4 ms·

I ran the numbers for going in the other direction. Suppose that: * The 99 slower gears are massless. * the fastest gear weighs 10 grams * The fas

I ran the numbers for going in the other direction. Suppose that: * The 99 slower gears are massless. * the fastest gear weighs 10 grams * The fastest gear is a cylinder of radius of 2cm * The slowest gear completes 1 rotation every 3 days. * Relativity only applies when I want it to. We have: * The fastest gear has an angular velocity of: w=10^95 radians/second = 10^95 s^-1 //Since radians are unitless * The fastest gear has a moment of intertia given by: I=mr^2 / 2 = 10^-2 * 4*10^-4 / 2 kg*m^2 * The fastest gear has a rotational energy of: E=Iw^2 / 2= 10^-2 * 4*10^-4 / 2 kg*m^2 * 10^190 s^-2 / 2 E = 10^184 kg*m^2*s^-2 * Ignoring all of relativity except E=mc^2, we have m = E/c^2 m ~ 10^184 kg*m^2*s^-2 / 10 ^ 17 m^2*s^2 m ~ 10^167 kg * Acknowledging more of relativity now, we have: Schwarzschild radius = 2Gm / c^2 Schwarzschild radius ~ 10^-12 m^3*kg^-1*s^-2 * 10^167 kg * 10^-17 m^-2*s^2 Schwarzschild radius ~ 10^138 m Giving us a black hole large enough to fit about 112 googol observable universes. I'm sure that I'm missing some relativistic effects here, but I don't even know how to begin to approach that.

spectramax 7y ago
What would be the torque required to spin the slowest gear 1/3 rotation per day? Let's assume the same radius: 2cm.
- hnuser123456 7y ago
  I would say you could ballpark it as something north of 10^100 Nm or ft-lb If you want to make these figures a little more reasonable, we could try to spin the slow gear only once in 100 years, in which case you can take about 5 whole powers of 10 off these numbers
hnuser123456 7y ago
>Giving us a black hole large enough to fit about 112 googol observable universes. *Giving us a black hole with a radius large enough to fit about 112 googol observable universes stacked end-to-end in a line. ~Triple the number of zeros for the number of universes to fit in its 3D volume. 112 x (4/3) x pi followed by 336 zeroes, more or less.
- hnuser123456 7y ago
  Doing some more order of magnitude approximations, in order to scale this black hole down so it were the size of our observable universe, define "the diameter of the universe in millimeters" (approx 10^30) as mmU. There are mmU * mmU * mmU * 0.5 mmU universes in the diameter of this black hole. For each millimeter in this scaled-down black hole, there are mmU groups of mmU universes, with another half layer of nesting, but then you can divide that by 1000. To fit the number of universes in the whole volume in a line end to end, repeat this process again, except without the extra half layer and division by 1000.
cjslep 7y ago
For relativistic effects I think you'll need to apply the Lorentz factor [0] [0] https://en.m.wikipedia.org/wiki/Lorentz_factor https://en.m.wikipedia.org/wiki/Lorentz_factor

spectramax 7y ago

What would be the torque required to spin the slowest gear 1/3 rotation per day? Let's assume the same radius: 2cm.

hnuser123456 7y ago
I would say you could ballpark it as something north of 10^100 Nm or ft-lb If you want to make these figures a little more reasonable, we could try to spin the slow gear only once in 100 years, in which case you can take about 5 whole powers of 10 off these numbers

hnuser123456 7y ago

>Giving us a black hole large enough to fit about 112 googol observable universes. *Giving us a black hole with a radius large enough to fit about 112 googol observable universes stacked end-to-end in a line. ~Triple the number of zeros for the number of universes to fit in its 3D volume. 112 x (4/3) x pi followed by 336 zeroes, more or less.

hnuser123456 7y ago
Doing some more order of magnitude approximations, in order to scale this black hole down so it were the size of our observable universe, define "the diameter of the universe in millimeters" (approx 10^30) as mmU. There are mmU * mmU * mmU * 0.5 mmU universes in the diameter of this black hole. For each millimeter in this scaled-down black hole, there are mmU groups of mmU universes, with another half layer of nesting, but then you can divide that by 1000. To fit the number of universes in the whole volume in a line end to end, repeat this process again, except without the extra half layer and division by 1000.

cjslep 7y ago

For relativistic effects I think you'll need to apply the Lorentz factor [0] [0] https://en.m.wikipedia.org/wiki/Lorentz_factor https://en.m.wikipedia.org/wiki/Lorentz_factor