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

[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.

 

[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.