Circular Region: $(a_{1}t_{1_{f}} \leq s_{proj} \leq a_{1}t_{1_{f}} + r \delta )$

The azimuth angle at any point of the circular region is determined with the following.

$$\theta = \gamma_{1} + \frac{s_{proj} - a_{1}t_{1_{f}}}{r}$$

The end of the first clothoid occurs when $t=t_{1_{f}}$. This point is labeled $(x_{1_{f}},y_{1_{f}})$ and is depicted in figure 2.4.

$$\begin{eqnarray*} x_{1_{f}} & = & a_{1} \ensuremath\int_{0}^{t_{1_{f}}} \cos(\f... ...1} \ensuremath\int_{0}^{t_{1_{f}}} \sin(\frac{\pi}{2} u^{2})\,du \end{eqnarray*}$$

The center of the circular region can now be found from the following relationship.

$$\begin{eqnarray*} x_{c} & = & x_{1_{f}} - r \sin\gamma_{1} \\ y_{c} & = & y_{1_{f}} + r \cos\gamma_{1} \end{eqnarray*}$$

The positions in the circular region are then determined with the following. Note that $z$ is calculated in the same manner as before, as it varies linearly with arclength along the curve.

$$x = x_{c} + r \sin\theta$$ (2.31)

$$y = y_{c} - r \cos\theta$$ (2.32)

$$z = s_{proj} \tan\phi$$ (2.33)

Derivatives with respect to arclength will need to be taken to determine the remaining functions. Since the position vector is expressed in terms of $\theta$, the chain rule is applied.

$$\begin{eqnarray*} \ensuremath \frac{d\theta}{ds} & = & \ensuremath \frac{d\thet... ...\ensuremath \frac{ds_{proj}}{ds} \\ & = & \frac{1}{r} \cos\phi \end{eqnarray*}$$

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

$$f_{x} = \ensuremath \frac{dx}{ds}$$

$$= \ensuremath \frac{d\left( x_{c} + r \sin\theta \right)}{d\theta} \ensuremath \frac{d\theta}{ds}$$

$$= \cos\theta \cos\phi$$ (2.34)

$$f_{y} = \ensuremath \frac{dy}{ds}$$

$$= \ensuremath \frac{d\left( y_{c} - r \cos\theta \right)}{d\theta} \ensuremath \frac{d\theta}{ds}$$

$$= \sin\theta \cos\phi$$ (2.35)

$$f_{z} = \ensuremath \frac{dz}{ds}$$

$$= \ensuremath \frac{d(s_{proj} \tan\phi)}{ds_{proj}} \ensuremath \frac{ds_{proj}}{ds}$$

$$= \sin\phi$$ (2.36)

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

$$\begin{eqnarray*} r_{x} & = & \ensuremath \frac{df_{x}}{ds} \\ & = & \ensurem... ...z}}{ds} \\ & = & \ensuremath \frac{d\sin\phi}{ds} \\ & = & 0 \end{eqnarray*}$$

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

$$r_{x} = -\sin\theta$$ (2.37)

$$r_{y} = \cos\theta$$ (2.38)

$$r_{z} = 0$$ (2.39)

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

$$k = \frac{1}{r} \cos^{2}\phi$$ (2.40)

Note the similarity of the above to equation 2.30. The curvature of a clothoid, $\frac{\pi t}{a}$ has been replaced by the curvature of a circle, $\frac{1}{r}$. The $\cos^{2}\phi$ term exists in both curvature equations, due to the linearly varying elevation along the curve.