Physical constant (anomaly)

We can demonstrate a curious geometrical relationship between the units (kg, m, s, A) by selecting 2 dimensioned quantities, here are chosen r, v (we can choose others) such that

${\displaystyle kg={\frac {r^{4}}{v}},\;m={\frac {r^{9}}{v^{5}}},\;s={\frac {r^{9}}{v^{6}}},\;A={\frac {v^{3}}{r^{6}}}}$

The units (kg, m, s, A) remain independent of each (i.e.: the kg cannot be replaced by the m or the s ...), and so we still have 4 independent units, however if 3 (or more) units are combined together, in a specific ratio, they can cancel.

${\displaystyle f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}={\frac {({\frac {r^{4}}{v}})^{9}({\frac {r^{9}}{v^{6}}})^{11}}{({\frac {r^{9}}{v^{5}}})^{15}}}=1}$

This f(X) embeds the units kg, m, s but itself is dimensionless, units = 1 (i.e.: it is a mathematical structure).

If we assign these SI units to the dimensioned quantities r, v;

units: ${\displaystyle r=({\frac {kg\;m}{s}})^{1/4}}$ (r⁴ are the units for momentum)

units: ${\displaystyle v={\frac {m}{s}}}$

units: ${\displaystyle f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}}$ units = 1

Mass

${\displaystyle {\frac {r^{4}}{v}}=({\frac {kg\;m}{s}})\;({\frac {s}{m}})=kg}$

Length (${\displaystyle f_{X}}$ = 1)

${\displaystyle m={\frac {r^{9}}{v^{5}}}}$

${\displaystyle (r^{9})^{4}={\frac {kg^{9}\;m^{9}}{s^{9}}}}$

${\displaystyle ({\frac {1}{v^{5}}})^{4}={\frac {s^{20}}{m^{20}}}}$

${\displaystyle ({\frac {r^{9}}{v^{5}}})^{4}={\frac {kg^{9}s^{11}}{m^{11}}}=m^{4}{\frac {kg^{9}s^{11}}{m^{15}}}=m^{4}f_{X}=m^{4}}$

Time

${\displaystyle s={\frac {r^{9}}{v^{6}}}}$

${\displaystyle (r^{9})^{4}={\frac {kg^{9}\;m^{9}}{s^{9}}}}$

${\displaystyle ({\frac {1}{v^{6}}})^{4}={\frac {s^{24}}{m^{24}}}}$

${\displaystyle ({\frac {r^{9}}{v^{6}}})^{4}={\frac {kg^{9}s^{15}}{m^{15}}}=s^{4}{\frac {kg^{9}s^{11}}{m^{15}}}=s^{4}f_{X}=s^{4}}$

We can also construct a unit-less structure using the ampere with length and time

${\displaystyle f_{X}={\frac {A^{3}m^{3}}{s}}={\frac {({\frac {v^{3}}{r^{6}}})^{3}({\frac {r^{9}}{v^{5}}})^{3}}{\frac {r^{9}}{v^{6}}}}=1}$

If we assign a numerical value θ to r (θ = 8) and to v (θ = 17), then we can assign a unit number θ to the SI units kg, m, s, A, K ^[3].

Table 1. unit relationship

attribute	SI equivalent	unit number θ	scalars
M (mass)	kg	8*4-17 = 15	${\displaystyle {\frac {r^{4}}{v}}}$
T (time)	s	8*9-17*6 = -30	${\displaystyle {\frac {r^{9}}{v^{6}}}}$
V (velocity)	m/s	17	${\displaystyle v}$
L (length)	m	8*9-17*5 = -13	${\displaystyle {\frac {r^{9}}{v^{5}}}}$
A (ampere)	A	17*3-8*6 = 3	${\displaystyle {\frac {v^{3}}{r^{6}}}}$
K (temperature)	K	17*4-8*6 = 20	${\displaystyle {\frac {v^{4}}{r^{6}}}}$

The Planck units are direct measures of the SI units; Planck mass in kg, Planck length in m, Planck time in s ... and so they are analogues to the attributes in the above table. The SI Planck units have numerical values, however to derive a mathematical relation between these SI units we cannot use numerical values, this is because numerical values are simply dimensionless frequencies of the SI unit itself, 299792458 could refer to the speed of light 299792458m/s or equally to the number of apples in a container (299792458 apples), numbers such as 299792458 carry no unit-specific information, and so the units are treated as independent by default. This therefore requires that to the number 299792458 is added a descriptive (the unit), which could be m/s or apples.

This inherent restriction can be resolved by assigning to each unit a geometrical object for which the geometry embeds the attribute (for example, the geometry of the time object T embeds the function time and so a descriptive unit s = seconds is not required). We may then combine these objects Lego-style to form more complex objects; from electrons to galaxies, while still retaining the underlying attributes (of mass M, wavelength L, frequency T ...). An apple has mass because its 'geometry' includes the geometrical object for mass.

Table 2. lists a set of attributes as the geometry of 2 dimensionless physical constants; the inverse fine structure constant alpha and Omega. As alpha and Omega are dimensionless (alpha = 137.035999139, Omega = 2.00713495...), so too are these geometrical objects.

The dimensioned variables (in this example we are using r and v) are unit specific. They are used to translate between the dimensionless geometrical objects MLTP... and local unit systems such as SI or imperial - i.e.: the numerical value for v will depend on whether we are using m/s or miles/hour or ...

Table 2. MLTVA Geometrical objects

attribute	geometrical object	numerical	unit number	dimensioned component
mass	${\displaystyle M=(1)}$	1	15	${\displaystyle {\frac {r^{4}}{v}}}$
time	${\displaystyle T=(\pi )}$	3.1415926535...	-30	${\displaystyle {\frac {r^{9}}{v^{6}}}}$
sqrt(momentum)	${\displaystyle P=(\Omega )}$	2.00713495...	16	${\displaystyle r^{2}}$
velocity	${\displaystyle V={\frac {2\pi P^{2}}{M}}=(2\pi \Omega ^{2})}$	25.3123819...	17	${\displaystyle v}$
length	${\displaystyle L=VT=(2\pi ^{2}\Omega ^{2})}$	79.5211931...	-13	${\displaystyle {\frac {r^{9}}{v^{5}}}}$
ampere	${\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}$	234.182607...	3	${\displaystyle {\frac {v^{3}}{r^{6}}}}$
temperature	${\displaystyle K={\frac {AV}{2\pi }}=({\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }})}$	943.425875...	20	${\displaystyle {\frac {v^{4}}{r^{6}}}}$

We can now solve those f(X) structures as dimensionless geometrical forms.

${\displaystyle f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}={\frac {(1)^{9}(\pi )^{11}}{(2\pi ^{2}\Omega ^{2})^{15}}}}$

The electron f_e (below) is one example of an f(X) structure

As this particular geometrical approach requires that the objects be interrelated (they are not independent of each other), this unit number relationship hypothesis can be easily tested. This is because, if these MLTVPA objects are natural Planck units, then they will be embedded within our dimensioned SI physical constants (G, h, c, e, m_e, K_B...).

We can use these text-book formulas to convert between Planck units and the common physical constants.

Table 3. Physical constant unit numbers

SI constant	geometrical analogue	unit number θ
Speed of light	c* = V	17
Planck constant	${\displaystyle h^{*}=2\pi MVL}$	15+17-13=19
Gravitational constant	${\displaystyle G^{*}={\frac {V^{2}L}{M}}}$	34-13-15=6
Elementary charge	${\displaystyle e^{*}=AT}$	3-30=-27
Boltzmann constant	${\displaystyle k_{B}^{*}={\frac {2\pi VM}{A}}}$	17+15-3=29
Vacuum permeability	${\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}}$	34+15+13-6=56

We can assign numerical values to alpha (the inverse fine structure constant) and Omega, and then use the dimensioned scalars (r, v) to convert from the MLTVA objects to their SI equivalents.

alpha α = 137.035999139

Omega Ω = 2.0071349496

For example, we can use scalar v to convert from dimensionless geometrical object V to dimensioned c.

scalar v = 11843707.905 m/s gives c = V*v = 25.3123819 * 11843707.905 m/s = 299792458 m/s (SI units)

scalar v = 7359.3232155 miles/s gives c = V*v = 186282 miles/s (imperial units)

r = 0.712562514304 ... (SI units)

As the scalars also carry the unit designation m/s or miles/hour ... (as well as an associated numerical value), they are dimensioned, and so we can apply the unit number relationship θ to them; scalar v (θ = 17), for r (θ = 8). We find that the unit number relationship is consistent regardless of the constants and the system of units used.

Table 4. Comparison θ; SI units and scalars

constant	θ from SI units	MLTVA	θ from r(8), v(17)
c	${\displaystyle {\frac {m}{s}}}$ (-13+30 = 17)	c* = ${\displaystyle V*v}$	17
h	${\displaystyle {\frac {kg\;m^{2}}{s}}}$ (15-26+30=19)	h* = ${\displaystyle 2\pi MVL*{\frac {r^{13}}{v^{5}}}}$	8*13-17*5=19
G	${\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}$ (-39-15+60=6)	G* = ${\displaystyle {\frac {V^{2}L}{M}}*{\frac {r^{5}}{v^{2}}}}$	8*5-17*2=6
e	${\displaystyle C=As}$ (3-30=-27)	e* = ${\displaystyle AT*{\frac {r^{3}}{v^{3}}}}$	8*3-17*3=-27
kB	${\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}$ (15-26+60-20=29)	kB* = ${\displaystyle {\frac {2\pi VM}{A}}*{\frac {r^{10}}{v^{3}}}}$	8*10-17*3=29
μ0	${\displaystyle {\frac {kg\;m}{s^{2}\;A^{2}}}}$ (15-13+60-6=56)	μ0* = ${\displaystyle {\frac {4\pi V^{2}M}{\alpha LA^{2}}}*r^{7}}$	8*7=56

This section uses the unit number relationship to formulate the constants (G, h, c, e, k_B) in terms of the 3 Planck units MTP and the fine structure constant alpha.

Initial constants: Note: for convenience α is assigned the value 137.035999139 (which is the CODATA 2014 inverse alpha).

α = 137.035999139, (dimensionless)

Ω = 2.007134949638 (dimensionless)

v = 11843707.9052487, unit u = 17

r = 0.71256251430467, unit u = 8

If we select MTP as the principal Planck units (they include a dimensionless geometrical component and 2 dimensioned variables; r and v);

${\displaystyle M=(1){\frac {r^{4}}{v}}}$, unit = 15 (Planck mass)

${\displaystyle T=(\pi ){\frac {r^{9}}{v^{6}}}}$, unit = -30 (Planck time)

${\displaystyle P=(\Omega )r^{2}}$, unit = 16 (sqrt of Planck momentum)

... then from MTP and alpha we can formulate VLA

${\displaystyle V={\frac {2\pi P^{2}}{M}}}$, unit = 17 (velocity c)

${\displaystyle L=VT}$, unit = -13 (Planck length)

${\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}}$, unit = 3 (Planck ampere)

... and from MTPVLA we can formulate the dimensioned fundamental physical constants using the textbook formulas for converting from Planck units and using the unit relationship (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20).

${\displaystyle c=V}$, unit = 17 (speed of light)

${\displaystyle \mu _{0}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}}$, unit = 56 (permeability of vacuum)

${\displaystyle h=2\pi MVL}$, unit = 19 (Planck's constant)

${\displaystyle e=AT}$, unit = -27 (elementary charge)

${\displaystyle G={\frac {V^{2}L}{M}}}$, unit = 6 (gravitational constant)

${\displaystyle k_{B}={\frac {2\pi VM}{A}}}$, unit = 29 (Boltzmann constant)

We can now construct the electron from magnetic monopoles AL and time T (AL units ampere-meter (ampere-length) are the units for a magnetic monopole).

${\displaystyle T=\pi {\frac {r^{9}}{v^{6}}},\;u^{-30}}$

${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}}$

${\displaystyle \psi ={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;units={\frac {(u^{-10})^{3}}{u^{-30}}}=1,scalars=({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}$

Both units and scalars cancel (units = scalars = 1), and so ψ (the formula for the electron) is dimensionless. We can solve the electron parameters; electron mass, wavelength, frequency, charge ... as the frequency of the Planck units, and this frequency is ψ. Our results (calculated) agree with CODATA 2014. This means that the formula ψ not only determines the frequency of the Planck units (and so the magnitude or duration of the electron parameters), but it also embeds those Planck units.

In other words, this formula ψ contains all the information needed to make the electron, and so by definition this formula ψ is the electron. However it is dimensionless (units = 1), and this means that the electron is a mathematical particle, not a physical particle. And if the electron is not a physical particle, then it is these electron parameters (wavelength, charge, mass ...), and not the electron itself, that we are measuring. The existence of the electron is inferred, it is not observed.

Solving the electron parameters using ψ

V = (2*pi*Omega^2*v) = 299792458m/s (V ⇔ speed of light)

T = (pi*r^9/v^6) = 0.53905178661e-43s (T ⇔ Planck time)

L = V*T = 0.1616036601e-34m (L ⇔ Planck length)

ψ = 4*pi^2*(2^6*3*π^2*α*Omega^5)^3 = 0.2389545307369 e23 (dimensionless)

1. Compton wavelength

λ_e = 2.4263102367e-12m (CODATA 2014)

λ_e = 2*π*L*ψ = 2.4263102386e-12m (calculated)

2. Electron mass

m_e = 9.10938356e-31kg (CODATA 2014)

M = (1*r^4/v) = 0.21767281758e-7kg (M ⇔ Planck mass)

ψ = 4*pi^2*(2^6*3*π^2*α*Omega^5)^3 = 0.2389545307369 e23 (dimensionless)

M/ψ = (1*r^4/v)/(4*pi^2*(2^6*3*π^2*α*Omega^5)^3) = (0.21767281758e-7/0.2389545307369e23)kg

m_e = M/ψ = 0.91093823211e-30kg (calculated)

3. Elementary charge

e = 1.6021766208e-19C (CODATA 2014)

A = 2^4*V^3/(alpha*P^3) = 0.2972212598e25C

e = A*T = 0.16021765130e-18C (calculated)

4. Rydberg constant

R = 10973731.568508/m (CODATA 2014)

${\displaystyle R=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}}\;u^{13}}$ = 10973731.568508/m (calculated)

In summary, we have a dimensionless geometrical mathematical electron formula ψ that resembles the formula for the volume of a torus or surface area of a 4-axis hypersphere (${\displaystyle 4\pi ^{2}(AL)^{3}}$), and that includes the information needed to make both the electron parameters and to make the Planck units. It can also be divided into 3 magnetic monopoles ${\displaystyle (AL)^{3}}$ and these suggest a potential 'quark' model for the electron as well as a rationale for electron spin.

In this section we solve the constants (G, h, c, e, k_B) using 2 dimensionless constants; the (inverse) fine structure constant and Omega, and 2 scalars (r, v) which have been assigned (best fit) numerical values consistent with the SI units. r can be defined in terms of μ₀ and v can be defined in terms of c, thus we may use either (α, Ω, r, v) or (α, Ω, c, μ₀) as the initial constants, however the (α, Ω, c, μ₀) set is more complicated to formulate, conversely (α, Ω, r, v) is more intuitive.

Table 5. Dimensioned constants (α, Ω, v, r)

constant	geometrical object	calculated (α, Ω, r, v)	CODATA 2014 (mean)[5]
Planck constant	${\displaystyle h^{*}=2\pi MVL=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}}}$	6.626069134e-34, u19	6.626070040e-34
Gravitational constant	${\displaystyle G^{*}={\frac {V^{2}L}{M}}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}}}$	6.67249719229e11, u6	6.67408e-11
Elementary charge	${\displaystyle e^{*}=AT={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}}}$	1.60217651130e-19, u-27	1.6021766208e-19
Boltzmann constant	${\displaystyle k_{B}^{*}={\frac {2\pi VM}{A}}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}}}$	1.37951014752e-23, u29	1.38064852e-23
Vacuum permeability	${\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7}}$	4π/10^7, u56	4π/10^7 (exact)

This section demonstrates why the geometrical Planck objects (M = 1, T = pi, P = Omega) could be natural Planck units, independent of any numbering system and of any system of units.

We can solve combinations of the constants (G, h, c, e, m_e, k_B) using only the 2 dimensionless constants (α, Ω).

The physical constants are used by science to describe our universe, and to solve them requires 4 numbers; 2 dimensionless constants (α, Ω) and 2 dimensioned scalars such as (r, v). However the universe itself only requires (α, Ω), the scalars are man-made artifacts, selected for convenience, and relevant only to the chosen system of units, such as SI. And so the actual sum universe itself could be dimensionless, composed only of dimensionless physical and mathematical constants.

For example, to derive h, c and e, which for science are 3 of the most essential constants, we can begin with a dimensionless formula that encodes these 3 constants, and then extract h, c and e from this formula (similar to what was done with the electron formula).

Here we have a dimensionless formula using (α, Ω) which encodes the geometrical h, c and e. We first break up this formula to extract the geometrical objects for h, c and e and then add the 2 scalars. The scalars chosen will depend on the system of units we will use. Mathematically both sides of the equation are still the same, nothing has been created or destroyed.

${\displaystyle {\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}={\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}={\frac {(2\pi MVL)^{3}}{(AT)^{13}(V)^{24}}}}$

${\displaystyle {\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})^{3}/({\frac {2^{7}\pi ^{4}\Omega ^{3}r^{3}}{\alpha v^{3}}})^{13}.(2\pi \Omega ^{2}v)^{24}}$ = 0.228 473 759... 10^-58, units = 1

${\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}$ 0.228 473 639... 10^-58, units = ${\displaystyle {\frac {kg^{3}s^{8}}{m^{18}A^{13}}}}$, units = 1 (15*3-30*8+13*18-3*13 = 0)

Each of the physical constants (G, h, c, e, m_e, k_B) has a unit number and 1 or 2 dimensioned scalars. We can then find combinations of these constants where the unit numbers θ and the scalars (r, v) will cancel, these combinations, which are unit-less (units = 1), will then return the same numerical value as the MLTVA object equivalents. This is because if the scalars have cancelled, and as the scalars embed the SI conversion values as well as the SI units, then these combinations are defaulting to the underlying MLTVA objects (the dimensioned SI components have cancelled leaving behind the dimensionless geometrical MLTVA objects).

This should therefore apply to any set of units, even extraterrestrial and non-human ones, suggesting that these MLTVA objects could candidates for the "natural units" as proposed by Max Planck. The precision of the results in following table can be used to verify this conjecture.

...ihre Bedeutung für alle Zeiten und für alle, auch außerirdische und außermenschliche Kulturen notwendig behalten und welche daher als »natürliche Maßeinheiten« bezeichnet werden können...

...These necessarily retain their meaning for all times and for all civilizations, even extraterrestrial and non-human ones, and can therefore be designated as "natural units"... -Max Planck ^[6][7]

Here we solve physical constant combinations using only α, Ω. As the scalars (v, r) have cancelled, we do not need to know their values or the associated units (because in all cases units = 1, scalars = 1). This means that column 1 does not equal column 2, rather column 1 is column 2. The precision of the results depends on the precision of the SI constants; combinations with G and k_B return the least precise values.

Note: the geometry ${\displaystyle \color {red}(\Omega ^{15})^{n}\color {black}}$ (integer n ≥ 0) is common to all ratios where units and scalars cancel (i.e.: only combinations with ${\displaystyle \Omega ^{0},\Omega ^{15},\Omega ^{30},\Omega ^{45}}$... will be dimensionless). However there is no Planck unit with a ${\displaystyle \Omega ^{15}}$ component (all constants are combinations of ${\displaystyle \Omega ^{2}}$ and ${\displaystyle \Omega ^{3}}$), and this suggests there is an underlying geometrical base-15.

Table 6. Dimensionless combinations (α, Ω)

CODATA 2014 (mean)	(α, Ω)	units uΘ = 1	scalars = 1
${\displaystyle {\frac {k_{B}ec}{h}}}$ = 1.000 8254	${\displaystyle {\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}}$ = 1.0	${\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1}$	${\displaystyle ({\frac {r^{10}}{v^{3}}})({\frac {r^{3}}{v^{3}}})(v)/({\frac {r^{13}}{v^{5}}})=1}$
${\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}$ 0.228 473 639... 10-58	${\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}$ 0.228 473 759... 10-58	${\displaystyle {\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1}$	${\displaystyle ({\frac {r^{13}}{v^{5}}})^{3}/({\frac {r^{3}}{v^{3}}})^{13}(v^{24})=1}$
${\displaystyle {\frac {c^{35}}{\mu _{0}^{9}R^{7}}}=}$ 0.326 103 528 6170... 10301	${\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=2^{295}\pi ^{157}3^{21}\alpha ^{26}\color {red}(\Omega ^{15})^{15}\color {black}=}$ 0.326 103 528 6170... 10301	${\displaystyle {\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1}$	${\displaystyle (v^{35})/(r^{7})^{9}({\frac {v^{5}}{r^{9}}})^{7}=1}$
${\displaystyle {\frac {c^{9}e^{4}}{m_{e}^{3}}}=}$ 0.170 514 342... 1092	${\displaystyle {\frac {(c^{*})^{9}(e^{*})^{4}}{(m_{e}^{*})^{3}}}=2^{97}\pi ^{49}3^{9}\alpha ^{5}(\color {red}\Omega ^{15})^{5}\color {black}=}$ 0.170 514 368... 1092	${\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1}$	${\displaystyle (v^{9})({\frac {r^{3}}{v^{3}}})^{4}/({\frac {r^{4}}{v}})^{3}=1}$
${\displaystyle {\frac {k_{B}}{e^{2}m_{e}c^{4}}}=}$ 73 095 507 858.	${\displaystyle {\frac {(k_{B}^{*})}{(e^{*})^{2}(m_{e}^{*})(c^{*})^{4}}}={\frac {3^{3}\alpha ^{6}}{2^{3}\pi ^{5}}}=}$ 73 035 235 897.	${\displaystyle {\frac {(u^{29})}{(u^{-27})^{2}(u^{15})(u^{17})^{4}}}=1}$	${\displaystyle ({\frac {r^{10}}{v^{3}}})/({\frac {r^{3}}{v^{3}}})^{2}({\frac {r^{4}}{v}})(v)^{4}=1}$
${\displaystyle {\frac {hc^{2}em_{p}}{G^{2}k_{B}}}=}$ 3.376 716	${\displaystyle {\frac {(h^{*})(c^{*})^{2}(e^{*})(m_{p}^{*})}{(G^{*})^{2}(k_{B}^{*})}}={\frac {2^{11}\pi ^{3}}{\alpha ^{2}}}=}$ 3.381 506	${\displaystyle {\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1}$	${\displaystyle ({\frac {r^{13}}{v^{5}}})v^{2}({\frac {r^{3}}{v^{3}}})({\frac {r^{4}}{v^{1}}})/({\frac {r^{5}}{v^{2}}})^{2}({\frac {r^{10}}{v^{3}}})=1}$

In this section we replace dimensionless Omega with dimensioned R. This has the advantage that Omega is an unknown constant, but R a very important and precisely measured constant.

We can use this to numerically solve the least precise dimensioned physical constants (G, h, e, m_e, k_B ...) in terms of the 3 most precise dimensioned physical constants); speed of light c (exact value), vacuum permeability μ₀ (exact value), Rydberg constant R (12-13 digits) and the dimensionless fine structure constant alpha.

R = 10973731.568508 (θ=13) (12-13 digit precision)

c = 299792458 (θ=17) (exact)

μ₀ = 4π/10⁷ (θ=56) (exact)

α = 137.035999139 (unit-less) (9-10 digit precision)

We first look for combinations in which the unit numbers are equal, and then add dimensionless numbers as required. For example;

${\displaystyle {(h^{*})}^{3}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}u^{19}}{v^{5}}})^{3}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}$

${\displaystyle {\frac {2\pi ^{10}{(\mu _{0}^{*})}^{3}}{3^{6}{(c^{*})}^{5}\alpha ^{13}{(R^{*})}^{2}}}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}$

Table 7. R, c, μ₀, α ... (CODATA 2014 mean)

constant	formula*	calculated	θ	CODATA 2014 mean [8]	Units
Planck constant	${\displaystyle {(h^{*})}^{3}={\frac {2\pi ^{10}{\mu _{0}}^{3}}{3^{6}{c}^{5}\alpha ^{13}{R}^{2}}}}$	h* = 6.626069134e-34	${\displaystyle {\frac {kg^{3}}{A^{6}s}}}$, 15*3-3*6+30 = 57	h = 6.626070040e-34	${\displaystyle {\frac {kg\;m^{2}}{s}}}$, θ = 15-13*2+30 = 19
Gravitational constant	${\displaystyle {(G^{*})}^{5}={\frac {\pi ^{3}{\mu _{0}}}{2^{20}3^{6}\alpha ^{11}{R}^{2}}}}$	G* = 6.67249719229e11	${\displaystyle {\frac {kg\;m^{3}}{A^{2}s^{2}}}}$, 15-13*3-3*2+30*2 = 30	G = 6.67408e-11	${\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}$, θ = -13*3-15+30*2 = 6
Elementary charge	${\displaystyle {(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}{c}^{4}\alpha ^{8}{R}}}}$	e* = 1.60217651130e-19	${\displaystyle {\frac {s^{3}}{m^{3}}}}$, -30*4+13*3 = -81	e = 1.6021766208e-19	${\displaystyle As}$, θ = 3-30 = -27
Boltzmann constant	${\displaystyle {(k_{B}^{*})}^{3}={\frac {\pi ^{5}{\mu _{0}}^{3}}{3^{3}2{c}^{4}\alpha ^{5}{R}}}}$	kB* = 1.37951014752e-23	${\displaystyle {\frac {kg^{3}}{s^{2}A^{6}}}}$, 15*3+30*2-3*6 = 87	kB = 1.38064852e-23	${\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}$, θ = 15-26+60-20 = 29
Electron mass	${\displaystyle {(m_{e}^{*})}^{3}={\frac {16\pi ^{10}{R}{\mu _{0}}^{3}}{3^{6}{c}^{8}\alpha ^{7}}}}$	me* = 9.10938231256e-31, u = 15	${\displaystyle {\frac {kg^{3}s^{2}}{m^{6}A^{6}}}}$, 15*3-30*2+13*6-3*6 = 45	me = 9.10938356e-31	${\displaystyle kg}$, θ = 15
	${\displaystyle {\frac {2^{3}\pi ^{5}({k_{B}}^{*})}{3^{3}\alpha ^{6}({e^{*}})^{2}({m_{e}}^{*})c^{4}}}}$	1.0	${\displaystyle {\frac {1}{m^{2}A^{2}K}}}$, 13*2-3*2-20 = 0	8π5kB/(27α6e2mec4) = 1.000 825	${\displaystyle {\frac {1}{m^{2}A^{2}K}}}$, θ = 0
Gyromagnetic ratio	${\displaystyle ({(\gamma _{e}^{*})/2\pi })^{3}={\frac {g_{e}^{3}3^{3}c^{4}}{2^{8}\pi ^{8}\alpha \mu _{0}^{3}R_{\infty }^{2}}}}$	(γe*/2π) = 28024.95355	${\displaystyle {\frac {m^{3}s^{2}A^{6}}{kg^{3}}}}$, -13*3-30*2+3*6-15*3 = -126	γe/2π = 28024.95164	${\displaystyle {\frac {A\;s}{kg}}}$, θ = -42
Planck length	${\displaystyle ({l_{p}^{*}})^{15}={\frac {\pi ^{22}{\mu _{0}}^{9}}{2^{35}3^{24}\alpha ^{49}c^{35}R^{8}}}}$	lp* = 0.161603660096e-34	${\displaystyle {\frac {kg^{9}s^{17}}{m^{18}A^{18}}}}$, 15*9-30*17+13*18-3*18 = -195	lp = 0.1616229e-34	${\displaystyle m}$, θ = -13
Planck mass	${\displaystyle ({m_{P}^{*}})^{15}={\frac {2^{25}\pi ^{13}{\mu _{0}}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}}}$	mP* = 0.217672817580e-7	u = ${\displaystyle {\frac {kg^{6}m^{3}}{s^{7}A^{12}}}}$, 15*6-13*3+30*7-3*12 = 225	mP = 0.2176470e-7	${\displaystyle kg}$, θ = 15
	${\displaystyle {\frac {2^{11}\pi ^{3}(G^{*})^{2}(k_{B}^{*})}{\alpha ^{2}(h^{*})(c^{*})^{2}(e^{*})(m_{P}^{*})}}}$	1.0	${\displaystyle {\frac {m^{4}}{kg^{3}s^{4}A\;K}}}$, -13*4-15*3+30*4-3-20 = 0	211π3G2kB/(α2hc2emP) = 1.001418	${\displaystyle {\frac {m^{4}}{kg^{3}s^{4}A\;K}}}$, θ = 0

For reference here we formulate the physical constants in terms of the Planck units M, T, P. These are proposed as candidates for natural Planck units because their geometries are very basic; M = 1, T = pi, P = Omega. This suggests that the attributes (of mass, length, time ... ) are built into these geometries; Planck mass is a point, Planck time is a rotation and P is a radius (from which higher geometries may emerge)... and so scalars (which carry the dimensioned component to solve the SI units) are not required (by the universe itself).

If this is correct, then particles do not exist at unit Planck time, but rather are events that occur over time, thus the quantum scale emerges from the Planck scale. This also means that quantum theories cannot be applied to the Planck scale because quantum states do not exist at the Planck scale.

The model uses individual Planck units as indivisible units, there is no half Planck mass or half Planck time .... therefore our observable physical universe begins at the Planck scale, however the existence of a rotation term (T = π) does suggest the possibility of a scale below the Planck scale, but we have no means to measure this. It is a philosophical argument.

${\displaystyle M=(1){\frac {r^{4}}{v}}}$

${\displaystyle T=(\pi ){\frac {r^{9}}{v^{6}}}}$

${\displaystyle P=(\Omega )r^{2}}$

Table 8. Physical constants in terms of (M, T, P, α)

SI constant	geometrical analogue	unit number θ
Speed of light	${\displaystyle c={\frac {2\pi P^{2}}{M}}}$	17
Planck constant	${\displaystyle h^{*}=2\pi MVL={\frac {8\pi ^{3}P^{4}T}{M}}}$	15+17-13=19
Gravitational constant	${\displaystyle G^{*}={\frac {V^{2}L}{M}}={\frac {8\pi ^{3}P^{6}T}{M^{4}}}}$	34-13-15=6
Elementary charge	${\displaystyle e^{*}=AT={\frac {2^{7}\pi ^{3}P^{3}T}{M^{3}\alpha }}}$	3-30=-27
Boltzmann constant	${\displaystyle k_{B}^{*}={\frac {2\pi VM}{A}}={\frac {M^{3}\alpha }{2^{5}\pi P}}}$	17+15-3=29
Vacuum permeability	${\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}={\frac {M^{6}\alpha }{2^{11}\pi ^{4}P^{4}T}}}$	34+15+13-6=56

The following is one of the most important formulas in physics; it describes the relationship between the fine structure constant and the dimensioned constants.

${\displaystyle \alpha ={\frac {2h}{\mu _{0}e^{2}c}}}$

However, if we replace the numerical (h, μ₀, e, c) with the geometrical (h, μ₀, e, c), we find that the equation collapses to give alpha;

${\displaystyle {\frac {2h}{\mu _{0}e^{2}c}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {red}\alpha \color {black}}$

Note also the units and scalars cancel

units = ${\displaystyle {\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1}$

scalars = ${\displaystyle ({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1}$

This is a good test of our model, both of the unit numbers thesis and the geometrical objects thesis, because this equation reduces to

${\displaystyle \alpha =2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})}$

${\displaystyle \alpha =\alpha }$

There is no uncertainty of measurement and the formula is well established as a key formula.

Omega is the second dimensionless constant used, there is a sqrt solution

${\displaystyle \Omega ={\sqrt {\left(\pi ^{e}e^{(1-e)}\right)}}}$ = 2.0071349543

This solution proposes that Omega is a mathematical constant (a construct of the naturally occurring mathematical constants pi and e). A sqrt solution is required because all mass related constants use Omega^2 and all charge constants use Omega^3; thus mass is always positive but charge can be plus or minus.

If so, this means that the model requires only 1 physical constant - the fine structure constant alpha. The rest are mathematical constants and these can be derived from integers.

The model is using CODATA 2014 values. This is because only 2 dimensioned physical constants can be assigned exact values, once 2 constants have been assigned values, then all other constants are defined by default. After CODATA 2014, 4 constants were assigned exact values which is problematic in terms of this model.

For comparison purposes, the following "8 constants" are chosen as they are "the most precisely known" of the CODATA 2014 constants (constants which can be formulated in terms of the Planck units). The 2014 constants (G, K_B) were known with low accuracy and so are not useful for comparison and so are not included here.

Note that although h and e both show slight divergence, when combined into the Von Klitzing constant (R_K = h/e^2), this divergence disappears.

The unit number relationship was used to tune (v, r, Omega) using c, μ₀, R. However this could only be possible if the dimensioned fundamental constants are not fundamental (they can be defined in terms of each other), and so these values may constitute evidence for a unit number relationship.

The "Calculated^*" values (column 2) were obtained using (α, Ω, r, v).

alpha = 137.035999139

Omega = 2.007134949638

v = 11843707.9052487

r = 0.71256251430467

M = r^4/v

T = pi*r^9/v^6

P = Omega*r^2

V = 2*pi*P^2/M

L = T*V

A = 16*V^3/(alpha*P^3)

ψ = 4*pi^2*(2^6*3*pi^2*alpha*Omega^5)^3 = 0.23895453074e23

c = V

μ₀ = 4*pi*V^2*M/(alpha*L*A^2)

R = (m_e)/(4*pi*L*alpha^2*M)

h = 2*pi*M*V*L

e = A*T

R_K = h/e^2

m_e = M/ψ

λ_e = 2*pi*L*ψ

We can construct a table of constants using these 3 geometries. Setting

${\displaystyle f(x)\;units=({\frac {L^{15}}{M^{9}T^{11}}})^{n}=1}$

i.e.: unit number θ = (-13*15) - (15*9) - (-30*11) = 0

${\displaystyle \color {red}i\color {black}=\pi ^{2}\Omega ^{15}}$, units = ${\displaystyle {\sqrt {f(x)}}}$ = 1 (unit number = 0, no scalars)

${\displaystyle \color {red}x\color {black}=\Omega {\frac {v}{r^{2}}}}$ , units = ${\displaystyle {\sqrt {\frac {L}{MT}}}}$ = u¹ = u (unit number = -13 -15 +30 = 2/2 = 1, with scalars v, r)

${\displaystyle \color {red}y\color {black}=\pi {\frac {r^{17}}{v^{8}}}}$ , units = ${\displaystyle M^{2}T}$ = 1, (unit number = 15*2 -30 = 0, with scalars v, r)

Note: The following suggests a numerical boundary to the values the SI constants can have.

${\displaystyle {\frac {v}{r^{2}}}=a^{1/3}={\frac {1}{t^{2/15}k^{1/5}}}={\frac {\sqrt {v}}{\sqrt {k}}}}$ ... = 23326079.1...; unit = u^1 = u

${\displaystyle {\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...}$ gives a range from 0.812997... x10^-59 to 0.123... x10⁶⁰

Note: 
1. The constants with unit numbers ${\displaystyle \theta }$ in the series ${\displaystyle \color {red}(\theta ^{15})^{n}\color {black}}$ have no Omega. This further suggests an underlying geometrical base-15.  

Table 10. Table of Constants

Constant	θ	Geometrical object (α, Ω, v, r)	Unit	Calculated	CODATA 2014
Time (Planck)	${\displaystyle \color {red}-30\color {black}}$	${\displaystyle T=\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}={\frac {\pi r^{9}}{v^{6}}}}$	${\displaystyle T}$	T = 5.390 517 866 e-44	tp = 5.391 247(60) e-44
Elementary charge	${\displaystyle \color {red}-27\color {black}}$	${\displaystyle e^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\pi \Omega ^{3}r^{3}}{v^{3}}}}$	${\displaystyle {\frac {L^{3/2}}{T^{1/2}M^{3/2}}}=AT}$	e* = 1.602 176 511 30 e-19	e = 1.602 176 620 8(98) e-19
Length (Planck)	${\displaystyle \color {red}-13\color {black}}$	${\displaystyle L=(2\pi )\color {red}{\frac {x^{\theta }i}{y}}\color {black}=(2\pi )\;{\frac {\pi \Omega ^{2}r^{9}}{v^{5}}}}$	${\displaystyle L}$	L = 0.161 603 660 096 e-34	lp = 0.161 622 9(38) e-34
Ampere	${\displaystyle \color {red}3\color {black}}$	${\displaystyle A=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}x^{\theta }\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{3}v^{3}}{r^{6}}}}$	${\displaystyle A={\frac {L^{3/2}}{M^{3/2}T^{3/2}}}}$	A = 0.297 221 e25	e/tp = 0.297 181 e25
Gravitational constant	${\displaystyle \color {red}6\color {black}}$	${\displaystyle G^{*}=(2^{3}\pi ^{3})\color {red}\color {red}x^{\theta }y\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{6}r^{5}}{v^{2}}}}$	${\displaystyle {\frac {L^{3}}{MT^{2}}}}$	G* = 6.672 497 192 29 e11	G = 6.674 08(31) e-11
Mass (Planck)	${\displaystyle \color {red}\color {red}15\color {black}}$	${\displaystyle M=\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}={\frac {r^{4}}{v}}}$	${\displaystyle M}$	M = .217 672 817 580 e-7	mP = .217 647 0(51) e-7
Velocity	${\displaystyle \color {red}\color {red}17\color {black}}$	${\displaystyle V=(2\pi )\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(2\pi )\;\Omega ^{2}v}$	${\displaystyle V={\frac {L}{T}}}$	V = 299 792 458	c = 299 792 458
Planck constant	${\displaystyle \color {red}19\color {black}}$	${\displaystyle h^{*}=(2^{3}\pi ^{3})\color {red}{\frac {x^{\theta }y^{3}}{i}}\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{4}r^{13}}{v^{5}}}}$	${\displaystyle {\frac {L^{2}M}{T}}}$	h* = 6.626 069 134 e-34	h = 6.626 070 040(81) e-34
Planck temperature	${\displaystyle \color {red}\color {red}20\color {black}}$	${\displaystyle {T_{p}}^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{5}v^{4}}{r^{6}}}}$	${\displaystyle {\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV}$	Tp* = 1.418 145 219 e32	Tp = 1.416 784(16) e32
Boltzmann constant	${\displaystyle \color {red}\color {red}29\color {black}}$	${\displaystyle {k_{B}}^{*}=({\frac {\alpha }{2^{5}\pi }})\color {red}{\frac {x^{\theta }y^{4}}{i^{2}}}\color {black}=({\frac {\alpha }{2^{5}\pi }})\;{\frac {r^{10}}{\Omega v^{3}}}}$	${\displaystyle {\frac {M^{5/2}T^{1/2}}{L^{1/2}}}={\frac {ML}{TA}}}$	kB* = 1.379 510 147 52 e-23	kB = 1.380 648 52(79) e-23
Vacuum permeability	${\displaystyle \color {red}56\color {black}}$	${\displaystyle {\mu _{0}}^{*}=({\frac {\alpha }{2^{11}\pi ^{4}}})\color {red}{\frac {x^{\theta }y^{7}}{i^{4}}}\color {black}=({\frac {\alpha }{2^{11}\pi ^{4}}})\;{\frac {r^{7}}{\pi \Omega ^{4}}}}$	${\displaystyle {\frac {M\;L}{T^{2}A^{2}}}}$	μ0* = 4π/10^7	μ0 = 4π/10^7