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

The vertical angle at any point on the circular region is determined with the following.

\begin{displaymath}
\phi = \gamma_{1} + \frac{s_{proj} - a_{1}t_{1_{f}}}{r}
\end{displaymath}

The end of the first clothoid occurs when $t=t_{1_{f}}$. The position at $t=t_{1_{f}}$ is found, and the center of the circular region is determined from it. This point, $(x_{c},z_{c})$, can be seen in figure 2.5.

\begin{eqnarray*}
x_{c} & = & a_{1} \ensuremath\int_{0}^{t_{1_{f}}} \cos(\frac{...
...frac{\pi}{2} u^{2})\,du - r \cos \left( \frac{\delta}{2} \right)
\end{eqnarray*}



The positions in the circular region are then determined with the following.
$\displaystyle x$ $\textstyle =$ $\displaystyle x_{c} + r \sin\phi$ (2.63)
$\displaystyle y$ $\textstyle =$ $\displaystyle s_{proj} \tan\alpha$ (2.64)
$\displaystyle z$ $\textstyle =$ $\displaystyle z_{c} + r \cos\phi$ (2.65)

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

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



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

$\displaystyle f_{x}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d\left( x_{c} + r \sin\phi \right)}{d\phi} \ensuremath \frac{d\phi}{ds}$  
  $\textstyle =$ $\displaystyle \cos\phi \cos\alpha$ (2.66)
$\displaystyle f_{y}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d(s_{proj} \tan\alpha)}{ds_{proj}} \ensuremath \frac{ds_{proj}}{ds}$  
  $\textstyle =$ $\displaystyle \sin\alpha$ (2.67)
$\displaystyle f_{z}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d\left( z_{c} + r \cos\phi \right)}{d\phi} \ensuremath \frac{d\phi}{ds}$  
  $\textstyle =$ $\displaystyle \sin\phi \cos\alpha$ (2.68)

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...
... \frac{d\phi}{ds} \\
& = & \frac{1}{r} \cos\phi \cos^{2}\alpha
\end{eqnarray*}



The unit vector in the direction of the radial vector must be determined.
$\displaystyle r_{x}$ $\textstyle =$ $\displaystyle -\sin\phi$ (2.69)
$\displaystyle r_{y}$ $\textstyle =$ $\displaystyle 0$ (2.70)
$\displaystyle r_{z}$ $\textstyle =$ $\displaystyle \cos\phi$ (2.71)

The curvature is the magnitude of the (non-unit) radial vector.
\begin{displaymath}
k = \frac{1}{r} \cos^{2}\alpha
\end{displaymath} (2.72)


next up previous contents
Next: Clothoid 2: Up: Loop Elements Previous: Clothoid 1:   Contents
Darla Weiss 2000-02-13