Modelling and Control of Blowing-Venting
Operations in Manned Submarines∗
R. Font,† J. Garćıa-Peláez,‡ J. A. Murillo,† and F. Periago†
May 25, 2021
Abstract
Motivated by the study of the potential use of blowing and venting operations of
ballast tanks in manned submarines as an alternative control system for manoeuvring,
we first propose a mathematical model for these operations. This model extends previ-
ous works where only blowing is considered. Then, the model is applied to the control of
an emergency manoeuvre by using only blowing and venting. To this end, we formulate
a suitable constrained, nonlinear, optimal control problem where controls are linked to
the variable aperture of blowing and venting valves of each of the tanks. The state
law is composed of a system of nonlinear differential equations where the equations
modelling blowing and venting processes are coupled with the Feldman, variable mass,
coefficient based hydrodynamic model for the equations of motion. In a second part,
we carry out a rigorous mathematical analysis of the model: existence of a solution
for both the state law and the optimal control problem is proved. Finally, we address
the numerical resolution of the optimal control problem by using a descent algorithm.
Numerical experiments seem to indicate that, indeed, an appropriate use of blowing
and venting operations may help in the control of an emergency manoeuvre.
Keywords: Ballast tanks, manned submarines, blowing-venting operations, optimal con-
trol, numerical simulations.
1 Introduction
Manned submarines are equipped with several ballast tanks distributed along its hull. When
filled with water, they contribute with the submarine mass allowing it to submerge. During
an unexpected event or emergence they act like a dispositive for emerging to the surface: air
is blown into the ballast tanks from very high pressure bottles expelling the water out of the
∗Work supported by projects 2989/10MAE from Navantia S. A. and 08720/PI/08 from Fundación Séneca,
Agencia de Ciencia y Tecnoloǵıa de la Región de Murcia (Programa de Generación de Conocimiento Cient́ıfico
de Excelencia, IIPCTRM 2007-10). The last author was supported by project MTM2007-62945 from Min-
isterio de Educación y Ciencia (Spain).†Departamento de Matemática Aplicada y Estad́ıstica, ETSI Industriales, Universidad Politécnica de
Cartagena, 30202 Cartagena, Spain - [email protected], [email protected], [email protected].‡Direction of Engineering, DICA, Navantia S. A., Cartagena, 30205, Spain - [email protected].
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1 INTRODUCTION 2
tanks. The submarine loss weight, its buoyancy is higher, and it can emerge quicker. In the
last years, several works have addressed these emergency rising manoeuvres (see [1, 2, 12, 13]
and the references given there). To fill the tanks with water again, air is vented out of the
ballast tanks. A valve located at the top of each of the tanks is opened, air escapes outside,
and water flows back into the tanks.
On the other hand, under certain circumstances, like gathering intelligence missions or
special operations, submarines may need to perform manoeuvres with very specific require-
ments. In these cases, submarines often perform small blowing and venting operations, not
because of an emergency, but to slightly modify the buoyancy of the vehicle. This way,
blowing and venting become a complementary tool for manoeuvring. These manoeuvres
are currently performed based exclusively on the operator experience and, due to the high
degree of accuracy required, would enormously benefit from the implementation of a control
system.
The issue of modelling the blowing of ballast tanks has been addressed, for instance,
in [2] and [13]. Up to the best knowledge of the authors, venting has not been addressed
so far, and most importantly, the coupling of both processes as a control system has not
been considered before. This is the first aim of the present work. In Section 2 we propose
a model for a coupled system of blowing-venting operations. For the particular case where
only blowing is considered, the model presented in this work was numerically compared in
[7] with the one proposed by Watt [13] giving similar results, but also showing the ability
to capture some phenomena that were overlooked by that last model. Then, these two
processes are coupled with the usual Feldman’s coefficient based hydrodynamic model for
the equations of motion (see [6] and more precisely [10]). We notice that, since the mass
of the submarine changes with blowing-venting operations, some of the parameters (e.g.,
moments and products of inertia, weight, mass, etc...) that remain constant in Feldman’s
model are, in our case, time dependent.
In a second part, we analyze the potential use of blowing-venting operations as a control
system in a typical emergency manoeuvre. To this end, we model such a manoeuvre as a
constrained, nonlinear, optimal control problem which has the aperture of blowing-venting
valves of each of the tanks as the control variables. The underlying state law is composed
of a nonlinear system of 3N + 12 ordinary differential equations (ODE’s), where N is the
number of ballast tanks.
The mathematical analysis of the model is presented in Section 4 and is divided into two
parts. First, the existence of solution for the state law is obtained by using the classical
existence theory for ODE’s. However, due to the size and complexity of the mathematical
model, checking the sufficient conditions for the existence of a solution is not an easy task.
In particular, we carry out a detailed analysis of the main properties of the time-dependent
mass matrix of the system, Subsection 4.1.1. Second, we also prove that the optimal control
problem is well-posed, that is, that there exists a solution to it (Theorem 4.3). This existence
result is obtained by using Filippov’s theory for Bolza-type optimal control problems. The
fact that controls appear in linear form simplifies the analysis.
At this point, it is convenient to comment on the choice of an open-loop control system
as an appropriate formulation for the problem considered in this work. Up to now, there
is not an available technology to implement a control system as the one described in this
2 MATHEMATICAL MODELLING 3
paper in a real manned submarine. In fact, the present study is motivated by the company
Navantia S.A. Shipyard (Spain) as a first step to analyze the effectiveness of blowing-venting
operations as a control system. In addition, the results provided by the formulation consid-
ered here lets naval architects fit several parameters of blowing-venting systems (such as size
of ballast tanks and/or blowing and venting valves) during the preliminary state of design
of a prototype. Of course, once the capabilities of blowing-venting operations are positively
evaluated by an open-loop control system, a second step would be the design of a closed-loop
control system to be able to correct errors in the model, for instance in the estimation of
the hydrodynamics coefficients, or the effects of disturbances acting on the surroundings of
the vehicle. We plan to address this issue in a future work.
Finally, for the numerical resolution of the optimal control problem we use a gradient
descent method as in [10]. The performance of the algorithm is illustrated through the
numerical simulation of an emergency manoeuvre. Numerical simulation results seem to
indicate that blowing-venting operations may help in a significant way to the control of this
type of manoeuvre.
To conclude it is important to point out that although the study of the present work has
been only applied to a typical emergency manoeuvre, the scope of the model and the tech-
niques developed here is not limited to this particular situation. Indeed, the approach pro-
posed in this paper can easily be extended to analyze the use of blowing-venting operations
as a depth controller in snorting condition, which is a relevant topic in naval engineering.
The stabilization, based on blowing and venting of ballast tanks, of other marine systems,
like offshore facilities or pontoon docks is another possible application.
2 Mathematical modelling
The three-dimensional equations of motion for an underwater vehicle are usually described
by using two coordinate frames: the moving coordinate frame which is fixed to the vehicle
and is called the body-fixed system, and the earth-fixed reference frame which is called the
world system. The position and orientation of the vehicle are described in the world system
while the linear and angular velocities are expressed in the body-fixed coordinate system.
These quantities are defined according to SNAME notation [8] as
η(t) =[ηT1 (t),η
T2 (t)
], η1(t) = [x(t), y(t), z(t)]
T, η2(t) = [φ(t), θ(t), ψ(t)]
T
ν(t) =[νT1 (t),ν
T2 (t)
], ν1(t) = [u(t), v(t), w(t)]
T, ν2(t) = [p(t), q(t), r(t)]
T.
Here t is the time variable, aT stands for the transpose of vector a, η1 denotes the position
of the vehicle in the world system, η2 is the orientation in the same reference system, ν1 is
the vector of linear velocities in the body-fixed frame (where as usual u is surge velocity, v
is sway velocity, and w is heave velocity), and finally ν2 is the vector of angular velocities
in the body-fixed reference system (p is roll rate, q is pitch rate, and r is yaw rate). See
Figure 1.
The mathematical model for the equations of motion is based on Gertler and Hagen’s [9]
six degree of freedom (DOF) submarine equations of motion, which were revised by Feld-
man [6]. Adapting these general equations to the particular characteristics of a prototype
2 MATHEMATICAL MODELLING 4
Figure 1: Variables and coordinate systems.
developed by the company Navantia S.A. Shipyard (Spain), a very similar coefficient based
hydrodynamic model was analyzed in [10]. This latter model will be the starting point for
the more general model that we will introduce in this section. Anyway, to make the paper
easier for readers we have collected these equations in Appendix A.
2.1 Blowing and venting model
As shown in Figure 2, the blowing/venting system is composed of ballast tank, pressure
bottle, blowing valve and venting valve. When the blowing valve is opened, air flows into
the tank increasing the pressure and forcing the water to flow out through the flood port
located at the bottom of the tank. When the venting valve is opened, air can flow out from
the tank. The model can be divided into four main parts:
• Air flow from pressure bottle.
• Air flow through venting valve.
• Water flow through flood port.
• Evolution of pressure in ballast tank.
2 MATHEMATICAL MODELLING 5
Figure 2: Blowing and venting processes.
Finally, in Subsection 2.1.5 we shall introduce a new set of variables that will control the
aperture of blowing and venting valves at each ballast tank, and shall summarize the whole
system.
Variables and symbols introduced in this section are summarized in Table 1.
2.1.1 Air flow from pressure bottle
When the blowing valve is opened, the air in the bottle is blown into the tank through
a nozzle. Pressure losses and heat transfer in the tube that connects the bottle and the
tank are, for the moment, neglected. However, as we will see later on, pressure losses can
be indirectly taken into account. Under the above conditions, we need to study the one
dimensional steady flow of an ideal compressible gas. This can be found in any classic text
on fluid mechanics (see for example [4]) but for completeness we include it here.
At the beginning of blowing, the flow is supersonic due to the high pressure difference
between bottle and tank. As air flows out of the bottle, this difference decreases and the
flow becomes subsonic at a certain time. This transition happens when pressures in bottle,
pF , and tank, pB , are such that (pFpB
)= Pc
where Pc =(γ+12
) γγ−1 , with γ the isentropic constant.
Supersonic flow: In this case, it is easy to calculate the mass flow rate at the nozzle
throat, where the Mach number is unity. Precisely,
ṁF = ρ∗av∗A = ρa,0a0
ρ∗aρa,0
a∗
a0A, (1)
2 MATHEMATICAL MODELLING 6
A area in nozzle throat (m2)
Ah outlet hole area (m2)
Av vent pipe section (m2)
Ch outlet hole coefficient
Htk ballast tank height (m)
hwc(t) height of water column in the tank (m)
mB(t) mass of air in ballast tank (kg)
mF (t) mass of air in pressure bottle (kg)
ṁF (t) mass flow rate from pressure bottle (kg/s)
ṁv(t) mass flow rate through venting valve (kg/s)
pB(t) pressure in ballast tank (Pa)
pext(t) pressure outside the venting valve (Pa)
pF (t) pressure in bottle (Pa)
pSEA(t) pressure outside the outlet hole (Pa)
qB(t) water flow through outlet hole (m3/s)
TB water temperature (K)
TF (t) temperature in pressure bottle (K)
VB0 initial air volume in ballast tank (m3)
VB(t) volume of air in ballast tank (m3)
VBB ballast tank volume (m3)
VF pressure bottle volume (m3)
(xb, yb, zb) location of geometrical center of ballast tank (m)
zh outlet hole distance from origin (m)
zt tank top distance from origin (m)
zv venting valve distance from origin (m)
γ isentropic constant
ρ density of water (kg/m3)
Table 1: Variables and symbols
where A is the area in the nozzle throat, a is the speed of sound, v is the air velocity, ρa is
the density of air, the asterisk means conditions wherein the Mach number is equal to unity
and the subscript 0 denotes stagnant conditions. These two conditions are related by(p0p∗
) γ−1γ
=
(ρa,0ρ∗a
)γ−1=(a0a∗
)2=T0T ∗
=γ + 1
2,
where T is the temperature.
Assuming that in this particular case stagnant conditions are the bottle conditions, using
the above relations, the ideal gas law ρa,0 =p0
RgT0and a0 =
√γRgT0, the mass flow rate is
ṁF (t) =ApF (t)√RgTF (t)
√γ
(2
γ + 1
) γ+1γ−1
, (2)
2 MATHEMATICAL MODELLING 7
where Rg is the gas constant for air and TF is the temperature in the bottle.
Subsonic flow: In this case, the mass flow rate is given by
ṁF = ρa,eMeaeAe = ρa,0a0ρa,eρa,0
aea0MeAe, (3)
where Ae is the exit area, in this case Ae = A, M is the Mach number and the subscript e
denotes conditions at the exit. The stagnant and static conditions are related by
T0T
=
(ρa,0ρa
)γ−1=
(p0p
) γ−1γ
= 1 +γ − 1
2M2.
Particularizing for p0 = pF and p = pe = pB , the Mach number at the exit is given by
Me =
√√√√ 2γ − 1
[(pFpB
) γ−1γ
− 1
]. (4)
Substituting (4) into (3) and proceeding as in the supersonic case, we get
ṁF (t) =ApF (t)√RgTF (t)
√√√√ 2γγ − 1
[(pB(t)
pF (t)
) 2γ
−(pB(t)
pF (t)
) γ+1γ
]. (5)
The mass flow rate from the bottle can then be expressed for both the subsonic and
supersonic cases as
ṁF (t) =ApF (t)√RgTF (t)
· µ(pB(t), pF (t)) (6)
with µ(pB(t), pF (t)) = µmax = cte ifpF (t)pB(t)
≥(γ+12
) γγ−1 .
As the pressure in the bottle drops, the temperature increases. If the heat transmission
is neglected, however, the process can be considered to be adiabatic. As mentioned in [2],
experimental results sustain this assumption. Let mF0, pF0 be the initial mass and pressure
in the bottle and VF the bottle volume. Under the above assumptions the momentary
pressure and temperature are given by
pF (t) =
(mF (t)
mF0
)γpF0, TF (t) =
pF (t)VFRgmF (t)
. (7)
Hence,
pF (t)√RgTF (t)
=
√mF (t)
γ+1pF0mγF0VF
.
Thus the final equation for the mass flow rate is
ṁF (t) = A
(mF (t)
γ+1pF0mγF0VF
) 12
µ(pB(t),mF (t)) (8)
2 MATHEMATICAL MODELLING 8
with
µ(pB ,mF ) =
√γ(
2γ+1
) γ+1γ−1
, Pc ≤pFpB√√√√√√ 2γγ − 1
pBpF0
(mFmF0
)γ 2γ−
pBpF0
(mFmF0
)γ
γ+1γ
, 1 < pFpB
< Pc
0,pFpB≤ 1
(9)
Since the initial mass flow rate depends only on the initial conditions in the bottle, it
can be considered as a constant with value
ṁF (0) = A
(γ
(2
γ + 1
) γ+1γ−1 pF0mF0
VF
) 12
. (10)
This initial mass flow rate has been measured for several blowing intensities. Let ṁFmaxbe this measured maximum mass flow rate. Instead of using the numerical value of area A
when computing (8), we obtain it from (10) and the measured value for ṁFmax, that is,
A = ṁFmax
((2
γ + 1
)− γ+1γ−1 VFγpF0mF0
) 12
. (11)
By doing so, we ensure that the initial mass flow rate calculated coincides with the real
measured value. This way, although pressure losses are not considered in the model, they
are indirectly taken into account.
2.1.2 Air flow through venting valve
Equations for the air flow from the tank can be obtained analogously to the ones used for the
mass flow from the bottle. In this case, the pressures to evaluate are the pressure in ballast
tank, pB , and the pressure outside the vent valve pipe system, pext = patm + ρg(z + zv −xb sin θ), where patm is the atmospheric pressure, z is the vehicle depth, zv is the vertical
distance between the origin of the body–fixed frame and the venting valve, and the term
xb sin θ models the tank height variation with the vehicle pitch. Thus, (z + zv − xb sin θ) isthe depth at which the exit of the venting system is located. Therefore, similarly to (6), we
have
ṁv(t) = µ (pB , pext)AvpB(t)
RgTB(t), (12)
where Av is the vent pipe section and
µ = µmax ifpextpB≤(pextpB
)crit
.
Although the geometric details of the venting system are not available and therefore
pressure losses can not be directly calculated, an evaluation of µ (pB , pext) for several values
2 MATHEMATICAL MODELLING 9
Figure 3: Curve fitting result for µ(Π) considering pressure losses.
of the pextpB ratio taking into account the different pressure losses along the system has been
provided. In particular, the critical pressure ratio is found to be(pextpB
)crit
= 0.5184 and
µmax = 0.3550. Let Π =pextpB
be the pressure ratio. Using a least squares fit of this data
(see Figure 3), the following expression for µ is obtained:
µ (Π) =
µmax if Π ≤ Πcrit
−66.97Π2 − 52.70Π + 119.70Π3 − 171.30Π2 − 75.65Π + 294.60
if Πcrit < Π < 1
0 if Π ≥ 1.
(13)
The variation in the mass of air in the ballast tank is the difference between the mass
flow rate from the bottle and the mass flow rate through the venting valve. This way, by
(12) and (13),
ṁB(t) = −ṁF (t)− µ(Π(t))AvpB(t)
RgTB(t). (14)
2.1.3 Water flow through flood port
The difference between the tank and outside pressure forces the water to flow in or out from
the tank trough the flood port located at the bottom. Let us assume, in the first place, that
the pressure in ballast tank is greater than the outside pressure. Then, the water flows out
from the tank. A detailed analysis of a draining tank filled with an ideal fluid can be found
in [4]. The pressure directly above the flood port is pB(t) + ρghwc(t), where hwc(t) is the
height of the water column in the tank. Assuming that the tank has a regular shape, the
2 MATHEMATICAL MODELLING 10
height of the water column can be expressed as
hwc(t) =
(1− ∆m(t)
∆mmax
)Htk,
where Htk is the ballast tank height, ∆m(t) is the mass loss in the tank and ∆mmax is the
maximum value of the mass loss (see Section 2.2 for more details).
The Bernoulli equation applied at both sides of the port gives
pB(t) + ρghwc(t)−∆ploss = pSEA(t) +1
2ρv2h(t),
where vh and ∆ploss are, respectively, the velocity and pressure losses in the outlet hole. Both
major and minor losses have been calculated by a separate model and can be compressed
into the one single term
∆ploss = ζh1
2ρv2h(t).
ζh is estimated to be ζh = 2.5.
This way the velocity in the flood port is given by
vh(t) =
√2(pB(t) + ρghwc(t)− pSEA(t))
ρ(1 + ζh)
and the volume flow from the ballast tank by
qB(t) = ChAhvh(t) = CnAh
√2(pB(t) + ρghwc(t)− pSEA(t))
ρ(1 + ζh), (15)
where
pSEA = patm + ρg(z + zh − xb sin θ).
Here zh is the outlet hole location, Ah is the outlet hole area and Ch is a coefficient that
takes into account that, since the outlet hole is actually a grid, the effective area is smaller
than Ah.
If pressure in ballast tank is lower than the outside pressure, then water flows into the
tank. In this case, the expression for the volume flow is (15), but with a negative value.
2.1.4 Evolution of pressure in ballast tank
When the blowing valve is opened the air is blown into the tank at a very high velocity,
rapidly mixing with water. This promotes good heat transfer from the water to the ex-
panding air and we can work under the assumption that the air will immediately adopt
the temperature in the tank. The process can then be considered to be isothermal. As
mentioned in [2], experimental results sustain this assumption.
From the ideal gas law it follows that
ṗB(t) =ṁB(t)RgTB − V̇B(t)pB(t)
VB(t).
2 MATHEMATICAL MODELLING 11
The volume occupied by air increases (and in the same quantity) as the volume occupied
by water decreases. Thus, the rate at which it changes, V̇B , will be the volume flow out of
the ballast tank, qB , given by (15). Using the perfect gas equation, the momentary volume
of air in the tank is VB(t) =mB(t)RgTB
pB(t). The variation in tank pressure is then
ṗB(t)−pB(t)
mB(t)ṁB(t) = −
p2B(t)qB(t)
mB(t)RgTB. (16)
At the mathematical and numerical levels, the presence of the square root in the equation
(15) generates serious difficulties. Indeed, if the term inside the square root vanishes, then
the gradient blows-up. To overcome this difficulty we approximate the square root near the
origin (x ∈ [0, ξ]) by a fourth-order polynomial as follows. Consider the polynomial
P (x) = a0 + a1x+ a2x2 + a3x
3 + a4x4.
The coefficients of this polynomial are determined in such a way that the following conditions
hold:
(a) P (0) = 0
(b) P (ξ) =√ξ
(c) Ṗ (0) = 0
(d) Ṗ (ξ) = 1√ξ
(e)∫ ξ0P (x) dx =
∫ ξ0
√xdx.
Notice that conditions (a)-(d) are continuity conditions of P and its derivative Ṗ at the
extremes x = 0, ξ, and condition (e) has the effect of preserving the average volume flow.
Numerical tests showed that ξ = 1 gives a good approximation for the square root in
(15). We obtain
a0 = a1 = 0, a2 = 8.75, a3 = −14 and a4 = 6.25.
To sum up, we consider the new function
P (x) =
−√−x if x < −ξ
−P (−x) if −ξ ≤ x ≤ 0P (x) if 0 < x ≤ ξ√x if x > ξ
and replace qB (t) by its approximation qB (t) defined as
qB (t) = CnAhP
(2 (pB(t) + ρghwc(t)− pSEA(t))
ρ(1 + ζh)
). (17)
2.1.5 Blowing and venting controlled system
Once a model for blowing an venting operations has been presented, our next goal is to use
such a system as a control mechanism to improve the manoeuvrability of the vehicle. To
this end, we introduce a new set of variables: the control variables of the blowing-venting
system. Let si, si ∈ L∞ (0, tf ; [0, 1]) denote, respectively, the grade of aperture of blowing
2 MATHEMATICAL MODELLING 12
MBT 2 MBT 3 MBT 4 MBT 5
Ah 0.191 0.191 0.191 0.191
Av 0.0177 0.0177 0.0177 0.0177
Ch 0.7 0.7 0.7 0.7
Htk 5 5 6.2 6.2
pF0 2.5 · 107 2.5 · 107 2.5 · 107 2.5 · 107
TF0 293 293 293 293
VB0 0.001 0.001 0.001 0.001
VBB 21.4 21.4 22.9 22.9
VF 0.8 0.8 0.8 0.8
xb −28.6 −28.6 23 23yb 1.2 −1.2 1.7 −1.7zb 0.595 0.595 0.975 0.975
zh −2.895 −2.895 −3.897 −3.897zt 2.105 2.105 2.303 2.303
zv 2.705 2.705 2.705 2.705
Table 2: Ballast tanks characteristics.
and venting valves of the i−th ballast tank during the time interval [0, tf ]. Thus, from (8),(14) and (16) we obtain the final equations governing the controlled evolution of the mass
of air in any pressure bottle (mFi(t)), the mass of air in the corresponding tank (mBi(t)),
and its pressure (pBi(t)). We obtain
ṁFi(t) = si(t)A
(mFi(t)
γ+1pF0mγF0VF
) 12
µi(pBi(t),mFi(t)) (18)
ṁBi(t) + ṁFi(t) = −µi(Π(t))si(t)AvpBi(t)√
RgTB, (19)
mBi(t)
pBi(t)ṗBi(t)− ṁBi(t) = −
pBi(t)qBi(t)
RgTB. (20)
The above formulation assumes that the flow trough the valves varies linearly with their
aperture, si(t), si(t). Of course, once the control system were implemented in a real vehicle
this assumption should be adapted to the particular characteristics of the chosen valves.
The values of all the required geometrical parameters for the four tanks considered here
are summarized in Table 2.
2 MATHEMATICAL MODELLING 13
2.2 Coupling of blowing-venting system with a variable mass model
for the equations of motion
As water flows in or out of the tanks there will be mass variations located at several points of
the vehicle. Since the equations of motion assume the mass of the submarine to be constant,
it is necessary to identify which terms, formerly constant, will become time dependent due
to its dependance with mass. We will need to write the following properties as a function
of the amount of water in the tanks:
• Mass (m).
• Weight (W ).
• Moments and products of inertia (Ix, Iy, Iz, Ixy, . . . ).
• Location of the center of gravity (xG, yG, zG).
Let us assume there are N ballast tanks with geometrical centers located at points (xbi,
ybi, zbi) (where the subscript i denotes the i−th ballast tank). Let m0 be the initial massof the submarine (with all tanks completely filled with water) and ∆mi the mass loss in the
i−th tank. It is 0 when the tank is completely filled with water, and reaches its maximumvalue when it empties.
The volume of water that has left the tank is equal to the volume occupied by air except
for the initial air volume in the tank, VB0, which depends on the initial mass of air in
the tank, mB0, and the initial depth. The mass loss in the i−th tank can be obtained bymultiplying this volume by the density of water, ρ, that is,
∆mi(t) = ρ (VBi(t)− VB0) = ρ(mBi(t)RgTB
pBi(t)− VB0
). (21)
The momentary mass of the submarine is then
m(t) = m0 −N∑i=1
∆mi(t), (22)
and the weight
W (t) = g
(m0 −
N∑i=1
∆mi(t)
), (23)
with g the acceleration due to gravity.
Although the vehicle buoyancy B is usually assumed to be constant, it is really a function
of the vehicle depth. Indeed, as the depth increases, the outside pressure compresses the
vehicle and therefore its volume decreases. Let B0 be the buoyancy at zero depth. The
buoyancy is assumed to be linearly dependent with respect to the vehicle depth in the form
B = B0
(1− 0.015
300z
)(24)
which means that the buoyancy decreases by a 1.5 percent a depth of 300 meters.
2 MATHEMATICAL MODELLING 14
Figure 4: Location of the mass loss.
To find expressions for the location of the center of gravity and moments and products
of inertia we will work under the assumption that the mass loss in each tank occurs at a
point, (xbi, ybi, zmli(t)), where zmli(t) is the height at which mass loss happens for each
tank. It varies, as shown in Figure 4, from the top of the tank, when it is completely filled,
to its geometric center, zbi, when it is completely empty. This variation is assumed to be
linear so that
zmli(t) = zti −(zti − zbi)∆mi(t)
∆mi,max,
where zti is the location of the tank top and ∆mi,max is the maximum value of the mass
loss, i.e, the value when the tank is completely empty.
Let Ix0, Iy0, Iz0, Ixy0, Ixz0, Iyz0 and xG0, yG0, zG0 be respectively the initial (all tanks
completely filled) moments and products of inertia and coordinates of the center of gravity.
Then the moments and products of inertia are given by
Ix(t) =
∫ (y2 + z2
)dm = Ix0 −
N∑i=1
(y2bi + zmli(t)
2)
∆mi(t),
Iy(t) =
∫ (x2 + z2
)dm = Iy0 −
N∑i=1
(x2bi + zmli(t)
2)
∆mi(t),
Iz(t) =
∫ (x2 + y2
)dm = Iz0 −
N∑i=1
(x2bi + y
2bi
)∆mi(t),
Ixy(t) =
∫xy dm = Ixy0 −
N∑i=1
xbiybi∆mi(t),
Ixz(t) =
∫xz dm = Ixz0 −
N∑i=1
xbizmli(t)∆mi(t),
Iyz(t) =
∫yz dm = Iyz0 −
N∑i=1
ybizmli(t)∆mi(t),
(25)
and the coordinates of the center of gravity by
2 MATHEMATICAL MODELLING 15
xG(t) =1
m0 −∑Ni=1 ∆mi(t)
(m0xG0 −
∑Ni=1 xbi∆mi(t)
)yG(t) =
1
m0 −∑Ni=1 ∆mi(t)
(m0yG0 −
∑Ni=1 ybi∆mi(t)
)zG(t) =
1
m0 −∑Ni=1 ∆mi(t)
(m0zG0 −
∑Ni=1 zmli(t)∆mi(t)
) (26)
Substituting (22)−(26) into the equations of motion completes our variable mass model.
2.3 Complete model in compact form
Summarizing, the state variables of the system can be expressed in vector form as
x(t) =[[mFi(t),mBi(t), pBi(t)]1≤i≤N ,η(t)
T ,ν(t)T]T
where N is the number of ballast tanks, η = [x, y, z, φ, θ, ψ]T
is the vector of positions and
Euler angles and ν = [u, v, w, p, q, r]T
are linear and angular velocities. This means that the
vector state x belongs to R3N+12.
Moreover we write the control vector as
u(t) = [si(t), si(t)]T1≤i≤N
si and si being, respectively, the aperture of blowing and venting valves of the i−th tank.Therefore, u ∈ R2N .
Finally, the state law is composed of equations (18)-(20) for the blowing-venting system,
the six kinematic equations as given in Appendix A.1, and the six equations of the hydro-
dynamic forces and moments (see Appendix A.2, where the time variable parameters as
described in Subsection 2.2 have been taken into account). In compact form we write these
equations as
A(x(t))ẋ(t) = f (t,x(t),u(t)) , (27)
where A(x(t)) ∈ M(3N+12)×(3N+12) and f (t,x(t),u(t)) ∈ R3N+12 for every t > 0. InSection 4 we will analyze in detail the structure of A and f . Nevertheless, at this point, it
is convenient to comment on the explicit dependence of f with respect to the time variable
t. Since we plan to analyze the potential use of blowing-venting as a control system for
manoeuvrability, deflection of bow plane (δb(t)), deflection of stern plane (δs(t)), deflection
of rudder (δr(t)) and propeller speed (n(t)), that typically are the elements used for the
manoeuvrability of the submarine, will be fixed to some convenient values during the whole
time of manoeuvre. More precisely, we assume that δb, δs, δr, n ∈ L∞(0, tf ;K), where
K =
[−5π
36,
5π
36
]×[−5π
36,
5π
36
]×[−7π
36,
7π
36
]× [0, 2.5].
Therefore, the explicit dependence of f on time is linked to these four functions (see Sub-
sections 5.2 and A.2 for more details on this passage).
3 FORMULATION OF THE CONTROL PROBLEM 16
3 Formulation of the control problem
Next, we plan to analyze the potential use of blowing-venting operations as a control mech-
anism in a typical emergency rising manoeuvre where the submarine must reach surface
quickly while keeping its stability. We are now in a position to formulate the manoeuvra-
bility control problem. Given an initial state x (0) = x0 and a desired final target xtf , the
goal is to calculate the vector of control u = u(t), which is able to draw our system from
the initial state x0 to (or near to) the final one xtf in a given time tf , also minimizing a
cost functional. In mathematical terms we have the Bolza-type problem
Minimize in u : J (u) = Φ (x(tf ),xtf ) +
∫ tf0
F (t,x (t)) dt
subject to
A(x(t))ẋ(t) = f(t,x(t),u(t))
x(0) = x0 ∈ Ω
0 ≤ si(t), si(t) ≤ 1, 1 ≤ i ≤ N, and x (t) ∈ Ω,
(Ptf )
where Ω stands for the set of constraints for the state variable. Typically
Φ(x(tf ),x
tf)
=
3N+12∑j=1
αj
(xj(tf )− x
tfj
)2(28)
with αj > 0 penalty parameters, and
F (t,x(t)) =
3N+12∑j=1
βj (xj(t)− xj (t))2 (29)
with βj > 0 also weight parameters and x(t) = [xj(t)] a desired trajectory.
The set Ω, which models the constraints on the state variable, has the following structure.
For the variables entering in the blowing-venting model, since the outside pressure at a
certain depth is the sum of the atmospheric pressure and the weight of the water column
above the submarine, even if the pressure in the ballast tank is slightly lower than the outside
pressure and the vehicle is close to the surface, we can safely assume that the pressure in
the tank will always be greater than the atmospheric pressure, that is pBi ≥ p−B = patm.Although an upper bound can not be so easily obtained, it is easy to see that the pressure in
the tank will always take finite values, which justifies the assumption pBi ≤ p+B < +∞. It isalso immediate to see that the upper bound for the mass of air in the tank and bottle is the
initial mass of air in the bottle. By hypothesis, there will always be a residual amount of
air in the tanks, mB0 . Therefore mBi ≥ m−B = mB0 > 0. As we stated before, the pressurein the tank will not drop below patm. Since the air will flow due to the pressure difference
between bottle and tank, the pressure in the bottle will have the same lower bound. This
way, using the perfect gas equation and (7) a lower bound for the air mass in the bottle can
be obtained, mFi ≥ m−F > 0. Summarizing, we are able to assume the constraints0 < m−F ≤ mFi ≤ m
+F < +∞
0 < m−B ≤ mBi ≤ m+B < +∞
0 < p−B ≤ pBi ≤ p+B < +∞.
(30)
4 MATHEMATICAL ANALYSIS 17
As for Euler angles, since we are dealing with a manned submarine, typically
−π4< φ <
π
4, −π
4< θ <
π
4, 0 < ψ < 2π.
Due to the bounded nature of ocean, the position components (x, y, z) are also limited to
some bounded rectangle. Finally, the physics of the problem also imposes a constraint on
the rest of components (i.e, linear (u, v, w) and angular (p, q, r) velocities).
To sum up, we can assume that Ω is a bounded rectangle.
4 Mathematical analysis
Once the mathematical model for the blowing-venting procedure has been established and
the associated control problem has been formulated as a Bolza-type problem, then we will
analyze it to provide an existence result. That is, we will show that, under suitable assump-
tions, for any admissible initial state x0 ∈ Ω, there exists a control u(t) minimizing the costfunctional in (Ptf ).
4.1 Well-posedness of the state law
We start our analysis by proving that for any initial state x0 ∈ Ω and any admissible controlu(t) there exists a unique solution of (27) starting from x0, and also satisfying x(t) ∈ Ω,defined in some interval 0 ≤ t ≤ tf where tf = tf (x0) only depends on the initial condition.However, since the state law is expressed in implicit form, we can not apply standard results.
To overcome this difficulty we will show that the matrix-valued map A is smooth and takes
nonsingular values, that is, the state law can be rewritten in explicit form as
ẋ(t) = A(x(t))−1f(t,x(t),u(t)) (31)
4.1.1 A is nonsingular-valued
It is clear that for any x ∈ Ω, the matrix A(x) has the form
A(x) =
BV(x) 03N×6 03N×606×3N I6 06×606×3N 06×6 M(x)
(32)where BV(x) is the submatrix associated to the blowing-venting equations and M(x) is the
so called (variable) inertia matrix. Furthermore, BV(x) is a 3N × 3N matrix structured in3× 3 diagonal blocks
BV(x) =
BV1(x) · · · 03×3... . . . ...03×3 · · · BVN (x)
(33)where, for 1 ≤ i ≤ N ,
BVi(x) =
1 0 01 1 00 −1 mBipBi
(34)
4 MATHEMATICAL ANALYSIS 18
with det BVi(x) =mBipBi≥ m
−B
p+B> 0 by (30). Thus BVi(x) is nonsingular with
BVi(x)−1 =
1 0 0−1 1 0−pBimBi
pBimBi
pBimBi
(35)and, therefore, the full matrix BV(x) is also nonsingular for every admissible state x ∈ Ωwith
BV(x)−1 =
BV1(x)−1 · · · 03×3
.... . .
...
03×3 · · · BVN (x)−1
. (36)It just remains to prove that the inertia matrix M is invertible. From the dynamic
equations of motion and (22)-(26) we have that M(x) has the form
M(x) = Mv(x) + Mc (37)
where
Mv(x) =
[m(x)I3 −S(x)S(x) I(x)
](38)
is the variable part of the matrix, with
S(x) =
0 −m(x)zG(x) m(x)yG(x)m(x)zG(x) 0 −m(x)xG(x)−m(x)yG(x) m(x)xG(x) 0
and
I(x) =
Ix(x) −Ixy(x) −Izx(x)
−Ixy(x) Iy(x) −Iyz(x)
−Izx(x) −Iyz(x) Iz(x)
the inertia tensor, and Mc is the so-called added inertia matrix
−ρ2 l3X ′u̇ 0 0 0 0 0
0 −ρ2 l3Y ′v̇ 0 −
ρ2 l
4Y ′ṗ 0 −ρ2 l
4Y ′ṙ
0 0 −ρ2 l3Z ′ẇ 0 −
ρ2 l
4Z ′q̇ 0
0 −ρ2 l4K ′v̇ 0 −
ρ2 l
5K ′ṗ 0 −ρ2 l
5K ′ṙ
0 0 −ρ2 l4M ′ẇ 0 −
ρ2 l
5M ′q̇ 0
0 −ρ2 l4N ′v̇ 0 −
ρ2 l
5N ′ṗ 0 −ρ2 l
5N ′ṙ
. (39)
It is usual in the literature on dynamics of submerged vehicles (see[8], Property 2.4, for
instance) to assume that Mc is a symmetric and positive definite matrix (and therefore
invertible). However, experimental values of the non-dimensional hydrodynamic coefficients
4 MATHEMATICAL ANALYSIS 19
reported by Navantia showed that this is not a realistic assumption in all cases. In partic-
ular the inertia matrix used in our numerical experiments, based in the experimental data
provided by Navantia, is not symmetric, but it is invertible (see Table 5 in the Appendix).
Let M0 be the rigid-body inertia matrix of the submarine with all the ballast tanks
completely filled with water, i.e.,
M0 =
m0I3 m0
0 zG0 −yG0−zG0 0 xG0yG0 −xG0 0
m0
0 −zG0 yG0zG0 0 −xG0−yG0 xG0 0
I0
. (40)
This matrix is symmetric and usually is assumed to be positive definite ([8], Property 2.2).
Therefore whenever matrices Mc and M0 are assumed to be symmetric and positive definite,
the inertia matrix M0 + Mc is invertible, since it is also symmetric and positive definite.
Otherwise the invertibility of M0 + Mc must be checked for any particular model.
Finally, nonsingularity of M(x) follows from the next Lemma.
Lemma 4.1 Let D ∈Mm×m be an invertible matrix and let ∆ ∈Mm×m be a square matrixsuch that
‖D−1∆‖ = sup{|D−1∆y| : |y| = 1
}< 1
where | · | denotes the Euclidean norm in Rm. Then D + ∆ is invertible with
(D + ∆)−1
=
∞∑n=0
(−1)n(D−1∆
)nD−1
Computing the operator norm of a matrix is not in general an easy task, but next Lemma
provides a very useful estimate.
Lemma 4.2 Let D = [dij ] ∈Mm×m be a square matrix. Then
‖D‖ ≤√∑
i,j
d2ij . (41)
The right-hand side in (41) is called the Frobenius norm of D, ‖D‖F .
Let x ∈ Ω be an admissible state. Since M(x) = (M0 + Mc) + (Mv(x)−M0) withM0 + Mc invertible, by Lemma 4.1, it suffices to show that∥∥∥(M0 + Mc)−1 (Mv(x)−M0)∥∥∥ < 1 (42)to have the nonsingularity of M(x). Furthermore, Lemma 4.1 also ensures that
M(x)−1 =
∞∑n=0
(−1)n(
(M0 + Mc)−1
(Mv(x)−M0))n
(M0 + Mc)−1. (43)
4 MATHEMATICAL ANALYSIS 20
Summarizing, M(x) is invertible, for any state x ∈ Ω, whenever M0 + Mc is invertibleand the variation of mass due to the blowing-venting manoeuvres is small enough. More
precisely, if
∆m+ <λmin√
3 + 4N∑Ni=1 |ci|2 +
92
∑Ni=1 |ci|4
(44)
holds, where ∆m+ =∑Ni=1 ∆mi,max = ρ
∑Ni=1(VBBi − VB0) is the maximum variation of
mass, λmin > 0 is the smallest singular value of the inertia matrix M0 + Mc and ci is the
geometrical center of the i−th ballast tank, then A is nonsingular-valued.
Theorem 4.1 If the matrix M0 + Mc is invertible and the inequality (44) is satisfied, then
the matrix-valued map A : Ω −→M(12+3N)×(12+3N) takes nonsingular values, that is, A(x)is invertible for any x ∈ Ω with
A(x)−1 =
BV(x)−1 03N×6 03N×606×3N I6 06×606×3N 06×6 M(x)
−1
.Moreover, the map x ; A(x)−1 is continuously differentiable.
Proof. To show that A(x) is invertible for any x ∈ Ω, it suffices to show that (42) is satisfieduniformly with respect to the state variable. We have∥∥∥(M0 + Mc)−1 (Mv(x)−M0)∥∥∥ ≤ ∥∥∥(M0 + Mc)−1∥∥∥ ‖Mv(x)−M0‖
=1
λmin‖Mv(x)−M0‖ (45)
by properties of the matrix norm (see Appendix A in [11], for instance). Let us now obtain
estimates for the supremum of the Frobenius norm of the variable matrix. Firstly, from (21)∣∣∣∣∣N∑i=1
∆mi(x)
∣∣∣∣∣ ≤N∑i=1
∆mi,max = ρ
N∑i=1
(VBBi − VB0). (46)
Let us now estimate the variation of the coordinates of the center of gravity∣∣∣∣∣N∑i=1
xbi∆mi(x)
∣∣∣∣∣ ≤ (∆m+)N∑i=1
|xbi|, (47)
∣∣∣∣∣N∑i=1
ybi∆mi(x)
∣∣∣∣∣ ≤ (∆m+)N∑i=1
|ybi| (48)
and
∣∣∣∣∣N∑i=1
zmli(x)∆mi(x)
∣∣∣∣∣ ≤ (∆m+)N∑i=1
|zbi| (49)
with (xbi, ybi, zbi) the coordinates of the geometrical center of the i−th tank (see Figure 4).Finally, from (25) and (46)
|Ix(x)− Ix0| =
∣∣∣∣∣N∑i=1
(y2bi + z
2mli
)∆mi(x)
∣∣∣∣∣ ≤ (∆m+)N∑i=1
(y2bi + z
2bi
)(50)
4 MATHEMATICAL ANALYSIS 21
and, in a similar way,
|Iy(x)− Iy0| ≤(∆m+
) N∑i=1
(x2bi + z
2bi
), |Iz(x)− Iz0| ≤
(∆m+
) N∑i=1
(x2bi + y
2bi
)(51)
and
|Ixy(x)| ≤(∆m+
) N∑i=1
|xbi| |ybi|, |Ixz(x)| ≤(∆m+
) N∑i=1
|xbi||zbi| (52)
|Iyz(x)| ≤(∆m+
) N∑i=1
|ybi| |zbi|. (53)
Hence, from Lemma 4.2, and estimates (46)-(53), it follows that, for any x ∈ Ω,
‖Mv(x)−M0‖ ≤(∆m+
)√√√√3 + 4N ( N∑i=1
x2bi + y2bi + z
2bi
)+
9
2
(N∑i=1
x2bi + y2bi + z
2bi
)2(54)
where we have also used the convexity inequality(∑N
i=1 ai
)2≤ N
(∑Ni=1 a
2i
)and also that
ab ≤ (a2 + b2)/2. Combining this inequality with (44), we have the desired inequality (42).Finally, from (36) and (43) it follows that x ; A(x)−1 is continuosly differentiable and the
proof is complete.
Remark 1 For the particular data of the prototype considered in our numerical experiments,
M0 + Mc is invertible. Moreover, λmin = 1.54 · 1012 and hence
(∆m+)
√3 + 4N
(∑Ni=1 x
2bi + y
2bi + z
2bi
)+ 92
(∑Ni=1 x
2bi + y
2bi + z
2bi
)2λmin
= 1.73 · 10−4. (55)
4.1.2 The state law is well-posed
Let K = [0, 1]2N be. An admissible control of the problem will be an essentially boundedmap taking values in K, that is, L∞(0,+∞;K) is the space of admissible controls. Givenu ∈ L∞(0,+∞;K), our aim in this section is to show that the system (31) has a solutionfor any initial state x0. We begin by recalling the classical theory on this subject. For the
details we refer the reader to [11, Appendix C].
By a (Carathéodory) solution we mean an absolutely continuous function t; x(t) ∈ Ω,defined on some interval I = [0, tf ], which satisfies the integral equation
x (t) = x0 +
∫ t0
g(s,x(s)) ds, for every t ∈ [0, tf ] ,
where we write g : I × Ω −→ R3N+12, g(t,x) = A(x)−1f(t,x,u(t)) for simplicity.As it is well-known, if g satisfies conditions (H1)-(H4) below, then we can ensure the
existence and uniqueness of a maximal solution of (31) for any initial state:
(H1) for each x ∈ Ω, the function g (·,x) : I → RN is measurable,
4 MATHEMATICAL ANALYSIS 22
(H2) for each t ∈ I, the function g (t, ·) : Ω→ RN is continuous,
(H3) g is locally Lipschitz with respect to x, that is, for each x0 ∈ Ω there exist a realnumber ε > 0 and a locally integrable function α : I → R+ such that the ball Bε
(x0)
of radius ε centered at x0 is contained in Ω and
|g (t,x)− g (t,y)| ≤ α (t) |x− y| for every t ∈ I and x,y ∈Bε(x0), (56)
(H4) g is locally integrable with respect to t, that is, for each x0 ∈ Ω there exist a locallyintegrable function β : I → R+ such that∣∣g (t,x0)∣∣ ≤ β (t) a.e. t ∈ I. (57)
Our main result in this section follows:
Theorem 4.2 For each x0 ∈ Ω and each u ∈ L∞ (R+;K) there exists a unique maximalsolution of (31) defined in [0, tf ] , where tf = tf (x
0) only depends on the initial state x0 and
is uniform with respect to the control u ∈ L∞ (R+;K) .
Proof. To begin with, we notice that thanks to Theorem 4.1 we may restrict the analysis
that follows to the right-hand side of the state law (27). Let us see that conditions (H1)-(H4)
above hold. Since the time variable t only appears in the control functions (si (t) , si (t)) ,
1 ≤ i ≤ N , and, by hypothesis, these functions belong to L∞ (R+; [0, 1]) , it is clear that foreach x ∈ Ω, the function t; f (t,x) is measurable.
As regards the continuity of the function f (t, ·), only the equations modelling the airflow from pressure bottles need of a detailed analysis to check that the pass from supersonic
flow to subsonic one is continuous. A direct computation shows that this is so whenever
mFi ≥ m−F > 0 and pBi ≥ p−B > 0.
As for condition (H4), from the form in which controls appear in the state law it follows
that for each x0 ∈ Ω the components fi(t,x0
), 1 ≤ i ≤ 3N + 12, of f
(t,x0
)are uniformly
bounded with respect to t, that is, (57) holds for a constant function β (t) = β and what is
more important, this constant is uniform with respect to u ∈ L∞ (R+;K) .Next we analyze the local Lipschitz condition of f (t,x). Consider firstly the equations
fi, i = 1 + 3j, 0 ≤ j ≤ N − 1, which model air flow from bottles. Taking into account theset of constraints (30), a direct computation shows that fi (t, ·) ∈ W 1,∞ (Ω) and thereforethey are Lipschitz. As before it is important to notice that the estimates on the partial
derivatives of fi (t, ·) are uniform with respect to t, that is, there exists L > 0 such that∣∣∣∣ ∂fi∂xj (t,x)∣∣∣∣ ≤ L for every x ∈ Ω and uniformly w.r.t. t ≥ 0. (58)
Similarly, functions fi, i = 2 + 3j, 0 ≤ j ≤ N − 1, which appear in the equations modellingair flow through venting valve are continuous and satisfy an estimate as in (58). Thus,
they are Lipschitz. As before, the Lipschitz constant is uniform with respect to the control
variable.
Consider now functions f3+3j , 0 ≤ j ≤ N−1, which appear in the equations for evolutionof pressure in ballast tanks. In this case, when pressure in a ballast tank equals outside
4 MATHEMATICAL ANALYSIS 23
pressure, the derivatives of velocities in the corresponding flood port blows-up because of
the presence of the square root. To avoid this singularity we have approximated the square
root as shown in Subsection 2.1.4. As a conclusion, once again we obtain that these maps
are Lipschitz.
The components fi, 3N+1 ≤ i ≤ 3N+6, only include the transformation matrix betweenbody and world reference frames. Taking into account the constraints on Euler angles, it is
clear that fi ∈ C∞ (Ω) , 3N + 1 ≤ i ≤ 3N + 6, and therefore they are Lipschitz. Notice thatthe time variable does not appear in these functions.
As for the remaining fi, 3N + 7 ≤ i ≤ 3N + 12, these components include: (a) polyno-mial terms and terms in the form of absolute values; all of them are Lipschitz, (b) terms
like xj√x2j + x
2k and |xj |
√x2j + x
2k for some 18 ≤ j ≤ 24. Since these functions are contin-
uous and the discontinuities of its derivatives are of a finite jump, they are also Lipschiz.
(c) Since our model is mass variable, it is necessary to look carefully at the terms includ-
ing mass m, weight W, center of gravity (xG, yG, zG) and moments and product of inertia
Ix, Iy, Iz, Ixy, · · · , because now they depend on some components of the state variable. Tak-ing into account the constraints (30) it is clear that these components are also Lipschitz.
Finally, the product of Lipschitz functions is also Lipschitz. As before, these components
fi do not include control variable u (t) and therefore the corresponding Lipschitz constants
are independent of u (t) .
This analysis lets us conclude that for each x0 ∈ Ω and u ∈L∞ (R+;K) there existsa maximal time tf = tf (x
0,u) and a unique maximal solution defined in[0, tf
(x0,u
)].
Looking at the proof of this existence result (see [11, Th. 36, pp. 347–351]) we realize that
tf depends on u through the functions α (t) = α (u (t)) and β (t) = β (u (t)) which appear in
(56) and (57). However, as shown before these functions α and β can be chosen uniformly
w.r.t. u ∈ L∞ (R+;K). As a conclusion, tf = tf(x0)
only depends on the initial condition.
This completes the proof.
4.2 Existence of a solution for the optimal control problem(Ptf)
Fix x0 ∈ Ω and let tf = tf(x0)
be the maximal time, as given by Theorem 4.2, for which
system (31) is well-posed. The goal of this section is to prove the following existence result:
Theorem 4.3 Let x0 ∈ Ω and let 0 < tf = tf(x0)< +∞ be as above. Then, there exists,
at least, one solution to (Ptf ).
Proof. The proof follows as a consequence of the classical Filippov existence theorem
for Bolza-type optimal control problems (see [3, Th. 9.3.i, p. 314]). Indeed, let us see that
the sufficient conditions for the existence of a solution hold. Due to the constrains on the
state variable, the set A = [0, tf ] × Ω is compact. Similarly, since the set K = [0, 1]2N forcontrol constraints is compact, the set A × K is also compact. In addition, the functionΦ : A×A → R defined by (28) is continuous. Continuity also holds for the functions F andg both defined on A×K.
Notice also that, by Theorem 4.2, the set of admissible solutions for (Ptf ) is not empty.
Finally, we must check that for each (t,x) ∈ A the orientator field
Q (t,x) ={(z0; z
)∈ R1+3N+12 : z0 ≥ F (t,x) , z = g (t,x,u) , with u ∈ K
}
5 NUMERICAL ANALYSIS 24
is convex. This convexity easily follows from the facts that the control variable u appears
in a linear form in the state law and the set K is convex.
5 Numerical analysis
5.1 Algorithm of minimization
There are several optimization methods which can be applied to solve (Ptf ). Due to the
complexity of the state law and the large number of variables involved in the problem, it is
quite reasonable to use a gradient descent method with projection. Briefly, the scheme of
this method consists of the following main steps:
1. Initialization of the control input u0.
2. For k ≥ 0, iteration until convergence (e.g.∣∣J (uk+1)− J (uk)∣∣ ≤ ε ∣∣J (u0)∣∣ , with
ε > 0 a suitable tolerance) as follows:
2.1 we consider the vector
vk+1 = uk − λ∇J(uk)
where λ > 0 is a fixed step parameter, and ∇J(uk)
is the gradient of the cost function.
2.2 Since vk+1 may be not admissible, we compute its orthogonal projection onto the
admissibility set K, the unit rectangle in R8, that is,
uk+1 = PK(vk+1
)where
uk+1j = min(max
(0,vk+1j
), 1).
The crucial step is the computation of the gradient ∇J(uk). This can be obtained by
using the adjoint method which is described next:
1. Given the control uk, k ≥ 0, solve the state equation{A (x (t)) ẋ (t) = f
(t,x (t) ,uk (t)
)x (0) = xk (0)
to obtain the state xk+1 (t) .
2. With the pair(uk (t) ,xk+1 (t)
), solve the linear backward equation for the adjoint
state p (t) A(xk+1 (t)
)Tṗ (t) = −∇xF
(t,xk+1 (t) ,uk (t)
)−[∇xf
(t,xk+1 (t) ,uk (t)
)]Tp (t)
A(xk+1 (tf )
)Tp (tf ) = ∇xΦ
(xk+1 (tf ) ,x
tf)
where∇x is the gradient with respect to the state variable x. Thus, we obtain pk+1 (t) .
5 NUMERICAL ANALYSIS 25
3. Finally,
∇J(uk)
= ∇uF(t,xk+1 (t) ,uk (t)
)+[∇uf
(t,xk+1 (t) ,uk (t)
)]Tpk+1 (t)
where now ∇u is the gradient with respect to u.
Here AT stands for the transpose of A. We refer to [5] for more details on this method.
5.2 A numerical experiment
In order to test the proposed models, this section shows the results of the numerical sim-
ulation of an emergency rising manoeuvre. These results are used to analyze both the
mathematical properties and the possible real-world applications of the proposed scheme.
At each iteration of the gradient algorithm, the numerical resolutions of the state and ad-
joint state equations have been carried out by using the ODE45 Matlab function, which is
a one-step solver based on an explicit Runge-Kutta method.
On an emergency situation, like on board fire or flood, submarines may need to rise
to the surface quickly. The typical protocol for this situation is to use the control planes
to pitch the nose of the submarine up, increase the speed, and blow the ballast tanks to
reduce the weight of the submarine and drive it to the surface with buoyancy. As analyzed
in [1, 13], small to medium size submarines exhibit a roll instability during these kind of
manoeuvres. Full scale trials showed that submarines rose with roll angles up to 25 degrees.
It is also known that if the the submarine emerges with a high roll angle, it may experiment
large roll oscillations on the surface. Of course, this is an undesirable situation, particularly
if the operators are attending to the original problem that required the emergency rise.
For these reasons, it might be interesting to check if this situation can be prevented by
using our control algorithm. To this end, an emergency rising manoeuvre has been simulated
for two different scenarios:
1) A standard manoeuvre. Initial depth is 100 m, initial speed is 2 m/s. At t = 0 the
stern and bow planes are set to −20 and 20 degrees respectively and the propeller isset to 150 rpm. At the same time, ballast tanks 2–5 are simultaneously blown with
half the maximum intensity (this corresponds to si = 0.5 in our model). Vent valves
remain closed (si = 0) throughout all the simulation. Simulation ends when submarine
reaches a depth of 10 m (an arbitrarily low value for which we can assume that the
vehicle has reached the surface). To sum up:
δr(t) = 0
δs(t) = −20δb(t) = 20
n(t) = 2.5
si(t) = 0.5, i = 1 . . . 4
si(t) = 0, i = 1 . . . 4
∀ t
We note that, although we refer to this scenario as standard manoeuvre, the constant
value for the deflection of control planes is of course a simplification of what would be
done in real operation.
5 NUMERICAL ANALYSIS 26
2) Same manoeuvre, with the control algorithm acting from t = 0 to t = 30 s using
Scenario 1 as initialization and looking to achieve three main objectives:
• Submarine must rise in a similar time as it does in the standard manoeuvre.• Rising pitch angle must be around 20 degrees and never above 25 degrees.• Roll angle must be as close as possible to zero throughout all the simulation.
To this end, the following set of parameters is used:
x(0) =(
[mF0i ,mB0i , pB0i ]1≤i≤4 , 0, 0, 100, 0, 0, 0, 2, 0, 0, 0, 0, 0)
tf = 30 s
ztf = xtf15 = 75, θ
tf = xtf17 = 10.
φ(t) = x16(t) = 0 ∀ t ∈ [0, 30].α15 = α17 = 1, αj = 0, j 6= 15, 17β16 = 5, βj = 0, j 6= 16λ = 0.01 ε = 10−4
where the initial mass of air in the bottles is mF0i = 237.8376 kg, the initial mass of
air in the tanks is mB0i = 0.0126 kg and the initial pressure in the tanks is pB0i =
1.0846 · 106 Pa. After t = 30 s, simulation continues with fixed controls (values equalto those used in Scenario 1) until the submarine reaches a depth of 10 m.
Results for Scenario 1 (dashed lines) and Scenario 2 (solid lines) are shown in figures 5 to
9. Figure 5 shows vehicle depth, pitch angle and roll angle for both scenarios. For Scenario 2,
pressure, mass of air in the tank and mass of air in the bottle are plotted for each of the tanks
on Figure 6. The rest of state variables have not been included since they are not directly
relevant for this particular manoeuvre. Comparison between dashed (standard manoeuvre)
and solid (optimal controls) lines shows that the three objectives have been achieved for
Scenario 2: the rising time is only a few seconds greater than in the standard manoeuvre,
the final pitch angle is close to 20 degrees and the roll angle has been significantly reduced
with respect to the standard manoeuvre. Indeed, Scenario 1 exhibits roll angles in the range
of 3–4 degrees during most of the simulation time while in Scenario 2, after an initial peak
of 2 degrees, the roll angle is kept within extremely low values during the whole t = [0, 30]
interval.
The optimal (solid lines) and standard (dashed lines) controls are shown in Figure 7
(blowing valves) and Figure 8 (venting valves). Although the convenience of the use of
venting during an emergency rise can be arguable and is indeed not an usual practice, it is
used here to demonstrate the algorithm capabilities. Very similar, although slightly worse
results, were obtained by using only the blowing valves as control.
As we can see in Figure 7, the rolling moment is compensated by blowing more ballast
from the starboard tanks. It may surprise that tanks 2 and 4 are being vented while there
is no blowing air yet, but this seems to allow a smoother transition when the two valves
are simultaneously open. As we said before, the use of venting valves seems to improve the
results obtained by only blowing the tanks.
The value of the cost function at each iteration is plotted in Figure 9. As we see, the
algorithm shows exponential convergence. As is usual in this type of algorithm, results
5 NUMERICAL ANALYSIS 27
Figure 5: Depth, pitch and roll angle for scenarios 1 (dashed line) and 2 (solid line).
5 NUMERICAL ANALYSIS 28
Figure 6: Pressure (solid line), air mass in bottle (dashed line) and air mass in tank (dash-dot
line) for tanks 2−5.
Figure 7: Blowing valve aperture for standard (dashed lines) and optimal (solid lines) con-
trols.
5 NUMERICAL ANALYSIS 29
Figure 8: Vent valve aperture for standard (dashed lines) and optimal (solid lines) controls.
Figure 9: Cost function at each iteration.
A APPENDIX: EQUATIONS OF MOTION 30
depend on the initialization. Different initializations were tested obtaining different optimal
controls. This seems to show the existence of several local and/or global minima.
A Appendix: equations of motion
A.1 Kinematic equations
φ̇ = p+ r cos(φ) tan(θ) + q sin(φ) tan(θ)
θ̇ = q cos(φ)− r sin(φ)
ψ̇ =r cos(φ) + q sin(φ)
cos(θ)
ẋ = u cos(φ) cos(ψ) + v(sin(φ) sin(θ) cos(ψ)− cos(φ) sin(ψ))+ w(sin(φ) sin(ψ) + cos(φ) sin(θ) cos(ψ))
ẏ = u cos(θ) sin(ψ) + v(cos(φ) cos(ψ) + sin(φ) sin(θ) sin(ψ))
+ w(cos(φ) sin(θ) sin(ψ)− sin(φ) cos(ψ))ż = −u sin(θ) + v cos(θ) sin(φ) + w cos(θ) cos(φ)
A.2 Dynamic equations
AXIAL FORCE EQUATION:
m[u̇− vr + wq − xG(q2 + r2) + yG(pq − ṙ) + zG(pr + q̇)]
=ρ
2l4[X
′
qqq2 +X
′
rrr2 +X
′
rprp+X′
q|q|q|q|] +ρ
2l3[X
′
u̇u̇+X′
vrvr +X′
wqwq]
+ρ
2l2[X
′
uuu2 +X
′
vvv2 +X
′
www2 +X
′
w|w|w|w|]
+ρ
2l2[X
′
δrδru2δ2r +X
′
δsδsu2δ2s +X
′
δbδbu2δ2b ]− (W −B) sin(θ) + ρT (1− tp)
LATERAL FORCE EQUATION:
m[v̇ − wp+ ur − yG(r2 + p2) + zG(qr − ṗ) + xG(qp+ ṙ)]
=ρ
2l4[Y
′
ṙ ṙ + Y′
ṗ ṗ+ Y′
r|r|r|r|+ Y′
pqpq]
+ρ
2l3[Y
′
rur + Y′
pup+ Y′
v̇ v̇ + Y′
wpwp]
+ρ
2l2[Y
′
∗u2 + Y
′
vuv + Y′
v|v|Nv|(v2 + w2)
12 |]
+ρ
2l2[Y
′
δru2δr + Y
′
δrηu2δr(η −
1
C)C]
+ρ
2l2Y
′
vwNvw + (W −B) cos(θ) sin(φ)
A APPENDIX: EQUATIONS OF MOTION 31
NORMAL FORCE EQUATION:
m[ẇ − uq + vp− zG(p2 + q2) + xG(rp− q̇) + yG(rq + ṗ)]
=ρ
2l4[Z
′
q̇ q̇ + Z′
q|q|q|q|+ Z′
rrr2] +
ρ
2l3[Z
′
ẇẇ + Z′
quq + Z′
vpvp+ Z′
vrvr]
+ρ
2l2[Z
′
∗u2 + Z
′
wuw + Z′
vvv2] +
ρ
2l2[Z
′
|w|u|w|+ Z′
wwN |w| (v2 + w2)12 ]
+ρ
2l2[Z′
δsu2δs + Z
′
δbu2δb + Z
′
δsηu2δs
(η − 1
C
)C
]+(W −B) cos(θ) cos(φ)
ROLLING MOMENT EQUATION:
Ixṗ+ (Iz − Iy)qr − Izxṙ − Izxpq + Iyzr2 − Iyzq2 + Ixypr − Ixy q̇myGẇ −myGuq +myGvp−mzGv̇ +mzGwp−mzGur
=ρ
2l5K
′
ṗṗ+ρ
2l5K
′
ṙ ṙ +ρ
2l5K
′
qrqr +ρ
2l5K
′
p|p|p|p|+ρ
2l5K
′
r|r|r|r|
+ρ
2l4K
′
pup+ρ
2l4K
′
rur +ρ
2l4K
′
v̇ v̇ +ρ
2l4K
′
wpwp
+ρ
2l3K
′
∗u2 +
ρ
2l3K
′
vuv +ρ
2l3K
′
v|v|v|v|+ρ
2l3K
′
δru2δr − ρQ
+(YGW − YBB) cos(θ) cos(φ)− (ZGW − ZBB) cos(θ) sin(φ)
PITCHING MOMENT EQUATION:
Iy q̇ + (Ix − Iz)rp− (ṗ+ qr)Ixy + (p2 − r2)Izx + (qp− ṙ)Iyz+m[zG(u̇− vr + wq)− xG(ẇ − uq + vp)]
=ρ
2l5[M
′
q̇ q̇ +M′
rprp+M′
q|q|q|q|+M′
rrr2] +
ρ
2l4[M
′
ẇẇ +M′
quq +M′
vrvr]
+ρ
2l3[M′
∗u2 +M
′
wuw +M′
vvv2 +M
′
w|w|Nw∣∣∣(v2 + w2) 12 ∣∣∣]
+ρ
2l3[M
′
vwvw +M′
|w|u|w|+M′
ww|w(v2 + w2)12 |]
+ρ
2l3[M′
δsu2δs +M
′
δbu2δb +M
′
δsηu2δs
(η − 1
C
)C
]−(xGW − xBB) cos(θ) cos(φ)− (zGW − zBB) sin(θ)
YAWING MOMENT EQUATION:
Iz ṙ + (Iy − Ix)pq − (q̇ + rp)Iyz + (q2 − p2)Ixy + (rq − ṗ)Izx+m[xG(v̇ − wp+ ur)− yG(u̇− vr + wq)]
=ρ
2l5[N
′
ṙ ṙ +N′
r|r|r|r|+N′
ṗṗ+N′
pqpq] +ρ
2l4[N
′
pup+N′
rur +N′
v̇ v̇]
+ρ
2l3[N
′
∗u2 +N
′
vuv +N′
v|v|Nv∣∣∣(v2 + w2) 12 ∣∣∣]
+ρ
2l3[N′
δru2δr +N
′
δrηu2δr
(η − 1
C
)C
]+ρ
2l3N
′
vwNvw
+(xGW − xBB) cos(θ) sin(φ) + (yGW − yBB) sin(θ)
The value of the hydrodynamic coefficients present in the above equations have been
obtained experimentally by using a scale model.
A APPENDIX: EQUATIONS OF MOTION 32
l 67 m m0 2352 · 103 kg B0 23073 · 103 NXG0 0 m YG0 0 m ZG0 −0.264 mXB 0 m YB 0 m ZB −0.595 mIx0 16390 ·103 kgm2 Iy0 659770 · 103 kgm2 Iz0 659770 · 103 kgm2
Ixy0 0 kgm2 Ixz0 0 kgm
2 Iyz0 0 kgm2
Table 3: Submarine parameters
KT0 0.5382 KTJ −0.3811KTJ2 −0.1645 KTJ3 0KTJ4 0 η 1.09431
C 0.8976 KQ0 0.0701
KQJ −0.0273 KQJ2 −0.0345KQJ3 0 KQJ4 0
tp 0.148 wf 0.344
D 3.821 m
Table 4: Propeller model coefficients
A APPENDIX: EQUATIONS OF MOTION 33
X ′u̇ −0.46 · 10−3 Y ′ṗ −0.3 · 10−3 Z ′q̇ −0.07 · 10−3
X ′rp 0.6 · 10−3 Y ′ṙ −0.83 · 10−3 Z ′ẇ −14.40 · 10−3
X ′qq 1.42 · 10−3 Y ′v̇ −18.37 · 10−3 Z ′vp −18.37 · 10−3
X ′q|q| 0 Y′p −3.05 · 10−3 Z ′q −6.99 · 10−3
X ′rr 2.08 · 10−3 Y ′pq −0.07 · 10−3 Z ′q|q| −3.98 · 10−3
X ′vr 22.37 · 10−3 Y ′wp 14.40 · 10−3 Z ′rr −3.96 · 10−3
X ′uu −1.124 · 10−3 Y ′r 7 · 10−3 Z ′vr −45.13 · 10−3
X ′vv 17.46 · 10−3 Y ′r|r| −4.65 · 10−3 Z ′vv 0.185
X ′ww 7.75 · 10−3 Y ′v −61.36 · 10−3 Z ′w −20.28 · 10−3
X ′w|w| 0 Y′v|v| −28.14 · 10
−3 Z ′|w| −1.99 · 10−3
X ′wq −13.16 · 10−3 Y ′δr −0.83 · 10−3 Z ′ww 20.00 · 10−3
X ′w|q| 0 Y′δrη
0.67 · 10−3 Z ′δs −5.12 · 10−3
X ′δrδr −3.90 · 10−3 Y ′∗ 0 Z
′δsη
−0.45 · 10−3
X ′δsδs −1.19 · 10−3 Y ′vwN 0 Z
′δb
−5.12 · 10−3
X ′δbδb −2.99 · 10−3 Y ′v|v|N −0.1162 Z
′∗ −0.3 · 10−3
X ′δbδb −2.99 · 10−3 Y ′v|v|N −0.1162 Z
′∗ −0.3 · 10−3
M ′q̇ −0.98 · 10−3 N ′ṗ −0.03 · 10−3 Z ′w|w|R 0M ′ẇ −1.39 · 10−3 N ′ṙ −1.20 · 10−3 K ′ṗ −0.07 · 10−3
M ′rp 1.12 · 10−3 N ′v̇ 0.68 · 10−3 K ′ṙ −0.05 · 10−3
M ′q −3.89 · 10−3 N ′p −0.68 · 10−3 K ′v̇ −0.60 · 10−3
M ′q|q| −3.03 · 10−3 N ′pq −0.91 · 10−3 K ′p −0.62 · 10−3
M ′rr −1.19 · 10−3 N ′r −4.83 · 10−3 K ′p|p| −0.30 · 10−3
M ′vr −15.73 · 10−3 N ′r|r| 2.09 · 10−3 K ′wp 0.30 · 10−3
M ′vv 34.61 · 10−3 N ′v −18.64 · 10−3 K ′qr −0.21 · 10−3
M ′w 4.78 · 10−3 N ′v|v|R 18.50 · 10−3 K ′r 0.26 · 10−3
M ′|w| −0.36 · 10−3 N ′δr −3.53 · 10
−3 K ′r|r| −0.19 · 10−3
M ′w|w|R −6.74 · 10−3 N ′δrη −0.34 · 10
−3 K ′v −2.87 · 10−3
M ′ww 0.45 · 10−3 N ′∗ 0 K ′v|v| −2.14 · 10−3
M ′δs −2.31 · 10−3 N ′vwN −0.303 K ′δr 0.13 · 10
−3
M ′δsη −0.38 · 10−3 K ′∗ 0
M ′δb 0.94 · 10−3
M ′∗ 0.02 · 10−3
Table 5: Non dimensional hydrodynamic coefficients.
REFERENCES 34
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1 Introduction2 Mathematical modelling2.1 Blowing and venting model2.1.1 Air flow from pressure bottle2.1.2 Air flow through venting valve2.1.3 Water flow through flood port2.1.4 Evolution of pressure in ballast tank2.1.5 Blowing and venting controlled system
2.2 Coupling of blowing-venting system with a variable mass model for the equations of motion2.3 Complete model in compact form
3 Formulation of the control problem4 Mathematical analysis4.1 Well-posedness of the state law4.1.1 A is nonsingular-valued4.1.2 The state law is well-posed
4.2 Existence of a solution for the optimal control problem (Ptf)
5 Numerical analysis5.1 Algorithm of minimization5.2 A numerical experiment
A Appendix: equations of motionA.1 Kinematic equationsA.2 Dynamic equations
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