Control What is the DP Current? Nils Albert Jenssen Kongsberg Maritime AS – Kongsberg, Norway October 17-18, 2006 Return to Session Directory What is the DP Current? Nils Albert Jenssen Kongsberg, Norway WORLD CLASS – through people, technology and dedication © KONGSBERG November 16, 2006 1 Mathematical Vessel Model State variables: u , v, r vessel velocity surge, sway and rate of turn u cE , v cE current velocity north and east, assumed constant or slowly varying xE , yE vessel position north and east ψ vessel heading © KONGSBERG November 16, 2006 2 Mathematical Vessel Model u& = ( M y vr − Fcx (u r , v r ) + Ftx + Fwx + Fx ) / M x v&r = (− M x ur − Fcy (u r , v r ) + Fty + Fwy + Fy ) / M y r& = (− M c (u r , v r ) + M t + M w + M ) / M Ψ where ur = u − uc v r = v − vc and Mx My Mψ Fcx Fcy Mc Ftx Fty Mt Fwx Fwy Mw Fx Fy M © KONGSBERG November 16, 2006 u c = u cE cosψ + vcE sinψ Relative velocity vc = vcE cosψ − u cE sinψ vessel mass (included added mass) current load (unknown) resulting thruster forces (may be measured) wind load (may be measured) represents all other forces (unknown) 3 Mathematical Vessel Model Two unknown variables (parameters) (Fcx , Fcy , Mc) - current (Fx , Fy , M) – others Not feasible to establish a mathematical model of (Fx , Fy , M) – neither based on simple inputs such as e.g. wave height measurements from wave radar or buoy – nor describing the dynamics mathematically in such a way that it can be distinguished from effects of current – but, (Fcx , Fcy , Mc) may be This means that in the context of Kalman filtering current and waves must be aggregated to a common unknown force - the “DP current” © KONGSBERG November 16, 2006 4 Vessel Characteristics Wind & Current Cu rre n t lo ad c o e fficie n ts Fcx (ur , vr ) = Cx (α )(u + v ) 2 r 2 r 30 20 10 Fcy (ur , vr ) = C y (α )(u + v ) 2 r 0 2 r -10 -20 -30 Current -40 M c (ur , vr ) = Cψ (α )(u + v ) 2 r S urge [tf.s ^2/ m^2] Sway[tf.s ^2/m^2] Yaw 1.0e-002*[tf.s ^2/ m^2] 40 2 r -50 -60 -70 -80 vr α = tan ( ) ur -90 -100 −1 -110 -120 0 10 20 30 Bow 40 50 60 70 80 90 100 Curre n t a n gl e [ de g ] 110 120 1 30 1 40 150 160 Angle of attack 170 180 Stern W in d lo ad c oe ffic ie nts 0,200 Cx , Cy and Cψ are the force characteristics, i.e. force and moment as a function of angle of attack S urge [tf.s^2/ m^2] Sway[tf.s^2/m^2] Yaw 1.0e-002*[tf.s ^2/ m^2] 0,150 0,100 0,050 0,000 -0,050 -0,100 -0,150 Wind -0,200 -0,250 -0,300 -0,350 -0,400 -0,450 -0,500 -0,550 0 © KONGSBERG November 16, 2006 10 20 30 40 50 60 70 80 90 100 Wi nd a ng le [d e g] 110 120 1 30 1 40 150 160 170 180 5 Vessel Characteristics Wave drift Wave -drift lo ad c o e ffic ie nts , s way 0 0.0 [de g] 15.0 [de g] 30.0 [de g] 45.0 [de g] 60.0 [de g] 75.0 [de g] 90.0 [de g] 105.0 [deg] 120.0 [deg] 135.0 [deg] -10 -20 -30 -40 [tf/m^2] -50 -60 -70 -80 -90 -100 -110 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00 1,10 1,20 1,30 1,40 Wave frequency [rad/sec] © KONGSBERG November 16, 2006 6 Load Conditions wind 30 knots, current 1 knot, waves 5 m Hs Lateral Longitudinal forces Latteral forces Longitudinal waves 60 0 0 wind 40 60 90 120 150 180 -60 Tonnes Tonnes 20 30 0 0 30 60 90 120 150 180 -20 -120 -40 current -180 -60 Angle of attack Angle of attack Turning moment Moment 4000 Ttonnes*m 2000 0 0 30 60 90 120 150 180 -2000 -4000 Angle of attack © KONGSBERG November 16, 2006 7 Load Conditions wind 30 knots, current 1 knot, waves 1 m Hs Lateral Longitudinal forces Longitudinal forces Longitudinal 4000 60 current wind 40 2000 Tonnes Tonnes 20 0 0 30 60 90 120 150 180 0 0 30 60 90 120 150 180 -20 -40 -2000 waves -4000 -60 Angle of attack Angle of attack Moment Longitudinal forces 0 0 30 60 90 120 150 180 Tonnes -60 -120 -180 Angle of attack © KONGSBERG November 16, 2006 8 Some Observations The shape of current and wave characteristics is similar especially with respect to the longitudinal and lateral forces At high beam seas wind loads and wave loads are quite similar in strength At ahead sea the wave loads are modest Latteral forces 0 0 30 60 90 120 150 180 wind Tonnes -60 -120 current waves -180 Angle of attack © KONGSBERG November 16, 2006 9 Discussion Is the best approach just to consider forces as totally unknown without any relation to physics? I.e. just an unknown force with three independent components (Fx , Fy , M) Should the current load be measured? Should there be any relationship between these components similar to the figures shown? How reliable are the load characteristics? Would the use of such relations introduce principal incorrect couplings between the different degrees of freedom? There may be two approaches; assuming three totally unknown external force components, or assuming the external forces to be modelled as current load characteristics (or any other for that matter) © KONGSBERG November 16, 2006 10 Advantages if load characteristics were known Rotating the vessel would not affect the station keeping since we could calculate the external load at any angle of attack and compensate for it Moving towards or with the current would not have any impact since we could use nonlinear decoupling to virtually make the vessel move in vacuum. The fact is that we are not really dealing with current loads, but with residual forces not covered by inputs to our mathematical model © KONGSBERG November 16, 2006 11 Modelling Errors Thruster set-point – feedback error Bias error of 25% (of full pitch) in the pitch feedback for a main propeller Result – Artificial ahead DP current of about 2.5 knots Erroneous thruster – The large artificial current result of the quadratic nature of drag. If there would have been a real current of 1 knot, the resulting DP current would have been below 3 knots © KONGSBERG November 16, 2006 12 Modelling Errors Wind Sensor Error Scenario: Real Current Current 1 knot from 0 deg Wind Wind 20 knots from 30 deg Scaling error 50% Observations: DP current is heavily corrupted 3 knots from a direction almost opposite to the wind © KONGSBERG November 16, 2006 Wind DP Current DP Current 13 Modelling Errors Wave Drift Forces Scenario: Current 1 knot from 0 deg Real Current DP Current Waves Wind Wind 20 knots from 30 deg Waves 5 m from 345 deg Observations: Current speed is overestimated (about 3 knots with direction 340 deg) Wind © KONGSBERG November 16, 2006 DP Current 14 Station Keeping Capability Waves set to zero © KONGSBERG November 16, 2006 Correct wave setting 15 Heading Change Wave drift Scenario: Heading change +15 deg towards wind Real Current DP Current Waves Observations: Wind DP current does not change significantly No significant position excursion End heading change Wind Start heading change © KONGSBERG November 16, 2006 DP Current 16 DP Current and Control DP current may be utilised in several ways: Method 1: Not directly used for control at all. Instead an external integrator shall secure zero mean positioning offset Method 2: Used for feedback of non-modelled external forces Method 3: Used for nonlinear decoupling making the vessel virtually behave as if moving in vacuum Method 2 Method 3 Fcx = C x (α )(u cE 2 + vcE 2 ) Fcx (u r , v r ) = C x (α )(u r2 + v r2 ) Fcy = C y (α )(u cE 2 + vcE 2 ) Fcy (u r , v r ) = C y (α )(u r2 + v r2 ) M c = Cψ (α )(u cE 2 + vcE 2 ) + mc M c (u r , v r ) = Cψ (α )(u r2 + v r2 ) + mc α = tan −1 ( vc ) uc © KONGSBERG November 16, 2006 α = tan −1 ( vr ) ur 17 Robustness to modelling error Stability limit 0.04 0.04 Stability limit 0.03 0.03 0.02 0.02 0.01 0.01 0 mag Axis -0.01 0 mag Axis -0.01 -0.02 -0.02 -0.03 -0.03 -0.04 -0.08 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 Real Axis Method 2 0 -0.04 -0.05 -0.04 -0.03 -0.02 Real Axis -0.01 0 0.01 Method 3 Growing modelling error, Kalman filter Growing modelling error, control © KONGSBERG November 16, 2006 18 Conclusion The DP current is to be considered an expression of all nonmodelled phenomena in the mathematical DP model We have seen that sensor errors (anemometers as well as thruster set-point – feedback problems) and unknown environmental loads such as wave drift will cause the DP current to grow This may be a nuisance to the DPO but does not destroy system stability © KONGSBERG November 16, 2006 19

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