Reducing yawing in a kayak with a rudder

Question

Disclaimer

I would first like to thank in advance the people for taking the time to read this question. Please, note that this is not either homework or an open question: I just wondered about the behaviour of a physical principle and tried to solve it by hand, but got stuck.

Setting

I have been wondering about how a simple rudder in a kayak works. In order to clarify the scenario a bit, a kayak is a type of boat which moves thanks to a paddler rowing on both sides of it alternatively. This alternation would produce a big yaw movement (defining the Z axis as the vertical one, parallel to gravity) each stroke. Since this movement would increase the drag of the hull in the water when trying to row in a straight line, sprint kayaks have a rudder at the stern in order to minimise it. In order to make things easier, we will assume that it is a single-place kayak and the paddler is sat in the middle of it, so that the center of mass lies just in the center of the kayak and the underwater depth of the hull is distributed uniformly along it.

In a first approach, we can assume that the movement of the paddle in the water will not be perfectly parallel to the hull but rather at an angle $\theta$. This way, having our reference frame at the very center of the kayak (and paddler) (see picture), we can decompose the force applied by the paddler into two orthogonal components in $x$ and $y$ ($F_x(t)$ and $F_y(t)$) We can even assume that these forces are sinusoids, since paddling is a periodic movement and we can decompose it by doing some Fourier analysis.

The rudder consists on an airfoil oriented in the direction of displacement. Let $S$ be the area of the rudder, then it produces a force given by:

$$D_r(t) = \frac{1}{2} \rho S C_D v^2(t)$$

For a movement perpendicular to the direction of displacement (necessary to overcome the yaw motion induced by the paddler), $C_D$ will take a large value ($~1.1$), so we expect this force to be relatively large.

For this problem, I will ignore the drag produced by the hull in the water when yawing, since it will not affect the final result much.

The problem

I want to know how much the kayak will yaw $\theta(t)$ and where the center of rotation is.

My attempt at solving this problem

Let $t=0$ be the very first instant in which the paddler starts applying a force. At this moment, the velocity of the rudder in the aforementioned equation will equal $0$, so we can conclude that the kayak will inevitably yaw (even if it's less that without the rudder).

In order to obtain the yaw angle, I though a good way would be calculating the torque generated by $F_y$ and $D_r$, and applying:

$\sum \tau(t) = I \alpha(t)$

For $I$ the moment of inertia and $\alpha = \frac{d^2 \theta}{dt^2}$ the angular acceleration. The problem with this approach is that I do not know where the center of rotation is, so I cannot directly obtain an expression for the torques and solve the differential equation. Moreover, another difficulty is that we must divide $F$ into two orthogonal components $F_r$ and $F_\theta$ with respect to the center of rotation in order to calculate the torque applied by the paddler.

In an attempt to gain some intuition on this problem, I decided to analyse the limit cases. First, imagine there was no rudder. In that case, we would have $D_r=0$, so the only torque would be produced by the paddler. The center of rotation (assuming the center of mass is in the middle of the kayak and so in the paddler) would be the center of the kayak. Obtaining the angular acceleration and the yaw angle would be trivial.

Now, for the case of an infinitely big rudder $S\to \infty$, $D_r$ would always be large enough for compensating for the force applied by the paddle. Hence, the center of rotation would be infinitely close to the rudder, and the amplitude of yaw would be at a minimum.

Due to continuity, we know that the center of rotation will lie in between the rudder and the paddler, and its position will depend on both $F$ and $D_r$. Now, here comes the tricky part: $D_r$ also depends on $v$, which depends on both $\alpha$ and the distance from the rudder to the center of rotation.

I find myself relatively close to the final solution, since I can more or less have certain intuition about the differential equation I will have to solve, but I am not able to find the center of rotation for analysing the torques and obtaining $\alpha$. (Or, maybe, there's a different path to solve this problem)

Any help on this will be highly appreciated! Thank you in advance.

Edit: I would like to know the reason this question got downvoted in order to correct it

Answer 1

I want to know how much the kayak will yaw $~\psi(t)~$ and where the center of rotation is.

I) the equation of motion

$${\frac {d}{d\tau}}\beta \left( \tau \right) -{\frac {{}{\it Fy }-m \left( {\frac {d}{d\tau}}\psi \left( \tau \right) \right) v \left( \tau \right) }{mv \left( \tau \right) }} =0$$

$${\frac {d^{2}}{d{\tau}^{2}}}\psi \left( \tau \right) -{\frac {{\it Fy} \,L}{\theta}} =0$$

$${\frac {d}{d\tau}}v \left( \tau \right) -{\frac {{\it Fx}{}}{m }} =0$$

where

• $v~$ velocity at the center of mass
• $\psi~$ the yaw angle
• $\beta~$ side slip angle (angle between the velocity and the x axis)
• $\theta~$ Inertia of the kayak z direction

II)

you can obtain analytical solution of the EOM's and get $~v(\tau)~,\beta(\tau)~,\psi(\tau)~$

III)

to obtain the center of rotation $~C_R~$, you need two velocities at the x-axis, at point $~L_1$ and $~L2~$

$$\vec v_1=v(\tau)\,\begin{bmatrix} \cos(\beta(\tau) \\ \sin(\beta(\tau) \\ \end{bmatrix} +\dot\psi(\tau)\, \begin{bmatrix} 0 \\ L1 \\ \end{bmatrix}$$

$$\vec v_2=v(\tau)\,\begin{bmatrix} \cos(\beta(\tau) \\ \sin(\beta(\tau) \\ \end{bmatrix} +\dot\psi(\tau)\, \begin{bmatrix} 0 \\ -L2 \\ \end{bmatrix}$$

the center of rotation is the intersection point between the two lines that are perpendicular to the velocities at point $~L_1~$ and $~L_2$, thus:

$$\vec r_{L1}=\begin{bmatrix} 0 \\ L_1 \\ \end{bmatrix}+\mu\,\begin{bmatrix} v_{1y} \\ -v_{1x} \\ \end{bmatrix}$$

$$\vec r_{L2}=\begin{bmatrix} 0 \\ -L_2 \\ \end{bmatrix}+\lambda\,\begin{bmatrix} v_{2y} \\ -v_{2x} \\ \end{bmatrix}$$

with $~\vec r_{L1}=\vec r_{L2}~$ you obtain $~\mu~$ and $~\lambda$

and for $~F_x=A\,\cos(\omega\tau)~,F_y=A\,\sin(\omega\,\tau)~$

$$\mu=\lambda=\frac{\theta\,\omega}{A\,L(\cos(\omega\,\tau)-1)}$$

notice that at each time you obtain other center of rotation