= Solution
For the unit-speed <curve>, the <unit tangent vector>, <curvature of a space curve>, <principal normal vector> and <binormal vector> are
$$
\boxed{t=\gamma',\qquad \kappa=|\gamma''|,\qquad
p=\frac{\gamma''}{|\gamma''|},\qquad b=t\times p.}
$$
Differentiating $t\cdot t=1$ gives $t\cdot\gamma''=0$, so $t,p,b$ form a positively oriented <orthonormal basis>. The hypothesis $\gamma''\ne0$ makes the <principal normal vector> well-defined.
To determine the sign of the <torsion of a space curve> without relying on a pictorial handedness convention, take $R>0$, $h\ne0$, $L=\sqrt{R^2+h^2}$ and use the unit-speed <circular helix>
$$
\gamma_h(s)=\bigl(R\cos(s/L),\,R\sin(s/L),\,hs/L\bigr).
$$
Writing $u=s/L$, direct <differentiation> gives its <Frenet frame>:
$$
\begin{aligned}
t&=\frac1L(-R\sin u,R\cos u,h),&
p&=(-\cos u,-\sin u,0),\\
b&=\frac1L(h\sin u,-h\cos u,R),&
b'&=\frac h{L^2}(\cos u,\sin u,0).
\end{aligned}
$$
Consequently
$$
\boxed{\kappa=\frac R{R^2+h^2},\qquad
\tau=-b'\cdot p=\frac h{R^2+h^2}.}
$$
At $s=0$ the marked point is $(R,0,0)$, with $t=(0,R/L,h/L)$, $p=(-1,0,0)$, $b=(0,-h/L,R/L)$ and $b'=(h/L^2,0,0)$. For $h>0$, $b'$ points opposite to $p$ and the <torsion of a space curve> is positive. Replacing $h$ by $-h$ reflects the <circular helix> across the horizontal plane and makes the <torsion of a space curve> negative; then $b'$ points in the same direction as $p$. The direction of increasing $s$ is fixed in both panels.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-3-helices.png]
{title=Unit-speed circular helices of opposite torsion, with tangent, principal normal, binormal and binormal derivative at the marked point}
Back to article page
