User:27 is the best number/sandbox

$10^{506}-10^{253}-1$ $d\sin \theta =m_{1}\lambda _{1}=5\times 400\times 10^{-9}=2.0\times 10^{-6}\ {\text{m}}$ $\sin \theta ={\frac {2.0\times 10^{-6}}{5.555\dots \times 10^{-6}}}=0.36$ $\theta =\arcsin(0.36)\approx 21.1^{\circ }$

BAGGED

Consonant chart for American English

		Labial	Dental	Alveolar	Post- alveolar	Palatal	Velar	Uvular	Glottal
Nasal		m	n				ng /ŋ/
Stop	Voiceless	p	t				k		/ʔ/
	Voiced	b	d				ɡ
Affricate	Voiceless				ch /t͡ʃ/
	Voiced				j /d͡ʒ/
Tap or flap			r /ɾ/
Fricative	Voiceless	f	th /θ/	s	sh /ʃ/		(ch /x/)		h
	Voiced	v	th /ð/	z	j /ʒ/
Approximant	Voiceless	(wh /ʍ/)					(wh /ʍ/)
	Voiced	w		l	r /ɹ̠/	y /j/	(w)

A better metric system

Length

Let's start by defining the speed of light, c, and the Planck length, ℓ_P: $$c=299\,792\,458\,\mathrm {m/s}$$ $$\ell _{\mathrm {P} }={\sqrt {\frac {\hbar G}{c^{3}}}}=1.616\,255\times 10^{-35}\,\mathrm {m}$$ Of course, the notation of using decimal is quite arbitrary; a far better base is binary because it only has two digits. $${\textrm {DEC}}\,299\,792\,458\,={\textrm {BIN}}\,1\,0001\,1101\,1110\,0111\,1000\,0100\,1010=\mathrm {11\,DE7\,84A} _{\textrm {HEX}}$$$${\textrm {DEC}}\,1.616\,255\times 10^{-35}={\textrm {HEX}}\,\mathrm {1.280\,34E\,7E4\,4A79} \times 2^{-76}$$ Of course, the notation that I want to use (where "down" is 0 and "up" is 1) can't really be done in LaTeX, so we'll have to use hex for now.

The speed of light is the maximum speed of causality, so let's set it equal to 1:$$c=1$$1... what? Well, it's a fraction (more specifically, 100%), so it's actually dimensionless. In fact, I want all units in this new metric system to be dimensionless whenever possible.

For reference (on a human scale), a typical walking speed is about $1.42\,{\textrm {m/s}}$, which is about $4.73661\,{\textrm {n}}c$ (the two numbers in this sentence are in decimal).

Since the Planck length is apparently the smallest length that can be measured, it also makes sense to make it equal to 1:

$\ell _{\mathrm {P} }=1\,\ell$

Here, l is a new unit that stands for length. But... uh, oh! We can't have both of them equal to one! What should we do?

First, let's convert c to scientific notation (in hex):$$\mathrm {11\,DE7\,84A} =1.1\mathrm {DE} \,784\mathrm {A} \times 10^{7}=1.1\mathrm {DE} \,784\mathrm {A} \times 2^{1\mathrm {C} }$$Remember, this is in hex, so that's sixteen to the power of seven.

Time

Now, we need the ratio of the Planck speed ($1{\frac {\ell _{\mathrm {P} }}{t_{\mathrm {P} }}}$) to the speed of light (in decimal):$${\begin{aligned}{\frac {\frac {\ell _{\mathrm {P} }}{t_{\mathrm {P} }}}{c}}&={\frac {\frac {1.616\,255\times 10^{-35}\,\mathrm {m} }{5.391\,247\times 10^{-44}\,\mathrm {s} }}{{\frac {299\,792\,458\,\mathrm {m} }{1\,\mathrm {s} }}{}}}\\&=1\end{aligned}}$$This should not be too surprising because the Planck time is defined as how long it would take something traveling at the speed of light to travel one Planck length. More importantly, to establish a scale, we need the ratio of all possible lengths from the Planck length to the speed of light. Doing that, we get...$${\begin{aligned}{\frac {\ell _{\mathrm {P} }}{c_{\textrm {length}}}}&={\frac {1.616\,255\times 10^{-35}\,\mathrm {m} }{299\,792\,458\,\mathrm {m} }}\\&=5.391\,246\,36\times 10^{-44}\end{aligned}}$$...which is exactly the value of the Planck time. Inverting that number results in 1.85485865×10⁴³. In other words, we have found another definition for the speed of light (remember, we want it to be 1):$$c=299\,792\,458\,\mathrm {m/s} =1.854\,858\,65\times 10^{43}\,\ell _{\textrm {P}}/{\textrm {s}}$$Solving, we get:$${\begin{alignedat}{2}t_{\mathrm {P} }&=5.391\,247\times 10^{-44}\,{\textrm {s}}\\{\textrm {Let}}\,t_{\mathrm {P} }&=1{\textrm {:}}\\1&=5.391\,247\times 10^{-44}\,{\textrm {s}}\\5.391\,247\times 10^{-44}\,{\textrm {s}}&=1\\1\,{\textrm {s}}&={\frac {1}{5.391\,247\times 10^{-44}}}\\\end{alignedat}}$$One second is 1.85485865×10⁴³ t_P. Using all this information, we can now finally let $c=1$:$$c=299\,792\,458\,\mathrm {m/s} =1.854\,858\,65\times 10^{43}\,\ell _{\textrm {P}}/t_{\textrm {P}}=1$$

Mass

Did you think I was going to stop there? Heck no, we gotta do all seven SI units!

The SI unit of mass is the kilogram for some reason (and not the gram), and a Planck mass feels very arbitrary. Let's instead use the least massive thing so far: The electron is the lightest particle with a non-zero rest mass whose mass has been measured, and it is (9.1093837139±28)×10⁻³¹ kg. Well, I can't call it the Planck mass, so I'll call it ${\textrm {m}}_{\textrm {e}}$:

${\textrm {m}}_{\textrm {e}}=9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}$

Just as before, we will ignore the errors and define the value exactly. Technically, particles can exist with zero mass (like the photon), but that wouldn't be very useful, now would it? Let's set ${\textrm {m}}_{\textrm {e}}$ to be $1\,{\textrm {m}}$, where ${\textrm {m}}$ is the unit of mass.

Temperature

This is the first unit that has both a lower and an upper bound! The SI unit is Kelvin, so let's use absolute zero (coldest temperature possible) as the minimum and the Planck temperature as the maximum (because our models of temperatures break down):

$${\text{Range of new metric temperature scale:}}\,0\,{\textrm {K}}<T\leq 1.416\,784\times 10^{32}\,{\textrm {K}}$$

Let's set ${\textrm {t}}=1.416\,784\times 10^{32}\,{\textrm {K}}$, where $t$ is the unit of temperature. Let's look at some temperatures:

Description	Temperature
	$^{\circ }{\textrm {C}}$	${\textrm {K}}$	${\textrm {t}}$
Absolute zero	−273.15	0	0
Temperature of the cosmic microwave background	−270.42452	2.72548	1.92371×10−32
Freezing point of water	0	273.15	1.9280×10−30
Room temperature	23.89	297.04	2.0966×10−30
Human body temperature	37.0±0.5	309.6–310.6	2.19×10−30
Boiling point of water	100	373.15	2.6338×10−30
Surface temperature of the Sun	5500	5772	4.074×10−29
Core temperature of the Sun	15700000		1.11×10−25
Planck temperature	1.416784×1032		1

Charge

There's already the elementary charge, so I'll use that. This way, a proton's charge can still be +1.
Except... We actually care a lot more about the electron, especially in chemistry. If an ion gains an electron, their charge... decreases? Isn't that confusing? Well, that ends today! $$1\,{\textrm {I}}=-e,$$ ...where e is the fundamental unit of charge.

Amount of substance

This is probably the dumbest SI unit because it's literally just a number. It equals the number of atoms in 12 g of carbon-12. If we're going to be as objective as possibly, then we'll instead define the new mole to be the number of atoms in 1 m of hydrogen, the most common element in the universe.

But this isn't actually possible because a hydrogen atom is larger than an electron. However, I do see the usefulness of the mole, so let's use an example problem:

A scientist has discovered that $1.000\,{\textrm {kg}}$ of a substance is 98% carbon and 2% hydrogen. How many atoms exist in this substance?

Normally, one would use Avogadro's number to solve for this: $$1.000\,{\textrm {kg}}=1000.\,{\textrm {g}}$$ $$m_{\text{C}}=(1000\,{\textrm {g}})(0.98)=980\,{\textrm {g}}$$$$m_{\text{H}}=(1000\,{\textrm {g}})(0.02)=20\,{\textrm {g}}$$ $$n_{\text{C}}={\frac {980\,{\cancel {\textrm {g}}}}{12.01\,{\cancel {\textrm {g}}}/{\textrm {mol}}}}\approx 81.66\,{\textrm {mol}}$$

$$n_{\text{H}}={\frac {20\,{\cancel {\textrm {g}}}}{1.008\,{\cancel {\textrm {g}}}/{\textrm {mol}}}}\approx 19.84\,{\textrm {mol}}$$$${\text{atoms of C}}={\Bigl (}81.66\,{\cancel {\textrm {mol}}}{\Bigr )}{\Bigl (}6.022\times 10^{23}\,{\frac {\textrm {atoms}}{\cancel {\textrm {mol}}}}{\Bigr )}\approx 4.91\times 10^{25}\,{\text{atoms}}$$

$${\text{atoms of H}}={\Bigl (}19.84\,{\cancel {\textrm {mol}}}{\Bigr )}{\Bigl (}6.022\times 10^{23}\,{\frac {\textrm {atoms}}{\cancel {\textrm {mol}}}}{\Bigr )}\approx 1.19\times 10^{25}\,{\text{atoms}}$$$${\text{Total atoms}}=(4.91\times 10^{25})+(1.19\times 10^{25})=6.10\times 10^{25}\,{\text{atoms}}$$

But... How many atoms are there really in 12 grams of carbon-12? Well, I suppose we could calculate the mass of the carbon-12 atom (let's be as exact as possible, no more "assume the mass of the proton and neutron are the same and ignore the electron" shenanigans. Carbon-12 has:

• 6 protons
• 6 neutrons
• 6 electrons

Now, recall that protons and neutrons are not fundamental particles (but electrons are). They are made of quarks:

• 1 proton has two up quarks and one down quark.
• 1 neutron has one up quark and two down quarks.
• Electrons are fundamental particles.

Here is the mass of each:

• $m_{q_{u}}=2.2\,{\frac {\textrm {MeV}}{c^{2}}}$
• $m_{q_{d}}=4.7\,{\frac {\textrm {MeV}}{c^{2}}}$
• ${\textrm {m}}_{\textrm {e}}=9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}$

$$1\,{\frac {\textrm {MeV}}{c^{2}}}=1\,{\frac {10^{6}\,{\textrm {eV}}}{c^{2}}}=1{\frac {10^{6}\,{\textrm {eV}}}{(299\,792\,458\,{\textrm {m/s}})^{2}}}=1{\frac {10^{6}(1.602\ 176\ 634\times 10^{-19}\,{\text{C}})\times 1\,{\text{V}}}{(299\,792\,458\,{\textrm {m/s}})^{2}}}=1.782\ 661\ 92\times 10^{-30}\;{\textrm {kg}}$$So,$$1\,{\frac {\textrm {MeV}}{c^{2}}}=1.782\ 661\ 92\times 10^{-30}\;{\textrm {kg}}$$

Here is the mass of each, this time in SI units:

• $m_{q_{u}}=(2.2)(1.782\ 661\ 92\times 10^{-30}\,{\textrm {kg}})=3.921\ 856\ 224\times 10^{-30}\,{\textrm {kg}}$
• $m_{q_{d}}=(4.7)(1.782\ 661\ 92\times 10^{-30}\,{\textrm {kg}})=8.378\ 511\ 024\times 10^{-30}\,{\textrm {kg}}$
• ${\textrm {m}}_{\textrm {e}}=9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}$

Now, for protons and neutrons:

• ${\begin{alignedat}{2}m_{p}&=2(3.921\ 856\ 224\times 10^{-30}\,{\textrm {kg}})+8.378\ 511\ 024\times 10^{-30}\,{\textrm {kg}}\\&=7.843\ 712\ 448\times 10^{-30}\,{\textrm {kg}}+8.378\ 511\ 024\times 10^{-30}\,{\textrm {kg}}\\&=16.222\ 223\ 472\times 10^{-30}\,{\textrm {kg}}\end{alignedat}}$

Wait, that isn't right! Well, that's because protons and neutrons also have gluons holding the quarks together. Here are their actual masses:

• $m_{p}=1.672\ 621\ 925\ 95\times 10^{-27}\,{\textrm {kg}}$
• $m_{n}=1.674\ 927\ 498\ 04\times 10^{-27}\,{\textrm {kg}}$
• $m_{\textrm {e}}=9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}$

Now, we can find the mass of one atom of carbon-12:$${\begin{aligned}m_{{}^{12}{\text{C}}}&=6(1.672\ 621\ 925\ 95\times 10^{-27}\,{\textrm {kg}})+6(1.674\ 927\ 498\ 04\times 10^{-27}\,{\textrm {kg}})+6(9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}})\\&=6{\Bigl [}(1.672\ 621\ 925\ 95\times 10^{-27}\,{\textrm {kg}})+(1.674\ 927\ 498\ 04\times 10^{-27}\,{\textrm {kg}})+(9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}){\Bigr ]}\\&=6(3.348\ 460\ 36\times 10^{-27}\,{\textrm {kg}})\\&=2.009\ 076\ 22\times 10^{-26}\,{\textrm {kg}}\end{aligned}}$$Now, this is the mass of ONE atom of carbon-12. The number of atoms in twelve grams of carbon-12 turns out to be 5.972894348×10²³ atoms.

I guess an equivalent would be the 32-bit integer limit plus one: 2147483648, or $2^{31}$, as a base number for now. Just as the mole is defined as the number of atoms in 12 g of ¹²
C, I can make the new mole equal to the number of atoms in $2^{31}\,{\textrm {m}}$ of ¹
H:

$mm_{\text{H}}=1.673\ 823\ 37\times 10^{-27}\ {\textrm {kg}}\,{\text{(using Avogadro's number)}}$

${\begin{aligned}m_{\textrm {H}}&=1.672\ 621\ 925\ 95\times 10^{-27}\ {\textrm {kg}}\\&{\underline {+9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}}\\&1.673\ 532\ 86\times 10^{-27}\ {\textrm {kg}}\end{aligned}}$

The second method simply adds the masses of the proton and the electron and is what I will use. Technically, since the proton and electron are charged particles, a hydrogen atom also has a photon, but it has zero rest mass, so it will be ignored (for now). We can now set up a relationship:$${\frac {1\ {\text{atom}}}{1.673\ 532\ 86\times 10^{-27}\ {\textrm {kg}}}}={\frac {x}{2^{31}\cdot 9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}}$$$${\frac {(1\ {\text{atom}})2^{31}\cdot 9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}{1.673\ 532\ 86\times 10^{-27}\ {\textrm {kg}}}}={\frac {x{\cancel {2^{31}\cdot 9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}}}{\cancel {2^{31}\cdot 9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}}}$$$$x={\frac {(1\ {\text{atom}})2^{31}\cdot 9.109\ 383\ 7139\times 10^{-31}\ {\textrm {kg}}}{1.673\ 532\ 86\times 10^{-27}\ {\textrm {kg}}}}$$$$x={\frac {(1\ {\text{atom}})2^{31}\cdot 5.443\ 205\ 76\times 10^{-31}\ {\cancel {\textrm {kg}}}}{10^{-27}\ {\cancel {\textrm {kg}}}}}$$$$x=2^{31}\cdot 5.443\ 205\ 76\times 10^{-4}\ {\text{atoms}}$$$$x=1\,168\,919.536\,229\,94\ {\text{atoms}}$$That is not comparable enough. However, if we multiply by 2³¹ three more times, we get ${\textstyle x=5.390\ 689\ 88\times 10^{24}\ {\text{atoms}}}$, or ${\textstyle x=(2^{31})^{4}\cdot 5.443\ 205\ 76\times 10^{-4}\ {\text{atoms}}=2^{124}\cdot 5.443\ 205\ 76\times 10^{-4}\ {\text{atoms}}}$.

Luminosity

The SI unit of luminosity is the candela. So uh, I've never actually learned about the candela nor have ever used it in school. But apparently, it's tuned at 555 nm, because green is where human eye sensitivity is at its peak? Here's the actual definition:

The candela [...] is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency 540×10¹² Hz, K_cd, to be 683 when expressed in the unit lm W⁻¹, which is equal to cd sr W⁻¹, or cd sr kg⁻¹ m⁻² s³, where the kilogram, metre and second are defined in terms of h, c and Δν_Cs.

So, uh, there is already a "dimensionless" version of luminosity, which I found on Luminous efficacy#Photopic vision, where the efficiency is already measured as a fraction. 555 nm feels arbitrary, but at this point, I don't feel like I understand the candela enough to optimize it and make it dimensionless.

Angle

So, while the SI doesn't consider the angle one of its seven base units, I think I can still optimize it. We often use degrees (based on SHCN 360) and radians (based on the half-circle constant π), but the most objective way to do this is to make one full turn equal to... well, one, and angles can once again be measured as fractions and percentages. This would make half a turn equal to $0.5\,\tau$.

Conversions table

Here is a summary of the units we have used so far:

Quantity	SI	New metric	Conversion factor
Speed	Metre per second	Speed of light	$c=299\,792\,458\ {\textrm {m/s}}$
Length	Metre	Planck length	$\ell _{\mathrm {P} }=1.616\,255\times 10^{-35}\,\mathrm {m}$
Time	Second	Planck time	$t_{\mathrm {P} }=5.391\,247\times 10^{-44}\,{\textrm {s}}$
Mass	Kilogram	Mass of the electron	$m_{\textrm {e}}=9.109\,383\,7139\times 10^{-31}\,{\textrm {kg}}$
Temperature	Kelvin	Upper: Planck temperature	$1\ {\textrm {T}}=1.416784\times 10^{32}\ {\textrm {K}}\ {\text{such that}}\ 1\ \Delta {\textrm {T}}={\frac {1}{1.416784\times 10^{32}}}\ \Delta {\textrm {K}}$
		Lower: Absolute zero
Amount of substance	Mole	New mole	${\textstyle N_{1}=5.390\ 689\ 88\times 10^{24}\ {\text{atoms}}\approx 8.951\,451\,14\cdot N_{0}}$
Luminosity	Candela	"New" candela	$1\ {\textrm {cd}}_{\textrm {new}}=683.002\ {\textrm {cd}}$
Angle	Radian	Turn	$1\ \tau =2\pi \ {\textrm {rad}}$

Now, I'll have to convert all of these numbers to binary.