My (Overcomplicated) Coherent System of Measurements

July 23, 2020 · 18:25 pm · Metrology

There are many measurement systems, and the ones by Système international and the imperial system are arguably two of the most famous. There are also the furlong-firkin-fortnight (FFF) system and the CCC system by Jan Misali, both of which are sarcastic systems made as a joke. I have here a system that falls into the latter category that aims to be as difficult to use as possible. Here are the base units:

Barye (Ba)(Ba)0.1Pa=0.1kgms20.1 Pa = 0.1 \frac{kg}{m \cdot s^2}
gallon (gal)(gal)231in3=0.003785411784m3231 in^3 = 0.003785411784 m^3
standard gravity (g)(g)9.80665ms29.80665 \frac{m}{s^2}
vacuum permittivity (ϵ0)(\epsilon_0)8.8541878128×1012C2s2kgm3\approx 8.8541878128 \times 10^{-12} \frac{C^2 \cdot s^2}{kg \cdot m^3}1
specific heat of water at 25°C (c)(c)4179.6m2s2K\approx 4179.6 \frac{m^2}{s^2 \cdot K}1

Let’s try to derive the units for some common dimensions.


We know volume is represented by galgal, so gal30.1558491279m\sqrt[3]{gal} \approx 0.1558491279 m is our unit of length.


We can’t use BaBa nor ϵ0\epsilon_0, since those have kgkg and CC, respectively, that needs to be removed by multiplying its own reciprocal, which cancels itself out. That leaves us with galgal and gg to be manipulated. Time can be represented by seconds, so we have

s=s2=s2mm=s2mm33=s2mm36\begin{aligned} s &= \sqrt{s^2} \\ &= \sqrt{\frac{s^2}{m} \cdot m} \\ &= \sqrt{\frac{s^2}{m} \sqrt[3]{m^3}} \\ &= \sqrt{\frac{s^2}{m}} \sqrt[6]{m^3} \\ \end{aligned}

We can replace s2m\frac{s^2}{m} with g1g^{-1} and m3m^3 with galgal to get

g1gal60.1260642227s.\sqrt{g^{-1}} \sqrt[6]{gal} \approx 0.1260642227 s.


Velocity is acceleration times time, so

gg1gal6=ggal61.2362677099ms\begin{aligned} g \cdot \sqrt{g^{-1}} \sqrt[6]{gal} &= \sqrt{g} \sqrt[6]{gal} \\ &\approx 1.2362677099 \frac{m}{s} \\ \end{aligned}


kg=kgms2ms2=kgms2m2s2m\begin{aligned} kg &= \frac{kg}{m \cdot s^2} m \cdot s^2 \\ &= \frac{kg}{m \cdot s^2} m^2 \cdot \frac{s^2}{m} \\ \end{aligned}

We can replace kgms2\frac{kg}{m \cdot s^2} and m2m^2 with BaBa and gal23\sqrt[3]{gal^2}, respectively, to get

Bag1gal232.4767836812×104kgBa \cdot g^{-1} \sqrt[3]{gal^2} \approx 2.4767836812 \times 10^{-4} kg


Force is mass times acceleration, so

Bag1gal23g=Bagal230.002428895068N\begin{aligned} Ba \cdot g^{-1} \sqrt[3]{gal^2} \cdot g &= Ba \sqrt[3]{gal^2} \\ &\approx 0.002428895068 N \\ \end{aligned}


Momentum is mass times velocity, so

Bag1gal23ggal6=Bag1gal563.06196768983×104kgms\begin{aligned} Ba \cdot g^{-1} \sqrt[3]{gal^2} \sqrt{g} \sqrt[6]{gal} &= Ba \sqrt{g^{-1}} \sqrt[6]{gal^5} \\ &\approx 3.06196768983 \times 10^{-4} \frac{kg \cdot m}{s} \\ \end{aligned}


Energy has the same units as work, which is force times distance, so

Bagal23gal3=Bagal3.785411784×104Nm\begin{aligned} Ba \sqrt[3]{gal^2} \cdot \sqrt[3]{gal} &= Ba \cdot gal \\ &\approx 3.785411784 \times 10^{-4} N \cdot m \\ \end{aligned}

Electric Charge

C=C2=C2s2kgm3kgm3s2=C2s2kgm3kgmsmsm\begin{aligned} C &= \sqrt{C^2} \\ &= \sqrt{\frac{C^2 \cdot s^2}{kg \cdot m^3} \cdot \frac{kg \cdot m^3}{s^2}} \\ &= \sqrt{\frac{C^2 \cdot s^2}{kg \cdot m^3} \cdot \frac{kg \cdot m}{s} \cdot \frac{m}{s} \cdot m} \\ \end{aligned}

We can replace in ϵ0\epsilon_0, momentum, velocity, and length accordingly:

ϵ0Bag1gal56ggal6gal3=ϵ0Bagal56gal6gal3=ϵ0Bagal43=ϵ0Bagal232.285509959336463587506×108C\begin{aligned} \sqrt{\epsilon_0 \cdot Ba \sqrt{g^{-1}} \sqrt[6]{gal^5} \cdot \sqrt{g} \sqrt[6]{gal} \cdot \sqrt[3]{gal}} &= \sqrt{\epsilon_0 \cdot Ba \sqrt[6]{gal^5} \cdot \sqrt[6]{gal} \cdot \sqrt[3]{gal}} \\ &= \sqrt{\epsilon_0 \cdot Ba \cdot \sqrt[3]{gal^4}} \\ &= \sqrt{\epsilon_0 \cdot Ba} \sqrt[3]{gal^2} \\ &\approx 2.285509959336463587506 \times 10^{-8} C\\ \end{aligned}

Electric Field

Electric Field is force per charge, so

Bagal23ϵ0Bagal23=Baϵ0106273.65935983474NC\begin{aligned} \frac{Ba \sqrt[3]{gal^2}}{\sqrt{\epsilon_0 \cdot Ba} \sqrt[3]{gal^2}} &= \sqrt{\frac{Ba}{\epsilon_0}} \\ &\approx 106273.65935983474 \frac{N}{C}\\ \end{aligned}


Voltage is energy per charge, so

Bagalϵ0Bagal23=Baϵ0gal316562.657137136224V\begin{aligned} \frac{Ba \cdot gal}{\sqrt{\epsilon_0 \cdot Ba} \sqrt[3]{gal^2}} &= \sqrt{\frac{Ba}{\epsilon_0}} \sqrt[3]{gal} \\ &\approx 16562.657137136224 V \\ \end{aligned}


Capacitance is charge per voltage, so

ϵ0Bagal23Baϵ0gal3=ϵ0gal31.37991744948×1012F\begin{aligned} \frac{\sqrt{\epsilon_0 \cdot Ba} \sqrt[3]{gal^2}}{\sqrt{\frac{Ba}{\epsilon_0}} \sqrt[3]{gal}} &= \epsilon_0 \sqrt[3]{gal} \\ &\approx 1.37991744948 \times 10^{-12} F \\ \end{aligned}


Current is charge per time, so

ϵ0Bagal23g1gal6=ϵ0Bagalg1.81297271306×107A\begin{aligned} \frac{\sqrt{\epsilon_0 \cdot Ba} \sqrt[3]{gal^2}}{\sqrt{g^{-1}} \sqrt[6]{gal}} &= \sqrt{\epsilon_0 \cdot Ba \cdot gal \cdot g} \\ &\approx 1.81297271306 \times 10^{-7} A \\ \end{aligned}


Resistance has the same units as vacuum permittivity divided by velocity, so

ϵ0ggal67.1620311209×1012Ω\frac{\epsilon_0}{\sqrt{g} \sqrt[6]{gal}} \approx 7.1620311209 \times 10^{-12} \Omega


K=s2Km2m2s2\begin{aligned} K &= \frac{s^2 \cdot K}{m^2} \cdot \frac{m^2}{s^2} \\ \end{aligned}

We can replace in cc and velocity accordingly:

c1ggal62.95786130251×104Kc^{-1} \sqrt{g} \sqrt[6]{gal} \approx 2.95786130251 \times 10^{-4} K

What about luminous intensity and substance amount?

I don’t know how luminous intensity works so I can’t list an overcomplicated version of it. But feel free to use candela.

For substance amount, I don’t know any established units that are in the range of Avogadro’s Number that are ridicuous, so I’ll just say the Baker’s Dozen (13)(13) works.

  1. Technically, these don’t have a defined exact value, but I couldn’t think of any other unit or constant that was as complex.

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