equivalent to oberth maneuver?

[metastable]

riddle me this...

are the following 2 oberth maneuvers equivalent (ignoring the weight of the cable and friction)

in the following diagram, the vehicle (in vacuum) accelerates from gravity pulling down the water tank until the tank reaches the flat section at the bottom of the tunnel, at which point the vehicle applies a mechanical impulse to the tether sufficient to stop the tank, which transfers 100% of the mechanical impulse plus the kinetic energy of the tank to the vehicle.



&

in the following diagram, the vehicle and tank (in vacuum) accelerates from gravity until the vehicle & tank reaches the flat section at the bottom of the tunnel, at which point the vehicle applies a mechanical impulse to the tether sufficient to stop the tank, which transfers 100% of the mechanical impulse plus the kinetic energy of the tank to the vehicle, the vehicle then coasts back to the surface on a second ramp.



assuming the masses of tank and vehicle and the depth of the ramp are the same, will the energetic requirements and final vehicle velocity at the surface be the same?

Where:

M_t = 240 = Trailer Tank + Water Mass ( K i l o g r a m s )
M_v = 10^3 = Passenger Vehicle Mass ( K i l o g r a m s )
T_d = 30 = Free Fall Duration ( S e c o n d s )
V_i = 0 = Initial Velocity Before Free Fall ( m / s )
G_e = 9.80 = Gravity Acceleration at Earth’s Surface ( m / s^2 )
W_i = 1.28 ∗ 10^7 = Mechanical Impulse Energy ( J o u l e s ) That Brings Trailer to 0 Velocity At Ramp Bottom

W_i = M_t * V_i * T_d * G_e + ( M_t * T_d^2 * G_e^2 + M_t * V_i^2 ) / 2 + ( M_t^2 * ( V_i + T_d * G_e )^2 ) / ( 2 * M_v )

W_i = 240*0*30*9.80665+(240*30^2*9.80665^2+240*0^2)/(2)+(240^2*(0+30*9.80665)^2)/(2*1000)

W_i = 1.28 ∗ 10^7 Joules

(this assumes vertical ramp, but as long as it is the same depth it works out the same energy regardless of slope)

[metastable]

I suspect the first vehicle will take less impulse energy on the tether to stop the tank, and so its final velocity will be lower, however, if the first vehicle first uses its impulse on the tether till the tank stops, and then uses some more impulse energy on the ground (same total impulse energy as the 2nd vehicle), then I suspect both final impulse energies and velocities will be the same (the final vehicle kinetic energy being greater than the impulse energy used on account of the tank PE->KE transferring through the tether to the vehicle).

[metastable]

So I've done some calculations for both vehicles and it seems both are equivalent (as suspected). In other words it isn't necessary for the vehicle to drop into a gravitational well to perform an Oberth Maneuver.... only the reaction mass needs to.

For example with:

A = Tank Mass kg = 1000kg
B = Vehicle Mass kg = 1000kg
D = Gravity m/s^2 = 9.8m/s^2
E = Height m = 1000m

For the second vehicle (both vehicle and reaction mass drop into well):
K = Comparison Vehicle Velocity at Surface = 242.48m/s
L = Comparison Vehicle Vehicle KE at Surface = 29400002 J
M = Comparison Vehicle Mechanical Impulse = 19600001 J

For the first vehicle (only the reaction mass drops into well):
H = Vehicle Final Velocity After Impulse = F+G = 197.98989m/s
I = Mechanical Impulse = ((1/2)AF^2)+(((1/2)BG^2)) = 9800000 J
J = Vehicle Kinetic Energy After Impulse = (1/2)BH^2 = 19599998 J
N = Difference in Mechanical Impulse = M-I = 19600001 J -9800000 J = 9800001 J
P = Vehicle Kinetic Energy After Impulse + Difference in Mechanical Impulse Used on Ground = J+N = 19599998 + 9800000 = 29399998J
Q = P = Vehicle Kinetic Energy on Surface After Mechanical Impulse on Tether and After Mechanical Impulse on Ground, Same Total Impulse as M = 29399998 J

Comparison
L =~ Q

In other words as long as the reaction mass drops into the well (whether the vehicle enters the well or not), and the same total impulse is used (entirely on the cable with the second vehicle or combination cable/ground with first vehicle), the final vehicle kinetic energy and velocity at the surface is the same.

The first vehicle is more practical since the passengers don't need to experience negative and positive G forces, there doesn't need to be 2 separate ramps to depth, and the same ramp can be used to accelerate the vehicle in any direction.

Only the empty tank is lifted... the water is emptied into an underground chamber where it is heated with geothermal and run through a turbine and steam pipe to the surface, enabling power generation both as the water tank descends and ascends.

Assuming 70% recovery of the vehicle's kinetic energy at the destination via regen braking, more energy is recoverable from the vehicle than was used for the mechanical impulse.