Zac  Manchester Harvard  University Recursive  Inertia  Estimation  with   Semidefinite  Programming Mason  Peck Cornell  University 

Why? 1 

Gyrostat  Dynamics 2 Inertia Rotor  Torque External  Torque Angular  Velocity Rotor  Momentum J ˙ ! + !⇥(J! + ⇢) + ˙ ⇢ = ⌧ 

A  Least-­‐Squares  Problem 3 j = ⇥ J11 J22 J33 J12 J13 J23 ⇤T 

A  Least-­‐Squares  Problem 4 J! = G(!)j j = ⇥ J11 J22 J33 J12 J13 J23 ⇤T 

A  Least-­‐Squares  Problem 5 J! = G(!)j G( ˙ !) + !⇥G(!) j = ⌧ ˙ ⇢ !⇥⇢ j = ⇥ J11 J22 J33 J12 J13 J23 ⇤T 

A  Least-­‐Squares  Problem 6 J! = G(!)j H y G( ˙ !) + !⇥G(!) j = ⌧ ˙ ⇢ !⇥⇢ j = ⇥ J11 J22 J33 J12 J13 J23 ⇤T 

A  Least-­‐Squares  Problem 7 J! = G(!)j H y G( ˙ !) + !⇥G(!) j = ⌧ ˙ ⇢ !⇥⇢ H(!, ˙ !)j = y(!, ⇢, ˙ ⇢, ⌧) j = ⇥ J11 J22 J33 J12 J13 J23 ⇤T 

Some  Least-­‐Squares  Problems… 8 H(!, ˙ !)j = y(!, ⇢, ˙ ⇢, ⌧) 

Some  Least-­‐Squares  Problems… 9 H(!, ˙ !)j = y(!, ⇢, ˙ ⇢, ⌧) We  don’t  have  measurements  of  this 

Some  Least-­‐Squares  Problems… 10 H(!, ˙ !)j = y(!, ⇢, ˙ ⇢, ⌧) We  don’t  have  measurements  of  this This  doesn’t  necessarily  form  a  valid  inertia  matrix 

Principle  of  Least  Action 11 L(q, ˙ q) = T V S = Z tf t0 L (q(t), ˙ q(t)) dt = 0 

12 Discrete  Mechanics = N X k=0 Z tk+1 tk L(q, ˙ q) dt S = Z tf t0 L(q, ˙ q) dt 

13 Discrete  Mechanics = N X k=0 Z tk+1 tk L(q, ˙ q) dt S = Z tf t0 L(q, ˙ q) dt ⇡ N X k=0 L ✓ qk, qk+1 qk h ◆ h 

14 Discrete  Mechanics = N X k=0 Z tk+1 tk L(q, ˙ q) dt S = Z tf t0 L(q, ˙ q) dt Sd = N X k=0 Ld(qk, qk+1) = 0 ⇡ N X k=0 L ✓ qk, qk+1 qk h ◆ h 

15 Discrete  Gyrostat  Equation fk = 2 6 6 6 4 k q 1 T k k 3 7 7 7 5 H( k , k+1)j = y( k , k+1 , ⇢k , ⇢k+1 , ⌧k+1) 

16 What  Makes  A  “Valid”  Inertia  Matrix? 

17 J 2 S3 What  Makes  A  “Valid”  Inertia  Matrix? 

What  Makes  A  “Valid”  Inertia  Matrix? 18 J 2 S3 J > 0 x T Jx > 0 8 x 6= 0 

19 J 2 S3 J > 0 Jii  Jkk + J`` x T Jx > 0 8 x 6= 0 What  Makes  A  “Valid”  Inertia  Matrix? 

Semidefinite  Programming  (SDP) 20 minimize x cT x subject to F0 + n X i =1 xiFi 0  x z z y 0 

21 Schur Complement  A B BT C 0 =) ( C > 0 A BC 1BT 0 

22 SDP  Inertia  Estimation  Formulation minimize ⇥ · · · 0 · · · 1 ⇤  i s subject to 8 > > > > > > > > < > > > > > > > > :  s ( Hj y ) T ( Hj y ) I 0 J > 0 J11 + J22 J33 0 J11 + J33 J22 0 J22 + J33 J11 0 

0 20 40 60 80 -0.2 -0.1 0 0.1 1 0 20 40 60 80 -0.1 -0.05 0 0.05 2 0 20 40 60 80 Time (s) -0.04 -0.02 0 0.02 3 0 20 40 60 80 -0.05 0 0.05 0.1 1 0 20 40 60 80 -0.05 0 0.05 0.1 2 0 20 40 60 80 Time (s) -0.05 0 0.05 0.1 3 23 Slewing  Spacecraft  Example 

0 20 40 60 80 10-9 10-7 10-5 10-3 10-1 J 11 SDP Variational SDP Finite Diff. Momentum Based 0 20 40 60 80 10-9 10-7 10-5 10-3 10-1 J 22 0 20 40 60 80 Time (s) 10-9 10-7 10-5 10-3 10-1 J 33 0 20 40 60 80 10-9 10-7 10-5 10-3 10-1 J 12 0 20 40 60 80 10-9 10-7 10-5 10-3 10-1 J 13 0 20 40 60 80 Time (s) 10-9 10-7 10-5 10-3 10-1 J 23 24 Slewing  Spacecraft  Example 

0 10 20 30 40 50 60 -0.2 0 0.2 1 0 10 20 30 40 50 60 -0.2 -0.1 0 0.1 2 0 10 20 30 40 50 60 Time (s) 0.76 0.77 0.78 0.79 3 0 10 20 30 40 50 60 -0.1 0 0.1 1 0 10 20 30 40 50 60 -0.1 0 0.1 2 0 10 20 30 40 50 60 Time (s) -0.1 0 0.1 3 25 Spinning  Spacecraft  Example 

0 10 20 30 40 50 60 10-8 10-6 10-4 10-2 100 J 12 0 10 20 30 40 50 60 10-8 10-6 10-4 10-2 100 J 13 0 10 20 30 40 50 60 Time (s) 10-8 10-6 10-4 10-2 100 J 23 0 10 20 30 40 50 60 10-8 10-6 10-4 10-2 100 J 11 SDP Variational SDP Finite Diff. Momentum Based 0 10 20 30 40 50 60 10-8 10-6 10-4 10-2 100 J 22 0 10 20 30 40 50 60 Time (s) 10-8 10-6 10-4 10-2 100 J 33 26 Spinning  Spacecraft  Example 

27 Conclusions • SDP  formulation  guarantees  that  a  valid  inertia  is  returned • Discrete  mechanics  formulation  eliminates  noise   amplification  problems • A  priori  knowledge  can  be  included  in  the  estimator  to   improve  convergence 

Questions? 28 [email protected]