GTOC 9, Multiple Space Debris Rendezvous Trajectory Design in the J2 environment

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1 GTOC 9, Multiple Space Debis Rendezvous Tajectoy Design in the J envionment Macus Hallmann, Makus Schlottee, Ansga Heidecke, Maco Sagliano Fedeico Fumenti, Volke Maiwald, René Schwaz Institute of Space Systems, DLR, Robet-Hooke-St. 7, 359 Bemen (Gemany) May, 7 Abstact This pape discusses the methods used by the team fom the Geman Aeospace Cente (DLR) fo solving the 9th Global Tajectoy Optimization Competition (GTOC) poblem. The GTOC is an event taking place evey yea lasting oughly one month duing which the best aeospace enginees and mathematicians wold wide challenge themselves to solve a nealy-impossible poblem of tajectoy design. Intoduction The pape is oganized as follows: section ecaps the elevant figues of the poblem statement; section 3 points out how the oveall stategy was developed; section focuses on the combinatoial pat of the poblem and 5 on the tansfe between two debis. Finally, in sections and 7 we discuss the esults and daw some conclusions. Macus.Hallmann@dl.de Poblem Statement The task was to design a scenaio with n missions which collect a given set of 3 space debis on Sun-synchonous Low Eath Obits. The following cost function has to be minimized: J = n i= c i + α(m i m dy ) () whee c i is the base cost (inceasing linealy duing the competition time fame fom 5 MEUR to 55 MEUR). Each spacecaft initial mass m is the sum of dy mass, popellant mass and N times the deobit package mass: m = m dy + m p + Nm de, with m de = 3kg. The α paamete is set to be. MEUR/kg. In ode to contol the spacecaft five impulsive manoeuves ae allowed duing the debis to debis tansfe. In addition to an impulsive manoeuve at depatue and at aival. The oveall time between two successive debis endezvous, within the same mission, must not exceed 3 days. The deobit package deploy-

2 ment takes 5 days. That esults in a maximum tansfe time of 5 days. The time between two missions must be at least 3 days. And the mission must take place between 37 MJD and 9 MJD. The adius of peicente p is constained to be smalle than m = km. The spacecaft dynamics is descibed by the following set of Odinay Diffeential Equations: ẍ = µx 3 ( + 3 J ÿ = µy 3 ( + 3 J z = µz 3 ( + 3 J ( eq ( eq ( eq ) ( 5 ( z ) )) ) ( 5 ( z ) )) ) (3 5 ( z ) )) () which descibes a Kepleian motion petubed by an oblate Eath. The obital elements of the space debis ae given fo a cetain epoch and ae popagated via a moe simplified model than equation (): ) Ω = 3 ( J eq ncosi p ) ω = ( 3J eq n ( p 5cos i ) (3) It can be seen that the ascending node is the obital element which encountes the most vaiations caused by J. That will have an impact on ou oveall stategy. Fo moe details on the poblem statement efe to the GTOC 9 poblem statement [] o visit 3 Oveall Stategy The poblem we have to solve can be classified as Time Dependent Taveling Salesman dot [deg/day] height (km) Figue : Ω ove height Poblem, with a nested optimal contol poblem fo each tansfe. One way to solve the combinatoial pat would be to explicitly evaluate all possible combinations. That appoach is only applicable fo small dimensions. In ou case we have 3 debis to sot fo the best sequence (which gives 3! pemutations) and than we have to chop them into n missions, which causes anothe dimension. Even with the most poweful computes and the smatest appoach to calculate the V fo tansfe fom one debis to the next one it would take yeas to detemine the entie tee. Section deals about how we solved the combinatoial pat of the poblem. Befoe we look into the tansfe we analyze the design space to get some easonable boundaies fo ou design vaiables like tansfe time, delta v ange, numbe of missions we need and so on. The fist impotant question to answe is how ae the debis pieces spead out egading inclination i, eccenticity e and semi majo axis a. Figue shows the ange of the obital elements fo the debis pieces. It can be seen that the obits ae nealy cicula, inclination anges fom 9 to and the obital height (o a eq ) goes fom km to 9 km. These elements do not change ove

3 Numbe of debis Numbe of debis Numbe of debis a- eq [km] Inclination [deg] e Figue : Debis obital elements time fo the dynamic model which is used fo the debis. If we only conside the change in inclination and semimajo axis duing a tansfe, the poblem can be teated as a simple Taveling Salesman Poblem (TSP). The V which is needed to tavel fom debis A to debis B is the sum of the inclination change V inc plus the V sma fo the Hohmann tansfe: V inc = V sin((i A i B )/) V sma = µ/ ( k/( + k) ) + µ/( k)( /( + k)) V AB = V inc + V sma Whee k is the atio between a A and a B, as- can set up a cost matix which gives us the cost fo going fom one debis to the next debis, ignoing the phasing in tue anomaly and ight ascension, and we can apply a genetic algoithm implemented in Matlab to find the optimal oute. Figue 3 shows the esult of one un with 5 populations and 5 iteations. The total distance we get is aound 5 m/s, so an aveage V of.7 m/s fo one tansfe. In theoy and with no constaints on the mission time, this esult would equal to a J value below MEUR. In pactice we have to fulfill the yeas mission time constaint and the 5 day tansfe time constaint. That foces us to invest some easonable V fo changing the ight ascension. That can be done in two ways: Total Distance = 5.9, Iteation = 7775 a diect plane change Inclination [deg] a- eq [km] Figue 3: Optimal Path in the i and a TSP suming cicula obits. With that equations we an indiect plane change via a change in semi majo axis Let us compae the V fo both cases. If all othe elements besides Ω ae the same we can use a simila equation like we used fo the inclination change fo the diect plane change: V Ω = V sin((ω A Ω B )/) () Like the inclination change its a quite cost intensive maneuve. If thee is enough time 3

4 available an indiect tansfe maybe cheape. An impotant figue to look at is the change in ight ascension fo the debis which is plotted in figue ove height (with an inclination equal to 9 deg). Out of that we can see that the obits ae difting between. deg/day and. deg/day. Depending on ou initial height value we can get a diffeential dift aound.5 deg/day. Assuming a day tansfe time we could ovecome a delta of deg in Ω. Fo the indiect change we pefom two Hohmann like tansfes, the fist one to each the desied dift obit, the second one to get the semi majo axis of the aival debis. Let us compae some use cases to see if the dift appoach is bette o not. Ou obital elements fo both debis ae: a = km + eq, e =, i = 9deg. So we only conside a change in Ω. On Figue we can see that the of the depating and aival debis. The inclination change we can eithe pefom befoe o afte the dift change maneuves. And the choice has an impact on the equied V fo the dift change as it can be seen when looking at equation 3, cause its a function of inclination. With that thoughts we have a good set up fo ou combinatoial poblem, which will be discussed late. V aveage [m/s] Numbe of Stats V [m/s] diect indiect 5 days indiect days indiect 5 days indiect days indiect 5 days [deg] Figue : V ove Ω V needed fo a diect change goes nealy linea with Ω (dashed line). The othe cuves in the figue ae epesenting an indiect tansfe fo diffeent tansfe times (5:5:5 days). We can conclude that fo tansfe times lage than 5 days it is always bette to make an indiect tansfe. And that does even not take into account that we may save some V cause of diffeent semi majo axis and inclination Figue 5: J ove n and V aveage The next inteesting question to look at is how many missions we may need (do we stack one launche full as possible) and what s the J looks like. Assuming a ange of diffeent aveage tansfe V s and numbe of missions(n) we get figue 5, which shows the J function in MEUR. It can be seen that we should not use the maximum populsion available, cause if we analyze the J value fo an aveage V of 3 m/sec, the J value would be fo 9 stats 9. MEUR, stats 7. MEUR and afte that it goes up again. Although we would have times the base cost fo the launche instead of only 9 times, the used popellant mass goes in quadatic. And if we use the total allowed 5 kg we ae not optimal. So we have to educe the total allowed fuel o V pe mission by % to %, depending on how many missions we need. The issue is that we do not

5 know that befoe hand. Combinatoial Poblem In section 3 we set up ou stategy how to tackle the poblem. We can estimate the cost o V to pefom a debis to debis tansfe. The issue is, the poblem is time vaiant, cause ou cost matix depends on at which epoch we pefom the tansfe. O using the TSP syntax its not a city to city outing poblem, its a boat to boat one, whee the boats ae sailing aound. Maximum allowed tansfe time [days] Numbe of missions Figue : Maximum tansfe time ove n Befoe we go into detail of the gaph implementation we have to look a bit deepe into the handling of the tansfe time. We have seen fo the dift change maneuve, it would be bette to have a lage tansfe time available. So it looks obvious to use the maximum allowed 5 days. But if we use that fo all of the tansfes, we ae ending up having a total mission time which is lage than the allowed yeas. And that is even dependent on how many missions we need. Figue shows the maximum allowed tansfe times ove the numbe of missions we need, which is again not known befoe we have solved the gaph poblem. In fact we can only use aveage tansfe times between 5 and 9 days. In that case the next question is, if we let the tansfe time be a design vaiable in the combinatoial poblem o if we keep it to an aveage fixed value. We decided to keep it fixed, cause the pemutation space is lage aleady. That appoach allows us also to use a look up table fo all possible tansfes with a day gid size, which is pecalculated and loaded into the gaph algoithm. In the cost matix we can also handle the adius of peiapsis constaint, by setting the V = fo all tansfes whee p < pm. Fo solving the outing we tied to use the same genetic algoithm, which was deliveing good esults fo the inclination-sma outing poblem, on the time dependent one, but it didn t bought any good esults. Instead we developed a gaph algoithm which uses a cetain beam width. Each mission can be epesented as a gaph o tee (see figue 7). Fo the fist mission we have 3 nodes o debis as an option to stat fom. Keeping in mind that the J-function doesn t give any penalty at which debis we stat ou mission, it s a fee design paamete. So ou initial beam width would be 3. Than we can calculate fo each of that 3 debis possible debis tansfes, cause each debis should be only visited ones. That gives us 3 times possible options. Thee ae many methods to exploe such kind of tees o gaphs: Depth Fist Seach Beadth Fist Seach Beam Seach Geedy Seach 5

6 In ode to ovecome that we can apply a beam seach. Instead of only taveling along one path we select k best options and take them with us. Depending on the beam width k the computational time of couse gows in that case. In ou case we have an initial beam width of 3 fo the fist mission. On the next level we would have 3 = 5 maximum beam width and so on. We limited ou beam width fo the fist uns to and got aleady good esults fo J aound. The entie idea behind that method is, that we can not look into the futue, so we have to take also some bad paths with us to hope that they tun into the golden paths in the end. At each level we check each possible solution fo uniqueness. That can also be explained when looking again at figue 7, the sequence -3- is equal to 3--, because fo the next level we would have the same node, in that case node, to stat fom and the left ove debis-set is the same. So we only take the best sequence out of that two, because ou beam width is limited and we want as many diffeent pemutations to find the golden path. The gaph algoithm is implemented in matlab and takes oughly h computation time on a Intel Xeon CPU E3 3.5GHz, with a beam width set to. When we un the tee afte the fist mission we have the advantage that the left nodes ae getting less and the computation time goes down. With the 5 kg popellant we can oughly achieve a maximum V of 5 m/s. But with the esults we had fom ou qualitative Jfunction analysis we set ou maximum to 5 m/s and we also made uns with lowe values. Fo the fist mission we wee able to pefom 3 tansfes. But the beam width at that point was only aound to. So in that case we don t have enough pemutations fo the next 3 Figue 7: Mission Gaph The Depth Fist Algoithm tavels along the left o ight side of the tee. In ou case it would just visit the debis eithe in the sequence ::3 o 3:-:. We could add some backtacking if we get stuck somewhee o if we apply some heuistic, like aveage V is getting to woth along the cuent path. Beadth Fist in the opposite exploes the tee fist hoizontal and than continues to the next level with all possible pemutations.in ou case like mentioned aleady a full beadth fist seach would not be possible cause the pemutation would be to lage. Geedy seach is like depth fist, but it makes a decision on some heuistic which path to take. The easiest implementation is to make the decision on the next shotest (lowest V ) path. Thats the fist method we used to find a pope sequence and it bought esults fo J aound 5. The Geedy seach has the typical tee dawback that the best cookies ae eaten fist and the bad ones ae left ove in the end, and we still have to eat them, cause we have to collect the entie set and not only a subset. And that s something we obseved when unning a geedy seach on the tee. The last tansfes had a quite high V and that caused a high numbe of missions and a esulting high J Value.

7 missions. So instead of taking the maximum tansfe solution we took the one with a highe beam width. Ou algoithm is setup in a way that the numbe of tansfes is the same fo all beams, that may not be the optimal choice and some futhe investigation may be pefomed to see if a fee numbe of tansfes bings a significant impovement. Ou final sequence will be discussed in the Results section 5 Tansfe Poblem Befoe we solve the debis to debis final tansfe we e-optimize the sequence coming out of the beam seach algoithm. Like we discussed in section aleady we kept the tansfe time fo all tansfes to a fixed value. That will be e-optimized using the local optimize fmincon in matlab. The Cost function in that case is the sum of the V s fo all tansfes in that paticula mission. Ou design vaiables ae the tansfe times. The uppe bound is set to 5 days, constained by the poblem statement. The lowe bound to day, cause we may need some time fo the final phasing of the tue anomaly, which we have ignoed completely so fa. In the combinatoial pat we used a fixed gid size fo the tansfe time (e.g. always 7 days fo the fist missions and than 9 o fo the emaining, depending how many debis is left to collect). We also have to intoduce an inequality constaint that the sum of the tansfe times is not lage than the old sum of the fixed tansfe times (That means we can only shift some tansfe time fom one tansfe to the othe). Fo the e-optimize we will not use a look up table fo the V, we will calculate it duing the fmincon call. That gives us the advantage of having a eal value fo the tansfe time. With that appoach we saved between % and 5 % V pe mission. Fo each tansfe we know the following infomation: depatue epoch aival epoch tansfe time estimated V We used again matlab fmincon with an inteio point method to tackle that poblem. The contol paametes ae the times between maneuves and the 5 maneuves itself in Catesian fom. The cost function is quite easy in ou case, it s just the sum of all 5 V s we applied. We have 3 deep space maneuves, one at depatue and one at aival. The moe demanding pat is the constaint function we pass to fmincon. Hee we integate the equation of motion between the maneuves until we each ou final state. Than the final state should equal the aival debis state at that time. We have a global paamete in ode to activate o deactivate the constaints, and we can choose between the catesian state vecto, kepleian elements, o a mixtue, o a subset. Anothe inequality constaint is also in the game, because we have to take cae that the sum off all tansfe times between maneuves is not lage than the tansfe time we got out of the tee, othewise we would mess up the following tansfes. When using an ode-solve in fmincon we have the issue that integation eos we get fom the ode solve may distub the Jacobian o Hessian. And that was the case fo the fist uns we pefomed and we had to investigate which ode solve bings easonable esults. We came to the conclusion that a fixed 7

8 Inclination [deg] Epoch [days] RAAN [deg] Epoch [days] Semimajo axis [km] Epoch [days] Figue : Evolution of obital elements fo mission step size is moe stable than a vaiable step size solve. In the end we used a RK implemented in c++ with a step size of 5 sec(the RK needed sec step size), cause the matlab implementation was to slow. An inteesting obsevation was also when using the debis dynamic model fist and eunning the optimize with the spacecaft ode we achieved faste and bette esults. Fo the initial guess we set the fist and last two maneuves to the Hohmann like maneuves coming out of ou dift stategy. The maneuve in the middle is set to. We didn t touched the inclination change o phasing change and left that fo the optimize to solve. The fist and last tansfe times whee set to a half obital peiod. And the emaining two tansfe times whee chopped up equally (kind off mid couse maneuve). What we found out is that scaling ou state vecto x, the constaints and the cost function all close to is cucial fo success. Although you could assume that this techniques should be handled by optimizes automatically, that seems not to be the case. With the RK the algoithm took oughly 5 minutes on a Intel Xeon CPU E3 3.5GHz. And all tansfes conveged pope. The achieved V was even lowe than the estimated one out of the tee seach, cause in the tee seach we added scala the inclination change and the dift maneuve. In pactice they can be combined and the optimize seems to have taken cae of that. Results Numbe of tansfes Mission ID Figue 9: Mission Gaph Ou submitted solution had a total of n = missions and a pefomance index J = 99.5MEUR. Figue 9 shows the numbe of tansfes we have pe mission. Fo the fist mission the evolution of Ω, a and i ae plotted ove time (see figue ). It can be seen that mainly the inclination and semi majo axis was changed by the V and the ight ascension just difts along to the next taget.

9 7 Conclusion Fo the combinatoial pat genetic algoithms ae suitable when the poblem is time invaiant. But fo time vaiant poblems gaph algoithms seem to be the bette choice. The Beam seach algoithm bought easonable esults, but still suffes a bit fom the geedy effect, that we have the good sequences in the beginning and we obtain bad once in the end. One option to impove that may be to select some feasible continuation beams andomly. In the tansfe poblem we may use a multiple shooting method to get id off ou ode-integation issue. Refeences [] Daio Izzo. Gtoc 9 poblem statement. 7. 9