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

[metastable]

Would the slope riding vehicle use less energy to cover the same ground distance in the same time in the following manner:

Both ground vehicle and slope vehicle initially have the same mass.

Suppose 1/2 of the mass of the slope vehicle is an apparatus consisting of a linear motor, push rod, and a light-weight tank filled with a heavy amount of water.

The slope riding vehicle and ground vehicle are both accelerated to 100mph with outside means. The ground vehicle covers 10 miles at constant 100mph. The slope vehicle, locked to rails, enters a 45 degree downward slope to a depth of 2.5 miles. At the bottom of the slope the linear motor launches the tank off the back of vehicle with enough force that the tank is at rest on the tracks. A valve opens on the tank dumping out the water on the tracks and the lightweight tank is later retrieved. The rest of the vehicle, now with 1/2 the original mass, continues down the flat track and then up another 45 degree ramp and then does a series of S turns before crossing the finish line at the same time and velocity as the ground vehicle. Did the slope riding vehicle use less energy?

 

[jbriggs444]

Would the slope riding vehicle use less energy to cover the same ground distance in the same time in the following manner:

Both ground vehicle and slope vehicle initially have the same mass.

Suppose 1/2 of the mass of the slope vehicle is an apparatus consisting of a linear motor, push rod, and a light-weight tank filled with a heavy amount of water.

The slope riding vehicle and ground vehicle are both accelerated to 100mph with outside means. The ground vehicle covers 10 miles at constant 100mph. The slope vehicle, locked to rails, enters a 45 degree downward slope to a depth of 2.5 miles. At the bottom of the slope the linear motor launches the tank off the back of vehicle with enough force that the tank is at rest on the tracks. A valve opens on the tank dumping out the water on the tracks and the lightweight tank is later retrieved. The rest of the vehicle, now with 1/2 the original mass, continues down the flat track and then up another 45 degree ramp and then does a series of S turns before crossing the finish line at the same time and velocity as the ground vehicle. Did the slope riding vehicle use less energy?

No. It consumed potential energy in the form of the water in the tank.

 

[metastable]

Thank you. I meant can it potentially use less electrical energy to cover the same ground distance in the same time with the right parameters.

 

[jbriggs444]

Thank you. I meant can it potentially use less electrical energy to cover the same ground distance in the same time with the right parameters.

If your goal is to use less electrical energy, there are less Rube Goldberg ways of going about it. Use a gasoline engine, for instance.

 

[metastable]

I’m assuming the water will later evaporate so it won’t take any fossil fuels or electrical power to lift this water. If water can do the same thing as gasoline, well, I think it’s more renewable.

 

[jbriggs444]

I’m assuming the water will later evaporate so it won’t take any fossil fuels or electrical power to lift this water.

It is still not a fair comparison. If you want to add a hydropower generating system for use by the sub-surface vehicle, fair play demands that you let the surface car use it as well.

 

[metastable]

The surface vehicle can use hydropower to charge the batteries of its electrical wheel driven system.

 

[jbriggs444]

The surface vehicle can use hydropower to charge the batteries of its electrical wheel driven system.

And the result is that the surface car wins. It spends less energy on drag and both cars harvest the same energy from hydropower.

I am out of the discussion. The conservation of energy argument is solid. The various suggestions for defeating it amount to unsupported claims for perpetual motion.

 

[metastable]

I thought it wasn’t perpetual motion if it uses gravitational potential energy for propulsion instead of electrical energy. The goal of the slope riding vehicle is derive as much propulsion as possible from gravitational potential energy and as little as possible from electrical energy.

 

[metastable]

For example if the vehicle is 99% ballast mass, unless I am mistaken it won’t take comparatively very much electrical energy to make the remaining 1% that comes out of the tunnel achieve very high velocities.

 

[Dale]

Trying to simplify the problem as much as possible. ...

Ok, you are really making this more complicated than needed. The Oberth effect says that if you burn a given amount of rocket fuel at a low speed then you gain less KE than if you burn that same amount of fuel at a high speed.

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.

Ignore trying to match the starting and ending times, the Oberth effect says nothing about times. Don’t have different paths, the Oberth effect says nothing about paths.

 

[metastable]

Dale, what you suggest is a perfectly reasonable way to simplify the problem of calculating a land based Oberth maneuver.

Before I do the calculation, I wanted to ask if the following maneuver still counts as an “Oberth” maneuver, because the goal is to not use any form of fuel on the vehicle except water.

Lets assume the water weighs 20,000lbs and the passenger section of the slope vehicle weighs 200lbs. I want to go 10 miles as fast as possible, without using any gasoline or electricity — only water.

If we can assume the frontal area of the tank + vehicle on the way down the ramp terminal velocity is 600mph and the 200lb + 20,000lb tank vehicle was using electric regenerative braking the whole way down the 45 degree ramp to a depth of 2.5 miles to charge a bank of capacitors, the vehicle might store many watt hours of energy in the capacitor from regen by the time it reaches the bottom of the ramp if it reduces the speed of the tank from say 600+mph to 500mph with the 20,000lb water tank.

Once it reaches the bottom it releases all the stored regen energy in one push with a linear motor against the 20,000lb 500mph tank. It almost certainly has enough kinetic energy to make it back up to the surface on the second ramp, and it wasn’t powered by any electricity or fossil fuels, just water.

In this case would the maneuver at the bottom of the ramp with the pushrod, linear motor and stored regen energy still be considered an “Oberth” maneuver?

 

[Dale]

I wanted to ask if the following maneuver still counts as an “Oberth” maneuver ... using electric regenerative braking

I don’t know how regenerative braking could fit in. The goal is to use gravity to get to a high speed before using the rocket. Regenerative braking defeats that. Also, regenerative braking transforms mechanical energy into internal energy; there is no equivalent for a rocket.

 

[metastable]

According to this article:

https://www.google.com/amp/s/electrek.co/2018/04/24/regenerative-braking-how-it-works/amp/

^Using electric regen braking under optimal conditions up to 70% of a braking vehicle’s kinetic energy can be converted to thrust

So as far as the amount of energy my rocket can have, I will look at how much kinetic energy can be recovered by decelerating a 20,200lb vehicle from 600mph to 500mph at the bottom of a 2.5 mile deep vertical drop.

Next I will consider the 70% of the lost kinetic energy to be recovered kinetic energy which can be converted to thrust from this process my “rocket fuel.”

Next I will do a comparison of using this rocket fuel (at up to 9g passenger 200lb vehicle acceleration) via a push rod by a 200lb vehicle section against a 0mph 20,000lb (pinewood derby rocket at top of ramp) vehicle section or a 500mph 20,000lb (pinewood derby rocket at middle ramp) vehicle section.

I am using a certain amount of rocket fuel to push off of a 0mph 20,000lb tank or a 500mph 20,000lb tank with a 200lb passenger vehicle.

 

[Dale]

Ok, but if you are using regenerative braking then I wouldn’t call the result an Oberth maneuver.

 

[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