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I've been researching a land vehicle concept that takes advantage of a maneuver more commonly used in spacecraft propulsion-- the oberth maneuver, and an underground tunnel is the perfect environment for it.

I hope it can be built one day because it uses only sustainable fuel- namely a combination of gravitational potential energy, solar, and geothermal.

It requires an underground tunnel, magnetic levitation in a vacuum, like the hyperloop, but uses a different form of propulsion, and the tunnel has to be deep.

In essence it's shaped like a giant underground skateboard ramp, with an electric motor that pushes off a moving water tank instead of the ground to go faster than an airplane.

I’ve done some back of the calculations to compare a Tesla Model S vs a land based oberth maneuver vehicle of the same mass & available energy. These are rough numbers just to give an idea of the potential performance and compare with an automobile.

Long story short with it’s lithium battery pack partially charged up with 45kWh of energy, at roughly 70mph the 2250kg Tesla can do 135 miles in about 115.2 minutes (3mi/kWh). At 13.3 cents per kWh, it costs $5.98 for the energy. The ratio of kinetic energy to total kWH consumed with the Tesla on this trip is about 0.67%. (2250kg @ 70mph has 0.306kWh kinetic energy vs 45kWh consumed on the trip)

The land based oberth maneuver vehicle also weighs 2250kg and has 45kWh energy and follows a ramp (maglev in vacuum) which is 4.412km deep (equivalent to 30 second freefall in vacuum on Earth). It consists of a 1350kg passenger section (16.22kWh gravitational potential energy), 750kg water (9.01kWh gravitational potential energy), and a 150kg tank (1.8kWh gravitational potential energy) and 18.03kWh of electromechanical potential stored in capacitors in the track. The vehicle coasts down the ramp reaching about 294m/s at the bottom (657mph). The water tank is ahead of the passenger section on a long tether. On the flat section at the bottom of the ramp, the passenger section "reels in" then releases the tank with the 18.03kWh electromechanical potential, bringing the tank to a halt on the tracks (from conservation of momentum), transferring all its kinetic energy plus the mechanical impulse to the passenger section. After the mechanical impulse at the bottom of the ramp between the tank and passenger section, the 1350kg passenger section is traveling 1096mph, can go 135 miles in 7.38 minutes (about 15.6x faster). At $0.004/gallon the water costed $0.79 (less than 1/7th the energy cost). The ratio of kinetic energy to total kWH consumed with the land based oberth maneuver vehicle is about 100% (~99.3% more of the vehicle's potential energy was converted to kinetic energy).

After traveling up a second ramp back to the surface, the passenger section still has 28.8kWh of kinetic energy (which is all the mechanical impulse + the kinetic energy the water tank had at the bottom of the ramp). The water is emptied from the tank at the bottom of the ramp and only the empty tank is lifted (the water is left to evaporate, and rock temp increases with depth). Factoring regen braking with 70% kinetic-to-kinetic efficiency at the destination, and using some of the recovered energy to recharge the capacitors, and some of the energy lift the empty tank, there is still a 0.36kWh excess of recovered energy from the regen above and beyond the energy used to push the vehicle at the ramp bottom. The excess energy comes from the lowered gravitational potential energy of the water left to evaporate at the bottom of the tunnel. It isn't perpetual motion because the vehicle requires gravitational potential energy to move itself forward, and geothermal and solar energy to lift the water out of the tunnel.

In summary the hypothetical land based oberth maneuver vehicle can theoretically go 15.6x faster, for less than 1/7th the energy cost, with the same amount of energy and mass as a Tesla Model S.

Suppose we compare the land based oberth maneuver to simply using a linear accelerator in a vacuum tunnel on the surface-- but using the same amount of water (750kg + 150kg tank) to move a heavier 40000kg passenger/freight section (shipping container) instead of 1350kg passenger section in the previous example.

In this case with more freight, from conservation of momentum, the mechanical impulse that stops the same mass water tank is only 11.06kWh instead of 18.03kWh in the original example. The gravitational potential energy of the water + tank at the surface is still the same - roughly 9.8kWh.

At the bottom of the 4.412km ramp after the mechanical impulse the 40000kg load has 502.7kWh of kinetic energy and is traveling 672.9mph. Compared to the push of 11.06kWh and the 9.8kWh gravitational potential of the water, most of the kinetic energy at the bottom of the ramp comes from the gravitational potential energy of the load itself.

To get the same load up to the same velocity with linear accelerators you'd need a lot more stored electrical energy-- at least 502.7kWh factoring 0 losses.

Factoring the 30% unrecoverable energy loss after regen braking the linear accelerated load at the same 70% kinetic to kinetic efficiency--

The energy irrecoverably lost using the linear acceleration would be about 150.81kWh.

With the land based oberth maneuver, if we factor capturing recovering 70% of the kinetic energy at the surface after the load climbs a second ramp to the surface, and using all of it to lift the empty trailer AND the water, then it takes an additional outside energy input of 6.56kWh to get the water out of the tunnel and recharge the capacitors.

So in the scenario with a 40000kg load, to get the same speed with linear accelerators requires electrical storage/loss of 502.7kWh/150.81kWh for the same performance vs 11.06kWh/6.56kWh storage/loss with the land based oberth maneuver (95.66% less energy lost per load).

The water at the bottom of the ramp would be emptied into and evaporate from a separate chamber connected to the surface at atmospheric pressure. In many places the heat of the rocks at 4.4km depth is more than sufficient to boil water & generate steam.

Since evaporation is proportional to surface area and temperature, I had imagined it being emptied into shallow but wide pools, at a depth where the rock temperature is significant.

As a side note, I've calculated the negative G's experienced by the passenger can be reduced to comfortable levels while still achieving the same top speed by using a less steep ramp to the same depth, and the acceleration G's during the pull can be reduced to comfortable levels by using a longer tether.

Where:

M_t = 240 = Trailer Tank + Water Mass ( K i l o g r a m s )
M_v = 10^3 = Passenger Vehicle Mass ( K i l o g r a m s )
T_d = 30 = Free Fall Duration ( S e c o n d s )
V_i = 0 = Initial Velocity Before Free Fall ( m / s )
G_e = 9.80 = Gravity Acceleration at Earth’s Surface ( m / s^2 )
W_i = 1.28 ∗ 10^7 = Mechanical Impulse Energy ( J o u l e s ) That Brings Trailer to 0 Velocity At Ramp Bottom

W_i = M_t * V_i * T_d * G_e + ( M_t * T_d^2 * G_e^2 + M_t * V_i^2 ) / 2 + ( M_t^2 * ( V_i + T_d * G_e )^2 ) / ( 2 * M_v )

W_i = 240*0*30*9.80665+(240*30^2*9.80665^2+240*0^2)/(2)+(240^2*(0+30*9.80665)^2)/(2*1000)

W_i = 1.28 ∗ 10^7 Joules

(this assumes vertical ramp, but as long as it is the same depth it works out the same energy regardless of slope)

The question is does the cost of digging the hole at such a depth justify the time and energy savings for long distance transit?
 

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So all you need is loads of 4.4 km deep holes and a load of sick bags for the passengers.

If I want to pop to the shops do I still need to go down the hole and back?
 

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The question is does the cost of digging the hole at such a depth justify the time and energy savings
What about the energy used in digging the hole, and lugging all the spoil back up to the surface? (That's a whole lot of potential energy.)
Also you'll have to be continuously pumping to maintain the vacuum, even if the hole is perfectly airtight, because you keep chucking tonnes of water into it to evaporate. (Which also means you will partially lose your vacuum for a time after each transit.)

Sounds to me like a troll subject - but it's quite interesting as a thought experiment :)
 

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Discussion Starter #6
Also you'll have to be continuously pumping to maintain the vacuum, even if the hole is perfectly airtight, because you keep chucking tonnes of water into it to evaporate. (Which also means you will partially lose your vacuum for a time after each transit.)
This is the crucial part... the water is not emptied into the vacuum, but rather a separate chamber at atmospheric pressure connected to the surface. Geothermal energy then turns the water to steam, like a man-made geyser.



What about the energy used in digging the hole, and lugging all the spoil back up to the surface? (That's a whole lot of potential energy.)
Since the vehicle would travel at similar speed to an airplane, the energy expended digging the hole would have to be compared to using airplanes and aviation fuel for the same number of loads over the lifetime of the hole. Even comparing to linear acceleration at surface level, the hole saves considerable energy per load for the same airplane like performance.

So all you need is loads of 4.4 km deep holes and a load of sick bags for the passengers.
I think if the slope of the ramp is shallow enough (but to the same depth for the same performance), and if the tether is long enough, the positive and negative g-forces on the passengers can be reduced to reasonable levels.
 

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This is the crucial part... the water is not emptied into the vacuum, but rather a separate chamber at atmospheric pressure connected to the surface. Geothermal energy then turns the water to steam, like a man-made geyser.
So the 'load' passes.the tank after reeling it in and leaves it at the bottom?
 

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I've enjoyed reading this - thank you for posting it. It's an interesting thought experiment, and I enjoy a good physics puzzle.

You've designed a very inefficient system to extract geothermal energy. Gravitational potential energy (GPE) of the water is turned into kinetic energy (KE) firstly in the water itself, and thence into additional energy for the payload. The water remains at the bottom, and relies on geothermal energy to raise it back to the surface. The GPE of the water has been used, but now the water at the bottom must be removed to keep the system in a steady state. This is achieved by using geothermal energy. All the GPE produced by the fall of the water has to rely on geothermal energy to raise it back to the surface. It would be far more efficient to build a binary geothermal energy plant, and turn the geothermal energy directly into electrical energy.

Your argument becomes a bit fuzzy on the 'linear accelerators alone'. For an 'on the surface' version, you should be considering a like-for-like 1350kg payload being accelerated to a maximum velocity (at 675 mph this would require 17kWh, at 1125 mph it would require 40 kWh), and considering the losses in the generation and regeneration of this energy (at your estimate of 70% efficiency for each process, this would be a loss of 8kWh or 20kWh respectively.) This energy could be sourced from that geothermal energy plant that you'd built much more cheaply than a tunnel and a cavern!

The Oberth manouvre is important for spacecraft because they are isolated. All their propellant has, itself, got to be propelled, and the trick of gaining KE by leaving your propellant gases in a GPE well, such as that of Jupiter, increases its effectiveness. On the surface of the Earth, however, our vehicles generally don't use rocket or jet propulsion.

Though I am being critical, and do think this particular idea is bit a silly, I think it's great that you've thought it through and been prepared to present. Please don't feel patronised if I say 'well done!'

: o )
 

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Years ago at school I remember solving an applied maths problem, which came up with a slightly surprising answer!

Assume you create a frictionless hole between 2 points on Earth's surface, using a perfectly straight hole. You place your frictionless carriage stationary at one end of the hole, adn let go. How long does it take to arrive at the other end? All teh usual assumptions made, Earth is a perfect sphere, etc etc.

I forget the precise answer, but I think it was in the region of 2 to 3 hours. The surprising bit was that the period of this oscillation was constant, and independent of how far apart A & B were! So you could travel from UK to Australia in the same time, assuming yo have the boring technology !!! (Fat chance ot that!).

So using this design, you could drill a hole from London to Edinburgh, create a vacuum inside it & have a frictionless maglev rail. Journey times would be quite acceptable, and the fuel costs relatively small. Theoretically they could be zero, if it's possible to use permanent magnets to create the lift, rather like that floating biro desktop-toy thing I keep meaning to get myself.

Sadly, in the real world, all you'll need is a minor earthquake to break a seal somewhere, and disaster will ensue, followed by court cases & a total loss of confidence in the whole system.
 

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Discussion Starter #12 (Edited)
Your argument becomes a bit fuzzy on the 'linear accelerators alone'.
Let me try to clear up the comparison with linear accelerators...

Suppose we want to accelerate a 40000kg shipping container to 672.91mph horizontal speed in a maglev vacuum tunnel.

The two options are a surface linear accelerator in a straight line, or using a tunnel that first drops to 4412.99 meters depth, continues horizontally for some distance, and then climbs a second 4412.99 meter ramp back to the surface.

Both systems have capacitors in the track with enough storage that if called upon they would able to transfer 502.73kWh of mechanical energy to the container.

In this scenario, we aren't using any geothermal energy to reset the system for the next container, but rather buying electricity from the power company for $0.13/kWh to reset each system.

At the surface destination, regen braking is used to capture the surface kinetic energy of each container. The regen braking has 70% kinetic to kinetic efficiency which means 70% of the container's surface kinetic energy can be recaptured and transferred to the next container cycle. It also means the "70%" regen braking efficiency also factors all of the energy conversion losses incurred during the acceleration of the next container, because 70% of the surface kinetic energy of the first container will be directly transferred to the next container.

With the linear accelerator, since the target speed 672.91mph is 300.81m/s, and the kinetic energy of a 40000kg container traveling 300.81m/s is 502.73kWh, it requires all the energy stored in the track capacitors to get the container up to speed.

At the destination, the regen braking is able to capture 70% of the linear accelerated container's 502.73kWh kinetic energy, which is 351.91kWh. Since the next container also needs 502.73kWh, it means we have to buy 150.82kWh from the power company to make up for the difference and reset the system, which costs $19.60 at $0.13/kWh.

The 40000kg land based oberth maneuver container has a 150kg tank + tether filled with 750kg of water.

The water, tank and tether have 10.82kWh of gravitational potential energy at the top of the ramp. The shipping container itself has 480.9kWh of gravitational potential energy at the top of the ramp. It also takes a transfer of 11.06kWh of mechanical energy from the capacitors used as an impulse between the container and the tank/tether to cause the water tank to come to a stop at the bottom of the ramp.

After the 11.06kWh mechanical impulse at the bottom of the ramp, the 40000kg container is traveling at the 672.91mph target speed (300.81m/s) with 502.73kWh of kinetic energy, which is all of the GPE from the container, tank, water, and tether + the energy from the mechanical impulse.

After climbing a second ramp back to the surface, the container still has 21.88kWh of kinetic energy, which equals all of the 11.06kWh mechanical impulse energy plus all of the 10.82kWh of gravitational potential energy the tank/tether had on the surface.

After regen braking at the surface at 70% kinetic to kinetic efficiency, 15.31kWh of the container's surface 21.88kWh kinetic energy is recaptured.

From the recaptured 15.31kWh, 11.06kWh is used to recharge the capacitors to their initial state, leaving 4.25kWh.

Since it will take 10.82kWh to lift the water, tank, and tether out of the tunnel, all of the remaining 4.25kWh from regen is used to partially lift the water, tank and tether out of the tunnel.

Since it will take a further 6.57kWh from the power company to finish the job of lifting the water, tank and tether out of the tunnel, at $0.13/kWh it costs $0.85 to reset the system.

Comparing the energy cost from the power company of resetting each system for the next container, the linear accelerator costed $19.60 to reset the system compared to $0.85 for the land based oberth maneuver, without using any geothermal and the same price per kWh to reset both systems.

So for the same top speed performance and without using any geothermal power in the scenario, the energy costs are 23.05x higher when using a linear accelerator at ground level compared to the land based oberth maneuver in an underground tunnel.
 

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If you can get an air tight EV into a vacuum tunnel with a horizontally flat road, then it'd use virtually no energy to get to its destination, assuming the cabin air lasted long enough for the occupants. The minuscule energy it uses could be had from your geothermal scheme, avoiding all the mining effort to drill a deep scary tunnel.

The same problem will befall both parties; if there is a fault they will all suffocate.

Apart from that minor safety concern, maybe the reverse scenario is a good idea; use solar power to generate hydrogen fuel and use rockets to get where you are going?
 

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the energy costs are 23.05x higher
That's a dodgy bit of selling. 23 x b*ggerall is still not much.
With a 7+ minute transit a trip every 10 minutes is the most you'll get realistically, which is 144 per day. So the energy saving is $2,700 a day. By the time you add up staff costs, power for handling at each end and payments on the capital cost (which will be billions) that cost saving is going to be pissing in the ocean - and a hugely risky project to boot.
 

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Discussion Starter #16
If they're going to be building hyperloops underground anyway, what % more does it cost to dig the tunnel a bit deeper?
 

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Discussion Starter #20
My last calculation ignored the monetary value of gaining ready access to the geothermal resource which would be present along the entire route.
 
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