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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)

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?

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^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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