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

[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.