Let $I\subset\reals$ be an interval, and let $\gamma:I\to\reals^3$ be an arclength parameterization of an oriented space curve, assumed to be regular, and free of points of inflection. We interpret $\gamma(t)$ as the trajectory of a particle moving through 3-dimensional space. Let $T(t), N(t), B(t)$ denote the corresponding moving trihedron. The speed of this particle is given by$$ v(t) = \Vert \gamma'(t) \Vert.$$

The quantity$$ \kappa(t) = \frac{\Vert T'(t)\Vert}{v(t)} = \frac{\Vert \gamma'(t)\times \gamma''(t)\Vert} {\Vert \gamma'(t)\Vert^3}$$ is called the curvature of the space curve. It is invariant with respect to reparameterization, and is therefore a measure of an intrinsic property of the curve, a real number geometrically assigned to the point $\gamma(t)$ . If one parameterizes the curve with respect to the arclength $s$ , one gets the more concise relation that$$ \kappa(s) = \frac{1\cdot\Vert\gamma''(s)\Vert\cdot\sin\frac{\pi}{2}}{1^3} = \Vert\gamma''(s)\Vert.$$

Physically, curvature may be conceived as the ratio of the normal acceleration of a particle to the particle's speed. This ratio measures the degree to which the curve deviates from the straight line at a particular point. Indeed, one can show that of all the circles passing through $\gamma(t)$ and lying on the osculating plane, the one of radius $1/\kappa(t)$ serves as the best approximation to the space curve at the point $\gamma(t)$ .

To treat curvature analytically, we take the derivative of the relation$$ \gamma'(t) = v(t) T(t)$$ This yields the following decomposition of the acceleration vector:$$ \gamma''(t) = v'(t) T(t) + v(t) T'(t) = v(t) \left\{ (\log v)'(t)\, T(t) + \kappa(t)\, N(t)\right\}.$$ Thus, to change speed, one needs to apply acceleration along the tangent vector; to change heading the acceleration must be applied along the normal.