# B Land Based Oberth Manuever?

#### metastable

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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#### Dale

Mentor
I think that the only thing worse than wall-of-text posts is wall-of-math posts. Please

b) use standard variables and the common form of equations whenever possible

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#### Ibix

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.

#### Ibix

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?

#### metastable

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.

#### Dale

Mentor
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.

#### metastable

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.

#### Dale

Mentor
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.

#### Dale

Mentor
After a private discussion and a brief tutorial on LaTeX, @metastable will provide an improved analysis and question. So thread reopened.

#### metastable

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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#### metastable

To calculate the terminal velocity of the vehicle:

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

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.

#### jbriggs444

Homework Helper
[...]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.

#### metastable

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 $N$.

#### jbriggs444

Homework Helper
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.

#### metastable

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:

#### jbriggs444

Homework Helper
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?

#### metastable

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.

#### jbriggs444

Homework Helper
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.

#### metastable

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

#### rcgldr

Homework Helper
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.

#### russ_watters

Mentor
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.

#### metastable

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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#### russ_watters

Mentor
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.

#### metastable

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.

"Land Based Oberth Manuever?"

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