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

[jbriggs444]

I don’t think it takes any chemical energy in this case to accelerate the air because the acceleration of the slope vehicle down slope is from gravity.

Nope. That does not help. Now you are trying to design a perpetual motion machine by ignoring buoyancy.

 

[metastable]

If buoyancy is determined to be a factor with compressed air it can be a chemical rocket with a certain energy.

 

[jbriggs444]

If buoyancy is determined to be a factor with compressed air it can be a chemical rocket with a certain energy.

So now you are artificially limiting the efficiency of the surface vehicle so that you can claim a performance advantage for the sub-surface vehicle.

Chemical powered exhaust velocity is much higher than vehicle velocity for all reasonable earthbound vehicles. It follows that chemical rocket propulsion is much less energy-efficient than wheeled propulsion for such vehicles. You are better off running the rocket exhaust through a turbine and using the harvested energy to turn the wheels.

Some useful information may be found here.

 

[metastable]

Trying to simplify the problem as much as possible. Suppose I want to cover 10 miles on the surface at 100mph. I design a wheel driven car that goes 10 miles at 100mph with a certain amount of chemical or electrical energy -- the car is accelerated with outside means initially to 100mph before the start of the 10 mile straight track (flying start), and does not brake when crossing the finish line, and it gets a certain efficiency.

Next I design a track where the same frontal area, drag coefficient and initial mass car (locked to rails) is also first accelerated by outside means to 100mph, where it then enters a 45 degree downward slope to a depth of 2.5 miles, the track flattens out for a distance, and then climbs out via another 45 degree angle track to the point on the surface that is 10 surface miles from the starting point (smooth curves between the different angled sections). This vehicle has to expend enough energy via its rocket nozzle immediately after entering the flat section at the bottom of the tunnel that it crosses the finish line at 100mph -- but there is the added complication that it needs to cross the finish line at the same time as the other car that went constant speed on the surface. Does this second vehicle need to ride over an above ground hump to slow itself down enough to cross the finish line at the same time as the constant speed vehicle?

 

[jbriggs444]

Trying to simplify the problem as much as possible. Suppose I want to cover 10 miles on the surface at 100mph. I design a wheel driven car that goes 10 miles at 100mph with a certain amount of chemical or electrical energy -- the car is accelerated with outside means initially to 100mph before the start of the 10 mile straight track (flying start), and does not brake when crossing the finish line, and it gets a certain efficiency.

Next I design a track where the same frontal area, drag coefficient and initial mass car (locked to rails) is also first accelerated by outside means to 100mph, where it then enters a 45 degree downward slope to a depth of 2.5 miles, the track flattens out for a distance, and then climbs out via another 45 degree angle track to the point on the surface that is 10 surface miles from the starting point (smooth curves between the different angled sections). This vehicle has to expend enough energy via its rocket nozzle immediately after entering the flat section at the bottom of the tunnel that it crosses the finish line at 100mph -- but there is the added complication that it needs to cross the finish line at the same time as the other car that went constant speed on the surface. Does this second vehicle need to ride over an above ground hump to slow itself down enough to cross the finish line at the same time as the constant speed vehicle?

So the surface car runs at constant 100 mph speed through the entire track and completes the course in precisely six minutes. The energy budget for the surface car is dictated by air resistance. Elsewhere you have assumed quadratic drag for this.

The sub-surface car covers a total distance of 2.5√2+5.0+2.5√22.52+5.0+2.52 ~= 12 miles. Accordingly, its average speed must be 120 mph. Given quadratic drag, this means that the sub-surface car dissipates about 44% more energy to air resistance than does the surface vehicle.

If we assume an exhaust velocity of about 5000 miles per hour and a vehicle velocity of 100 mph, the energy in the rocket fuel would be delivered about 98% to the exhaust and about 2% to the vehicle.

At 120 mph, we are improving things slightly. Now we get about 97.7% to the exhaust and 2.3% to the vehicle. But that's not enough to make up for quadratic drag. [2.5 miles of vertical drop is enough to get much faster speeds. But quadratic drag grows faster than Oberth can catch up].

None of that was part of your immediate question though. You wanted to know about the hump at the end.

There is nothing that stops us from turning the track upside down and having the vehicle run inverted at the start and end of the trip so that it can shift from straight-and-level to downslope and from upslope to straight-and-level. That detail is uninteresting from the point of view of an energy analysis.

 

[metastable]

That detail is uninteresting from the point of view of an energy analysis.

Except that since it's already a tunnel it would be a relatively minor energy and capital investment for it to be a vacuum as well.

 

[jbriggs444]

Except that since it's already a tunnel it would be a relatively minor energy and capital investment for it to be a vacuum as well.

So now you finally have a scenario that works.

In vacuum, there is no air resistance. The energy budget for both surface and sub-surface trips is exactly zero. You get a win on transit time by putting a dip in the trajectory. And you can discard the rocket motor.

 

[metastable]

The sub-surface car covers a total distance of 2.5√2+5.0+2.5√22.52+5.0+2.52 ~= 12 miles. Accordingly, its average speed must be 120 mph

I thought with a 2.5 mile drop on a 45 degree ramp starting with 100mph at the top, the average speed would be considerably faster than 120mph even without a vacuum

 

[jbriggs444]

I thought with a 2.5 mile drop on a 45 degree ramp dstarting with 100mph at the top, the average speed would be considerably faster than 120mph even without a vacuum

You specified equal transit times.

 

[metastable]

Equal transit times, but the below ground portion must have (2) 45 degree ramps that each reach a depth below ground of 2.5 miles, so I was asking if it had to go on a big above ground journey to take the trip in the same time as the constant speed car.

 

[jbriggs444]

Equal transit times, but the below ground porting must have (2) 45 degree ramps that each reach a depth below ground of 2.5 miles

Divide distance by time and you have average speed. You specified equal time and you specified the track length and layout. There is no wiggle room. You effectively specified 120 mph.

 

[metastable]

Gravity will cause the vehicle to accelerate to much greater speeds than 120mph on the ramps and flat portion.

 

[jbriggs444]

Gravity will cause the vehicle to accelerate to much greater speeds than 120mph on the ramps and flat portion.

It is not my problem if you over-specify the scenario.

 

[metastable]

I'm sorry if my wording wasn't very clearly phrased. I meant, because of the ramps, the ramp vehicle is going to take the journey in less time, so what elements above ground if any do I have to add to the track so both vehicles cross the finish line at the same speed at the same time, while the second vehicle can only use rocket power at the flat section on the bottom of the tunnel.

 

[jbriggs444]

I'm sorry if my wording wasn't very clearly phrased. I meant, because of the ramps, its going to take the journey in less time, so what elements above ground if any do I have to add to the track so both vehicles cross the finish line at the same speed at the same time, while the second vehicle can only use rocket power at the flat section on the bottom of the tunnel.

Huh? What? If you want to burn some time, just put in a loop.

It won't make the trip any more efficient. Quadratic drag still trumps Oberth.

 

[metastable]

Quadratic drag still trumps Oberth.

How can this always be true when in the video at 4:07, the marble that takes the "low road" then climbs back to the original height covers a longer path at a higher speed with the same amount of energy. So if I make the tunnel a vacuum can I go anywhere on land in the least time for the least energy by riding down 45 degree ramps to straight sections of track in vacuum tunnels which are ~ 2.5 miles deep?

 

[metastable]

This is the pertinent still frame:

 

[jbriggs444]

How can this always be true when in the video at 4:07, the marble that takes the "low road" then climbs back to the original height covers a longer path at a higher speed with the same amount of energy. So if I make the tunnel a vacuum can I go anywhere on land in the least time for the least energy by riding down 45 degree ramps to straight sections of track in vacuum tunnels which are ~ 2.5 miles deep?

For the n'th time, that's not Oberth and it's not saving any energy.

That scenario involves a [nearly] lossless situation. The energy use for both high road and load road is nearly zero. If one were to do careful measurements however, one would find that the low road consumed more energy than the high road.

If you cover a longer path in a shorter time, you will have a higher average speed.

If you travel a longer path at a higher average speed you will tend to experience more drag and dissipate more energy.

Roughly speaking, Oberth buys you a linear increase in propulsion efficiency. Double the speed and you double the efficiency. Roughly speaking, quadratic drag costs a cubic increase in energy dissipation. Double the track length for a fixed transit time and you pay eight times the energy. Reduce the transit time and you pay even more.

 

[metastable]

If the vehicle is very massive for a given frontal area, wouldn’t the drag possibly become insignificant compared to the kinetic energy at the bottom of the first ramp?

 

[jbriggs444]

If the vehicle is very massive for a given frontal area, wouldn’t the drag possibly become insignificant compared to the kinetic energy at the bottom of the first ramp?

Irrelevant. The kinetic energy at the bottom of the first ramp must be repaid on the climb back up.

Recall your goal. Minimize expended energy. Neither the kinetic energy you happen to have at the midpoint of the journey nor the corresponding potential energy you do not have there enter into the calculation of total energy expended.

 

[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.