Clothoid 2: $(a_{1}t_{1_{f}} + r \delta \leq s_{proj} \leq l_{proj} )$

The end of the loop has zero curvature, and is where $t=0$. Therefore, $t$ ranges from $t=t_{2_{f}}$ at the beginning of clothoid 2, to $t=0$ at the end of clothoid 2.

$$t = t_{2_{f}} - \frac{s_{proj} - (a_{1}t_{1_{f}} + r \delta)}{a_{2}}$$

The vertical angle through clothoid 2 is found from the following.

$$\phi = \frac{\pi t^{2}}{2}$$

The coordinates at the beginning of clothoid 2 are found from the coordinates of the center of the circular region of the loop.

$$\begin{eqnarray*} x_{2_{0}} & = & x_{c} - r \sin \left( \frac{\delta}{2} \right... ... z_{2_{0}} & = & z_{c} + r \cos \left( \frac{\delta}{2} \right) \end{eqnarray*}$$

The end of the clothoid is calculated from $t=t_{2_{f}}$, and is translated to begin at $(x_{2_{0}},z_{2_{0}})$, as depicted in figure 2.5.

$$\begin{eqnarray*} x_{2_{f}} & = & x_{2_{0}} + a_{2} \ensuremath\int_{0}^{t_{2_{... ...2} \ensuremath\int_{0}^{t_{2_{f}}} \sin(\frac{\pi}{2} u^{2})\,du \end{eqnarray*}$$

The position of all points along the clothoid can now be calculated relative to the zero curvature end. The $y$ component is determined the same way as for the previous sections.

$$x = x_{2_{f}} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du$$ (2.73)

$$y = s_{proj} \tan\alpha$$ (2.74)

$$z = z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du$$ (2.75)

Derivatives with respect to arclength of the position vector must be taken. The required chain rule relationships are first determined.

$$\begin{eqnarray*} \ensuremath \frac{dt}{ds} & = & -\frac{1}{a_{2}} \cos\alpha \... ...nsuremath \frac{d\phi}{ds} & = & -\frac{\pi t}{a_{2}} \cos\alpha \end{eqnarray*}$$

The forward vector is calculated as the derivative of the position vector with respect to arclength.

$$f_{x} = \ensuremath \frac{d\left[ x_{2_{f}} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du \right]}{dt} \ensuremath \frac{dt}{ds}$$

$$= \cos\phi \cos\alpha$$ (2.76)

$$f_{y} = \ensuremath \frac{d(s_{proj} \tan\alpha)}{ds_{proj}} \ensuremath \frac{ds_{proj}}{ds}$$

$$= \sin\alpha$$ (2.77)

$$f_{z} = \ensuremath \frac{d\left[ z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right]}{dt} \ensuremath \frac{dt}{ds}$$

$$= -\sin\phi \cos\alpha$$ (2.78)

The radial vector is calculated as the derivative of the forward vector with respect to arclength.

$$\begin{eqnarray*} r_{x} & = & \ensuremath \frac{d\left( \cos\phi \cos\alpha \ri... ...\phi}{ds} \\ & = & \frac{\pi t}{a_{2}} \cos\phi \cos^{2}\alpha \end{eqnarray*}$$

The unit vector in the direction of the radial vector must be determined.

$$r_{x} = \sin\phi$$ (2.79)

$$r_{y} = 0$$ (2.80)

$$r_{z} = \cos\phi$$ (2.81)

The curvature is the magnitude of the (non-unit) radial vector.

$$k = \frac{\pi t}{a_{2}} \cos^{2}\alpha$$ (2.82)