Tangential/normal basis #rkt
Consider a particle moving with position vector →r and corresponding velocity →v and acceleration →a. The tangential/normal basis ˆet,ˆen,ˆeb is:
Tangential/normal basis vectors. #rkt‑eb
ˆet=ˆvtangential basis vectorˆen=˙ˆet‖
Derivation
The tangential basis vector \hat{e}_t points tangential to the path, the normal basis vector \hat{e}_n points perpendicular (normal) to the path towards the instantaneous center of curvature, and the binormal basis vector \hat{e}_b completes the right-handed basis.
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Tangential/normal basis associated with movement around a curve in 3D. Observe that the velocity \vec{v} is always in the \hat{e}_t direction and that the acceleration \vec{a} always lies in the \hat{e}_t,\hat{e}_n plane (the osculating plane). The center of curvature and osculating circle are defined below. #rkt‑fb
Path length #rkt‑sl
The distance along the path is s, which can be found by integrating the speed v with respect to time. Correspondingly, the speed is the derivative of path length.
Path length s and speed v. #rkt‑es
\begin{aligned} v &= \dot{s} \\ s &= \int_0^t v(\tau) \,d\tau \end{aligned}
Derivation
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Path length s and speed v shown for motion back and forth along a curve. Observe that the path length is always increasing, even when moving “backwards”, as speed cannot be negative. For ease of visualization, this figure resets the path length to zero when the particle returns to the starting point. #rkt‑fs
When differentiating geometric quantities related to paths, it is often convenient to differentiate with respect to path length rather than with respect to time. These derivatives are related by:
Derivatives with respect to path length s and time t. #rkt‑ea
\begin{aligned} \frac{d}{dt} &= v \frac{d}{ds} \\ \frac{d}{ds} &= \frac{1}{v} \frac{d}{dt} \end{aligned}
Derivation
Curvature and torsion #rkt‑sk
Curvature and torsion. #rkt‑ek
To better understand the geometry of the tangential/normal basis, we can use the curvature \kappa to describe the curving of the path, and the torsion \tau to describe the rotation of the basis about the path. These quantities are defined by:
\begin{aligned} \kappa &= \frac{d\hat{e}_t}{ds} \cdot \hat{e}_n = \frac{1}{v} \dot{\hat{e}}_t \cdot \hat{e}_n & &\text{curvature} \\ \rho &= \frac{1}{\kappa} & &\text{radius of curvature} \\ \tau &= -\frac{d\hat{e}_b}{ds} \cdot \hat{e}_n = -\frac{1}{v} \dot{\hat{e}}_b \cdot \hat{e}_n & &\text{torsion} \\ \sigma &= \frac{1}{\tau} & &\text{radius of torsion} \end{aligned}
Derivation
The radius of curvature \rho is the radius of equivalent circular motion, and the torsion determines the rate of rotation of the osculating plane, as described below in Section #rkt-so.
Basis derivatives and angular velocity #rkt‑sd
As the point P moves along its path, the associated tangential/normal basis rotates with an angular velocity vector \omega given by:
Angular velocity of the tangential/normal basis. #rkt‑ew
\begin{aligned} \vec{\omega} &= v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b \end{aligned}
Derivation
Knowing the angular velocity vector of the tangential/normal basis allows us to easily compute the time derivatives of each tangential/normal basis vector, as follows:
Tangential/normal basis vector derivatives. #rkt‑ed
\begin{aligned} \dot{\hat{e}}_t &= \phantom{-v\kappa\,\hat{e}_t - } v\kappa\,\hat{e}_n \\ \dot{\hat{e}}_n &= -v\kappa\,\hat{e}_t \phantom{ - v\kappa\,\hat{e}_n + } + v\tau\,\hat{e}_b \\ \dot{\hat{e}}_b &=\phantom{-v\kappa\,\hat{e}_t } - v\tau\,\hat{e}_n \end{aligned}
Derivation
Notation note
The tangential/normal basis is also called the Frenet–Serret frame after Jean Frédéric Frenet and Joseph Alfred Serret, who discovered it independently around 1850. The equations #rkt-ed for the basis derivatives are often called the Frenet-Serret formulas, typically written in terms of s derivatives: \begin{aligned} \frac{d\hat{e}_t}{ds} &= \phantom{-\kappa\,\hat{e}_t - } \kappa\,\hat{e}_n \\ \frac{d\hat{e}_n}{ds} &= -\kappa\,\hat{e}_t \phantom{ - \kappa\,\hat{e}_n + } + \tau\,\hat{e}_b \\ \frac{d\hat{e}_b}{ds} &=\phantom{-\kappa\,\hat{e}_t } - \tau\,\hat{e}_n. \end{aligned}
If we divide the angular velocity vector #rkt-ew by v then we obtain the vector \frac{1}{v} \vec{\omega} = \tau\,\hat{e}_t + \kappa\,\hat{e}_b, which is known as the Darboux vector after its discoverer, Jean Gaston Darboux.
Kinematic relations #rkt‑sk
While the motion of a point P along a path defines the tangential/normal basis, we can also use this basis to express the kinematics of P itself, giving the following expressions for velocity and acceleration.
Velocity and acceleration in tangential/normal basis. #rkt‑ev
\begin{aligned} \vec{v} &= \dot{s} \, \hat{e}_t \\ \vec{a} &= \ddot{s} \, \hat{e}_t + \frac{\dot{s}^2}{\rho} \hat{e}_n \end{aligned}
The above formula shows that the normal acceleration component a_n is determined by the radius of curvature. We can therefore also find the radius of curvature from knowing the normal acceleration:
Derivation
Radius of curvature \rho for velocity \vec{v} and acceleration \vec{a} with angle \theta between them. #rkt‑er
\rho = \frac{v^2}{a_n} = \frac{v^2}{a\sin\theta}
Derivation
Movement: | circle | var-circle | ellipse | arc |
trefoil | eight | comet | pendulum | |
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Origin: | O_1 | O_2 |
Velocity and acceleration in the tangential/normal basis. Note that the tangential/normal basis does not depend on the choice of origin or the position vector, in contrast to the polar basis. #rkt‑ft
Tangent lines and osculating circles #rkt‑so
Given a point P moving along a complex path, at a given instant of time we can match different components of the point's motion with successively more complex geometric shapes:
match | geometric object | name |
---|---|---|
\vec{r} | point | position |
\vec{r},\vec{v} | line | tangent line |
\vec{r},\vec{v},\vec{a} | circle | osculating circle |
The osculating circle is the instantaneous equivalent circular path. That is, a particle traveling on the osculating circle with the same location P, speed \dot{s}, and tangential acceleration \ddot{s} as our particle would have matching velocity and acceleration vectors.
Osculating circle. #rkt‑eo
\text{center $C = \vec{r} + \rho \hat{e}_n$,} \qquad \text{radius $\rho$,} \qquad \text{normal $\hat{e}_b$}
Derivation
The osculating circle lies in the \hat{e}_t,\hat{e}_n plane, which is thus called the osculating plane. This plane has normal vector \hat{e}_b, and rotates about \hat{e}_t with a rotation rate determined by the torsion \tau, as we see from the expression #rkt-ed for the derivative of \hat{e}_b.
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Match phase: | 0% |
Tangent lines and osculating circles, showing matching of velocity or velocity and acceleration, respectively. #rkt‑fo
Curves and curvature #rkt‑sc
Given a parametric curve, its curvature can be directly evaluated with:
Curvature of parametric curve \vec{r}(u) in 3D. #rkt‑ec
\kappa = \frac{\|\vec{r}' \times \vec{r}''\|}{\|\vec{r}'\|^3}
Derivation
Curvature of parametric curve x = x(u), y = y(u) in 2D. #rkt‑e2
\kappa = \frac{|x''y' - y''x'|}{(x'^2 + y'^2)^{3/2}}
Derivation