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

The end of the corkscrew has zero curvature, and is where $t=0$. Therefore, $t$ ranges from $t=t_{1_{f}}$ at the beginning of clothoid 2, to $t=0$ at the end of clothoid 2.

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

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

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

The coordinates at the end of the corkscrew are found from the coordinates of the center of the circular region of the corkscrew.

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

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.

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

The position vector is now rotated downward $\lambda$ radians about the $x$ axis.

$$x = x_{f} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du$$ (2.104)

$$y = s_{proj} \tan\alpha \cos\lambda - \left[ z_{f} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right] \sin\lambda$$ (2.105)

$$z = s_{proj} \tan\alpha \sin\lambda + \left[ z_{f} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right] \cos\lambda$$ (2.106)

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.

$$\begin{eqnarray*} f_{x} & = & \ensuremath \frac{d\left[ x_{f} - a_{2} \ensurema... ...uremath \frac{dt}{ds} \nonumber \\ & = & -\sin\phi \cos\alpha \end{eqnarray*}$$

This vector is then rotated $\lambda$ radians about the $x$ axis.

$$f_{x} = \cos\phi \cos\alpha$$ (2.107)

$$f_{y} = \sin\alpha \cos\lambda + \sin\phi \cos\alpha \sin\lambda$$ (2.108)

$$f_{z} = \sin\alpha \sin\lambda - \sin\phi \cos\alpha \cos\lambda$$ (2.109)

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.

$$\begin{eqnarray*} r_{x} & = & \sin\phi \\ r_{y} & = & 0 \\ r_{z} & = & \cos\phi \end{eqnarray*}$$

This vector is then rotated $\lambda$ radians about the $x$ axis.

$$r_{x} = \sin\phi$$ (2.110)

$$r_{y} = -\cos\phi \sin\lambda$$ (2.111)

$$r_{z} = \cos\phi \cos\lambda$$ (2.112)

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

$$k = \frac{\pi t}{a_{2}} \cos^{2}\alpha$$ (2.113)