# Control What is the DP Current? Nils Albert Jenssen

```Control
What is the DP Current?
Nils Albert Jenssen
Kongsberg Maritime AS – Kongsberg, Norway
October 17-18, 2006
What is the DP Current?
Nils Albert Jenssen
Kongsberg, Norway
WORLD CLASS – through people, technology and dedication
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
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
u c = u cE cosψ + vcE sinψ
Relative velocity
vc = vcE cosψ − u cE sinψ
resulting thruster forces (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”
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
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
6
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
7
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
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
Latteral forces
0
0
30
60
90
120
150
180
wind
Tonnes
-60
-120
current
waves
-180
Angle of attack
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)
10
 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
11
Modelling Errors
Thruster set-point – feedback error
 Bias error of 25% (of full pitch) in the pitch feedback for a main
propeller
 Result
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
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
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
knots with direction 340
deg)
Wind
DP Current
14
Station Keeping Capability
Waves set to zero
Correct wave setting
15
Wave drift
Scenario:
towards wind
Real Current
DP Current
Waves
Observations:
Wind
 DP current does not change
significantly
 No significant position
excursion
Wind
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
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
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
19
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