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

[jbriggs444]

The potential efficiency increase I am studying which I have referred to as an "oberth maneuver" can be referenced here:

https://en.wikipedia.org/wiki/Oberth_effect

"It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's kinetic energy is balanced by a relative decrease in the kinetic energy the exhaust is left with (the kinetic energy of the exhaust may still increase, but it does not increase as much"

Right. And for this to help in the case at hand, you need to be picking up energy from somewhere.

You can't extract potential energy by moving air around. The extra energy required to expel it at depth will defeat the gains from compressing it and carrying it downward. You could get potential energy from rocks. But that'll just wind up filling in the hole you dug for the road to run in.

And guess what -- now we're discussing perpetual motion.

 

[metastable]

"So if a spacecraft is on a parabolic flyby of Jupiter with a periapsis velocity of 50 km/s and performs a 5 km/s burn, it turns out that the final velocity change at great distance is 22.9 km/s, giving a multiplication of the burn by 4.58 times."

https://en.wikipedia.org/wiki/Oberth_effect

 

[jbriggs444]

"So if a spacecraft is on a parabolic flyby of Jupiter with a periapsis velocity of 50 km/s and performs a 5 km/s burn, it turns out that the final velocity change at great distance is 22.9 km/s, giving a multiplication of the burn by 4.58 times."

https://en.wikipedia.org/wiki/Oberth_effect

And it's not magic. And it conserves energy. But it leaves the fuel which started with high potential energy and high kinetic energy down lower in the gravitational well with reduced potential energy and reduced kinetic energy.

You cannot pull that same pet trick using compressed air in an atmosphere.

 

[Vanadium 50]

I am having a hard time figuring what the question is and how it relates to the Oberth Effect. I don't understand the purported equivalence of pressurized air and gravitational slingshots. Further, this looks an awful lot like a perpetual motion machine.

 

[jbriggs444]

The theory as I understand it:

The vehicle draws in still air at ambient pressure and temperature at ground level at the start of its trip. It uses engine power to compress this air and embarks on a path that dips below ground level, picking up speed as it moves.

At the bottom of the dip, an exhaust valve is opened and the compressed air is released, functioning as a rocket motor. The expectation is for a large efficiency multiplier.

Indeed, there might be a slight efficiency multiplier compared to an approach that used a compressed air rocket motor (also known as a low bypass jet engine) on level ground. But that is the wrong baseline to compare against. Rocket motors (or low bypass jet engines) pushing backward on high speed exhaust gasses are hideously inefficient compared to ordinary motors pushing backward on the ground.

 

[metastable]

I am having a hard time figuring what the question is and how it relates to the Oberth Effect. I don't understand the purported equivalence of pressurized air and gravitational slingshots

If I’m not mistaken, the “oberth effect” is a completely separate effect from a “gravitational slingshot.”

“Not to be confused with Slingshot maneuver.”

“Therefore, the larger the v at the time of the burn, the greater the final kinetic energy, and the higher the final velocity.

The effect becomes more pronounced the closer to the central body, or more generally, the deeper in the gravitational field potential the burn occurs, since the velocity is higher there.”

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

 

[jbriggs444]

If I’m not mistaken, the “oberth effect” is a completely separate effect from a “gravitational slingshot.”

Indeed so. Which means that you are not harvesting kinetic or gravitational potential energy from a third body. Which closes down that particular loophole in an energy conservation argument.

 

[metastable]

The unknown here is the efficiency of an electric motor used to drive a pump to compress air into the tank, and at what speed such a setup would be most efficient, taking into account all of the energy losses.

I found this reference about air compression efficiency:

https://web.archive.org/web/20080911042043/http://www.efcf.com/reports/E18.pdf

 

[jbriggs444]

I found this reference about air compression efficiency:

Make life easy. Assume 100% efficiency at compressing air and in exhausting it.

There is still no free lunch. Any trajectory other than the straight and level, constant speed one is going to involve an expenditure of additional energy to deal with wind resistance. With no energy sources in sight, the required energy cannot be supplied. And that dooms the plan.

 

[metastable]

Any trajectory other than the straight and level, constant speed one is going to involve an expenditure of additional energy to deal with wind resistance. With no energy sources in sight, the required energy cannot be supplied. And that dooms the plan.

Assuming this is true then how does the marble in the video reach the other side quicker? (the one that spends time at lower altitude)

 

[jbriggs444]

Assuming this is true then how does the marble in the video reach the other side quicker? (the one that spends time at lower altitude)

Remember what problem you are solving. You are trying to optimize energy expenditure for a fixed transit time under quadratic drag. If you have been paying attention, the version of the problem that I have been taking pains to address is one with a running start and a running stop. That is the problem with an optimal solution that cannot be improved upon.

Optimizing transit time with zero energy input, negligible resistance and a start from rest is a different problem. Brachistochrone

If one were trying to optimize transit time with fixed energy input (elevation difference between start and finish), quadratic drag and a start from rest, one would expect a curve with an initial dip, a near constant slope and a final rise -- a flattened Brachistochrone curve.

 

[russ_watters]

There is still no free lunch. Any trajectory other than the straight and level, constant speed one is going to involve an expenditure of additional energy to deal with wind resistance. With no energy sources in sight, the required energy cannot be supplied.

While that's true, I'm not sure it provides an apples-to-apples comparison. There's an assumption I haven't seen discussed, which is that the vehicle reaches point B with zero velocity (otherwise the scenarios don't achieve the same result). The vehicle on the straight and level has to expend energy braking whereas the vehicle on the curved path does not. All that is required to save energy is for the additional drag on the curved one to be less than the energy lost in braking the level one.

 

[jbriggs444]

While that's true, I'm not sure it provides an apples-to-apples comparison. There's an assumption I haven't seen discussed, which is that the vehicle reaches point B with zero velocity (otherwise the scenarios don't achieve the same result). The vehicle on the straight and level has to expend energy braking whereas the vehicle on the curved path does not. All that is required to save energy is for the additional drag on the curved one to be less than the energy lost in braking the level one.

I thought I'd mentioned such an assumption in passing. You are correct, of course, that the rules for the start and end of the trip will affect the optimal road slope at the start and end of the trip.

 

[metastable]

I wasn’t assuming either vehicle brakes. I wanted them both crossing the starting line and the finishing line at the same velocity (flying start and stop) and also both completes its own course in the same amount of time. The ground level vehicle should be at constant speed, and the parabola riding vehicle should expend just enough energy to get to the finish line in the same time, and cross the finish line at the same velocity as the baseline vehicle.

 

[russ_watters]

I wasn’t assuming either vehicle brakes. I wanted them both crossing the starting line and the finishing line at the same velocity (flying start and stop) and also both completes its own course in the same amount of time. The ground level vehicle should be at constant speed, and the parabola riding vehicle should expend just enough energy to get to the finish line in the same time, and cross the finish line at the same velocity as the baseline vehicle.

These are conflicting constraints. If they start and end at the same speed, the curve-riding vehicle gets there first and if they take the same time the straight line vehicle has a higher starting and ending speed. The only way to correct that is with additional acceleration and/or deceleration.

 

[metastable]

These are conflicting constraints. If they start and end at the same speed, the curve-riding vehicle gets there first and if they take the same time the straight line vehicle has a higher starting and ending speed. The only way to correct that is with additional acceleration and/or deceleration.

I thought the wind drag might prevent the parabola riding vehicle from exiting with enough $V$ to cross the line at the same velocity as the baseline vehicle, unless it uses energy along the way, particularly very close to or at periapsis.

 

[russ_watters]

I thought the wind drag might prevent the parabola riding vehicle from exiting with enough $V$ to cross the line at the same velocity as the baseline vehicle, unless it uses energy along the way, particularly very close to or at periapsis.

It will, and you can make that boost whatever you want to get whatever exit speed you want, in this case equal to the starting speed. But that isn't where the problem lies. The problem is with equal starting and ending speeds (all four), the curved track vehicle reaches point B first.

 

[metastable]

I thought it would stand to reason that there would be some downward slope that would be too shallow for the slope riding vehicle to beat the surface vehicle with no energy usage, because of its wind drag, so wouldn’t there also be some slope where they would cross at the same time, with the same start / finish velocities, while the surface vehicle goes constant speed? This is the scenario where I wanted to see if the slope riding vehicle can use less energy.

 

[jbriggs444]

I thought it would stand to reason that there would be some downward slope that would be too shallow for the slope riding vehicle to beat the surface vehicle with no energy usage, because of its wind drag, so wouldn’t there also be some slope where they would cross at the same time, with the same start / finish velocities, while the surface vehicle goes constant speed? This is the scenario where I wanted to see if the slope riding vehicle can use less energy.

So your scenario now is entirely different from the one you started the thread with, correct? No air compressors or rocket nozzles anywhere in sight.

You want a "surface" vehicle on a level track with a running start, constant speed from start to finish and a final speed equal to the starting speed. The motor runs at its maximum rated output from start to finish.

You want a "slope" vehicle on a track with the same endpoints, same starting speed and same ending speed, same transit time but reduced energy usage.

The total distance traversed by the "slope" vehicle is greater than that traversed by the "surface" vehicle. The speed with which that distance is traversed must be greater than the speed of the "surface" vehicle. Accordingly, the energy dissipated to wind drag will be greater. Yet the energy available to dissipate is lower.

The answer is clear: No, it cannot be done.

 

[metastable]

So your scenario now is entirely different from the one you started the thread with, correct? No air compressors or rocket nozzles anywhere in sight.

You want a "surface" vehicle on a level track with a running start, constant speed from start to finish and a final speed equal to the starting speed. The motor runs at its maximum rated output from start to finish.

You want a "slope" vehicle on a track with the same endpoints, same starting speed and same ending speed, same transit time but reduced energy usage.

The total distance traversed by the "slope" vehicle is greater than that traversed by the "surface" vehicle. The speed with which that distance is traversed must be greater than the speed of the "surface" vehicle. Accordingly, the energy dissipated to wind drag will be greater. Yet the energy available to dissipate is lower.

The answer is clear: No, it cannot be done.

So if I understand correctly, you’d say the slope vehicle can go A to B in less time with the same energy, but not the same time with less energy?

 

[jbriggs444]

So if I understand correctly, you’d say the slope vehicle can go A to B in less time with the same energy, but not the same time with less energy?

No. The slope vehicle loses. It takes more energy and gets there more slowly.

 

[metastable]

No. The slope vehicle loses. It takes more energy and gets there more slowly.

So if that’s true how does the marble in the video that takes the lower ramp get to the other side quicker with the same initial height and same ending height?

4min07sec:

 

[jbriggs444]

So if that’s true how does the marble in the video that takes the lower ramp get to the other side quicker with the same initial height and same ending height?

Stop changing the scenario. I already answered that same question with this same answer.

 

[rcgldr]

Any trajectory other than the straight and level, constant speed one is going to involve an expenditure of additional energy to deal with wind resistance. With no energy sources in sight, the required energy cannot be supplied. And that dooms the plan.

Assuming this is true then how does the marble in the video reach the other side quicker? (the one that spends time at lower altitude).

The marble arrives sooner, but assuming no losses, it end ups with the same energy as the marble traveling along the straight path. With aerodynamic drag losses, the faster marble ends up with less energy (slower speed) once it returns to the original height. The "negative work" done by drag = force · distance, so the higher speed and higher drag path involves more "negative work".

For an example of energy consumed to maintain constant speed, assume there's a speed of maximum efficiency. For example, say a gasoline fueled car gets it's best fuel mileage at 45 mph, and is restricted to constant power output, so any change in speed will be due to a slope. If the initial == final speed is greater than 45 mph, then it would be more efficient to climb to a point where the speed is reduced to 45 mph, then at the end of the elevated straight, descend back to the initial speed. If the initial == final speed is less than 45 mph, then it's more efficient to descend to 45 mph, then climb back up at the end to the initial speed. Neither of theses cases are oberth effect.

Oberth effect works in space (zero velocity related losses) because a decrease in GPE coexists with an increase of KE of the remaining fuel, which could be compressed air in a tank. If operating in an atmosphere, the drag increases with the square of the speed, so when the GPE is decreased, the increase in speed and KE of the remaining fuel is less due to drag, and since the drag is increased by the square of the speed, a greater amount of thrust is required. Seems like there is probably an ideal speed for maximum efficiency, and the path could be used to increase or decrease speed to the ideal speed as noted above, taking the Oberth effect into account, but I haven't done the math.

 

[russ_watters]

I'm not quite following what you are saying here:

You want a "surface" vehicle on a level track with a running start, constant speed from start to finish and a final speed equal to the starting speed. The motor runs at its maximum rated output from start to finish.

You want a "slope" vehicle on a track with the same endpoints, same starting speed and same ending speed, same transit time but reduced energy usage.

[snip]

The answer is clear: No, it cannot be done.

[snip]
No. The slope vehicle loses. It takes more energy and gets there more slowly.

This doesn't appear clear to me. To start with, isn't the combination of these two scenarios physically impossible/contradictory?

Second, from the video of the balls rolling on the tracks it is clear to me that the deeper the curve the faster the transit from point A to point B for a case where drag is insignificant - and I don't think this is a different scenario from what you answered. I'm not convinced that non-zero drag instantly changes things -- I think there's a threshold of drag vs mass, on one side of which the curve helps and on the other side it hurts. But let's take this step by step...

Zero drag, flat vs curved. That's what the video shows, right? And the curved track ball gets to point B first, right?

 

[russ_watters]

The marble arrives sooner, but assuming no losses, it end ups with the same energy as the marble traveling along the straight path. With aerodynamic drag losses, the faster marble ends up with less energy (slower speed) once it returns to the original height. The "negative work" done by drag = force · distance, so the higher speed and higher drag path involves more "negative work".

Agreed; the marble following the curve arrives at point B sooner but needs a larger energy boost because of the higher speed and associated drag. Now what if we speed up the flat track marble/vehicle so that they arrive at the end at the same time. Which one uses less energy now? Assume the extra boost of the flat track vehicle is not recoverable and it brakes at the end to match the speed of the marble exiting the curved track.

We have competing constraints and it doesn't appear to me they are even being acknowledged as competing constraints, much less analyzed together to see how they affect the outcome.

 

[metastable]

Perhaps some reasonable constraints (serving to limit the possible solutions) would be as follows:

-both vehicles cross the starting line with the same initial velocity, and it is the same velocity as the electric vehicle’s peak velocity

-the parabola riding vehicle only uses tunnels, and these can be as deep as the deepest point underground yet reached by man

-perhaps even easier to solve the problem if we do it backwards: figure out how fast the parabola riding vehicle can cover a ground distance under ideal circumstances with the constraint of the tunnel depth and initial velocity, and then determine how much power and efficiency the baseline vehicle gets covering the same distance in a straight line on land

Agreed; the marble following the curve arrives at point B sooner but needs a larger energy boost because of the higher speed and associated drag. Now what if we speed up the flat track marble/vehicle so that they arrive at the end at the same time. Which one uses less energy now?

Yes I think it makes sense for ease of calculation to determine the performance of the curve riding vehicle first (with a flying start and flying finish of the same velocity, and a shallow enough parabola where the total time taken is the same amount of time that it would take to cover A to B on the surface at the constant initial velocity). Then once the curve and oberth boost that achieved these parameters are determined, the parameters of the baseline vehicle are determined that gets it from A to B on the surface in the same time at constant velocity.

 

[jbriggs444]

Then once the curve and oberth boost

There is no Oberth boost without rocket propulsion. You do not have rocket propulsion.

 

[metastable]

Will the compressed-air-powered parabola riding vehicle use less energy to reach the same distance in the same time as the standard electric vehicle?

There is no Oberth boost without rocket propulsion. You do not have rocket propulsion.

I thought I specified the curve riding vehicle uses a compressed air rocket for its boosts rather than wheel driven electric on the baseline vehicle.

 

[jbriggs444]

I thought I specified the curve riding vehicle uses a compressed air rocket for its boosts rather than wheel driven electric on the baseline vehicle.

And then you post a video of a marble on a track. Make up your mind, please.

 

[metastable]

And then you post a video of a marble on a track. Make up your mind, please.

I thought the marble on the track shows what happens to the vehicle without propulsion (since I don’t know the equation for the curve vehicle), but in the actual scenario we discuss propulsion is added so the vehicles have the same speed at the end as they had at the start. I see the goal at this point as finding a certain slope and oberth boost, if possible, that gives a consistent start and end velocity, and takes the same time from A to B as constant surface velocity, in order to compare with the wheel driven surface vehicle. The surface driven vehicle’s parameters will be determined after the curve riding vehicle’s parameters.

 

[jbriggs444]

I see the goal at this point as finding a certain slope and oberth boost, if possible, that gives a consistent start and end velocity, and takes the same time from A to B as constant surface velocity, in order to compare with the wheel driven surface vehicle.

Oberth gives no advantage relative to wheeled propulsion. There is no possibility of improvement on the constant speed powered surface trip.

Do the energy accounting. Any Earth-relative exhaust velocity other than zero will result in energy being wasted into the exhaust stream. An Earth-relative exhaust velocity of zero is equivalent to wheeled propulsion.

 

[metastable]

Oberth gives no advantage relative to wheeled propulsion.

I thought with wheeled propulsion, it takes increasing mechanical power to maintain constant thrust at increasing speeds, because the ground (reaction mass) is moving away from the vehicle faster as the vehicle travels faster. But with the rocket, the reaction mass (compressed air) has 0 velocity relative to the vehicle at the moment before it is expelled, so I thought it took less chemical energy to obtain the same kinetic energy when the rocket travels at very high speeds.

 

[jbriggs444]

But with the rocket, the reaction mass (compressed air) has 0 velocity relative to the vehicle at the moment before it is expelled, so I thought it took less chemical energy to obtain the same kinetic energy when the rocket travels at very high speeds.

You are picking up the air from the atmosphere where it starts with zero Earth-relative velocity. You are then accelerating the air to vehicle speed. That takes energy.

If you proceed to expel the air at an exhaust velocity less than the vehicle's Earth-relative speed, you have a net loss of momentum -- you are slowing your vehicle down. If you expel the air at more than the vehicle's Earth-relative speed, you have a net loss of efficiency compared to wheeled propulsion.

 

[metastable]

You are picking up the air from the atmosphere where it starts with zero Earth-relative velocity. You are then accelerating the air to vehicle speed. That takes energy.

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.