

dy/dt=θ

d²y²/dt²=dθ/dt=M/EI

d³y/dt³=dM/EI dt=V/EI

=2/[0.4233)(4/3)=345.359

April 3, 2005 @ 3:36 P.M. Date and Time Idol of the oppressor was installed at the Cathedral of the Immaculate Conception in Saint John: Divine Mercy Sunday.

EVIL=Energy (Shear)(Inertia)(Length)=(1)(2)(1/202)(1/2)=1/202=1/ Y=1/E=t Time is evil (Figure 9).

fluid-mechanics-cathedral-immaculate-conception

Figure 9: Cathedral of the immaculate conception in Saint John: Divine mercy Sunday.

Conjugate Beam Theory

The slope (t=1.0123 rads) and the deflection (E=0) is the shear in the real beam is the shear (58) and the moment (0) in the conjugate beam [1,2].

In the Conjugate Beam, the,

Shear V= Slope θ

and

Deflection δ= Moment M

so,

V=dM/dt =2=√3/sin θ

θ=dE/dt =1

δ=t =∫M/EI x dx =1-sin 1=Fd d=1682

M=1-sin1=Fd=0.1585

EαMc²

Derivative:

dE/dt= dM/dt × 2c

θ=V×2 ×δ

θ=2√3/sin θ ×δ

θ/δ=4.00=t=40% of a cycle

δ=θ/4

θ=1/2π

δ=0.1592/4=0.0398=1/251=1/T=t

δ=t=y=E

The Universal Beam

∫ BMD=∫( -1 to 2) t³-t²-t-2

=-[1.333-1]

=-0.333

=-1/c

s-t=-1/c

[|E||t|sin θ]-t=-1/c

t[ E sin θ-1]=-1/c

1[sinθ-1]=-0.333

sin θ=0.777

θ=2.98~c

sin θ-1=-1/c

1-sin θ=1/c

θ=0.7296 rads

41.8°/360°=116.1=M

1/c=E×t

1/(tc)=E

1/c ÷ (1/E)=E

E/c=t

E/c=1

E=3 (Figures 10 and 11).

fluid-mechanics-universal-diagram

Figure 10: Universal M/EI diagram

fluid-mechanics-bending-moment-diagrams

Figure 11: Load, shear, and bending moment diagrams.

Axial load

ð=PL/AE

2t-1=w=0

2t=1

t=1/2

Moment

t³-t²-t-2=0

(1/2)³-(1/2)²-(1/2)-2=26.25

δ=1=P(117.27)/[A (0.4233)}

P/A=σ=3.587

σ/Mp+= 3587/938=26.15

Work

WInternal=∫(0-L)=M²/2EI dx

(where M=moment is this particular case.)

W=(F x)²/2EI dx

=F ²(2/3x³)/(2EI) from 0 -L

F is the Superforce=2.667

=2.667² (2/3)(118)/2 (0.4233)(1/202)

=1/127.5=1/ρ.

And the Ether:

Shear Modulus G

1/G=2(1+υ)/E

=2(1.27)/0.4233)

=6=1/G

G=0.167= monoatomic gas γ

WI=K∫ (0-L) V²/(2GA) dx

1/ρ=K/A dx/(2(1/6))

K/A dx = 941

K for a square =1.2

1.2/A dx =941

A=x²

1.2 x-2 dx=941 dx

1.2 (-1/x)=941.

x=127.5=ρ

A=x²=127.5²=1625 (~1623 Navier-Stoke’s Solution)

ΣWInternal=WAxial+WBending+WShear+WTwist

ΣWInternal=4 (1/ρ)=3.13~π=t.

Real work

ΣWInternal=Mom. +V

=P/2(Average Mass +1)

(Note: Atomic Number of Element Te=58)

=2.667/2[58+1]

=1.334(59)=78.667=1/ρ

For a Virtual Displacenent, δ=1, the Total work done =0

ΣWExternal=ΣWInternal

ΣWTotal=4 ΣWInternal

F × d=(2.667)(118)=4/ρ=4/127.6=π

d=118=Mass of the Periodic Table of the Elements.

Conclusion

So we see that Statics and mechanics as applied to a force polygon provide useful insights into physical constants. In addition, we show that the physical and spiritual are inseparable. Energy methods are useful is understanding the Ether as a monoatomic gas.

References

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