Choice of (a) Hydrodynamic-Derivative Classification or (b) Steady Turning Radius: 25-Nav-A3 Hydrodynamics of Ships (II)_ Ship Motion, December 2019, Q6

Question 6: Choice of (a) Hydrodynamic-Derivative Classification or (b) Steady Turning Radius (10 marks; either part)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a): Classifying Linear Hydrodynamic Derivatives from Force-vs-Speed Curves

Given. Three characteristic force-vs-speed curves $F(u)$, each marked at two operating points: the origin $O$ ($u=0$) and a forward-speed point $A$ ($u>0$).

Figure 5 — three characteristic $F(u)$ curve shapes, each marked at O and A.

Find. For each of the three curves, classify the local slope (the linearized hydrodynamic derivative $dF/du$) at O and at A as zero, small, finite positive, or finite negative.

Approach. A linearized hydrodynamic derivative is, by definition, the local tangent slope of the underlying nonlinear force curve at the chosen operating point — so this reduces to reading the tangent slope directly off each curve at O and at A.

(i) First curve — flat-bottomed (near-parabolic) curve. At $O$ the curve sits at a smooth minimum (horizontal tangent): the derivative is zero. At $A$, on the steeply rising branch away from the minimum, the tangent has a clear upward slope: the derivative is finite positive.

(ii) Second curve — S-shaped (odd, near-cubic) curve. The curve is steep only far from the origin and flattens through an inflection that spans both marked points; at both $O$ and $A$ the tangent is shallow but not exactly horizontal: the derivative is small (small positive) at both points.

(iii) Third curve — curve with an interior maximum. At $O$ the curve is on its steep rising branch (clear positive tangent): the derivative is finite positive. Point $A$ sits exactly at the curve's peak (horizontal tangent at the maximum): the derivative there is zero.

Curve	Derivative at O	Derivative at A
(i) flat-bottomed / near-parabolic	zero	finite positive
(ii) S-shaped / near-cubic	small	small
(iii) curve with interior maximum	finite positive	zero

Concept & Theory

Topic. Linearized hydrodynamic derivatives depend on the operating point chosen — they are local tangent slopes of an inherently nonlinear force characteristic, not universal constants.

Key relations. $Y_v=\partial Y/\partial v\big|_{\text{operating point}}$ (and likewise for any force/moment derivative); a linear maneuvering model is only valid in a neighbourhood of the point at which each derivative was evaluated.

Why this works. Real hull hydrodynamic force characteristics (drag-like, lift-like, or curves with an interior extremum such as a maximum rudder force before stall) are all genuinely nonlinear in the governing velocity; "the" derivative reported for a hull is only ever a local linearization, and this question exercises reading that linearization directly off a force curve at two different, explicitly marked points.

Common pitfall. Assuming a derivative must be either strictly zero or a single "the" value for the whole curve — the same physical curve can have a zero derivative at one speed and a large one at another, which is exactly what curves (i) and (iii) illustrate.

Source. Lewandowski, The Dynamics of Marine Craft, Ch. 3 (hydrodynamic derivatives, linearization); Lewis (ed.), PNA Vol. III, Ch. 9.

Part (b): Steady Turning Radius from Linear Maneuvering Derivatives

Given. $\Delta=8000$ tonnes, $L=160$ m, $T=6$ m; nondimensional (prime-system) derivatives $Y'_v=-0.31$, $Y'_r=0.0822$, $N'_v=-0.1073$, $N'_r=-0.0871$, $Y'_\delta=0.0777$, $N'_\delta=-0.03$; rudder angle $\delta=35^\circ$; CG at the origin; seawater $\rho=1025$ kg/m³.

Find. The steady turning radius $R$.

Approach. At steady turn (constant sway velocity $v$, constant yaw rate $r$, all accelerations zero), the linear sway and yaw equations of motion reduce to two algebraic equations for the nondimensional $v',r'$ in terms of the rudder deflection; with the CG at the origin there is no $x_G$ coupling term, and the resulting yaw rate gives the turning radius directly as $R=L/|r'|$.

Nondimensionalize the mass. In the SNAME prime system, $m'=m/(0.5\rho L^2T)$, with $m=\Delta=8000$ tonnes $=8.00\times10^6$ kg: $$m'=\frac{8.00\times10^6}{0.5(1025)(160)^2(6)}=\frac{8.00\times10^6}{7.872\times10^7}=\boxed{0.1016}.$$
Steady-state sway/yaw equations. With all accelerations zero and $x_G=0$, the linear equations of motion reduce to $$Y'_v v' + (Y'_r-m')r' = -Y'_\delta\,\delta, \qquad N'_v v' + N'_r r' = -N'_\delta\,\delta.$$ Substituting the given derivatives and $\delta=35^\circ=0.6109$ rad: $$-0.31\,v' -0.0194\,r' = -0.0474, \qquad -0.1073\,v' -0.0871\,r' = 0.01833.$$
Solve the $2\times2$ system for $r'$. By Cramer's rule (determinant $=0.02492$): $$r'=\frac{(-0.31)(0.01833)-(-0.0474)(-0.1073)}{0.02492}=\boxed{-0.4324}$$ (and $v'=0.1802$, not required for the radius).
Turning radius. Since the nondimensional yaw rate is $r'=rL/U$, a steady turn has $r=U/R$, so $r'=L/R$, giving $$R=\frac{L}{|r'|}=\frac{160}{0.4324}=\boxed{370\ \text{m}}\qquad(R/L=2.31).$$

Quantity	Result
Nondimensional mass $m'$	0.1016
Nondimensional yaw rate $r'$	−0.432
Steady turning radius $R$	370 m ($R/L=2.31$)

Concept & Theory

Topic. Steady-state solution of the linear (prime-system) sway-yaw maneuvering equations to find turning radius.

Key relations. $m'=m/(0.5\rho L^2T)$; steady-state $Y'_v v'+(Y'_r-m')r'=-Y'_\delta\delta$, $N'_v v'+N'_r r'=-N'_\delta\delta$; $R=L/|r'|$.

Why this works. At a steady turn every acceleration term (and hence every added-mass derivative) drops out of the equations of motion identically, leaving a simple $2\times2$ linear system in $v',r'$ driven only by the rudder deflection; because $r'=rL/U=L/R$ by definition of the nondimensionalization, the yaw-rate solution converts directly to a turning radius with no further data needed.

Common pitfall. Forgetting to subtract $m'$ from $Y'_r$ in the sway equation (the "$m'u'r'$" centripetal-coupling term) — omitting it changes the coefficient matrix and gives a noticeably different (and wrong) turning radius, even though $m'$ never appears in the yaw equation itself (because $x_G=0$ removes it there).

Source. Lewandowski, The Dynamics of Marine Craft, Ch. 4–5 (linear maneuvering equations, steady turning); Lewis (ed.), PNA Vol. III, Ch. 9.

Question 6 of 6: Choice of (a) Hydrodynamic-Derivative Classification or (b) Steady Turning Radius

Question 6: Choice of (a) Hydrodynamic-Derivative Classification or (b) Steady Turning Radius (10 marks; either part)

Part (a): Classifying Linear Hydrodynamic Derivatives from Force-vs-Speed Curves

Part (b): Steady Turning Radius from Linear Maneuvering Derivatives