@Alpacaslapper You are correct it takes energy to make the tunnel, but depending how long the tunnel is used for this energy invested can potentially be recouped over time due to the energy efficiency of the system. Rate of evaporation can be increased by making the surface area of the pools as large as possible, for example, stacks of very wide, long, but shallow pools. Also increasing the temperature of the water hastens evaporation, and the rock surface temperature at 2.5 miles in the deepest gold mine in the world is ~150F. Increasing the # passengers per vehicle, increasing tunnel depth, increasing water volume and increasing regen energy harvest per tank can all increase the efficiency further. Because the vehicle consumes the gravitational potential energy of the water during the journey, and because it requires solar and geothermal energy to lift the water out of the tunnel, it isn’t perpetual motion, or free lunch.
This conversation wants me to introduce more and more entropy into the universe with every comment 
The velocity involved (>2km/s) likely rules out wheels, so magnetic pseudo-levitation in a vacuum might become necessary:
“ Mechanical constraint (pseudo-levitation)Edit
With a small amount of mechanical constraint for stability, achieving pseudo-levitation is a relatively straightforward process.
If two magnets are mechanically constrained along a single axis, for example, and arranged to repel each other strongly, this will act to levitate one of the magnets above the other.”
Always turn off the vision positioning when flying in fog, found this out the hard way also
Or cover them with tape
How would you apply brakes?
I think if the whole tunnel was a vacuum with pseudo magnetic levitation you can actually recover a lot (~70%) of the passenger vehicle’s kinetic energy with regen braking to use for something else at the destination. maybe charge your skateboard?
Your magnetic constraint system would also need to be able to adapt to various loads and position of the load depending on passengers and cargo, also the weight of the vehicle before and after dumping the water.
I think the only part that requires pseudo magnetic levitation might be for the passenger vehicle in transit in a vacuum tunnel at constant speed on or close to the surface because this is where there is the least dynamic loads.
(professor_shartsis)
I’m still not very clear on understanding the difference between the passenger vehicle pushing off the tank or off the ground when at the bottom of the ramp.
(professor_shartsis)
I do know that it takes proportionately more electrical power (even in a vacuum) for an electric vehicle to produce the same amount of thrust when it is traveling at higher ground speeds than when it is at lower ground speeds.
(physics forum moderator)
That is a good starting point for a discussion that avoids the wall-of-numbers effect.
If one pushes off the ground the interaction requires energy. Part of the energy goes into the vehicle. Part of the energy goes into the ground. You can calculate how much by looking at the work done by the interaction force for each.
The ground does not move. Force applied times distance moved is zero. No work is done.
The vehicle does move. Force applied times distance moved is non-zero. Work is done.100% of the energy applied in the interaction goes into the vehicle.
If one pushes off from a 500 mile per hour trailer, things are roughly similar. Energy is still involved. Part goes into the trailer, part into the vehicle. We can look at work done.
The trailer moves in a direction opposite to the force applied on it. The work done on the trailer is negative. It loses kinetic energy as a result.
The vehicle moves faster than before. Accordingly, the work done on the vehicle is increased relative to the push-off-from-the-ground case.More than 100% of the energy applied in the interaction goes into the vehicle. The excess is deducted from the kinetic energy of the trailer.
There is no free lunch. Energy is conserved in either case. The total increase in kinetic energy from the push-off is equal to the energy provided (by piston, muscles, batteries, engine or whatever).
Here’s a (not so) weird theory :
Ever thought of dampening wobbles on classic trucks with d30 or similar material? Like a d30 sleeve around the bushings.
Or a d30 core inside the bushing.
Way to kill spiky loads. Unless it’s a load from @Brenternet.
I’ll have you know my loads are smooth as butter. In fact, I am so smooth I slide uphill.
Can confirm… one of those things 
(professor_shartsis)
I was looking into how much of the kinetic energy in the vehicle can potentially be recovered at the destination. The following numbers likely sound absurd but I created them with the same aforementioned formulas.
If the “trailer” is 10^4 kg, the passenger car is 100 kg, no regen braking is used at the bottom of the 30 second vacuum free-fall ramp, the mechanical “push” off the trailer on the flat section on the bottom is 43.7 GJ–
-The passenger vehicle’s KE after returning to the surface is 44.1 GJ (12250 kilowatt hours)
-The passenger vehicle’s velocity after returning to the surface is 29.7 km/s
-The tank’s velocity at the bottom of the ramp is 0 m/s
-Assuming the passenger vehicle loses no KE along the journey (maglev in vacuum), and regen braking is used at 70% efficiency at the destination, 30.4 GJ is recoverable from the vehicle and available for a second impulse
-Assuming . 432 GJ is used to lift the trailer back to the surface, about 29.9 GJ is recoverable
-The total energy non-recoverably consumed while accelerating and decelerating the vehicle was 13.8 GJ
-In this case the vehicle’s kinetic energy on the surface (~ 44.1 GJ) is a factor of ~3.19 times larger than the total energy non-recoverably consumed while accelerating and decelerating the vehicle (~ 13.8 GJ)
(physics forum moderator)
Have you looked at the Carnot cycle and at Carnot’s theorem? It appears to me that the numbers for your system are near the limits. If you bring in a second stage, as is done in a combined cycle power plant, to recover some of the energy unused at/by the primary stage, you might get an even better efficiency, but I don’t see how you could do that practically, and I think that you’re already nearing the maximum.
I went ahead and made a downloadable excel spreadsheet calculator for doing land based oberth maneuver calculations:
Here we have a 100000kg passenger vehicle with a 20kg trailer containing 200kg of water, perched at the top of a vertical 30 second drop in a vacuum, with a curved section of track at the bottom leading to a flat section that leads to the destination where there is a second curved section leading to a second vertical section…
At the bottom of the ramp, the passenger vehicle exerts a mechanical impulse equivalent to 2.65kWh, pushing off the 658mph trailer instead of the ground, which brings the trailer to a halt on the tracks…
^Assuming the passenger vehicle is using maglev in vacuum, and therefore does no work while coasting, the passenger vehicle is traveling 659.55mph in a straight line at the bottom of the ramp after the mechanical impulse. Regen braking (@ 70% kinetic to kinetic conversion efficiency) of the passenger vehicle at its destination (after climbing a second vertical ramp to the surface) results in (after the empty trailer is lifted), a factor of 130% as much usable energy recovered from the vehicle than was exerted in the mechanical impulse at the bottom of the ramp, a net surplus if the geothermal and solar energy required to evaporate this water from under the ground is ignored. The surplus comes from the lowered gravitational potential energy of the 200kg water at the bottom of the tunnel. If the full trailer is lifted with the regen braking energy instead of geothermal and solar, then only 40% mechanical impulse energy is recoverable. If the full trailer is lifted from the recovered regen braking energy, then a total of 1.58kWh energy was “unrecoverably” consumed accelerating and decelerating the 100000kg vehicle to 659mph. If the 200kg water is left at the bottom of the tunnel and allowed to evaporate naturally, then an “excess” of 0.8kWh energy above and beyond the total mechanical impulse energy is obtained from regen braking the vehicle at 70% conversion efficiency at the destination, even after some of the recovered energy is used to lift the empty trailer. The kinetic energy of the passenger vehicle after the mechanical impulse at the bottom of the ramp is ~760x greater than total energy “unrecoverably” consumed, after lifting the full (not empty) trailer all the way back to the surface, using the recovered kinetic energy from the regen braking after the journey.
I’ve been waiting all my life with the idea of a flying longboard, something with propellers, height would be controlled with a handheld remote which would increase or decrease height as you push/pull the trigger, but a gps would keep it stable at a given height.
Moving front and back would be controlled by the weight distribution as a onewheel, same for steering.
I’ve seen development already for this device but thee mudsr be a forums for this kind of ideas at the moment.
Can’t wait to get on one of these devices
I’ve been wondering if its possible to attach one big motor to a dual vesc to push more amps.
are you talking about something like this? it already hurts too much when i fall from a few inches on top of the board. 
No cogging is not the same as no drag, I think. It can still have drag but it’s just smooth and uniform. Like @hummieee said, anytime you have permanent magnets moving over a conductor there will be eddy currents which create heat and that heat is taking rotational energy away from the system.
Is it more efficient to go for a higher kv or a lower gear ratio to achieve a higher top speed