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

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In summary, the author thinks that by using a vehicle that is powered by compressed air instead of an electric motor, the energy used to cover the same distance between points can be reduced.
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metastable
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I was out skateboarding the other day when I wondered if the efficiency of a vehicle attempting to cover the distance between point A and point B can be improved in the following manner. I put this thread in general relativity section since it involves gravity.

First I will share an example of the equations I would use to calculate the power consumption of a given standard land vehicle at a given speed in a given set of conditions.

peak mechanical power is 10746.218459832w

A = meters per second = XX.XXX
B = drag coefficient = 0.75
C = frontal area = 0.6m^2
D = fluid density of air = 1.225kg/m^3
E = wind drag force in watts
F = sine of 5% slope = sin(atan(5/100)) = 0.04993761694389223373491
G = acceleration of gravity = 9.80655m/s^2
H = vehicle mass in kg = 90.7184kg = 200lb / 2.20462lb/kg
I = mechanical watts required for constant speed up slope with no wind drag
J = mechanical watts required for constant speed up slope including wind drag
K = H * G * F
L = (1/2) * D * C * B

E = ((1/2) * D * C * (A^2) * B) * A

I = H * G * A * F

J = E + I

J = (((1/2) * D * C *(A^2) * B) * A) + (H * G * A * F)

J = (1/2) * D * C * B * A^3 + H * G * F * A

J = (L * A^3) + (K * A)

^this can be rearranged to:

A=(sqrt(3) * sqrt(27 * J^2 * L^4 + 4 * K^3 * L^3) + 9 * J * L^2)^(1 / 3) / (2^(1 / 3) * 3^(2 / 3) * L) - ((2 / 3)^(1 / 3) * K) / (sqrt(3) * sqrt(27 * J^2 * L^4 + 4 * K^3 * L^3) + 9 * J * L^2)^(1 / 3)

we know:

J = 10746.218459832w peak mechanical
L = 0.275625 = (1/2) * D * C * B
K = 44.42622815547907982077 = H * G * F

therefore:

A=(sqrt(3) * sqrt(27 * 10746.218459832^2 * 0.275625^4 + 4 * 44.42622815547907982077^3 * 0.275625^3) + 9 * 10746.218459832 * 0.275625^2)^(1 / 3) / (2^(1 / 3) * 3^(2 / 3) * 0.275625) - ((2 / 3)^(1 / 3) * 44.42622815547907982077) / (sqrt(3) * sqrt(27 * 10746.218459832^2 * 0.275625^4 + 4 * 44.42622815547907982077^3 * 0.275625^3) + 9 * 10746.218459832 * 0.275625^2)^(1 / 3)

A=32.32 meters per second

^therefore the peak velocity up slope is 32.32 meters per second


My question is would the following method potentially improve the energy efficiency of a given vehicle to cover the same distance between 2 points on land in the same time using less energy:

-First a track is constructed which consists of a series of parabolas (think of the trajectory of the "vomit comet" aircraft which is used for zero g astronaut training)

- The vehicle is modified so that, rather than its electric motor directly powering the wheels as in a standard automobile, its electric motor is used to force air into a high pressure tank

-The tank is connected to a compressed air thruster on the back of the vehicle, similarly to a reaction control system on a spacecraft

-The vehicle starts down the track, accelerating from gravity towards the bottom of the first parabola. Once it is almost at the bottom, it fires its compressed air thruster in a very short blast with just enough energy to surpass the next crest, and also in such a way that it eventually reaches point B in the same time as the standard vehicle.

For reference, unless mistaken I believe the parabola riding vehicle is taking advantage of an oberth manuever at the bottom of each parabola.

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?
 
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  • #2
I think that the only thing worse than wall-of-text posts is wall-of-math posts. Please

a) use LaTeX to format your math in a readable style

b) use standard variables and the common form of equations whenever possible
 
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  • #3
In addition to what Dale says, do not post any number to more than 3 significant figures. Unless you expect to be able to control the density of air to better than one percent, there is no point to more precision.
 
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  • #4
Would it be acceptable enough to use a downloadable spreadsheet format? (it strips off the extra zeros while retaining them in memory)
daf33cba802352e9f9581f5d01388274107a25ca.jpg

Spreadsheet: https://files.secureserver.net/0fT5HhdTQhpC24

6ad8f5964d87eaa70159d0a2704eaaa60300533c.jpg


Spreadsheet: https://files.secureserver.net/0fhemIK29ta5TN
 
  • #5
No. You are still not explaining in words what you are doing and why you think it is the right thing to do. If you cannot be bothered to do this, why would you expect me to be bothered to figure it out?
 
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  • #6
I thought it might work because:

"The gain in efficiency is explained by the Oberth effect, wherein the use of an engine at higher speeds generates greater mechanical energy than use at lower speeds."

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

I wasn't sure if other factors might offset the potential efficiency benefits of this technique.
 
  • #7
metastable said:
Would it be acceptable enough to use a downloadable spreadsheet format?
That depends what your goal is. If your goal is to simply to post it for a record to establish priority or something then it is fine. If the goal is to get people to actually evaluate your work and give you feedback then see post 2.
 
  • #8
I'm posting to share the method I believe is the easiest way I know to calculate the efficiency of an electric vehicle (so that others may more easily calculate as well), by entering the vehicle parameters into a spreadsheet which, to the best of my knowledge models the correctly applicable physics equations.
 
  • #9
That sounds like something more suited to a personal blog than a thread on PF. You don’t need any discussion from the community to accomplish that.

I encourage you to avail yourself of any of the many blogging sites for future posts of this nature. For posts where you want community discussion here, please follow the usual courtesy outlined in post 2.

Thread closed.
 
  • #10
After a private discussion and a brief tutorial on LaTeX, @metastable will provide an improved analysis and question. So thread reopened.
 
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  • #11
To begin I will post some details and equations used to describe the baseline electric vehicle (iron losses and rolling resistance ignored):

Frontal Area: 3 ##m^2##
Drag Coefficient: 0.02 ##C_D##
Air Density: 1.225 ##kg/m^3##
Mass: 300 ##lb##
Mass: 136.08 ##kg##
Tire Diameter: 500 ##mm##
BLDC Motor KV: 50 ##rpm/v## no load
Motor Resistance: 0.01 ##Ohm##
DC Battery Voltage: 110 ##V##
Battery Current Limit: 30 ##A##
Gear Ratio: 3.21:1 ##final##
Peak Velocity MPH: 100.08 ##mph##
Peak Velocity m/s: 44.74 ##m/s##
Electrical Power: 3300 ##W##
Peak Mechanical Power: 3291 ##W##
Efficiency @ Peak Velocity: 30.32 ##mi/kWh##

##I^2 \cdot R=W##

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

##F_D=1/2pv^2C_DA##

https://en.wikipedia.org/wiki/Drag_(physics)

##W=Fs##

https://en.wikipedia.org/wiki/Work_(physics)
 
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  • #12
To calculate the terminal velocity of the vehicle:

##V_t=\sqrt{2mg/pAC_d}##

https://en.wikipedia.org/wiki/Terminal_velocity
Using 136.08 ##kg## mass I get 190.55 ##m/s## or 426.24 ##mph## terminal velocity.

Interestingly if the mass of the vehicle is increased to 13600.08 ##kg##, I calculate 1905.02 ##m/s## or 4261.41 ##mph## terminal velocity.
 
  • #13
metastable said:
[...]terminal velocity[...]
You are interested in increasing efficiency -- how little energy you can consume to get a vehicle from point A to point B. Right?

For a typical wheeled vehicle, you can push it from point A to point B by applying a force significantly less than the vehicle's weight. Right? [The invention of the wheel was a good thing]

If part of your trajectory from point A to point B involves forward progress at terminal velocity then that portion of the journey is encountering resistance equal to the vehicle's weight. You can't win the game by throwing away energy that recklessly.
 
  • #14
jbriggs444 said:
For a typical wheeled vehicle, you can push it from point A to point B by applying a force significantly less than the vehicle's weight. Right? [The invention of the wheel was a good thing]

If I use:

##F_D=1/2pv^2C_DA##

https://en.wikipedia.org/wiki/Drag_(physics)

If I've done my calculations correctly, baseline electric vehicle at constant 100.08 ##mph## uses 73.55 ##N##.
 
  • #15
metastable said:
If I use:

##F_D=1/2pv^2C_DA##

https://en.wikipedia.org/wiki/Drag_(physics)

If I've done my calculations correctly, baseline electric vehicle at constant 100.08 ##mph## uses 73.55 ##N##.
Since the baseline vehicle masses 136 kg, its weight is approximately 1360 N. So yes, 73.55 is less than 1360 by a factor of about twenty to one.

You do not want to allow your vehicle to get anywhere near terminal velocity. Especially if is spending that portion of the journey not making forward progress.
 
  • #16
I feel as if this video at t=5m:01s relates to the problem at hand, but I'm not sure how to describe it with an equation:
 
  • #17
metastable said:
I feel as if this video at t=5m:01s relates to the problem at hand, but I'm not sure how to describe it with an equation:
Are you trying to optimize for time required or energy utilized?
 
  • #18
Ideally I'd like to see both vehicles get from point A to point B in the same time, but have the parabola riding vehicle use less energy.
 
  • #19
metastable said:
Ideally I'd like to see both vehicles get from point A to point B in the same time, but have the parabola riding vehicle use less energy.
The energy losses that you are budgeting for are motor, drive train and wind resistance, I believe.

Do the losses from the motor and drive train disappear if you turn off the motor and shift into neutral?
Do both cars start from a stand-still?
Is point A at the same altitude as point B?

The thrust of these questions is to set the stage for a possible "no free lunch" conservation of energy argument. The idea being that there is a certain minimum energy requirement. And that, perhaps, it can be approached arbitrarily closely with a simpler strategy.

In addition, being forced to live within a precomputed energy budget would rule out huge swaths of strategies.
 
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  • #20
jbriggs444 said:
Do both cars start from a stand-still?
Is point A at the same altitude as point B?

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
 
  • #21
metastable said:
I feel as if this video at t=5m:01s relates to the problem ...
What that video shows is that the ball that spends more time with reduced GPE (gravitational potential energy) spends that time at higher speed. If the path distance doesn't increase as much as the increase in speed, the ball's average speed is higher. This is clear in the first comparison, where there is a single drop at the start and a single rise at the end. What wasn't shown in the video is a path where the tops of the peaks are higher than the straight line path, where part of the time is at a higher GPE. An obvious example would be a single rise at the start, a straight line, and a single drop at the end. There's a path where the average GPE on the curved path is just decreased enough to compensate for the longer path, in which case the time from point A to point B would be the same. If the average GPE doesn't compensate for the longer path, the curved path takes longer. Note that this is a comparison of time, not of efficiency.

Getting back to the original question, unlike a rocket in space, an object operating in the atmosphere doesn't get the same increase in velocity (delta v), from the same impulse (force · time), due to drag, and the atmosphere reducing the "exit velocity" of the exhaust from the compressed air tank.

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.

Another example would be a jet engine based land vehicle. Since jet engines are more efficient at higher speeds, than a path that involves spending more time at the jet engines ideal speed (taking into account drag), should be more efficient.
 
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  • #22
jbriggs444 said:
The energy losses that you are budgeting for are motor, drive train and wind resistance, I believe.

Do the losses from the motor and drive train disappear if you turn off the motor and shift into neutral?
Do both cars start from a stand-still?
Is point A at the same altitude as point B?

The thrust of these questions is to set the stage for a possible "no free lunch" conservation of energy argument. The idea being that there is a certain minimum energy requirement. And that, perhaps, it can be approached arbitrarily closely with a simpler strategy.
To put a finer point on it, for most transportation from point A to point B (and often back to point A) type problems, the minimum energy requirement is zero. The billiards balls on the tracks in the video, for example; in all cases the energy required to get from one side to the other and back is just about zero. This can be useful to remember when, for example, asking the question: what is the maximum fuel efficiency possible for a car?

My first thought on seeing the OP (after oh-god-I'm-not-going-to-read-this-wall-of-math) was that this idea either does nothing (because no energy is required to do what is desired) or is looking for a perpetual motion machine.

In either case, I agree you need to very precisely define what you want to do and what you want to achieve. The answer is probably simple after you make those decisions.
 
  • #23
russ_watters said:
To put a finer point on it, for most transportation from point A to point B (and often back to point A) type problems, the minimum energy requirement is zero. This can be useful to remember when, for example, asking the question: what is the maximum fuel efficiency possible for a car?
Since the baseline vehicle is required to maintain its maximum possible constant speed at its optimal gear ratio (for achieving max constant velocity), and since it has a drag coefficient and there is an air density, the energy requirement for the baseline vehicle can't be zero, and at optimal gear ratio for maximum speed there is only one possible efficiency.
 
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  • #24
metastable said:
Since the baseline vehicle is required to maintain its maximum possible constant speed at optimal gear ratio with a drag coefficient and air density, so the energy requirement for the baseline vehicle can't be zero, and unless I'm mistaken it has only one possible efficiency.
Fair enough; if you define a "baseline vehicle" to have certain traits you are not allowed to optimize, then the minimum energy requirement for that vehicle is non-zero. To me, though, adding a bunch of arbitrary complexities to the scenario you are trying to analyze just gets in the way of what you want to know. If this is all you wanted to know:
Ideally I'd like to see both vehicles get from point A to point B in the same time, but have the parabola riding vehicle use less energy.
...then you really don't need any of that extra stuff. Since in the most basic case - the zero energy case - the vehicle riding the deepest curve gets to point B fastest, any attempt to add speed to the flat vehicle adds to the energy requirement. I do not think trying to quantify the energy requirement matters in answering the question or exploring similar questions.

The secondary question - regarding an Oberth Maneuver - hopefully you recognize by now that you don't even need it.
 
  • #25
russ_watters said:
the zero energy case - the vehicle riding the deepest curve gets to point B fastest
On this point I'm not so sure because as a result of the air drag during one descent, with zero energy it could possibly not make it all the way back to the surface under certain circumstances, but in the actual scenario we are expending energy along the way to return to the surface.
 
  • #26
metastable said:
On this point I'm not so sure because as a result of the air drag...
I said in the zero energy case. As in zero energy required, meaning no drag or other losses.

Whether it is true for other cases or not will likely depend on the case. I'm sure for shallow depths the curve will help and I suspect there is an optimal depth above which it starts doing more harm than good, but it will vary depending on the specifics of the case. That's why I don't think the exercise is very useful: the answer you get is as arbitrary as the scenario. That's why I think it is so important to focus on what you really want to know rather than putting so much effort into dealing with a bunch of arbitrary constraints, otherwise you may find that you pre-determined the answer to your question by applying constraints you shouldn't have, leading to a conclusion that is wrong or unhelpful. Again, I think the arbitrary constraints get in the way of what you really want to know.
 
  • #27
russ_watters said:
That's why I think it is so important to focus on what you really want to know

I want to know if there is another way to travel between 2 arbitrarily distant points on the surface (in the same amount of time), that is more efficient than using an electric wheel-driven vehicle of the same mass, frontal area and drag coefficient traveling in a straight line at constant speed along the surface.
 
  • #28
metastable said:
I want to know if there is a way to travel between 2 arbitrarily distant points on the surface in the same amount of time, that is more efficient than using an electric wheel-driven vehicle of the same mass, frontal area and drag coefficient traveling in a straight line along the surface.
You didn't define all the necessary constraints, so for fun I'll fill them in for you and say yes; the parabola accomplishes that. Now what?

I'm not trying to be coy or difficult here, I'm trying to break through what I see as a wall of unproductive thinking and try to see where this is going.
 
  • #29
If the answer is "yes" then I'm satisfied.
 
  • #30
metastable said:
If the answer is "yes" then I'm satisfied.
Fair enough...but does the title of the thread still have any relevance?
 
  • #31
Ideally I was hoping to be able to calculate how efficient each vehicle is (I can already calculate the baseline electric).
 
  • #32
metastable said:
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.
 
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  • #34
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.
 
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  • #35
jbriggs444 said:
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."
 
<h2>1. What is a compressed-air-powered parabola riding vehicle?</h2><p>A compressed-air-powered parabola riding vehicle is a type of vehicle that uses compressed air as its main source of energy to propel itself and ride along a parabolic path. It typically consists of a pressurized tank of compressed air, a propulsion system, and a parabola-shaped track or ramp.</p><h2>2. How does a compressed-air-powered parabola riding vehicle work?</h2><p>The vehicle works by releasing compressed air from the tank through a nozzle, which creates a thrust force that propels the vehicle forward. The parabola-shaped track or ramp provides the necessary curvature for the vehicle to follow a parabolic path, allowing it to reach high speeds and perform various maneuvers.</p><h2>3. What are the advantages of using compressed air as an energy source for this type of vehicle?</h2><p>Compressed air is a clean and renewable energy source, making it more environmentally friendly than fossil fuels. It also allows for quick refueling and can be easily stored, making it a practical option for powering vehicles. Additionally, the use of compressed air can result in lower operating costs compared to traditional fuel-powered vehicles.</p><h2>4. Are there any limitations or drawbacks to using compressed air as an energy source for this type of vehicle?</h2><p>One limitation is that compressed air has a lower energy density compared to traditional fuels, which means the vehicle may have a shorter range or require more frequent refueling. Another drawback is that the compression process can generate heat, which can affect the performance and efficiency of the vehicle.</p><h2>5. How can the energy efficiency of a compressed-air-powered parabola riding vehicle be improved?</h2><p>The energy efficiency of the vehicle can be improved by optimizing the design and materials used for the propulsion system, reducing friction between the vehicle and the track, and implementing regenerative braking systems to capture and reuse energy. Additionally, using renewable sources of energy to compress the air can also contribute to improving the overall energy efficiency of the vehicle.</p>

1. What is a compressed-air-powered parabola riding vehicle?

A compressed-air-powered parabola riding vehicle is a type of vehicle that uses compressed air as its main source of energy to propel itself and ride along a parabolic path. It typically consists of a pressurized tank of compressed air, a propulsion system, and a parabola-shaped track or ramp.

2. How does a compressed-air-powered parabola riding vehicle work?

The vehicle works by releasing compressed air from the tank through a nozzle, which creates a thrust force that propels the vehicle forward. The parabola-shaped track or ramp provides the necessary curvature for the vehicle to follow a parabolic path, allowing it to reach high speeds and perform various maneuvers.

3. What are the advantages of using compressed air as an energy source for this type of vehicle?

Compressed air is a clean and renewable energy source, making it more environmentally friendly than fossil fuels. It also allows for quick refueling and can be easily stored, making it a practical option for powering vehicles. Additionally, the use of compressed air can result in lower operating costs compared to traditional fuel-powered vehicles.

4. Are there any limitations or drawbacks to using compressed air as an energy source for this type of vehicle?

One limitation is that compressed air has a lower energy density compared to traditional fuels, which means the vehicle may have a shorter range or require more frequent refueling. Another drawback is that the compression process can generate heat, which can affect the performance and efficiency of the vehicle.

5. How can the energy efficiency of a compressed-air-powered parabola riding vehicle be improved?

The energy efficiency of the vehicle can be improved by optimizing the design and materials used for the propulsion system, reducing friction between the vehicle and the track, and implementing regenerative braking systems to capture and reuse energy. Additionally, using renewable sources of energy to compress the air can also contribute to improving the overall energy efficiency of the vehicle.

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