# One-Dimensional Mathematical Model for Kayak Propulsion
## 2 Mathematical Model
### Derivation of Relevant Quantities  
The constant force, or asymptotic solution (Equation (3)), is generic for any stroke model, which we will see later, and the constant \(K\) is zero if the kayak starts from rest. It can be observed that the described curve can be separated into an acceleration interval followed by an interval, where the velocity is constant. In this section, we assume that the first interval corresponds to a uniform acceleration followed by a uniform velocity interval, which in turn we will decompose into two intervals: one that we will call the stabilization interval followed by a uniform velocity interval. Likewise, we introduce a rule to approximate them in a simple way.  
The area under the velocity-time curve (Figure 1) corresponds to the distance covered by the kayak. We start by finding a point on the acceleration interval such that it approximates this distance by a uniform acceleration interval; it will therefore be a quadrature rule. Let \(A\) be such a point. Then, the following holds:  
\[v_{a}=a\cdot\sqrt{\frac{f}{d}}\qquad\text{ with }\qquad 0<a<1. \tag{6}\]  
This is the velocity reached at this moment, where \(a\) represents a fraction over the limiting velocity. Substituting this expression in (3), and clearing the time, we obtain the following:  
\[t_{a}=\frac{m}{f}\sqrt{\frac{f}{d}}tanh^{-1}(a)=\frac{m}{2f}\sqrt{\frac{f}{d} }\ln\frac{1+a}{1-a} \tag{7}\]  
where we have used the following identity \(tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\). The mean acceleration \(<a>\) between the initial instant and \(t_{a}\) writes as follows:  
\[<a>=\frac{v_{a}}{t_{a}}=\frac{a\cdot\sqrt{\frac{f}{d}}}{\frac{m}{2f}\sqrt{ \frac{f}{d}}\ln\frac{1+a}{1-a}}=\frac{2f}{m}\frac{a}{\ln\frac{1+a}{1-a}}. \tag{8}\]  
The cut-off point between the constant acceleration interval and the constant velocity displacement gives us the transition point between both intervals, and is obtained by equating both straight lines, that is, as follows:  
\[v_{limit}=<a>\cdot t_{accl}\qquad\rightarrow\qquad t_{accl}=\frac{v_{limit }}{<a>}=\frac{\ln\left(\frac{1+a}{1-a}\right)\cdot m}{a\cdot 2\cdot\sqrt{d \cdot f}}. \tag{9}\]  
The approximate distance based on the approximate mean acceleration is as follows:  
\[dist_{0}=\frac{1}{2}<a>t_{accl}^{2}=\frac{1}{2}\frac{v_{limit}^{2}}{<a>}= \cdots=\frac{1}{a\cdot 4}\frac{m}{d}\ln\left(\frac{1+a}{1-a}\right). \tag{10}\]  
This is opposed to the analytical distance, which is obtained by integrating the velocity, with the following results:  
\[dist_{1}=\int_{0}^{t_{accl}}v(t)dt=\cdots=\frac{m}{d}\ln\left(cosh\left( \frac{1}{2a}\ln\left(\frac{1+a}{1-a}\right)\right)\right). \tag{11}\]  
By equating both distances, we obtain an expression for the parameter and with this, the optimal quadrature point. The equation is the following:  
\[\frac{1}{a\cdot 4}\ln\left(\frac{1+a}{1-a}\right)=\ln\left(cosh\left(\frac{1}{2a }\ln\left(\frac{1+a}{1-a}\right)\right)\right) \tag{12}\]  
whose numerical resolution obtains \(a\sim\)0.68. Once estimated and with the kinematic magnitudes entered, we can obtain the expression for the average work as follows:\[<W>=m<a>dist_{0}=m\cdot\frac{2f}{m}\frac{a}{\ln\left(\frac{1+a}{1-a}\right)}\cdot \frac{1}{a\cdot 4}\frac{m}{d}\ln\left(\frac{1+a}{1-a}\right)=\cdots=\frac{mf}{2d} \tag{13}\]  
approximated from the rest up to \(t_{accel}\). Note that this could be written as follows:  
\[<W>=\frac{mf}{2d}=\frac{1}{2}m\left(\sqrt{\frac{f}{d}}\right)^{2}. \tag{14}\]  
The kayak kinetic energy, when the limiting velocity \(v_{limit}=\sqrt{f/d}\) is approximated, corresponds to the work required to accelerate the kayak from rest to that velocity, and is therefore independent of acceleration interval. The average power is written as follows:  
\[<P>=\frac{W}{t_{accel}}=\frac{\frac{mf}{2d}}{t_{accel}}=\cdots=\frac{f}{\ln \left(\frac{1+a}{1-a}\right)}\cdot v_{a}. \tag{15}\]