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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.

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

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

\begin{displaymath}
\phi = \frac{\pi t^{2}}{2}
\end{displaymath}

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.
$\displaystyle x$ $\textstyle =$ $\displaystyle x_{2_{f}} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du$ (2.73)
$\displaystyle y$ $\textstyle =$ $\displaystyle s_{proj} \tan\alpha$ (2.74)
$\displaystyle z$ $\textstyle =$ $\displaystyle z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du$ (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.

$\displaystyle f_{x}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d\left[ x_{2_{f}} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du \right]}{dt} \ensuremath \frac{dt}{ds}$  
  $\textstyle =$ $\displaystyle \cos\phi \cos\alpha$ (2.76)
$\displaystyle f_{y}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d(s_{proj} \tan\alpha)}{ds_{proj}} \ensuremath \frac{ds_{proj}}{ds}$  
  $\textstyle =$ $\displaystyle \sin\alpha$ (2.77)
$\displaystyle f_{z}$ $\textstyle =$ $\displaystyle \ensuremath \frac{d\left[ z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right]}{dt} \ensuremath \frac{dt}{ds}$  
  $\textstyle =$ $\displaystyle -\sin\phi \cos\alpha$ (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.
$\displaystyle r_{x}$ $\textstyle =$ $\displaystyle \sin\phi$ (2.79)
$\displaystyle r_{y}$ $\textstyle =$ $\displaystyle 0$ (2.80)
$\displaystyle r_{z}$ $\textstyle =$ $\displaystyle \cos\phi$ (2.81)

The curvature is the magnitude of the (non-unit) radial vector.
\begin{displaymath}
k = \frac{\pi t}{a_{2}} \cos^{2}\alpha
\end{displaymath} (2.82)


next up previous contents
Next: All regions Up: Loop Elements Previous: Circular Region:   Contents
Darla Weiss 2000-02-13