The person who made this did a one minute video on it answering this question (below).
The tl;dw version is he cannot move the last gear at all.
Hahaha! That was well done.
I was just wondering about that. Not surprising the last gear won’t move. I wonder if the material is strong enough to ever transmit movement to the last gear when driven as intended.
By the way, the LEGO construction incorporates a “compound planetary gear”, a means of obtaining high ratios because the ratio is the differential of the ring gears. Machinery’s Handbook has the mathematics if you care to look it up.
Chrysler used compound planetary gears in radar sets and tank transmissions during WWII The first was to create a high gear reduction with minimal backlash. The second was to create a large reduction in limited space.
There are also “split gears” to minimize backlash.
And thus you demonstrated why it’s necessary for a car to start in a very low gear and shift gears as you gain speed 
And then if you hook the output of the gear-down train into the input of the gear-up train you’ll have an over-center perpetual motion machine.
Wait, what??/?!!/!? 
Actually, that’s an interesting scenario. Would the end of the gear-up train recover the input torque of the initial drive? Would you see the output shaft of the final drive turning at the speed of the input motor?
In a perfect world, it seems like it should. In the real world, I suspect torque losses (lash, rolling resistance, etc.) would eat up the transmitted torque.
Just like in abstract mathematics, multiplication and division are perfectly reversible to infinite precision, but in real life you eventually round down to zero.
We saw in the video posted above (post 21) that you cannot turn the last gear. It seems the whole thing only works one way.
Assuming the whole mechanism is absolutely frictionless, it would work in either direction. However, the drive ratio means that any torque applied to the last gear is reduced by a factor of 10100 on its way to the first gear. Grab that last gear with a firm grip, and maybe you can put 10 lb-ft of torque on it. That means the first gear experiences a torque of just 10/(10100) lb-ft. Most folks are familiar with the basic “force equals mass times acceleration” rule in linear scenarios (like a jet accelerating down a runway), but there’s an analogous equation for rotating systems: “torque equals mass moment of inertia times angular acceleration.” That first gear has some non-zero rotating inertia, and when you apply a very tiny amount of torque to it, it will experience a very tiny amount of angular acceleration. Also, since every intermediate gear has its own inertia, each one will absorb some of the torque applied to it by accelerating, passing on only part of it to the next gear. So the net torque that actually gets applied to the first gear will be extremely small - much, much less than 10/(10100) lb-ft. So if there’s no friction, the mechanism definitely will work - you will be able to get the first gear rotating by applying a torque to the last gear. It’s just that it will take an extremely long time to accelerate that first gear enough to achieve an RPM that you could notice with your unaided eye.
Izzat with or without a treadmill?
I’ll let myself out.
I’m curious when this will start to visually break the concrete.
Never.
The first gear will be worn to a nub before the last gear starts to move.
And of course, “frictionless” is an ideal that’s never realized, either. We can build some devices with really, really low friction, but there’s “really really low”, and then there’s “less than 1/googol”.
So, we build two of these sets out of pure frictionless rigidium. I attach the “ends” of them together, so that both reduced gears are in contact. I then spin the first gear. Does the last gear, after being geared down and back up, spin instantly* and at the same rate that I’m spinning the first?
*ridgidium has perfect rigidity, which means the speed of sound in it is infinite.
Exactly what I was asking.
The reduction should be perfectly reversible by a second set of the geartrain connected in reverse. You should recover the full input torque, at the same rotation speed, on the final output shaft even if the entire middle of the combined geartrain appears completely immobile.
Friction, lash, and elasticity should probably prevent this from happening IRL.
Yeah, in a problem like this, it’s not just friction. All of the inconvenient real-world factors that we usually pretend are small enough to ignore, are amplified to the point that we can’t ignore them.
Should work fine, but reversing the two units - so that the reduced gears are at both ends of the entire assembly - will not. This is because the intermediate gears (particularly the ones at the middle of the entire assembly) will be spinning extremely fast, as described in my earlier post, and will require a lot of mechanical work from your end of the system in order to achieve those speeds.
I am not going to do the math, but I’m pretty sure you are severely understating the case, in that the intermediate gears would probably be breaking lightspeed by a pretty comfortable margin.
One thought - gears don’t only transmit rotation and torque, they also produce a force that pushes the gears apart. So as the torque approaches - so does the force pushing on the last gear’s axle.