Assignment 2, Part 1

Resources:

• Chapter 3,4,5,6 (Fossen, 2021)

---

This is the first part of a five-part assignment where you will create a high-fidelity simulation of a ship. The simulation model and its functionality will be built up step-by-step through these assignments

We assume that the geometrical shape of the ship is similar to a rectangular prism Where $L=161$ m, $B=21.8$ m, and $H=15.8$ m.

Problem 1: Rigid-Body Kinetics of a Rectangular Prism

Resources:

• Chapter 3
• Rigid-Body Kinetics

---

Goal: Find the rigid-body mass matrix $M_{RB}$ and the Coriolis matrix $C_{RB}$ of the ship (rectangular prism).

a)

The formula for the moment of inertia about the prism’s center of gravity is given by

$$I_z^{CG}={\int }_{\nu }(x^2+y^2)p_{m}d\nu$$

Which we can write as

$$\begin{array}{rl}{I}_z^{CG}& ={\rho }_m{\int }_{-H/2}^{H/2}{\int }_{-B/2}^{B/2}{\int }_{-L/2}^{L/2}{x}^2+y^{2}dxdydz\\ I_z^{CG}& =\frac{1}{12}m(L^2+B^2)\\ I_z^{CG}& =3.7544\cdot 10^{10}\end{array}$$

We know that $I_{xy}^{CG}=I_{yx}^{CG}=0$ since we take the integral of an odd function from $-a$ to $a$, which becomes zero due to symmetry.

b)

Let $r_{bg}^b=[x_g=-3.7,y_g=0,z_g=H/2]$. Using the Parallel Axis Theorem we can find the new inertia matrix

$$\begin{array}{rl}{I}_z^{CO}& =I_z^{CG}+m(x_g^2+y_g^2)\\ I_z^{CO}& =I_z^{CG}+m((-3.4{)}^2+0^2)=3.7777\cdot 10^{10}\end{array}$$

And the ratio is

$$I_{z,prism}^{CO}/I_{z,boat}^{CO}=1.7383$$

c)

Recall the Skew-Symmetric Matrix

Skew-Symmetric Matrix

The term $v^x$ is the skew-symmetric form of $v$

$$V=\left[\begin{array}{ccc}0& -c& b\\ c& 0& -a\\ -b& a& 0\end{array}\right]$$

^36cc79 Where

$$v=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$

---

Link to original

From equation 3.24, 3.29 and 3.30 in the book we can write

$$\begin{array}{rl}{M}_{RB}& =\left[\begin{array}{cc}m{I}_3& -mS(r_{bg}^b)\\ mS(r_{bg}^b)& I_g^b-m{S}^2(r_{bg}^b)\end{array}\right]\\ & \\ C_{RB}& =\left[\begin{array}{cc}mS({\omega }_{nb}^b)& -mS({\omega }_{nb}^b)S(r_{bg}^b)\\ mS(r_{bg}^b)S({\omega }_{nb}^b)& -mS(r_{bg}^b)S({\omega }_{nb}^b)S(r_{bg}^b)-S(I_g^b{\omega }_{nb}^b)\end{array}\right]\end{array}$$

Where $r_{bg}^b=[x_g=-3.7,y_g=0,z_g=H/2]$, $x_g$ is the distance from CG to CO, and $w_{nb}^b=[p,q,r{]}^T$ are the angular velocities. This gives us

$$S(r_{bg}^b)=\left[\begin{array}{ccc}0& -z_g& y_g\\ z_g& 0& -x_g\\ -y_g& x_g& 0\end{array}\right]$$

Since we’re only interested in the surge (x), sway (y) and yaw, which are respectively the first, second and last (sixth) state in the rigid-body mass matrix $M_{RB}$, we get:

$$M_{RB}=\left[\begin{array}{ccc}m& 0& -m{y}_g\\ 0& m& m{x}_g\\ -m{y}_g& m{x}_g& I_z^{CG}\end{array}\right]$$

We do the same for finding the rigid-body Coriolis and centripetal matrix, $C_{RB}$:

$$\begin{array}{rl}{C}_{RB}& =\left[\begin{array}{ccc}0& -mr& -mr{x}_g\\ mr& 0& -mr{y}_g\\ mr{x}_g& mr{y}_g& 0\end{array}\right]\\ C_{RB}& =m\left[\begin{array}{ccc}0& -r& -r{x}_g\\ r& 0& -r{y}_g\\ r{x}_g& r{y}_g& 0\end{array}\right]\end{array}$$

d)

We can see that $C_{RB}$ is a Skew-Symmetric Matrix. “This is a desired property when designing a nonlinear motion control system since the quadratic form ${\nu }^T{C}_{RB}\nu \triangleq 0$” (Fossen, page 67), meaning that the Coriolis effect does not change the kinetic energy in the system. A way to think about this is that a change in a state (mode) only redistributes the energy to other states.

e)

When irrotational ocean currents, $v_c=[u_c,v_c,w_c,0,0,0],$enter the equations of motion, the Coriolis matrix $C_{RB}$ is unaffected due to it not depending on linear velocity.

Problem 2: Hydrostatics of a Rectangular Prism

Resources:

• Chapter 4
• Hydrostatics

---

a)

For a floating vessel at rest, Archimedes stated that buoyancy and weight are in balance

$$mg=\rho g\nabla$$

Thus giving us the water displacement as

$$\nabla =\frac{m}{\rho }=1.6651\cdot 10^4{\text{ m}}^3$$

b)

The waterplane area $A_{wp}$ for a prism is constant (as long as it isn’t completely submerged), and is given by:

$$A_{wp}=LB=3509.8{\text{ m}}^2$$

The hydrostatic force $Z_{hs}$ in heave is given by the difference between the gravitational and the buoyancy forces (Eq. 4.12 Fossen)

$$\begin{array}{rl}{Z}_{hs}& =mg-\rho g(\nabla +\delta \nabla (z^n))\\ Z_{hs}& =-\rho g\delta \nabla (z^n)\end{array}$$

For small perturbations we can write $Z_{hs}$ as (Eq. 4.14 Fossen)

$$\begin{array}{rl}{Z}_{hs}& \approx -\rho g{A}_{wp}{z}^n\\ Z_{hs}& \approx -\rho gLB{z}^n\end{array}$$

c)

From equation 4.78 in Fossen, we can approximate the heave period as

$$\begin{array}{rl}{T}_3& \approx 2\pi \sqrt{\frac{2T}{g}}\\ T_3& \approx 2\pi \sqrt{\frac{2\cdot 4.74}{9.81}}s\approx 6.1766s\end{array}$$

d)

Computing the metacenter heights for this surface vessel can be done using Eq. 4.34 in Fossen, and is given by

$$\begin{array}{r}G{M}_T=B{M}_T-BG,G{M}_L=B{M}_L-BG\end{array}$$

Where $B{M}_T=\frac{I_T}{\nabla }$, $B{M}_L=\frac{I_L}{\nabla }$, and $BG$ is the height from the CB to the CG.

We start by finding $B{M}_T$ and $B{M}_L$ using the Munro-Smith formula, but since the term

$$\frac{6C_w^3}{(1+C_w)(1+2C_w)}=1$$

Where $C_w=\frac{A_{wp}}{LB}$, we get

$$\begin{array}{rl}B{M}_T& =\frac{I_T}{\nabla }=\frac{\frac{1}{12}{B}^{3}L}{\nabla }\\ B{M}_L& =\frac{I_L}{\nabla }=\frac{\frac{1}{12}B{L}^3}{\nabla }\end{array}$$

We can then find $BG=KG-KB$, where $KB$ is estimated using the Morrish formula:

$$\begin{array}{rl}BG& =KG-KB\\ BG& =\frac{H}{2}-\frac{1}{3}\left(\frac{5T}{2}-\frac{\nabla }{A_{wp}}\right)\end{array}$$

Combining this, we can get the final expressions for $G{M}_T$ and $G{M}_L$:

$$\begin{array}{rl}G{M}_T& =\frac{1}{12}\frac{B^{3}L}{\nabla }-\frac{H}{2}+\frac{1}{3}\left(\frac{5T}{2}-\frac{\nabla }{A_{wp}}\right)\approx 2.8162m\\ G{M}_L& =\frac{1}{12}\frac{L^{3}B}{\nabla }-\frac{H}{2}+\frac{1}{3}\left(\frac{5T}{2}-\frac{\nabla }{A_{wp}}\right)\approx 449.77m\end{array}$$

e)

According to definition 4.2 in Fossen, a floating vessel is said to be tranverse metacentrically stable if

$$G{M}_T\ge G{M}_{T,min}>0$$

And longitudinal metacentrically stable if

$$G{M}_L\ge G{M}_{L,min}>0$$

For most ships $G{M}_{T,min}$ is usually $0.5m$, while $G{M}_{L,min}$ is usually $100m$. We can therefore conclude that the prism is metacentrically stable.

Problem 3: Added Mass and Coriolis

Resources:

• Chapter 5
• Maneuvering Models

---

Given the following general $6\times 6$ matrix of added mass coefficients:

a)

$M_A$ can be rewritten for surge, sway and yaw as

$$M_A=-\left[\begin{array}{ccc}{X}_{\dot{u}}& 0& 0\\ 0& Y_{\dot{v}}& Y_{\dot{r}}\\ 0& Y_{\dot{r}}& N_{\dot{r}}\end{array}\right]$$

Since $N_{\dot{v}}=Y_{\dot{r}}$ (since the inertia for the yaw rate in y rate is equal to the inertia for the y rate in yaw rate), and since we can decouple the surge from steering dynamics due to xz-plane symmetry. Heave, pitch and roll modes are also neglected as we assume small values.

b)

From property 6.2 in Fossen, we know that (Eq. 6.46)

$$C_A(\nu )=\left[\begin{array}{cccccc}0& 0& 0& 0& -a_3& -a_2\\ 0& 0& 0& a_3& 0& -a_1\\ 0& 0& 0& -a_2& a_1& 0\\ 0& -a_3& a_2& 0& -b_3& b_2\\ a_3& 0& -a_1& b_3& 0& -b_1\\ -a_2& a_1& 0& -b_2& b_1& 0\end{array}\right]$$

Since we’re only interested in surge,sway and yaw (mode 1,2, and 6) we get:

$$C_A(\nu )=\left[\begin{array}{ccc}0& 0& a_2\\ 0& 0& -a_1\\ -a_2& a_1& 0\end{array}\right]$$

Where $a_1=X_{\dot{u}}u$, and $a_2=Y_{\dot{v}}v+Y_{\dot{r}}r$ (Eq. 6.47 → 6.52).

Problem 4: Implementing the Maneuvering Model in Matlab

Goal: Implement the model in Matlab. See the files ship.m and main.m.

a)

This was achieved by adding the added mass matrices, $M_A$ and $C_A$.

b)

With linear damping for all modes, nonlinear surge damping (Eq. 6.76, 6.77), and cross-flow drag (Eq. 6.87, 6.88).

c)

d)

Where

$$\begin{array}{r}{\tau }_{RB}=B_{i}u=\left[\begin{array}{cc}1-t_{thr}& x_{{\delta }_2}\\ 0& Y_{\delta }\\ 0& N_{\delta }\end{array}\right]\end{array}$$

And $C_{RB}$, $M_{RB}$ as described earlier in the report.