# B Land Based Oberth Manuever?

#### 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.

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#### Dale

Mentor
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

Science Advisor
Homework Helper
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.

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#### jbriggs444

Science Advisor
Homework Helper
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

metastable said:
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.
Wile E. Coyote said:
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

#### 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 ...

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#### 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

#### 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

#### 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.

#### metastable

The equations seem to give absurd results with very high passenger vehicle masses: #### sysprog

Obviously we can't see on your spreadsheet the equation or inequality expression that resulted in the apparently absurd result -- in addition to your having the ability to use $\LaTeX$ here to show your math readably, you also can use BBcode [code ] tags to show your code -- maybe you would care to show, cell by cell, your .xls code that generated the 'absurd' result $\dots$

#### Dale

Mentor
Excel code is almost impossible to debug.

#### Vanadium 50

Staff Emeritus
Science Advisor
Education Advisor
Excel code is almost impossible to debug.
A wall-O-numbers in Excel is still a wall-O-numbers.

#### metastable

I think its an error in excel rather than the formulas because it jumps from a reasonable answer when I keep adding zeros to suddenly a ridiculous answer. In other words, with one less $0$ at the end of the passenger vehicle mass I think it gives the correct answer still.

Suppose we have a $10^4$ $kg$ passenger vehicle with a $20$ $kg$ trailer containing $200$ $kg$ 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 $9.54$ $MJ$, pushing off the $294$ $m/s$ 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 $294$ $m/s$ in a straight line at the bottom of the ramp after the mechanical impulse. Assuming regen braking of the passenger vehicle at its destination (after climbing a second vertical ramp to the surface) results in $70$ % kinetic to kinetic conversion efficiency, and that only the empty trailer is lifted, a factor of $130$ % as much usable energy is recovered from the vehicle after it climbs to the surface than was exerted in the mechanical impulse at the bottom of the ramp, a net surplus if the geothermal energy required to evaporate this water is ignored. The surplus comes from the lowered gravitational potential energy of the $200$ $kg$ water at the bottom of the tunnel. If the full trailer is lifted with the regen braking energy instead of geothermal, then only $40$ % of the mechanical impulse energy is recoverable. If the full trailer is lifted from the recovered regen braking energy, then a total of $5.71$ $MJ$ energy was “unrecoverably” consumed accelerating and decelerating the $10^4$ $kg$ vehicle to $294$ $m/s$. If the $200$ $kg$ water is left at the bottom of the tunnel and allowed to evaporate naturally, then an “excess” of $2.88$ $MJ$ 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 using 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 a factor of $~760x$ greater than the total energy “unrecoverably” consumed when lifting the full (not empty) trailer all the way back to the surface, using the recovered kinetic energy via the regen braking after the journey.  Last edited:

#### metastable

First I will share an example of the equations I would use to calculate the power consumption of a given standard land vehicle at a given speed in a given set of conditions.
The ACME Corporation cannot guarantee the safety of a rider aboard an electric skateboard operated at a speed exceeding 100 mph100 mph100\ mph.
for comparison I will share a photo of, what I personally consider, a "standard car"-- quite literally the drivetrain of the fastest electric skateboard in the world, capable of 70+ $mph$: • sysprog

#### sysprog

for comparison I will share a photo of, what I personally consider, a "standard car"-- quite literally the drivetrain of the fastest electric skateboard in the world, capable of 70+ $mph$:

View attachment 246979
Please sign your organ donor card before riding that thing, kid; how do you deal with the snap (aka jounce)? Oh, and, I have a friend whose pet kitty needs a cornea transplant. On a slightly more technical note, umm -- WB cartoon speed was measured in frames per second (12 fps doubled to 24); not really mph -- Mr. Coyote's speed is forever less than that of the Road Runner, speed in this instance being considered as the derivative of distance with respect to time -- the reasoning behind the standard for the time required to complete 1 frame (not for the artist; for the viewer) is hinted at by the following inequality constraint: $1(frametime) < 1(blinkofaneye)$. Actually, it has more to do with human brain visual image processing time -- the frame rate is determined by how many frames per second are required to cause the visual system to report continuous movement instead of separate still images.

And I didn't mean to hijack your thread.

@metastable posted me, and I mostly begged off for now, regarding his to-me-somewhat-quizzical spreadsheet code. My at-a-glance response was that the source of the problem may/might be in his having exhausted the capacity of the sqrt function.

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#### metastable

@metastable posted me, and I mostly begged off for now, regarding his to-me-somewhat-quizzical spreadsheet code. My at-a-glance response was that the source of the problem may/might be in his having exhausted the capacity of the sqrt function.
For me, the most interesting part of the whole exercise was determining the formula for calculating how many Joules of Mechanical Impulse between the vehicle and the trailer will bring the trailer to a complete halt at the bottom of the ramp every time, depending on the Trailer Mass, Vehicle Mass, Free Fall Duration, Initial Velocity Before Free Fall, and the Gravitational Acceleration at Earth's Surface:

$W_i$ = $12.8*10^6$ = Mechanical Impulse Energy ($Joules$) That Brings Trailer to 0 Velocity At Ramp Bottom
$M_t$ = $240$ = Trailer Tank + Water Mass ($Kilograms$)
$M_v$ = $10^3$ = Passenger Vehicle Mass ($Kilograms$)
$T_d$ = $30$ = Free Fall Duration ($Seconds$)
$V_i$ = $0$ = Initial Velocity Before Free Fall ($m/s$)
$G_e$ = $9.80$ = Gravity Acceleration at Earth’s Surface ($m/s^2$)

$W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+(M_t^2(V_i+T_dG_e)^2)/(2M_v)$

#### sysprog

For me, the most interesting part of the whole exercise was determining the formula for calculating how many Joules of Mechanical Impulse between the vehicle and the trailer will bring the trailer to a complete halt at the bottom of the ramp every time, depending on the Trailer Mass, Vehicle Mass, Free Fall Duration, Initial Velocity Before Free Fall, and the Gravitational Acceleration at Earth's Surface:

$W_i$ = $12.8*10^6$ = Mechanical Impulse Energy ($Joules$) That Brings Trailer to 0 Velocity At Ramp Bottom
$M_t$ = $240$ = Trailer Tank + Water Mass ($Kilograms$)
$M_v$ = $10^3$ = Passenger Vehicle Mass ($Kilograms$)
$T_d$ = $30$ = Free Fall Duration ($Seconds$)
$V_i$ = $0$ = Initial Velocity Before Free Fall ($m/s$)
$G_e$ = $9.80$ = Gravity Acceleration at Earth’s Surface ($m/s^2$)

$W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+(M_t^2(V_i+T_dG_e)^2)/(2M_v)$
Please disambiguate the last-line expression -- in particular, the '/2' -- does that fractionate the entirety? -- it's ok to use $\frac 1 2$ and/or extra parens (instead of making me have to go with PEMDAS and the rest of my feeble memory of the order of operations) to keep it clear what you mean $\dots$

#### metastable

Please disambiguate the last-line expression -- in particular, the '/2' -- does that fractionate the entirety? -- it's ok to use 1212\frac 1 2 and/or extra parens (instead of making me have to go with PEMDAS and the rest of my feeble memory of the order of operations) to keep it clear what you mean
I'm not sure what you mean exactly, does this help?

240*0*30*9.80665+(240*30^2*9.80665^2+240*0^2)/(2)+(240^2*(0+30*9.80665)^2)/(2*1000) ### Want to reply to this thread?

"Land Based Oberth Manuever?"

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