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

[metastable]

Ideally I was hoping to be able to calculate how efficient each vehicle is (I can already calculate the baseline electric).

 

[russ_watters]

Ideally I was hoping to be able to calculate how efficient each vehicle is (I can already calculate the baseline electric).

Fair enough. This approach may have value, for example, when dealing with a mass transit system where you are entitled to fix the constraints (unlike for motor vehicles, where you can't). A train could have curved tunnels to reduce travel time without increasing energy input, for example.

 

[metastable]

Depth of deepest mine in the world: 2.5 $miles$

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

 

[jbriggs444]

Like @russ_watters, I have a concern that you are unwittingly designing a perpetual motion machine.

If one starts with a machine that runs over level ground with constant speed and a running start (and running finish) under quadratic drag it seems obvious that any variation in speed will increase the total energy requirements for the trip.

The only way I see that a compressed air rocket motor can provide a performance win is if it embodies a perpetual motion machine. [You could get a performance win by loading a rock into the car at the top and kicking it out the back at the bottom. But that's cheating -- you've harvested the gravitational potential energy of the rock]

The "parabola" shape that optimizes transit time in the zero energy case is a brachistochrone.

 

[metastable]

I have a concern that you are unwittingly designing a perpetual motion machine.

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"

"At very high speeds the mechanical power imparted to the rocket can exceed the total power liberated in the combustion of the propellant; this may also seem to violate conservation of energy. But the propellants in a fast-moving rocket carry energy not only chemically, but also in their own kinetic energy, which at speeds above a few kilometres per second exceed the chemical component. When these propellants are burned, some of this kinetic energy is transferred to the rocket along with the chemical energy released by burning. This can partly make up for what is extremely low efficiency early in the rocket's flight when it is moving only slowly. Most of the work done by a rocket early in flight is "invested" in the kinetic energy of the propellant not yet burned, part of which they will release later when they are burned."

 

[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?