71–80 of 115 posts
I'm surprised Oskar van Deventer (YT: OskarPuzzle, https://oskarvandeventer.nl/ ) hasn't done more than 10^9:1 reduction.
I think this calls for 10^10^2 reduction using a series of cycloidal drives (34x 1000:1 should do it just fine)
Earlier quoted context omitted.
But the construction of it ??! It's one thing to put the numbers down, we know they round out nicely, but in my head this needs a very minute level of construct engineering.. For example: If 1 gear(cog?) is 1/1000 of a cm out, would that not effect the ratio over this 'distance' ? Edit: I might just sound like an idiot right now, but when we get down (or up) to these numbers, i can't help but feel manufacturing numbe…
As in the video, they just reuse the same ratio gears in cascade?
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…
I bet the zero point fluctuation motion of the last gear dwarfs the deterministic motion.
If you made this in a physics engine could you get the last gear to spin in a reasonable time and not break the simulation?
In fact, I can simulate this in text form with no floating-point or number theory issues at all. Here's the simulation:
> For every 10^100 turns of the first gear, the last gear will move about one full turn.
Earlier quoted context omitted.
When all the slack has been rotated out, it would require a tremendous amount of force to rotate that last gear in reverse. After all, the system wants to amplify the gear speed from that end.
I think parent’s idea would work if the system was mirrored and duplicated. Then you could attach the two the them together with the stationary gears in the middle
Earlier quoted context omitted.
It's pretty simple: for each pair of gears that mesh with each other, one has ten times as many teeth as the other, which gives a ten to one reduction. If you chain 100 such reductions, you get 10^100 to one reduction.
But the construction of it ??! It's one thing to put the numbers down, we know they round out nicely, but in my head this needs a very minute level of construct engineering.. For example: If 1 gear(cog?) is 1/1000 of a cm out, would that not effect the ratio over this 'distance' ? Edit: I might just sound like an idiot right now, but when we get down (or up) to these numbers, i can't help but feel manufacturing numbe…
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…
What would be the torque required to spin the slowest gear 1/3 rotation per day? Let's assume the same radius: 2cm.
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
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…
*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.