How a continuous thrust orbit transfer can be compared with Hohmann transfer?

Question

Suppose that I have found the solution of an optimal continuous thrust planar orbit transfer from circular orbit A to orbit B by means of a numerical method. It's considered that the mission must be done in a fixed final time and the performance index is to minimize the magnitude of the commanded acceleration to the spacecraft. After solving the problem, I have the optimal time histories of the accelerations, tangential and normal to velocity vector.

• How can I compare the performance of the solution with an equal Hohmann transfer?
• Is that a good idea to compare a two impulse transfer with a continuous thrust one?
• Theoretically, I think that continuous thrust transfers must have better efficiency with respect to a limited two impulse transfer; is that correct?
• How can I calculate Delta-v for this problem; by simply integrating the magnitude of the acceleration command vector?

I am interested in comparing the efficiency of the continuous thrust acceleration command with the ideal Hohmann transfer regardless of the Isp of these two different types of propulsion systems. Actually, I want to know that continuous application of thrust is better or impulsive application, from theoretic and mathematical point of view.

Answer 1

How can I compare the performance of the solution with an equal Hohmann transfer?

There are multiple metrics you can use. One is $\Delta \text{v}$, the cumulative change in velocity. This will be a losing metric for a continuous thrust compared to impulsive burns.

Another metric is dollars: How much does it cost? As impulsive burns are impossible to achieve, making measuring their cost a bit difficult. What you can do is compare the cost of a vehicle that uses high thrust but low specific impulse chemical thrusters with the cost of a vehicle that uses low thrust but high specific impulse non-chemical thrusters.

Is that a good idea to compare a two impulse transfer with a continuous thrust one?

If the comparison is $\Delta \text{v}$, you lose. Non-impulsive burns (aka finite burns) inherently suffer from gravity loss, also known as gravity drag.

Theoretically, I think that continuous thrust transfers must have better efficiency with respect to a limited two impulse transfer; is that correct?

This depends on how one defines "efficiency". Using $\Delta\text{v}$ as the sole metric, a non-impulsive burn solution is going to be a losing proposition compared to an impulsive burn solution. However, $\Delta\text{v}$ is not necessarily a good metric. For one thing, there is no such thing as an engine that can provide impulsive burns.

This means that other metrics must necessarily come into play. Time, cost, complexity, risk, something else: These can all be good metrics. Combining those multiple metrics into a single "efficiency" metric is a game that goes under the name of multi-criteria decision making (MCDM). Written very sarcastically, one of the "nice" features of MCDM is that an analyst can twist the combinatorics used to arrive at the singular metric in many ways.

How can I calculate Delta-v for this problem; by simply integrating the magnitude of the acceleration command vector?

That's the correct approach regarding $\Delta\text{v}$. It's also a losing approach regarding low thrust / high specific impulse thrusters. You'll want some other metric such as a cost-benefit-risk analysis.

Answer 2

The Hohmann transfer is (usually) the most energy-efficient transfer between two ideal circular orbits. The idealised Hohmann transfer has two instantaneous burns, one at the periapsis, one at the apoapsis.

Never forget the Oberth effect:

the most energy-efficient method for a spacecraft to burn its fuel is at the lowest possible orbital periapsis, when its orbital velocity (and so, its kinetic energy) is greatest.

So we get more efficiency by burning our fuel when the ship and the fuel has its highest kinetic energy (relative to the primary we're orbiting). Of course, we also need some fuel for when we reach apoapsis so we can circularize the orbit at the larger radius.

Your continuous thrust manoeuvre wastes fuel by not burning it when it's got a lot of KE. Instead, you're using delta-vee to slow down the ship and to slow down all that heavy fuel you're carrying.

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The Hohmann isn't always the most efficient transfer. The bi-elliptic transfer

requires a lower amount of total delta-v than a Hohmann transfer when the ratio of final to initial semi-major axis is 11.94 or greater, depending on the intermediate semi-major axis chosen.

The long transfer time of the bi-elliptic transfer is a major drawback for this maneuver. It even becomes infinite for the bi-parabolic transfer limiting case. The Hohmann transfer takes less than half of the time because there is just one transfer half-ellipse.

Answer 3

Your question is interesting. Long time ago I published a paper about the advantages of continuous thrust strategy. For what concern the DeltaV, I simply make the ratio of the impulsive DeltaV (Hohmann) with respect to the DeltaV (real, continuous with optimised thrust orientation).

DeltaV efficiency (impulse/real) in the paper (for figure 3) is about 67% meaning that the real continuous DeltaV is about 50% (1/67%-1) higher than the Hohmann one.

Ref. ADVANTAGES OF A CONTINUOUS THRUST STRATEGY. FROM A GEOSYNCHRONOUS TRANSFER ORBIT,. USING HIGH SPECIFIC IMPULSE THRUSTERS. Christophe R. Koppel. https://issfd.org/ISSFD_1999/pdf/OC1_1.pdf