Clothoid 2: $(a_{1}t_{1_{f}} + r \delta \leq s \leq l )$

The end of the joint 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.

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

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

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

The coordinates at the beginning of clothoid 2 are initially found by determining the endpoint of the circular region. For the non-transformed (not yet negated and rotated) circle, this position is found from the following.

$$\begin{eqnarray*} x_{2_{0}} & = & x_{c} + r \sin(\gamma_{1} + \delta) \\ z_{2_{0}} & = & z_{c} - r \cos(\gamma_{1} + \delta) \end{eqnarray*}$$

If a downward joint is being calculated, $z$ must be negated. This position vector is then rotated $\phi_{0}$ radians about the $y$ axis.

$$\begin{eqnarray*} x_{2_{0}} & = & (x_{c} + r \sin(\gamma_{1} + \delta)) \cos\ph... ...\phi_{0} \pm (z_{c} - r \cos(\gamma_{1} + \delta)) \cos\phi_{0} \end{eqnarray*}$$

The position of clothoid 2 is determined by creating a clothoid which initially ends horizontally. The method used here is much like the method used for a curve.^2.10 First, the end of the clothoid is determined. See figure 2.11.

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

Now the positions within the clothoid are determined relative to the endpoint.

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

If a downward joint is being calculated, $z$ must be negated. The clothoid is then rotated $\phi_{f}$ radians about the $y$ axis to line it up with the adjacent elements. It is then translated to line up with the end of the circular region.

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

$$\mp \left[ z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right] \sin\phi_{f}$$ (2.137)

$$y = 0$$ (2.138)

$$z = z_{2_{0}} + \left[ x_{2_{f}} - a_{2} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{2})\,du \right] \sin\phi_{f}$$

$$\pm \left[ z_{2_{f}} + a_{2} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du \right] \cos\phi_{f}$$ (2.139)

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

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

The forward vector is calculated by taking the derivative of the position vector as it was first calculated. It will then be transformed in the same way as the position vector was, with the exception of translations, which do not affect the forward and radial vectors.

$$\begin{eqnarray*} f_{x} & = & \ensuremath \frac{d\left[ a_{1} x_{2_{f}} - a_{2}... ...du \right]}{dt} \ensuremath \frac{dt}{ds} \\ & = & -\sin\phi \end{eqnarray*}$$

The $z$ coordinate is negated if a downward joint is being calculated. The vector is then rotated $\phi_{f}$ radians about the $y$ axis.

$$f_{x} = \cos\phi \cos\phi_{f} \pm \sin\phi \sin\phi_{f}$$ (2.140)

$$f_{y} = 0$$ (2.141)

$$f_{z} = \cos\phi \sin\phi_{f} \mp \sin\phi \cos\phi_{f}$$ (2.142)

The radial vector is calculated as the derivative of the forward vector (with respect to arclength) as it is initially calculated.

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

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

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

The $z$ coordinate is negated if a downward joint is being calculated. The vector is then rotated $\phi_{f}$ radians about the $y$ axis. The result will still be a unit vector, as rotation does not alter magnitude.

$$r_{x} = \sin\phi \cos\phi_{f} \mp \cos\phi \sin\phi_{f}$$ (2.143)

$$r_{y} = 0$$ (2.144)

$$r_{z} = \sin\phi \sin\phi_{f} \pm \cos\phi \cos\phi_{f}$$ (2.145)

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

$$k = \frac{\pi t}{a_{2}}$$ (2.146)

^2.10 Refer to section 2.1.2.