• ## TandEM

A visualization of a solution for a Jupiter target

TandEM is an interplanetary trajectory with multiple fly-bys. The transcription proposed here uses one deep-space manouvre per leg, and a launch with the rocket Atlas 501, the performances being taken here. The global optimisation problems we propose draw inspiration from the work performed in April 2008 by the European Space Agency working group on mission analysis on the mission named TandEM. TandEM is an interplanetary mission aimed at reaching Titan and Enceladus (two moons of Saturn). The problems, written here as minimisation problems, are actually maximising the mass delivered to a high eccentricity final orbit. With respect to the other mgadsm problems proposed in these pages, the transcription of the TandEM problem is slightly different as it evaluates directly the spacecraft mass at departure, and then at the end, using an interpolation of the launcher capability table (declination and C3 vs mass). The fly-by sequence can be selected among a list of suggested ones. Each fly-by sequence creates a different problem instance. The wide bounds used are valid for all problem instances.

### The code

1. MATLAB: use the function tandem.m and pass to it the MGADSMproblem variable contained in tandem.mat after having selected the correct sequence in MGADSMproblem.sequence
2. C++: call the function "double tandem(const std::vector & x, double& tof, const int sequence_[])" provided in the GTOPtoolbox.
3. C++ (PaGMO): use the class pagmo::problem::tandem
4. Python 2.7 (PyGMO): use PyGMO.problem.tandem(prob_id,tof_c).obj_fun(x) to minimize -log(m_final)

### Problems Description

We propose two sets of 25 different instances:

1. Unconstrained continuous optimisation (solutions database)
2. Constrained continuous optimisation (solutions database) : A constraint on the total mission duration is considered: x(5)+x(6)+x(7)+x(8) < 3652.5. [In the C++ code c_ineq (output of the tandem function) needs to be negative]

### Box bounds

The box bounds on each of the decision vector variables are given below.

## Units

x(1) t0 5475 9132 MJD2000
x(2) Vinf 2.5 4.9 km/sec
x(3) u 0 1 n/a
x(4) v 0 1 n/a
x(5) T1 20 2500 days
x(6) T2 20 2500 days
x(7) T3 20 2500 days
x(8) T4 20 2500 days
x(9) eta1 0.01 0.99 days
x(10) eta3 0.01 0.99 n/a
x(11) eta3 0.01 0.99 n/a
x(12) eta4 0.01 0.99 n/a
x(13) r_p1 1.05 10 n/a
x(14) r_p2 1.05 10 n/a
x(15) r_p3 1.05 10 n/a
x(16) b_incl1 -pi pi n/a
x(17) b_incl2 -pi pi n/a
x(18) b_incl3 -pi pi n/a