Control What is the DP Current? Nils Albert Jenssen

Transcription

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
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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
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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)
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Mathematical Vessel Model
ƒ Two unknown variables (parameters)
(Fcx , Fcy , Mc) - current
(Fx , Fy , M) – others
ƒ 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
ƒ 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
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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
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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
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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
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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
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Some Observations
ƒ The shape of current and wave characteristics is similar
especially with respect to the longitudinal and lateral forces
ƒ At high beam seas wind loads and wave loads are quite
similar in strength
ƒ 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
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Discussion
ƒ 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)
ƒ Should the current load be measured?
ƒ Should there be any relationship between these components
similar to the figures shown?
ƒ How reliable are the load characteristics? Would the use of
such relations introduce principal incorrect couplings between
the different degrees of freedom?
ƒ 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
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Advantages if load characteristics were known
ƒ 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
ƒ 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.
ƒ 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
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Modelling Errors
Thruster set-point – feedback error
ƒ Bias error of 25% (of full pitch) in the pitch feedback for a main
propeller
ƒ 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
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Modelling Errors
Wind Sensor Error
Scenario:
Real Current
ƒ Current 1 knot from 0 deg
Wind
ƒ Wind 20 knots from 30 deg
ƒ Scaling error 50%
Observations:
ƒ DP current is heavily
corrupted
ƒ 3 knots from a direction
almost opposite to the wind
© KONGSBERG November 16, 2006
Wind
DP Current
DP Current
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Modelling Errors
Wave Drift Forces
Scenario:
ƒ Current 1 knot from 0 deg
Real Current
DP Current
Waves
Wind
ƒ Wind 20 knots from 30 deg
ƒ Waves 5 m from 345 deg
Observations:
ƒ Current speed is
overestimated (about 3
knots with direction 340
deg)
Wind
© KONGSBERG November 16, 2006
DP Current
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Station Keeping Capability
Waves set to zero
© KONGSBERG November 16, 2006
Correct wave setting
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Heading Change
Wave drift
Scenario:
ƒ Heading change +15 deg
towards wind
Real Current
DP Current
Waves
Observations:
Wind
ƒ DP current does not change
significantly
ƒ 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
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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
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Conclusion
ƒ The DP current is to be considered an expression of all nonmodelled phenomena in the mathematical DP model
ƒ 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
ƒ This may be a nuisance to the DPO but does not destroy
system stability
© KONGSBERG November 16, 2006
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