Clothoid 1: $(0 \leq s \leq a_{1}t_{1_{f}} )$

The equation for the arclength of a clothoid is used to determine $t$, the variable used for clothoid calculations. It is found in the same manner as it is for a loop and corkscrew, except the total arclength, instead of the projection, is used.^2.9

$$t = \frac{s}{a_{1}}$$

The vertical angle through the joint is defined as the angle between the forward vector and the $x$-$y$ plane. It is determined from the value of $t$ calculated above.

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

The position of points in clothoid 1 is calculated, initially beginning from horizontal, as shown in figure 2.11.

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

If a downward joint is being calculated, $z$ is negated. The clothoid is then rotated $\phi_{0}$ radians about the $y$ axis to line it up with adjacent elements. Refer to appendix B for an explanation of rotation transformations.

$$x = a_{1} \cos\phi_{0} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{... ...du \mp a_{1} \sin\phi_{0} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du$$ (2.117)

$$y = 0$$ (2.118)

$$z = a_{1} \sin\phi_{0} \ensuremath\int_{0}^{t} \cos(\frac{\pi}{2} u^{... ...du \pm a_{1} \cos\phi_{0} \ensuremath\int_{0}^{t} \sin(\frac{\pi}{2} u^{2})\,du$$ (2.119)

The calculation of the forward and radial vectors is nearly identical to the calculation for the curve. An abbreviated description will be shown here.

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

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

The forward vector is calculated by taking the derivative of the position vector (with respect to arclength) as it was first calculated. It will then be transformed in the same way as the position vector was.

$$\begin{eqnarray*} f_{x} & = & \ensuremath \frac{d\left[ a_{1} \ensuremath\int_{... ...)\,du \right]}{dt} \ensuremath \frac{dt}{ds} \\ & = & \sin\phi \end{eqnarray*}$$

The $z$ coordinate is negated if a downward joint is being calculated. The vector is then rotated $\phi_{0}$ radians about the $y$ axis.

$$f_{x} = \cos\phi \cos\phi_{0} \mp \sin\phi \sin\phi_{0}$$ (2.120)

$$f_{y} = 0$$ (2.121)

$$f_{z} = \cos\phi \sin\phi_{0} \pm \sin\phi \cos\phi_{0}$$ (2.122)

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

$$\begin{eqnarray*} r_{x} & = & \ensuremath \frac{d\left( \cos\phi \right)}{d\phi... ...uremath \frac{d\phi}{ds} \\ & = & \frac{\pi t}{a_{1}} \cos\phi \end{eqnarray*}$$

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 $\phi_{0}$ radians about the $y$ axis. The result will still be a unit vector, as rotation does not alter magnitude.

$$r_{x} = -\sin\phi \cos\phi_{0} \mp \cos\phi \sin\phi_{0}$$ (2.123)

$$r_{y} = 0$$ (2.124)

$$r_{z} = -\sin\phi \sin\phi_{0} \pm \cos\phi \cos\phi_{0}$$ (2.125)

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

$$k = \frac{\pi t}{a_{1}}$$ (2.126)

^2.9 Since the joint is a two-dimensional curve, the projection is the same as the total arclength.