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    Tangential and normal basis

    Normal basis

    Consider a particle moving with position vector \( \vec{r} \) and corresponding velocity \( \vec{v} \) and acceleration \( \vec{a} \). The tangential/normal basis \( \hat{e}_t,\hat{e}_n,\hat{e}_b \) is:

    Tangential/normal basis vectors. #rkt-eb
    $$ \begin{aligned}\hat{e}_t &= \hat{v}& &\text{tangential basis vector} \\\hat{e}_n &= \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|}= \frac{\operatorname{Comp}(\vec{a},\vec{v})}{\|\operatorname{Comp}(\vec{a},\vec{v})\|}& &\text{normal basis vector} \\\hat{e}_b &= \hat{e}_t \times \hat{e}_n& &\text{binormal basis vector} \\\end{aligned} $$

    These equations are definitions of the basis vectors, so the only thing to derive is the alternative formula for \( \hat{e}_n \). Using the definition of \( \hat{e}_t \) above and #rvc-eu, we see that

    $$ \dot{\hat{e}}_t = \dot{\hat{v}} = \frac{1}{v} \operatorname{Comp}(\dot{\vec{v}}, \vec{v}) = \frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v}). $$
    Normalizing both sides gives the desired expression:
    $$ \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|} = \frac{\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})}{ \left\|\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})\right\|} = \frac{\frac{1}{v} \operatorname{Comp}(\vec{a}, \vec{v})}{ \frac{1}{v} \|\operatorname{Comp}(\vec{a}, \vec{v})\|} = \frac{\operatorname{Comp}(\vec{a}, \vec{v})}{ \|\operatorname{Comp}(\vec{a}, \vec{v})\|}. $$

    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.

    Curvature and torsion

    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:

    $$ \begin{aligned}\kappa &= \frac{d\hat{e}_t}{ds} \cdot \hat{e}_n= \frac{1}{v} \dot{\hat{e}}_t \cdot \hat{e}_n & &\text{curvature} \\\rho &= \frac{1}{\kappa} & &\text{radius of curvature} \\\tau &= -\frac{d\hat{e}_b}{ds} \cdot \hat{e}_n= -\frac{1}{v} \dot{\hat{e}}_b \cdot \hat{e}_n & &\text{torsion} \\\sigma &= \frac{1}{\tau} & &\text{radius of torsion}\end{aligned} $$

    These equations are definitions, so we need only check that the two expressions for each of \(\kappa\) and \(\tau\) are equivalent. This follows immediately, however, from the chain-rule conversions #rkt-ea between \(d/ds\) and \(d/dt\).

    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 a section below.

    Basis derivatives and angular velocity

    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
    $$ \begin{aligned}\vec{\omega} &= v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b\end{aligned} $$

    We start by writing the angular velocity in the tangential/normal basis, giving:

    $$ \vec{\omega} = \omega_t\,\hat{e}_t + \omega_n\,\hat{e}_n+ \omega_b\,\hat{e}_b. $$
    Now the basis vector derivatives are given by the cross product by \( \vec{\omega} \) from #rkr-ew, so we can evaluate the expressions #rkt-ek for curvature and torsion to give:
    $$ \begin{aligned}\kappa &= \frac{1}{v} \dot{\hat{e}}_t \cdot \hat{e}_n \\&= \frac{1}{v} (\vec{\omega} \times \hat{e}_t) \cdot \hat{e}_n \\&= \frac{1}{v} (\omega_b\,\hat{e}_n - \omega_n\,\hat{e}_b) \cdot \hat{e}_n \\&= \frac{1}{v} \omega_b\end{aligned} $$
    and
    $$ \begin{aligned}\tau &= -\frac{1}{v} \dot{\hat{e}}_b \cdot \hat{e}_n \\&= -\frac{1}{v} (\vec{\omega} \times \hat{e}_b) \cdot \hat{e}_n \\&= -\frac{1}{v} (\omega_n\,\hat{e}_t - \omega_t\,\hat{e}_n) \cdot \hat{e}_n \\&= \frac{1}{v} \omega_t.\end{aligned} $$
    Rearranging the final expressions in each case gives \( \omega_b = v\kappa \) and \( \omega_t = v\tau \). From the definition #rkt-eb of \( \hat{e}_n \), we see that
    $$ \begin{aligned}\hat{e}_n &= \frac{\dot{\hat{e}}_t}{\|\dot{\hat{e}}_t\|} \\\hat{e}_n \cdot \hat{e}_b &= \frac{1}{\|\dot{\hat{e}}_t\|} (\vec{\omega} \times \hat{e}_t) \cdot \hat{e}_b \\0 &= \frac{1}{\|\dot{\hat{e}}_t\|} (\omega_b\,\hat{e}_n - \omega_n\,\hat{e}_b) \cdot \hat{e}_b \\&= - \frac{1}{\|\dot{\hat{e}}_t\|} \omega_n.\end{aligned} $$
    from which we conclude that \( \omega_n = 0 \), giving us all three components of \( \vec{\omega} \).

    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
    $$ \begin{aligned}\dot{\hat{e}}_t &= \phantom{-v\kappa\,\hat{e}_t - } v\kappa\,\hat{e}_n \\\dot{\hat{e}}_n &= -v\kappa\,\hat{e}_t \phantom{ - v\kappa\,\hat{e}_n + } + v\tau\,\hat{e}_b \\\dot{\hat{e}}_b &=\phantom{-v\kappa\,\hat{e}_t } - v\tau\,\hat{e}_n\end{aligned} $$

    We can use the expression #rkt-ew for \( \vec{\omega} \) together with #rkr-ew to find the basis vector derivatives:

    $$ \begin{aligned}\dot{\hat{e}}_t &= \vec{\omega} \times \hat{e}_t= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_t= v \kappa \,\hat{e}_n \\\dot{\hat{e}}_n &= \vec{\omega} \times \hat{e}_n= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_n= - v \kappa \,\hat{e}_t + v \tau \,\hat{e}_b \\\dot{\hat{e}}_b &= \vec{\omega} \times \hat{e}_b= (v\tau \,\hat{e}_t + v \kappa \,\hat{e}_b) \times \hat{e}_b= -v \tau \,\hat{e}_n.\end{aligned} $$

    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.

    Normal and tangential kinematics

    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-e
    $$ \begin{aligned}\vec{v} &= \dot{s} \, \hat{e}_t \\\vec{a} &= \ddot{s} \, \hat{e}_t + \frac{\dot{s}^2}{\rho} \hat{e}_n\end{aligned} $$

    From the definition #rkt-eb of \( \hat{e}_t \) we see that \( v\,\hat{e}_t = v\hat{v} = \vec{v} \), which is the first equation above. Differentiating this and using \( v = \dot{s} \) from #rkt-es gives

    $$ \begin{aligned}\vec{v} &= \dot{s}\,\hat{e}_t \\\vec{a} = \dot{\vec{v}}&= \ddot{s}\,\hat{e}_t + \dot{s}\,\dot{\hat{e}}_t \\&= \ddot{s}\,\hat{e}_t + \dot{s}(v \kappa \,\hat{e}_n) \\&= \ddot{s}\,\hat{e}_t + \frac{\dot{s}^2}{\rho} \,\hat{e}_n,\end{aligned} $$
    where we used the derivative #rkt-ed of \( \hat{e}_t \) in terms of the curvature \(\kappa\), and the definition #rkt-ek of the radius of curvature to give \( \kappa = 1/\rho \).

    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:

    Radius of curvature \(\rho\) for velocity \(\vec{v}\) and acceleration \(\vec{a}\) with angle \(\theta\) between them. #rkt-er
    $$ \rho = \frac{v^2}{a_n} = \frac{v^2}{|a\sin\theta|} $$

    From #rkt-ev we known that the normal component of the acceleration is given by \( a_n =\dot{s}^2/\rho \) and from the definition #rkt-es of path length we have \( \dot{s} = v \), so this equation can be rearranged to give \( \rho = v^2 / a_n \).

    Now because \( \vec{a} \) only has \( \hat{e}_t \) and \( \hat{e}_n \) components, the \( \hat{e}_n \) component is given by

    $$ \begin{aligned}a_n &= \|\operatorname{Comp}(\vec{a}, \hat{e}_t)\| \\&= \|\operatorname{Comp}(\vec{a}, \vec{v})\| \\&= |a \sin\theta|,\end{aligned} $$
    where we used the facts that \( \vec{e}_t \) is in the direction of \( \vec{v} \), that \( \operatorname{Comp}(\vec{a},\vec{v}) \) does not depend on the magnitude of \( \vec{v} \), and equation #rvv-em for the magnitude of the complementary projection.

    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.

    Tangent lines and osculating circles

    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
    $$ \text{center C} = \vec{r} + \rho \hat{e}_n,\qquad \text{radius} \; \rho,\qquad \text{normal to} \; \hat{e}_b $$

    Consider a particle \(P\) with position vector \( \vec{r}_P \) moving with velocity \( \vec{v}_P \) and acceleration \( \vec{a}_P \) , and let \( \hat{e}_t,\hat{e}_n,\hat{e}_b \)be the associated tangential/normal basis and \( \rho_P \) be the radius of curvature. Set \( \dot{s}_P = v_P \)and \( \ddot{s}_P = \vec{a}_P \cdot \hat{e}_t \) to be the speed and tangential acceleration of \(P\).

    Next define the osculating circle as above with center \( C = \vec{r}_P + \rho_P \) , radius \( \rho_P \), and normal vector \( \hat{e}_b \).

    Now take a second particle \(Q\) moving on the osculating circle so that its instantaneous position matches \(P\) and it has the same speed \( \dot{s}_Q = \dot{s}_P \) and tangential acceleration \( \ddot{s}_Q = \ddot{s}_P \). This means that the radius of the circle \( r_Q \) for \(Q\) is matches the osculating circle radius \( \rho_P \), so \( r_Q = \rho_P \).

    We want to prove that \( \vec{v}_Q = \vec{v}_P \) and \( \vec{a}_Q = \vec{a}_P \).

    We will use a polar coordinate system for \(Q\), centered at \(C\) and in the plane with normal vector \( \hat{e}_b \). At the current instant when the particles are at the same position, this means that \( \hat{e}_r = -\hat{e}_n \) and \( \hat{e}_\theta = \hat{e}_t \). Using the circular motion expressions #rke-ec we have that the velocity of \(Q\) is:

    $$ \begin{aligned} \vec{v}_Q &= v_Q \,\hat{e}_\theta \\ &= \dot{s}_Q \,\hat{e}_\theta \\ &= \dot{s}_P \,\hat{e}_t = \vec{v}_P. \\ \end{aligned} $$
    For a particle moving in a circle like \(Q\), the speed is given by \( \dot{s}_Q = v_Q = r_Q \omega_Q \), and differentiating this gives \( \ddot{s}_Q = r_Q \dot{\omega}_Q = r_Q \alpha_Q \). Using this, the acceleration of \(Q\) is:
    $$ \begin{aligned} \vec{a}_Q &= -\frac{v_Q^2}{\rho_Q} \,\hat{e}_r + \rho_Q \alpha_Q \,\hat{e}_\theta \\ &= -\frac{\dot{s}_Q^2}{\rho_Q} \,\hat{e}_r + \ddot{s}_Q \,\hat{e}_\theta \\ &= \frac{\dot{s}_P^2}{\rho_P} \,\hat{e}_n + \ddot{s}_Q \,\hat{e}_t \\ &= \vec{a}_Q. \end{aligned} $$
    We thus see that points \(P\) and \(Q\) have the same instantaneous velocity and acceleration. This also means that the tangential/normal basis computed at this instant for \(Q\) will match that of \(P\).

    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:

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    Tangent lines and osculating circles, showing matching of velocity or velocity and acceleration, respectively.

    Curvature on curves

    Given a parametric curve, its curvature can be directly evaluated with:

    Curvature of parametric curve \(\vec{r}(u)\) in 3D. #rkt-ec
    $$ \kappa = \frac{\|\vec{r}' \times \vec{r}''\|}{\|\vec{r}'\|^3} $$

    Because the tangential/normal basis vectors #rkt-eb are all normalized, it does not matter how fast the point moves along the line. We can thus evaluate the parametric curve as a function of time, giving \( \vec{r}(t) \). Now:

    $$ \begin{aligned}\|\vec{r}'' \times \vec{r}'\| &= \|\vec{a} \times \vec{v}\| \\&= a v \sin\theta \\&= \frac{v^3}{\rho},\end{aligned} $$
    where we used the cross product length formula #rvv-el and equation #rkt-er for the radius of curvature \( \rho \). By definition #rkt-ek the curvature is \( \kappa = 1/\rho \), so
    $$ \begin{aligned}\|\vec{r}'' \times \vec{r}'\|&= v^3 \kappa \\\kappa &= \frac{\|\vec{r}'' \times \vec{r}'\|}{\|\vec{r}'\|^3}.\end{aligned} $$

    While the above formula can be used in 2D by taking the third component to be zero, it can also be written in an explicitly 2D form:

    Curvature of parametric curve \(x = x(u)\), \(y = y(u)\) in 2D. #rkt-e2
    $$ \kappa = \frac{|x''y' - y''x'|}{(x'^2 + y'^2)^{3/2}} $$

    This equation is just #rkt-ec written in explicit 2D coordinates. To see this, we first take the position vector \( \vec{r}(u) \) to be

    $$ \vec{r}(u) = x(u)\,\hat\imath + y(u)\,\hat\jmath. $$
    Now \( r'(u) = \sqrt{(x'(u))^2 + (y'(u))^2} \) and
    $$ \begin{aligned}\vec{r}'' \times \vec{r}'&= (x''\,\hat\imath + y''\,\hat\jmath)\times (x'\,\hat\imath + y'\,\hat\jmath) \\&= (x''y' - y''x')\,\hat{k},\end{aligned} $$
    so evaluating #rkt-ec gives the desired expression.

    We can take this a step further, and obtain an expression for an explicitly defined function.

    Curvature of an explicitly defined function \(y = f(x).\) #rkt-e3
    $$ \kappa = \frac{|y''(x)|}{(1 + y'(x)^2)^{3/2}} $$

    Use equation #rkt‑e2 and parametrize the curve using \(x\) as the parametrization variable. Namely:

    $$ \begin{aligned}x = u, \quad \quad y = y(u) = y(x)\end{aligned} $$
    This yields a very elegant expression, as \( x'(u) = 1 \) and \( x''(u) = 0 \), which lets us arrive at the desired expression #rkt‑e3.