volumetric perpetual harmonic oscillator

8/17/2019 VOLUMETRIC PERPETUAL HARMONIC OSCILLATOR

1/44

G C

x( )2

2:=

Z c

2

ec

:= rs 100

98

:=RS rs:=

Pm 2( ) 7 RS⋅+[ ] PM⋅

7 RS⋅ 1 k 1−

7

2

+

⋅ 2( )+

:= Pn 2( ) 7 RS⋅+[ ] PM⋅

7 RS⋅ 1 k 1−

7

2

+

⋅ 2( )+

1 k 1−

7

2

+

⋅:=

0

∞

n

1−( )n

2 n⋅ 1+( )2∑

=Me C

1 x−( )7

7my⋅ 1

2

5−

⋅:= Mep PM k 1−

7

2 π⋅

10⋅

2

⋅:=

Dp C1 x+ k ⋅( )7

:= tp my C⋅7

:= Lp C tp⋅:= Mps Lp3 Dp⋅:=

DD 1

C3

1 2

5−

⋅

:= TT G

DD:= RU C TT⋅:= MU RU

3DD⋅:=

DATA SECTION : THE MATHEMATICAL SECTION IS BELOW THE FIRST HEADING (since

mathcad uses the data derived here to povide solution automatically.)

THE IMPORTANT ASPECT IS THAT NO MEASURED VALUES ARE USED EXCEPT THE

PLANCK'S AND COULOMB'S CONSTANT AND VELOCITY OF LIGHT ONLY FOR

COMPARSON.IT IS A HALLMARK OF SANKHYA THAT RECREATES NATURES DERIVATIONS THROUGH

AN AXIOMATIC PROCESS. THE PI AND CATALAN'S CONSTANT ARE DERIVED

THROUGH AXIOTIC PROCEDURES. I HAVE NOT EXPLAINED THE SYMBOLS HERE AS IT

IS IN THE BOOK. FURTHER ALL VALUES ARE ONLY RATIOS AND HENCE

DIMENSIONLESS.

x 1 2

2+ 1−

2:=

k 2

1

3:= C 10

2

x3

:= RS

0

100

n

2

100

n

∑=

:=Kx

101 x+

2

1

3⋅ 10

22−( )⋅

23

1−( ) 23⋅ 102⋅:=

Px 10

2 π⋅ 3⋅

3

:= PM Kx

Px C3⋅

:= my Kx

C6

:= Ne my C

2

⋅ 2 π⋅

7

2

⋅:= Lp my C

2

7⋅:=

i 0 200..:= A0

x

2:=

Ai 1+

1 1 Ai( )

2−−

2

Ai( )

2+

2:=

ec 1.60217733 10 19−

⋅:= KV 1

k 1−

3

:=h 6.6260755 10

34−⋅:=

c 299792458:=


2/44

The above values are derived constants for solving the formula below.

SIMULTANEOUSTHRESHOLD

2 π⋅( )2

k 1−

1−

1+ 1.0065838771=

7

ec G⋅ C⋅ 10⋅

1−

1.0065942766=

7 k 1−( )2

⋅

k 1−( ) 1⋅

1.8194473493=

PHOTOELECTRIC THRESHOLD

101 x+ k ⋅

C6

my⋅ 7⋅ 8⋅

1.0204081633=Mps

my C5

⋅

1

7⋅ 1.0204081633=

Mep Me−

Me Mee−

k 2

7⋅ 1.0204081633=

PM Pm−

Pn PM−

2

7⋅ 1.0204081633=

KV

7rs⋅

8.3014035528=10

1 x+k ⋅

Kx 7⋅ 8⋅ 1.0204081633=

Mps

Kx

C

7⋅ 1.0204081633=

Me0 9.1093897 10 31−

⋅:=Pm0 1.67262171 10 27−

×:=Pn0 1.67492728 10 27−

×:=

Measured

Mps Kx

C7⋅ rs⋅:=Mee 9.1093838239 10

31−×=Mee Me Mep Me−( )

k 2

7 RS⋅⋅−:=


3/44

PR 3.5714285714:=10 10⋅

7 7−( )−[ ] 7 7−( )−[ ]+ 3.5714285714=

SMC

SQCPR =

SQCSequential cycle =Simultaneous cycle = SMC

The first axiomatic increase in value is 1+1=2 units.The ratio 1/2 is the fundamental standing wave

ratio of a linear harmonic oscillator in a resonant state. The cube root of 2 = 1,259921 =k and k-1

= .255921. Therefore a volume will increase by 2 if the radius increases by k-1 =.259921.

Similarly a volume will increase by 7 if radius increases to two as 2 cubed = 8 volumes and 8

minus 1= 7 incremental volumes. Product of two simultaneous cycles = 10 x 10 = 100 divided bytwice the algebraic sum of sequential displacements in both forward and reverse directions of 7

(volumes) as 7 - (-7) =14. gives an oscillatory ratio PR. Perpetual interactive state or harmonic

oscillatory state requires a ratio of PR to be maintained continuously. When a number of

interactions occur within the same period as a single interactive cycle then it is considered to be a

simultaneous interaction with a common centre of action and displays mass / density

characteristics.

PLEASE NOTE THAT ALL VALUES ARE DERIVED FROM AN AXIOMATIC VALUE OF TWO

AND EVERY DERIVATION (SHOWN BELOW) EQUALS VALUES IN PHYSICS EXACTLY

_________________________________________________________________________________

VOLUMETRIC PERPETUAL HARMONIC OSCILLATOR

_________________________________________________________________________________

_________________________________________________________________________________

Background

In Physics , perpetual harmonic oscillation or motion is deemed to violate energy

conservation principles and hence is considered impossible. But Sankya logic based on

axioms and combinatorial mathematical procedures provides the proof for the existence

of perpetual oscillatory states in the continuum of space.

_________________________________________________________________________________ _________________________________________________________________________________

THE MATHEMATICAL SECTION

_________________________________________________________________________________


4/44

Pm0 1.67262171 10 27−

×=PM not known=Pn0 1.67492728 10 27−

×=Measured

in Physics

Pm 1.672621512 10 27−

×=PM 1.6744231791 10 27−

×=Pn 1.6749276458 10 27−

×=Derived =

As shown above the ratio of the mass decrease of Proton Pm from the coherent static mass

PM and the simultaneous increase of the Neutron Pn mass from the PM state is exactly equal

to RS. The Proton mass change is over 7 volumetric displacements while the Neutron is over

two volumetric displacements within the same oscillatory period of two cycles or the equivalent

of a half-wave / standing wave ratio. The values of PM , Pn and Pm are derived through axioms

and are not arbitrary, yet the dimensionless ratios match the experimentally measured values

of Pn0 & Pm0 precisely and within the tolerance values.

PROOFPM Pm−( )

Pn PM−( )

2

23

1−( )⋅ 1.0204081633=PR

2

7⋅ 1.0204081633=

102

102

2−

1.0204081633=RS 1.0204081633=RS

0

100

n

2

100

n

∑=

:=

An oscillatory rate of two changes of volume per cycle and a similar oscillatory rate of two

changes of radial distance per cycle should maintain a resonant harmonic cyclic period incomplete synchrony if both cyclic periods start at the same instant.. Since the radial distances

of both types of volumetric increments start from the same central point or common centre ofinteraction, the ratio of both increments will always be the same in every cycle as 2/7. Thereforethe product of PR and the ratio 2/7 should give the resonant ratio RS as the rate of decay100/(100-2) interactions per cycle = 1.020408163264 etc. as the transcendental sum of powerseries of harmonic decay of oscillatory periods, to infinite numbers that progressively reduce theintervals to infinitely small periods. It is a perpetually dynamic state.Therefore the coherentnuclear state can not decay or its decay time is infinite or eternal.

PR 3.5714285714=PM Pm−( )

Pn PM−( )3.5714285714=

The two particulate or fermion states of Proton ( Pm) and Neutron (Pn) oscillate around a central

coherent state PM that is new to Physics. Coherent states cannot be detected as a particle butonly as a resonance. The change in mass as PM minus Pm during the axpanding phase

compared to the change in mass as Pn minus PM during the compressive phase must equal PR

to maintain perpetual harmonic oscillation.

HADRONIC STATES


5/44

The above derivation indicates that leptonic states are area dependant flux densities displaying

charge characteristics that act as a simultaneous surface of interaction at the distance k, at the

same rate of 2 interactions per cycle but at distance k. Therefore, the variable states of Pm and

Mee must increase or decrease in the same ratio to maintain the standing wave ratio of 2 or half

wavelength. The Pn or Mep states must decay to maintain the resonant state. The decay must

take place at the maximum rate by converting the incremental potential to kinetic energy. Thereverse, compressive interactive state must convert the linear / angular momentum by

synchronising the flux density to a simultaneously interactive mass density, which will show a

reduction or apparent loss of flux density that equals the coupling constant. As all these factors

change within one cycle which is in a resonant harmonic state of half a cycle, the reactive

displacement never exceeds the interactive displacements and the resonant harmonic state

exists perpetually.

Me0 9.1093897 10 31−

×=Me not_predicted=Mep not_predicted= As Measured

in experiments

in Physics

Mee 9.1093838239 10 3−

×=Me 9.110233722 10 31−

×=Mep 9.1140580241 10 31−

×=Derived

PROOFMep Me−

Me Mee−

2

7 k ⋅⋅ 1.0204081633=

Mep Me−

Me Mee−

k 2

7⋅ 1.0204081633=

PR k ⋅ 4.4997180353=Mep Me−

Me Mee− 4.4997180353=

7 rs⋅ k ⋅

24.4997180353=

The same interactive process of perpetual oscillatory state also prevails at a relative distance

k from the centre, in the form of sequential interactive states displaying kinetic characteristics.

These interactions have a common surface of interaction that displays charge characteristics

or flux density. The three leptonic states, Mep is derived from the PM state, Me is derived

from the Andhatamishra or Planck mass state and Mee is the resonant balancing state which

equals the current measured value. The ratio of the mass decrease of Mep to Me and the

simultaneous increase of the Mee to Me mass is exactly equal to RS. The Mep to Me change

is over 7 area displacements while the Me to Mee change is over 2/k or k squared radial

displacements within the same oscillatory period of two cycles or the equivalents of a half-wave

/ standing wave ratio.

LEPTONIC STATES

PM

Pm1.0010771517=

Pn

PM1.0003012779=

Pn

Pm1.0013787541=

Pm0 Pm−

Pm1+ 1.0000001184=

Pn Pn0−

Pn1+ 1.0000002184=


6/44

When the perpetually oscillatory state is interrupted by an accelerative and out of phaseinteraction, interactive stresses created by the breakdown of the coherent state, transmigratein a wave form. When the rate of acceleration exceeds C, photons or stress quanta are

radiated in a holographic form as a particle. The wave characteristics exist as two states. Thesimultaneous phase represented C x and the sequential phase as C1-x transmigrate as a

wave and the product equals C. In a balanced state it is Cx / C1-x = Cx

3

equals the

impedance in space as shown below

IMPEDANCE IN100

k 1−( ) rs⋅ 377.0375659826=

Cx

C1 x−⋅

C1=

PM Px⋅

C

2 π⋅( )2

k 1−( )−

7

⋅ 6.6261986282 10

34−×= h 6.6260755 10

34−×=

PROOF

The magnetic state is created at a resonant oscillatory rate of two cycles per period in which themaximum coherence at maximum stress density is attained. The atomic mass number at which

perpetual magnetic resonance will be maintained is given below which is equal to Fe atomic mass

number and element number 26

Rp3

C1 x+( )

3

⋅

1−

2 rs⋅( )

1

31−

3

1+

55.8670172024= 55.8670172024

2 1

7+

26.0712746945=


7/44

Mps

PM Px⋅

Ne 7⋅

my

1.2665147955=

2 π⋅

1010

18⋅ my⋅

Ne 7⋅ 1.2666197678=RATIO

my C2

⋅ 2 π⋅( )

2k 1−( )−

72

⋅ 9.4659980402 10 35−

×= Ne 9.5287340542 10 35−

×=

Kx k 1−( )⋅

Px

1−.939⋅ 82.7086719379=

Px

k 1−( ) Kx⋅ .9396⋅

1 22

+( )2

1−

92.5301933599=

MW

MZ

h 6.6260755 10 34−

×=Lp 2 π⋅( )( )2

k 1−( )−⋅ 6.6261986282 10 34−

×=

PROOF10

2

2 π⋅( )2

1

k 3

⋅ 1.2665147955=HiddenMass

Hiddenenergy

Mps

PM Px⋅

my

Ne 7⋅⋅= 1.2665147955=

h

my4.9278416147 10

17×=

Ne 7⋅

my4.9605931212 10

17×=Hiddden dark energy

2 π⋅

1010

18⋅

1

6

925.4720588405=Mps

PM Px⋅ 6.2826645825 10

17×=hidden dark mass

The PM forms the neutral or balanced state between the limiting mass Mps (Planck) at the

highest density and the rest energy state of the Neutrino, Plancks constants and the Planck

length. The ratio between Mps and PM is angular converted to mass as potential in the form

of hidden 'dark matter'. "my" is the fundamental mass value in Sankhya and its energy value isequal to Plank's constant. Shown below, the potential driving ESP and Astrological factors is

shown to be only 0.266 of the measurable Planck value and is therefore hidden. But in terms of

the frequency spectrum it covers 1.3 E+17 interactions per second, which is a very large mass

interval driving the parapsychological phenomenon yet it cannot be detected by instrumentation

because it acts simultaneously- that is it is a change in potential but not discrete enough to

cause a change in frequency .


8/44

The foregoing algorithm is scale invariant & self similar. It can be applied to any field of

components that are in a coherent and confined state. The components that form space are in a

coherent, dynamic and perpetual state of balanced activity. That is the reason the interactive

formulation shown above work correctly.

Photons travel as a wave with a forward and bacward oscillatory motion and therefore the

loss in distance is one cycle (or 1 metre) in 36.6 years of travel time.( Pioneer Anomaly)

k 1−

c

yr

s⋅ 36.5497624716⋅

1=

The proof of the above is in an enigmatic behaviour of the nuclera spectrum where a the

perpetual oscillation is called the self energy of the Vacuum and its value is indicated by the

frequency of oscillation called the Lambshift, measured as 1057.862 Megacycles/sec.

The theoretical value is shown and the measured value is different for reasons.The existance

or the mass value of PM is not known in Physics because any measurent needs 7 plancks

constant of energy to register a frequency count and the PM is hidden within this value.

My = photon rest mass in Sankhya and the equivalence is shown below.

C

Pn PM( )−[ ]

PM( ) Pm−[ ]

1.0591998819 109

×=

Theoretical

1.0591998819 109

×

1 k 1−

7

2

+

1.0577415164 109

×= Experimental

_________________________________________________________________________________

REDSHIFT k 1−

7 10

6

⋅ 3.7131578556 104

×=

Derivation from axioms

x 1 2

2+ 1−

2:= x 0.6180339887=


9/44

Mee 9.1093838239 10 31−

×=Mee Me Mep Me−( ) k

2⋅

7 rs⋅−:=

Me C1 x−( )

7my

7⋅ 1

2

5−

⋅:=

Mep 9.1140580241 10 31−

×=Mep PM

7

k 1−

10

2 π⋅⋅

2:=

Pn 1.6749276458 10 27−

×=Pn

2( ) 7 rs⋅+[ ] PM⋅

7 rs⋅ 1 k 1−

7

2

+

⋅ 2( )+

1 k 1−

7

2

+

⋅:=

Pm 1.672621512 10 27−

×=Pm 2( ) 7 rs⋅+[ ] PM⋅

7 rs⋅ 1 k 1−

7

2

+

⋅ 2( )+

:=

PM 1.6744231791 10 27−

×=PM Kx

Px C3

⋅

:=

Px 20.9479860976=Px 10

2 π⋅ 3⋅

3

:=

Kx 0.9149879388=Kx 10

1 x+2

1

3⋅ 10

22−( )⋅

23

1−( ) 23⋅ 102⋅:=

C 2.9657596692 108

×=C 10

2

x3

:=

π

A100

2100

⋅

10=

Ai 1+

1 1 Ai( )

2−−

2

Ai( )

2+

2:=

A0

x

2:=i 0 100..:=


10/44

___________________________________________________________________________________

___________________________________________________________________________________

Pn 1.6749276458 10 27−

×= Pm 1.672621512 10 27−

×= PM 1.6744231791 10 27−

×=

Me 9.110233722 10 31−

×= Mee 9.1093838239 10 31−

×=

Pn0

Mee

Pn

Me− 0.1711300947=

Pm0

Mee

Pm

Me− 0.1715128546=

PM

Mee

PM

Me− 0.1714800482=

n 1 8..:=

Lxn

2( ) n rs⋅+[ ]

n rs⋅ 1 k 1−

n

2

+

⋅ 2( )+

1 k 1−

n

2

+

⋅ PM⋅:= = Pn

Ldn

2( ) n rs⋅+[ ]

n rs⋅ 1 k 1−

n

2

+

⋅ 2( )+

PM⋅:= = Pm

1 2 3 4 5 6 7 81.62 .10

27

1.64 .10 27

1.66 .10 27

1.68 .10 27

1.7 .10 27

1.72 .10 27

Lxn

Ldn

PM


11/44

n

___________________________________________________________________________________

n 1 10..:=

0 2 4 6 8 100

1 .10 30

2 .10 30

3 .10 30

Pn PM−( ) n⋅ rs⋅[ ] PM Pm−( ) 2⋅−

n

Ne Z⋅ k 1−( )⋅

rs13.6154988477=

1

x x+

Ne Z⋅ k 1−( )⋅

rs⋅ 11.0151699546=

Ne Z⋅ k 1−( )⋅

rs

1

x x+

Ne Z⋅ k 1−( )⋅

rs⋅− 2.600328893=

13.6 1.8146371585−

13.60.8665707972=

13.602( ) 1 .25− 13.602⋅( )− 1.822=


12/44

PM Pm−( ) 2⋅

Pn 2⋅ PM−( ) 7⋅ rs⋅ 3.0109651557 10

4−×=

Mep Me−

Me Mee−

2

7 k ⋅⋅ 1.0204081633=

Mep Me−

Me Mee−

PM Pm−( )

Pn 2⋅ PM−( )

4.1844424786 103

×=

Me Mep Me−( ) 2⋅

7 k ⋅ rs⋅− 9.1093838239 10

31−×= Mee 9.1093838239 10

3−×=

volume area and length sync

mass and flux density

2 and 7 with 1.020408Phn

Me

Mep Me−( ) 2

7 k ⋅ rs⋅⋅

PM Pm−( ) 2⋅

Pn n⋅ PM−( ) 7⋅ rs⋅

−

Me n( )⋅:=

Ph

Phn( )

0.9999067095

0.3450820841

0.126807209

0.0176697715

0.04781269110.0914676661

0.1226497911

0.1460363849

=

k 1−( ) 3−

56.947628372=

PM1 Mps

2 π⋅ 1017

⋅ Px⋅

:=

PM1

n

Mps

Px 2⋅ 1017⋅ 2n⋅ An⋅ 10⋅

:=

Mep1n

PM1n

7

k 1−

10

2 π⋅⋅

2:=


13/44

volume area and length syn

mass and flux density

2 and 7 with 1.020408Ph

n Me

Mep1n

Me−( ) 2

7 k ⋅ rs⋅⋅

PM1n Pm−( ) 2⋅

Pn PM1n−( ) 7⋅ rs⋅

−:=

0 2 4 6 89.1 .10

31

9.11 .10 31

9.12 .10 31

9.13 .10 31

Phn

n

h

ec4.1361270287 10

15−×= my

C2

2 π⋅( )2

k 1−( )−⋅

7⋅ 6.626198=

2 π⋅ 1018

⋅ h⋅

22.0816430113 10

15−×=

h

ec

1

2⋅ 2.0=

ec

1.38 10 23−⋅ 2⋅

5.8043478261 103

×=7

k 1−1

2⋅ 13.465627356=

PM C2

⋅

Kb1.066721439 10

13×= PM C

2⋅

9.1933675935 108

×=


14/44

1.8835327 10

28−

⋅ Kb k 1−( )2

⋅ Cx

⋅ 1.6063382333 10 19−

×=

k

1 1

343+

343⋅

1+

10⋅ C⋅ my C2

⋅ 2 π⋅( )2

k 1−( )−⋅⋅ 1.3806581776 10 23−

×=

my C2

⋅ 2 π⋅( )2

k 1−( )−⋅

7

6.6261986282 10

34−×=

my C2

⋅ 2 π⋅( )2

k 1−( )−⋅

7

6.6261986282 10

34−×=

7

k 1−( )Lp 2 π⋅( )2

k 1−( )−⋅ 6.6261986282 10 34−

×=

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C

3

C⋅

6.1795617215 1018

×=

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C

3

C2

⋅

2.0836353618 1010

×=

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C

3

C2

⋅

my C2

⋅ 2 π⋅( )2

k 1−( )−⋅

7⋅ 1.3806581776 10×=


15/44

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C

3

C⋅

my C2

⋅ 2 π⋅( )2

k 1−( )−⋅

7⋅ 4.0947003402 1×=

Lp2

G⋅

Mps7⋅ C

2⋅

2 π⋅( )2

k 1−( )−

7⋅ 6.6261986282 10

34−×=

PM Pm−

Pn PM−

2

7⋅ 1.0204081633=

Mep Me−

Me Mee−

k 2

7⋅ 1.0204081633=

7 rs⋅ 2+

7 rs⋅ 1 k 1−

7

2

+

⋅

2+

Kx

Px C3

⋅

⋅ 1.672621512 10 27−

×= Pm 1.6726=

7 rs⋅ 2+

7 rs⋅ 1 k 1−

7

2

+

⋅

2+

1 k 1−7

2

+

⋅ Kx

Px C3

⋅

⋅ 1.6749276458 10 27−×=

Kx

Px C3

⋅

k 1−

7

2

⋅ 2 π⋅

10

2

⋅ 9.1140580241 10 31−

×= Mep 9.1140580241 10 31−

×=

C1 x−( )

7

C6

Kx

7⋅ 1

2

5−

⋅ 9.110233722 10 31−

×= Me 9.110233722 10 31−

×=

Kx k 1− 2

⋅ 2 π⋅

2

⋅


16/44

Px C3

⋅ 7 10

Me1.0004197809=

my C⋅ C2

⋅

Px1.6744231791 10

−×=

PM Pt

2

⋅Rp

3G⋅

1=

Kx 0.9149879388= Px 20.9479860976= KV 56.9476283=

π 2⋅

100.6283185307=

2 π⋅( )

k 1− 24.1734377024=

This QuickSheet can be used to create a parametric surface

plot of a unit sphere.

X φ θ,( ) sin φ( ) sin θ( )⋅:=0 φ≤ π≤

Y φ θ,( ) sin φ( ) cos θ( )⋅:=0 θ≤ 2 π⋅≤

Z φ θ,( ) cos φ( ):=


17/44

X Y, Z,( )

KV

7rs⋅

Kb6.012642923 10

23×=

7

k 1−

3

KV

7 8⋅ 1.0204829⋅

2−1−

−

n 1 8..:=

n

2

3

1

1.587401052

2.0800838231

2.5198420998

2.9240177382

3.3019272489

3.65930571

4

= 1

n

2

3

3−

1

4

9

16

25

36

49

64

=

656.3 10 9−

⋅

656.1 10 9−

⋅

1.000=

C

656.1 10 9−

⋅ C1 x+

⋅

8.85038=

6.561 10 6−

⋅ C

2.4177207157 10

14

⋅( )

−

5.33432412=

ec

C h⋅( )

C

k 2

103

⋅( )− 6.283801213 10

5×= 2 π⋅ 10

5⋅

k 2

103

⋅

k 1−

100

⋅ 1.0288087687=


18/44

23

1−

k 3

rs⋅ 3.5714285714=ExpP

ComP3.5714285714=

10

logec

h

logc

656.1 10 9−

⋅

−

1−

1.88992484=

h

ecC⋅

6.561 10 7−

⋅( ) 1.8696477257=

log 1 k 1−( )2

+ 1

Pt⋅

1−1+

4.3928=

2

log 1 k 1−( )2

+ 1

Pt⋅

1−

1+

1.000304=

656.3 10 9−

⋅

656.1 10 9−

⋅ ,

=

Pn

Pm1.0013787541=

Pn0

Pm01.=

PM 1.0003045377⋅

Pm1.0013820174=

Pn

PM

1.0003012779=

10

log 1 k 1−( )2

+ 1

Pt⋅

1−

1+

log 2( )⋅

1.0003045377=

10

logC

656.1 1⋅

log 2( )

log C

656.1 10 9−

⋅

log

C

656.3 10 9−

⋅

−

1.3236649962 10

4−×=

log C

656.3 10 9−

⋅ log 2( )

48.=c

656.1 10 9−

⋅

4.5693104405 1014

×=


19/44

Kx

Px C3

⋅

Kx

Px c3

⋅

− 5.331878286 10 29−

×=

1

C3

1

c3

−

Kx

Px⋅

Mep

Kx

C3

Kx

c3

−

Mep1.225492661 10

3×=

my C3

⋅ Kx

c3

−

GePM Mee+( )− 4.7776831516 10

30−×=

4000 PM⋅ Na⋅ 4.0333505538=

my C3

⋅ 3.5075793477 10 26−

×=Kx

Px c

3

⋅

my C3

⋅

Px−

5.331878286− 10 29−

×=

73

343=103

1 103

×=Sahasrara =thousandfold

63

216= Ajna= motivating

53

125=Visuddha=purifying

5 8+ 1+ 14=43

64= Anahata= livingforce

5 5+ 1+ 2+ 1+33

27=Manipuraka = fullmagnetic

1 9+ 10=23

8=Swadhisthana =independant

5 3+ 8=13

1=k 3

2=Moola=root

ExpP ExpM⋅

ComP ComM⋅

3

0.5=

ExpP

ComP

2

rs⋅ 7=

ExpM k ⋅ 3.5714285714=


20/44

Kx

Px C3

⋅

1.6744231791 10 27−

×=

Mee 7 rs⋅

k 2

⋅ −

7 rs⋅

k 2

−

Me 7 rs⋅

k 2

1−

⋅

7 rs⋅

2Pn⋅ Pm+

7 rs⋅

2

1+

1.6744231791 10 27−

×=

Mee 7 rs⋅

k 2

⋅ −7 rs⋅

21+

PM⋅ 7.6545059615 10 27−

×=

7 rs⋅

k 2

Me⋅ Me− =PM PM 7 rs⋅

2⋅+ 7.6545059615 10

27−×=

7 rs⋅

2Pn⋅ Pm+ 7.6545059615 10

27−×=

Mep Me−

Me Mee− 4.49=

PM Pm−

Pn PM− 3.5714285714=

Lp k 1−( )⋅ 4.39152098 10 36−

×=

h

2 π⋅( )3

k 1−( ) 2⋅ π⋅ 1−

4.3914393767 10 3−

×=2 π⋅( )3

248.0502134424=

Pm Ne 8⋅ k 1−( )⋅−

Pm00.9999997632=

Pn Ne

k 1−−

Pn0

0.99999999=Pn Pn0−

Ne

k 1−

0.9979458885=


21/44

Kx

Px0.0436790408=

7 rs⋅

2Pn⋅ Pm+

7 rs⋅

21+

C3

⋅ 0.0436790408=

Kx

Px

k 1−

7

2 π⋅

10⋅

2

⋅ 2.3774952294 10 5−

×=

Mee 7 rs⋅

k 2

⋅ Mep−

7 rs⋅

k 2

1−

C 3( )

⋅ 2.3759275133 10−

×=

n 1 ..:=

Mps

Pn 1018

⋅ Px⋅

1

2⋅ 0.3140386162= An 2n

⋅

0.3128689301

0.3138363829

0.3140785261

0.3141390794

=

Mps

Mep 7

k 1−

10

2 π⋅

⋅

2

⋅ 1018

⋅ Px⋅

1

2⋅ 0.3141332291=

An 2

n⋅

0.31286893010.3138363829

0.3140785261

0.3141390794

=

log Dp( )

log 1

x3

154.1715269782= log Dp( )

log 2( )321.09700775=


22/44

70 0.0143876−(−

7


23/44

7

k 1−

3

1.9533036532 104

×=

1 1

73

+ 1.0029154519=

1

73

4.9⋅

1−1

73

4.95⋅

1−−

0.7070707071=

343

273.26 1−

1−

8.3081529089=

343 273.26−

343

1−1

k 1−− 1−

1−

2

3.7543677222=


24/44

n 3 4..:=

7 rs⋅ C2

⋅ 10⋅

2n

An

⋅ 10⋅ 2⋅ 1018

⋅

1−

1−

1−

-5.7425171218·103

5.3695760681·104

=

2 π⋅( )2

710

4⋅ 5.6397739435 10

4×=

7 k 1−( )⋅ 1.8194473493=

100

2 π⋅( )2

21− 0.2665147955=


25/44

π

A100

2100

⋅

10=

Kaivalya

MU

my

1

Lp3

2 x⋅( )6

⋅

−

MU

my

1−

7

k 1−

3

−

392.7616293442=

C3

6.022 1023

⋅ 7

k 1−

10

2 π⋅⋅

⋅

1.0106258319=

100

283.5714285714=

102

7 7−( )−[ ] 7⋅ 1.0204081633= Mps

Kx

C7⋅ RS⋅:=


26/44

KV 1

k 1−

3

:=Tama

102

7 7−( )−[ ][ ] 7 7−( )−[ ]+

3.5714285714=

0

100

n

1−( )n

2 n⋅ 1+( )2∑

=

0.915977847=

101 x+

2

1

3⋅

2

3

2

3

1−( )⋅

100 2−

100

⋅ 0.9149879388=

Rp k 1−

C1 x+

:=

0

100

n

2

100

n

∑=

1.0204081633=

1 1

49+ 1.0204081633=

1

100

n

1 22

+ 1−

2

n

∑=

1.6180339887=

1 1 2

2+ 1−

2

+ 1.6180339887=

0

100

n

0.6816901138( )n

∑=

3.1415926534=

Pt PM

Rp3 G⋅

1−

:= Pt 1.0803802741 10 3−

×=

C1 x+( )

3

KV⋅

1−

1.3179913508 10 43−

×=


27/44

Pm0 Pm−

Pm

C

10

⋅

3.5101237714=

Lp 2 π⋅( )2

⋅ Lp k 1−( )⋅−

Ne 7⋅ Lp k 1−( )⋅− 1=

n 1:= c 299792458:=

h

79.4658221429 10

35−×=

Rp

k 1−

C1 x+:= Rp 5.0890594006 10

15−

×=

KV 1

k 1−

3

:=

PM

NePx⋅

2 π⋅

10

2

⋅ 2⋅ rs⋅ 2.9657596692 108

×=

PM n⋅ Pm n⋅−

Pn n⋅ PM n⋅−

Mep n⋅ Me n⋅−

Me n⋅ Mee n⋅−

1−⋅ 0.793700526=


28/44

7

k 1−( ) 26⋅

1.035817489=

26 2 1

7+

⋅ 55.7142857143=2 rs⋅( )

1

31−

3

1+

1.0193425607=

55.867

2 1

7+

26.0712666667=

h C⋅

PM Px⋅

2 π⋅( )2

7−

7⋅

1−

3.836565253−=

Mee 9.1093838239 10

31−

×=

PM Px⋅

C

2 π⋅( )2

k 1−( )−

7

⋅ 6.6261986282 10

34−×=

Mee

Ne9.5599098181 10

3×=

1

k 0.793700526=

Mee

h C1 x−

⋅

0.7982972696=

h 6.6260755 10 34−

×=Planks constant

h C1 x−

⋅

k 9.0569303036 10

31−×=

PM Px⋅

Cx

k ⋅

2 π⋅( )2

k 1−( )−

7

⋅ 9.0570986028 10

31−×=

Cx

3Cx

C1 x−

100=

PM Px⋅Mee2 π⋅10

3

⋅ 9.5512037113 103×=


29/44

M

Pd PM

Rp3

:=

Ge

Dp

DD

1

3

DD⋅

PM

Rp3

:=

1

Pt

6

6.2883994818 1017

×==

Mps

PM Px⋅

Ne 7⋅

my−

1.3220714614 1017

×=

Dp

DD

1

3

DD⋅

PM

Rp3

1−

1.5042349611=

7my

h

C2

⋅ 39.2177677955=

1

k 1−( )2

c⋅

4.937378163 10 8−

×=

A10

210

⋅ 10⋅ 1

k 1−( )2

C⋅

+ 3.1415926542=π 3.1415926536=

7 rs⋅

23.5714285714=

Me Mee−

2

C2

ec⋅

−


30/44

10

1

3

7−

13.6058163144=

c 299792458:= n 1 3..:= k 1−

70.0371315786=

my C2

⋅ 2 π⋅( )2

⋅

72

9.5287340542 10 35−

×=

ec 1.602 10 19−

⋅:=

2 π⋅ 1017

⋅

107

G⋅ 4⋅ π⋅

3⋅ 1.0115452142=

2 π⋅

1010

18⋅ my⋅

1057.862 106

⋅ h

ec4.3754516108 10

6−×=

1057 106

⋅ h⋅

c2

13−

5.0439411733 1013

×=

C1 x+( )

3−C

2⋅

h 106

⋅

rs2

⋅ 1.0374127495 103

×=


31/44

Mep

Me1− 4.1978089734 10

4−×=

Me

Mee1−

1−

1.071820629 10

4×=


32/44


33/44

Kx rs⋅ 7⋅

4913.605⋅ 1.8146371585= Mps C⋅ 6.5356281344=

4.39 1014

⋅ h⋅

ec

13.605

1−

Kx⋅ rs⋅ 6.9956721976=

Kx rs⋅

7 13.605⋅

ec

h⋅ 4.3872858497 1014

×=

Mps C⋅

4913.605⋅

ec

h⋅ 4.3872858497 10

14×=

224577


34/44

1

15 60⋅ 1+ 1.0011111111=

4 π⋅3

103⋅ 4.1887902048 103×= 72

3.5=

n 1 8..:= m 1 8..:=

0 5 100

0.5

1

n

7 Kx⋅ rs⋅ 6.5356281344=

PM1n

PM=


35/44

1.0040409863

1.000945863

1.00017417

0.9999813769

0.9999331868

0.9999211397

0.999918128

0.9999173751

13.6 .28⋅ 3.808=

Phn

Mee−( ) Ne

25.640481881

4.1967287414

0.4336199355

-0.0290405127

-0.0849418215

-0.0942358728

-0.0962482202

-0.0967315501

=

Kb 1.380658 10 23−

⋅:=

6282 10 34−

×1

ec 2⋅ 3.1210986267 1018

×=

680635144 10 15−

×

nn 9.9395351411:=

nn

ec6.2044538958 10

19×=

KbRp

k 1−

C1 x+

:=

ec

1 x− 13.4752991801=


36/44

Kb2

⋅

Cx

k 1−( )2

⋅ 1.163458462 104

×=

n 1 3..:=

Kb 1.380658 10 23−

×=

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C⋅ 2.0836353618 1010

×=

k

1 1

343+

343⋅

1+

10⋅ 7⋅ C⋅ my C

2⋅ 2 π⋅( )

2k 1−( )−⋅

7⋅ 1.3806581776 10

23−×=

2⋅

C1 x−

2⋅ 1.1594835022 10

4×=

ec

Kb1.1603163129 10

4×=

1 1

k 1−+

103

1+ 1.0048473221=23 ec

h2⋅

C1.6304225462 10

6×=


37/44

15− h

ec 4.0947003402 10 15−×( )⋅

2

1.0203366523=

Kx

Px C3

⋅

k 1−

7

2

⋅

2 π⋅

10

2

⋅

9.1140580241 10

31−

×=

1512 10 27−

×

Pt PM

Rp3

G⋅

1−

:=

Pn 1.6749276458 10 27−×=

PM C3

⋅ 0.0436790408= Kx

Px0.0436790408=

Kx

PxPM C

3⋅− 0=

Kx rs⋅( ) 7⋅ Mps C⋅− 8.881784197 10 16−

×=

Kx 3⋅

7 10⋅ 2

⋅− =


38/44

Px k 1−( ) 2⋅ π⋅

Px

7

2.9925694425=7

72 rs 1.0204081633=

8.314 8−

8

1−

2

5.0475446513=


39/44

KV

5.04711.2834611397=

7

k 1−

3

137.02

1.0201506579=

1137.0375366202=

C1 x−

4 π⋅ 137.0433964307=

048316

2

x10

3⋅

1−3.0901699437 10

4−×=

ec

h2.4177207157 10

14×=

3607C

ec

h

1−

2

Pt⋅ 0.9754650967=

72 10 6−

× k 2

103

⋅

C

5.3524264574 10 6−

×=

ec

h2.4177207157 10

14×=


40/44

SwadhistanaMoola

15

ComM Mep Me−:=ComP Pn PM−:=

ExpM Me Mee−:=ExpP PM Pm−:=

2 4.3928808931 10

4−×( )

1.0003045377=08931 10

4−×

5377

1 k 1−( )2

+ 1

Pt⋅

1−1+

1.0010120099=

013784169

987.1794998356 Pt⋅ 1.0665292587=

1

.001012987987.1794998356=

0 9−

log

C

656.3 10 9−

⋅

log 2( )

−

1.001012987=

Pt 1+ 1.0010803803=

249

248.6829676837

− 1.1105910028 1014

×=829676837


41/44

Vijna Ajna

Sahasrara 103

1 103

×= 73

343=

Pd PM

Rp3

:=

14=Dp

DD

1

3DD

Pd⋅ 0.6647897608=

Na 6.022 1023

⋅:=

Ge Dp

DD

1

3DD

Pd⋅:=

1

5 100⋅1+ 1.004472136=

x

2 100⋅ 1+ 1.0030901699=

58.5016934481=1

1.3717421125=


42/44

.

96

Mps

C2

G⋅

Mps

C2

G⋅

k 1−( )⋅− 1.2504074747 10 35−

×=

2 π⋅( )3

2 π⋅ k 1−( )−

Lp⋅ 6.6261986282 10

34−×=

h 6.6260755 10 34−

×=

7180353

3.1883249263 10 30−

×

ep 3.1875600659 10 30−

×=

3.1883249263 10 30−

×=

Mep

1

9.1080482305 10 31−

×=


43/44

Px

Kx

7

k 1−

10

2 π⋅⋅

2

⋅

4.2061072832 10

4×=

7

k 1−

3

RS137.036−

1−

14.1697921538−=5

4

Kx 7⋅ rs⋅

Pn 1018

⋅ Px⋅

1

2 C⋅⋅ 0.3140386162=

Px

Kx 103

⋅ rs⋅

c3

⋅ Na−

Na3.8627538596 10

3−×=

Na

k 1−( ) 1000⋅ 2.3168573697 1021

×=

Kx

3

1

1− 0.2339694184−=


44/44

7

k 1− 26.931254713=

1)

1−

14.1213123948=

volumetric perpetual harmonic oscillator

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