Energy Efficiency of Compressed-Air-Powered Parabola Riding Vehicle?

[metastable]

Why not? The chemical reactions in a rocket use electromagnetic forces to accelerate mass.

The 200lb vehicle is also using EM forces to accelerate mass, so is it not considered a rocket and not considered an Oberth maneuver?

 

[Dale]

I explained why not in post 103.

 

[metastable]

I read that post, but changing a 20,000lb tank’s velocity from 600mph to 500mph is roughly 100 million joules, 70million recoverable. If this same energy is used to accelerate a 200lb mass from 0mph (pushing off of the ground not a 500mph tank), it gets to roughly 2700mph in a vacuum, if I’ve done my math properly.

 

[Dale]

Ok. How does that address my objections?

 

[metastable]

I thought you were saying that decelerating 20,000lbs from 600mph to 500mph to obtain the “rocket fuel” in the form of stored EM energy to later use on a reaction mass either moving 0mph or 500mph would be counterproductive to the goal of accelerating.

Let’s say I convert the 70million recoverable joules from dropping the 20,000lb tank down the 2.5mile shaft and decelerating from 600mph to 500mph into rocket fuel via electrolysis, and then I use the rocket fuel on the pinewood derby cars, is that acceptable? Then the only fuel is still water, but we’ve lost efficiency through the electrolysis.

 

[Dale]

That would be fine. As long as you don’t convert any of the mechanical energy to internal energy during the part that you want to call an “Oberth maneuver”

 

[metastable]

Here are my findings (though I am not sure I have done all the math right, but it's based on conservation of energy):

Vehicle: 200lbs
Reaction Mass: 20,000lbs Water Tank
Impulse Energy: 99,151,458 J

Ramp: 2.56 mile vertical shaft in vacuum (29sec @ 1g), with horizontal sections at top and bottom, connected by smooth curves. The flat section on top gives one of the vehicles a place to exert its impulse before entering the ramp, and the flat section on bottom gives both vehicles a place to go after exiting the ramp. One of the vehicles exerts its impulse before entering the top of the ramp, and the other accelerates from gravity on the ramp and then exerts its impulse immediately after entering the flat section at the bottom of the ramp.

Max velocity of "Impulse at Top" 200lb vehicle along the flat section at the bottom: 2422.43mph, 53,193,796 J
Max velocity of "Impulse at Bottom" 200lb vehicle along the flat section at the bottom: 5659.76mph, 290,372,128 J

 

[Dale]

Without having gone through the arithmetic myself, the general trend looks right to me. So what is your conclusion? Does the Oberth effect apply for land based vehicles? What are the requirements?

 

[metastable]

Well assuming my math was right, my first observations are it appears that it is more efficient in terms of kinetic energy gained for electrical energy consumed to push off of a large mass which has been gravitationally accelerated, rather than using the same amount of electrical energy to simply push off the Earth itself, as would be done with a standard car.

 

[metastable]

And according to this:

https://en.m.wikipedia.org/wiki/Mponeng_Gold_Mine
^The rock at 2.5 miles depth is 151F/66C, which should help with the evaporation after dumping the water. From what I understand, the pools would need a large surface area and an airflow path to the surface to evaporate most quickly.

 

[metastable]

Although my method gave what seems like could be a reasonable answer, I was uncertain on many of the steps:

PE 20,200lb tank + vehicle at 2.56 mile height in vacuum: = 370,531,290 J = 20,200lb @ 636.16mph @ 29 sec drop

KE 20,200lb tank + vehicle at bottom of 2.56 mile drop in vacuum: = 370,531,290 J = 20,200lb @ 636.16mph @ 29 sec drop

Electrical Impulse Energy = 99151458 J**

Kinetic Energy of Separation Between Tank and Vehicle Per Vehicle = (99151458 J /2) = 49,575,729 J

Kinetic Energy of 20,000lb Tank Before Separation (20,000lb @ 636.16mph) = 366862663 J

Kinetic Energy of 20,000lb Tank After Separation (366862663 J - 49,575,729 J) = 317286934 J

Velocity of 20,000lb tank after separation: 591.625mph (20,000lb @ 317286934 J)

Energy of Tank After Separation: 317286934 J

Kinetic Energy of 200lb vehicle after separation: (99151458 + 370,531,290 J - 317286934 J) = 152395814 J

Velocity of 200lb vehicle after separation (impulse at bottom): 4100.22mph (152395814 J @ 200lb)

---------------------------------------

Velocity of 200lb vehicle w/ Impulse at Top 49,575,729 J (1/2 Total Impulse Energy)= 2338.59mph

This is roughly 1.53 seconds per mile so in a 2.5mile vertical ramp 1.53s/mi * 2.5 miles * 9.8m/s^2 = 37.48m/s change in velocity on the ramp

37.48m/s is 83.84mph

2338.59mph + 83.84mph = 2422.43mph

Velocity of 200lb vehicle at bottom of ramp (impulse at top): 2422.43mph

Kinetic Energy of 200lb vehicle after separation: 53,193,796 J (200lb @ 2422.43mph)

---------------
Side Note:
**This 99151458 J electrical impulse energy is equivalent to (in a completely separate location and time) decelerating a separate 20,200lb vehicle from 636.16mph to 500mph, and obtaining the resulting impulse energy at 70% conversion efficiency

 

[Dale]

So, just some advice on working problems like this.

First, keep it as variables as long as possible. It will be much easier to find errors if you do most things algebraically and only substitute in numbers at the very end. Also, use standard variable names and clear definitions.

Second, use SI units exclusively. Particularly in a problem like this where you are in charge of all of the quantities, it makes things easier.

Third, use no more than 3 significant figures in any computation. If a given intermediate calculation requires more precision, then make sure to keep that part algebraic.

 

[metastable]

Conclusions

Using the same (202,000,000 J) amount of mechanical impulse energy at the top or bottom of the ramp, impulse at the top of the ramp resulted in 204,080,604.5 J KE at the bottom of the ramp while impulse at the bottom of the ramp resulted in 260,921,883.60 J KE at the bottom of the ramp, which is more than the total mechanical impulse energy.

Even after returning to the surface on a second ramp, the "impulse at bottom vehicle" had more kinetic energy than the total energy in the mechanical impulse, while the "impulse at top vehicle" had less.

The impulse at top vehicle did worse in terms of efficiency than a standard car pushing off the ground after returning to the surface, while the "impulse on bottom" vehicle did better.

---------------------

Vertical Drop Length: 4123.69 meters (29sec @ 1g - 2.56 miles)
Free-fall Velocity at Bottom of Ramp: 284.39 meters per second (636.16 miles per hour)
Water Tank Mass: 10000 kilograms
Passenger Vehicle Mass: 100 kilograms
Mechanical Impulse Energy: 202,000,000 J

-------------------

Impulse on Top (202,000,000 J):

m1 = 100kg
m2 = 10000kg
v1 = 2000m/s
v2 = 20m/s

m1v1 = m2v2

v2 = (m1v1)/m2

v2 = (100kg*2000m/s)/10000kg

v2 = 20m/s

m1 KE = (1/2)mv^2
m1 KE = (1/2)100kg*2000m/s^2
m1 KE = 200,000,000 J

m2 KE = (1/2)mv^2
m2 KE = (1/2)10000kg*20m/s^2
m2 KE = 2,000,000 J

m1 KE + m2 KE = 200,000,000 J + 2,000,000 J = 202,000,000 J = Mechanical Impulse Energy

Covering 4123.69 meters with initial 2000 meters per second velocity and 1 g acceleration takes 2.07 seconds.

2.07 seconds with 1 g acceleration is a velocity change of 20.3 meters per second

m1 velocity at ramp bottom with impulse on top: 2020.3m/s = 2000m/s + 20.3m/s

m1 KE (bottom of ramp, impulse on top) = (1/2)mv^2
m1 KE (bottom of ramp, impulse on top) = (1/2)100kg*2020.3m/s^2
m1 KE (bottom of ramp, impulse on top) = 204,080,604.5 J

-------------------

Impulse on Bottom (202,000,000 J):

Free-fall Velocity at Bottom of Ramp: 284.39 meters per second (636.16 miles per hour)

m1 = 100kg
m2 = 10000kg
v1 = 2000m/s
v2 = 20m/s

m1v1 = m2v2

v1 = (m2v2)/m1

v1 = (10000kg*20m/s)/100kg

v1 = 2000m/s

V1's actual velocity is 2000m/s + the 284.39 freefall velocity at the bottom of the ramp:

2000m/s + 284.39m/s = 2284.39m/s

m1 KE (bottom of ramp, impulse on top) = (1/2)mv^2
m1 KE (bottom of ramp, impulse on top) = (1/2)100kg*2284.39m/s^2
m1 KE (bottom of ramp, impulse on top) = 260,921,883.60 J

--------------------

Conclusions:

Max Velocity with no ramp: 2000m/s (4473.87mph @ ground)
Energy with 202,000,000 J Impulse on top: 200,000,000 J <----

Max Velocity "Standard Car" pushing off ground: 2009.97m/s (4496.19mph @ ground)
Energy with 202,000,000 J Impulse on top: 202,000,000 J <----

Max Velocity with impulse on top of ramp: 2020.3m/s (4519.28mph @ bottom)
Energy with 202,000,000 J Impulse on top: 204,080,604.5 J

Ground Level Exit Velocity with impulse on top of ramp: 2000.18m/s (4474.28mph @ ground)
Exit Energy with 202,000,000 J impulse on top of ramp: 200,036,002 J <----

Max Velocity with impulse at bottom of ramp: 2284.39m/s (5110.03mph @ bottom)
Energy with 202,000,000 J Impulse on bottom: 260,921,883.60 J

Ground Level Exit Velocity with impulse at bottom of ramp: 2266.64m/s (5070.33mph @ ground)
Exit Energy with 202,000,000 J impulse on bottom of ramp: 256,882,844 J <----

 

[Vanadium 50]

So, just some advice on working problems like this.

Followed by another wall-O-innumeracy.

@metastable, why do you bother asking questions when you ignore all the advice you get?

 

[metastable]

I used SI variables, standard variable names, and no more than 3 significant figures for the decimals, as instructed by @Dale

 

[A.T.]

...so is it not considered a rocket and not considered an Oberth maneuver?

Who cares how you call it?

 

[fresh_42]

I used SI variables, standard variable names, and no more than 3 significant figures for the decimals, as instructed by @Dale

Well, yes and no. The problem is, that your posts are difficult to read with all those numbers in it and especially without making use of LaTeX (cp. https://www.physicsforums.com/help/latexhelp/).

E.g.

Vertical Drop Length: 4123.69 meters (29sec @ 1g - 2.56 miles)
Free-fall Velocity at Bottom of Ramp: 284.39 meters per second (636.16 miles per hour)
Water Tank Mass: 10000 kilograms
Passenger Vehicle Mass: 100 kilograms
Mechanical Impulse Energy: 202,000,000 J

should read:
$L_d = |p_1-p_0|$ vertical drop from position $p_1$ to position $p_0$
$v_0$ velocity at $(0,0,0)$, the bottom of the ramp
$m_w$ mass of water tank
$m_v$ mass of vessel
$E_{kin}(p)$ kinetic energy at location $p$
$E_{pot}(p)$ potential energy at location $p$
followed by formulas. If the correct calculation is found, then we insert data, but only then. Btw. what is mechanical impulse energy?

This is only an example and instead of position, you could as well use time as parameter for the motions involved, or choose different coordinate system. I haven't followed the thread, only the last posts here which were about layout instead of physics. Of course you can write whatever and however you want. Unfortunately you cannot expect to communicate this way. For communication a certain layout (frame) is necessary, and each science has its own. In post #118 you failed to show how you calculated which quantity. It looks more like an Excel sheet of numbers and deliberate names where the calculations can only be seen in the edit box of a cell, only that we do not have this option.

The general way to present a problem is:

1. name all relevant quantities (in a way it is usually done, too)
2. add the data, i.e. figures, but do not use them, yet
3. start the calculation with those variables and explain what you thought and did
4. deduce a final answer in terms of variables
5. only now replace the variables with your data

Again, this is only meant as an advice how an efficient communication can be structured as a response on what this thread has become, not on what it was meant to be.

 

[metastable]

So have two “pinewood derby” vehicles both on a straight slope, both with identical rocket engines. One fires the engine at the top of the track and the other fires the engine in the middle. Determine the final KE of each at the end.

Thank you for all of the above input on my formatting, I will try to present everything from now on using the appropriate equations, variables, formatting and # of significant digits. (I obtained @Dale 's permission before posting this)

---------------------

Firstly a vertical drop depth of $29$ seconds at $9.8 m/s^2$ in a vacuum is chosen because it is roughly equivalent in depth to the world's current deepest goldmine ( $4.1km$ ) in hopes it can be achievable with present technology. There is a vertical drop, a flat section, followed by another vertical ramp, connected by smooth curved sections, in vacuum.

The drop depth, time and final velocity can be calculated with:

$a=(v_f−v_i)/Δt$
$a=2∗(Δd−v_i∗Δt)/Δt²$
$a=F/m$

where:

$a$ = the acceleration,
$v_i$ = initial velocity,
$v_f$ = final velocity,
$Δd$ = distance traveled during acceleration,
$Δt$ = acceleration time,
$F$ = is the net force acting on an object that accelerates,
$m$ = is the mass of this object.$Δd=4.12km$
$v_f=284m/s$
$m=m_{tan}+m_{pas}=10.1*10^3kg$
$m_{tan}:1*10^4 kg$
$m_{pas}:100kg$
$E_{impulse}:202MJ$

-------------------

The $E_{impulse}$ is calculated from a desired $2km/s$ velocity boost for the passenger vehicle, above and beyond its velocity from free fall at the bottom of the ramp from momentum conservation and kinetic energy equations:$m_1v_1=m_2v_2$
$v_1=(m_2v_2)/m_1$

$m_{1KE}=(1/2)mv^2$
$m_{2KE}=(1/2)mv^2$

---------
Results

Impulse on Top ( $202MJ$ ):

Covering $4.1km$ with initial $2km/s$ initial velocity and $9.8m/s$ acceleration takes $2.07$ seconds.

$2.07s$ with $9.8m/s$ acceleration is a velocity change of $20.3m/s$

$m_1$ velocity at ramp bottom with impulse on top: $2.02km/s$

$m_{1KE}$ (bottom of ramp, impulse on top) = $204MJ$

-------------------

Impulse on Bottom ( $202MJ$ ):

Free-fall Velocity at Bottom of Ramp: $284m/s$

$V_1$'s actual velocity is velocity obtained from the impulse in addition to the free-fall velocity at the bottom of the ramp:

$m_1$ velocity at ramp bottom with impulse on bottom = $2.28km/s$

$m_{1KE}$ (bottom of ramp, impulse on bottom) = $260MJ$

--------------------

Conclusions:

Max Velocity with no ramp: $2km/s$ @ ground
Energy with $202 MJ$ Impulse on top: $200 MJ$ <----

Max Velocity "Standard Car" pushing off ground: $2.009km/s$ @ ground
Energy with $202MJ$ Impulse on top: $202MJ$ <----

Max Velocity with impulse on top of ramp: $2.02km/s$ @ bottom
Energy with $202MJ$ Impulse on top: $204MJ$

Ground Level Exit Velocity with impulse on top of ramp: $2km/s$ @ ground
Exit Energy with $202MJ$ impulse on top of ramp: $200MJ$ <----

Max Velocity with impulse at bottom of ramp: $2.28km/s$ @ bottom
Energy with $202MJ$ Impulse on bottom: $260MJ$

Ground Level Exit Velocity with impulse at bottom of ramp: $2.26km/s$ @ ground
Exit Energy with $202MJ$ impulse on bottom of ramp: $256MJ$ <----------------
Conclusions

Using the same $202MJ$ amount of mechanical impulse energy at the top or bottom of the ramp, impulse at the top of the ramp resulted in $204MJ_{KE}$ at the bottom of the ramp while impulse at the bottom of the ramp resulted in $260MJ_{KE}$ at the bottom of the ramp.

After returning to the surface on a second ramp, the "impulse at bottom vehicle" had more kinetic energy than the total energy in the mechanical impulse, while the "impulse at top vehicle" had less.

The "impulse at top" vehicle did worse in terms of kinetic energy obtained per Joule of mechanical impulse energy than a "standard car" pushing off the ground after returning to the surface, while the "impulse on bottom" vehicle did better.

-----------------

Questions:

Are the conclusions correct? I'm having a hard time understanding what difference does it make if the passenger vehicles uses the mechanical impulse to push off the $284m/s$ tank at the bottom after free fall as opposed to pushing off the ground after free fall?

 

[metastable]

So what is your conclusion? Does the Oberth effect apply for land based vehicles? What are the requirements?

Btw. what is mechanical impulse energy?

This thread has covered a lot of territory, and the concept of the vehicle has changed slightly to simplify the equations, so here is my attempt to summarize the vehicle.

Short summary version:

- The 100kg passenger vehicle continues to its destination at more than 5000mph, having consumed no electricity or fossil fuels – fueled only by the gravitational potential energy of 10000+kg water at ground level

Long Version:

- The rider’s journey starts from ground level or slightly below ground level like entering a subway

- The downward ramp is a tunnel 45 degrees, 2.5 miles depth since its achievable with existing technology, the deepest gold mines are 2.5 miles depth

- The water is supplied from surface water resources, via gravity

- The rider enters a lightweight capsule (for example 100kg total passenger vehicle mass including rider)

- The aerodynamic tank is filled with water (for example 10000kg or ~2m edge cube equivalent)

- The tank + passenger vehicle are dropped down the ramp

- The tank has a “train” of small, lightweight roller cars with gear teeth in front of it and the passenger vehicle

- If the ramp were vertical and a vacuum, the tank + passenger vehicle reach ~636mph after ~2.5 miles

- There is a curved section of track at the bottom of the ramp leading to a horizontal section

- Once the tank and vehicle are at 636mph at the bottom of the ramp, regen braking kicks in decelerating the tank + vehicle to 500mph, and the energy is stored in capacitors in the track and utilized at 70% efficiency, a process which can produce a usable 100 million joule mechanical impulse

- The mechanical impulse is utilized to push the passenger vehicle off the 500mph 10000kg tank (not the ground), utilizing the “train” of gear teeth in front of the vehicle and co-moving with the tank

- If the mechanical impulse were instead 202 million joules (19kWh) pushing off the 636mph tank (which is all I have done calculations for so far – suppose more water and a deeper tunnel were used), even after traveling back to the surface, the passenger vehicle has 256 million joules KE (~5070mph) which is more kinetic energy than the total mechanical impulse. It is also more kinetic energy than a standard car would have (202 million joules - ~4496mph) from pushing off the ground with the same mechanical impulse at ground level, a difference of nearly 600mph. This is known in astronautics as an “Oberth maneuver”

- After the passenger vehicle pushes off the tank and returns to the surface via a second ramp, and is now traveling 5070mph on the surface, the tank no longer has sufficient kinetic energy to return all the way to the surface, so the water is instead dumped into a special chamber at the bottom of the tunnel, and the lightweight tank is retrieved with energy from harvesting more of the remaining kinetic energy in the tank at the bottom of the ramp

- The nearly 10,000kg water now sits in a high surface area chamber at the bottom of the tank and heats up because the rock at that depth can be 150f from geothermal, and so it evaporates back into the atmosphere via a special air channel dug to the surface, to be later recycled

- The 100kg passenger vehicle continues to its destination at more than 5000mph, having consumed no electricity or fossil fuels – fueled only by the gravitational potential energy of 10000kg water at ground level

- The velocity involved (>2km/s) likely rules out wheels, so magnetic pseudo-levitation in a vacuum might become necessary

https://en.m.wikipedia.org/wiki/Magnetic_levitation

 

[Dale]

fueled only by the gravitational potential energy of 10000+kg water at ground level

And whatever mechanism ejects the water during the “burn”.

 

[metastable]

And whatever mechanism ejects the water during the “burn”.

I was imaging the tank has lightweight roller cars in front with gear teeth on top, and an electric motor on the passenger vehicle or tank uses all the energy stored in the capacitors to accelerate the passenger vehicle at 9g by pushing off the linear arrangement of "gear teeth" which are rolling in front of the tank. the mechanical impulse would equate to 70% of the kinetic energy lost during the regen braking phase, based on the conversion efficiency of regen braking.

 

[Dale]

I was imaging the tank has ...

Sure, that is fine, I was just pointing out that you forgot to include it in the “fueled only by ...” list.

 

[metastable]

- The tank has a “train” of small, lightweight roller cars with gear teeth in front of it and the passenger vehicle

Ah sorry, I thought I included it but I wasn't very clear on how these would be used with an electric motor etc and the 9g rate of acceleration. 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.

 

[metastable]

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.

 

[jbriggs444]

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.

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).

 

[metastable]

Will the [...] parabola riding vehicle use less energy to reach the same distance in the same time as the standard electric 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.

Thank you, it is quite an interesting answer. So in summary are you saying it would be more logical for the passenger vehicle (considering the goals of such a vehicle) to push off the tank rather than the ground at the bottom of the ramp?

Also thanks to everyone who contributed to this thread and for also having great patience with me to refine the concept.

In summary, in terms of "electrical energy" or "fossil" fuel efficiency, the parabola riding vehicle uses less "electrical" or "fossil" fuel to get from point A to point B than a "standard car," because it uses a combination of gravitational potential energy, geothermal energy and solar energy to replace the electrical or fossil fuel, while reduction in passenger vehicle mass, reduction in rolling resistance via magnetic levitation, and reduction in wind resistance via vacuum tunnels can further improve the efficiency with which the gravitational, geothermal and solar fuel is used.

 

[jbriggs444]

Thank you, it is quite an interesting answer. So in summary are you saying it would be more logical for the passenger vehicle (considering the goals of such a vehicle) to push off the tank rather than the ground at the bottom of the ramp?

I am saying that it does not matter. The energy books balance either way.

If the passenger vehicle pushes off from the ground, more energy is available to be harvested by slowing down the trailer after the vehicle has moved on.

All that is achieved with this 2.5 mile Rube Goldberg arrangement is a sub-optimal way to harvest geothermal energy.

 

[metastable]

All that is achieved with this 2.5 mile Rube Goldberg arrangement is a sub-optimal way to harvest geothermal energy.

At some time in the future (in another thread) I’d like to conduct a study on the conversion efficiency of say a 2m edge length cube of water through a hydroelectric dam, then through the power distribution system, through a charger, through a battery, through a motor of an electric car versus 2m cube down 2.5 mile tunnel to capacitor to motor. There could be a difference in efficiency due to the difference in number of energy conversion steps as wells as differences in the efficiency of conversion for each step.

 

[sysprog]

I was out skateboarding the other day when I wondered if the efficiency of a vehicle attempting to cover the distance between point A and point B can be improved in the following manner.

If I've done my calculations correctly, baseline electric vehicle at constant 100.08 $mph$ uses 73.55 $N$.

The ACME Corporation cannot guarantee the safety of a rider aboard an electric skateboard operated at a speed exceeding $100\ mph$.

 

[metastable]

The ACME Corporation cannot guarantee the safety of a rider aboard an electric skateboard operated at a speed exceeding 100 $mph$.

My standard power source can't quite reach 100 $mph$... I'll have to compare the efficiency to solar/geothermal/nuclear power sources sometime.

 

[sysprog]

I guess maybe I should be a bit sorry for derailing your thread in the direction of the absurd -- I think that your question is legitimate and very well-articulated (even more well-articulated after @Dale discussed with you the use of $\LaTeX$ and you so well responded to the associated adjurements) -- but I also think that @jbriggs444 and others answered it well -- you know that you can't get more out than your doggies or other energy sources put in -- that pesky 2nd law and all that ...

 

[metastable]

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 $GJ$

-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 $GJ$)

 

[sysprog]

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 $GJ$

-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 $GJ$)

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.

 

[metastable]

Hopefully this post will be acceptable since I already posted the proper formulas in LaTex format and solutions to only 3 significant digits using SI variables:

It is an excel spreadsheet calculator I made today for performing "Land Based Oberth Maneuver" efficiency calculations:

Download Link: https://files.secureserver.net/0fpI5aSJ5iT3iF

Screenshot:

 

[sysprog]

Thanks for trying to stick within the PF guidelines. I've had a bit of trouble with that before myself. I think that I can safely (i.e. without fear of being incorrect) say that PF would not be such a great forum set without its moderators taking such good care and without the rest of its membership taking such good heed.