Lesson 5 · 30 min
Path Coordinates: Tangential and Normal
A driver does not think in compass directions. They think "forward" and "sideways toward the inside of the bend". Path coordinates formalize that view, and they make curved-path problems remarkably short.
Learning objectives
- Define the tangential and normal unit vectors \(\et\) and \(\en\), and draw them at any point of a path.
- Explain the radius of curvature \(\rho\), the center of curvature and the osculating circle.
- Write the velocity in path coordinates as \(\vvec = v\,\et\), and explain why it has no normal component.
- Find \(\et\) and \(\en\) in terms of \(\ihat\) and \(\jhat\) from the slope of the path and the direction of travel.
Axes that ride with the particle
In path coordinates (also called normal and tangential, or \(n\)–\(t\), coordinates) the origin is the particle itself, and the two axes move with it:
The path unit vectors
- \(\colT{\et}\), the tangential unit vector, is tangent to the path and points in the direction of motion.
- \(\colN{\en}\), the normal unit vector, is perpendicular to \(\et\) and points toward the center of curvature, on the concave (inner) side of the path.
Compare this with rectangular coordinates. There, \(\ihat\) and \(\jhat\) are fixed and the particle's coordinates change. Here, the particle is always at the origin of its own frame, and it is the unit vectors that change direction as the particle moves. That is the price of path coordinates; Lesson 6 shows it is also where the normal acceleration comes from.
Radius of curvature and the osculating circle
Near any point, a smooth curve looks like an arc of a circle. The circle that fits the curve best there (same tangent, and bending at the same rate) is the osculating circle, from the Latin for "kissing". Its radius is the radius of curvature \(\rho\) and its center is the center of curvature \(C\). The normal \(\en\) points from the particle toward \(C\).
- A circle of radius \(R\) has \(\rho = R\) everywhere.
- A straight line never bends: \(\rho = \infty\), and \(\en\) is undefined (any sideways direction will do, since it will be multiplied by zero).
- A tight bend has a small \(\rho\); a gentle one has a large \(\rho\). The curvature \(\kappa = 1/\rho\) measures how sharply the path bends.
- At an inflection point, where the path switches from bending one way to bending the other, \(\rho \to \infty\) and \(\en\) jumps to the other side of the path.
Lesson 7 gives formulas for \(\rho\). For now, the picture is what matters: at every point there is a best-fit circle, and \(\en\) points to its center.
Velocity in path coordinates
Measure distance along the path by the path coordinate \(s\), increasing in the direction of motion. Lesson 2 showed that \(\vvec\) is tangent to the path with magnitude \(v = ds/dt\). Since \(\et\) is defined as the unit tangent in the direction of motion,
Velocity in path coordinates
\[ \vvec = v\,\colT{\et}, \qquad v = \dot s = \frac{ds}{dt} \]The velocity has no normal component: \(v_t = v\) and \(v_n = 0\), always.
That is the great simplification of path coordinates: one number, the speed, describes the velocity completely. All the complexity moves into the acceleration, where \(\et\) turning produces a normal part. You will see it in Lesson 6.
Writing \(\et\) and \(\en\) with \(\ihat\) and \(\jhat\)
To connect the two systems you often need \(\et\) and \(\en\) in rectangular form. Let \(\psi\) be the angle from the \(+x\) axis to the direction of motion. Then
\[ \et = \cos\psi\,\ihat + \sin\psi\,\jhat, \qquad \en = \pm(-\sin\psi\,\ihat + \cos\psi\,\jhat) \]Both choices of sign are perpendicular to \(\et\); take the one that points to the concave side (\(+\) when the path turns counter-clockwise, that is, to the left of the direction of travel). If the velocity is known, \(\et = \vvec/v\).
Example 5.1 — Unit vectors on a parabolic guide
A bead slides along the wire \(y = x^2/4\) (metres) in the \(+x\) direction. Find \(\et\) and \(\en\) at \(x = 2\ \text{m}\). What changes if the bead moves in the \(-x\) direction instead?
Show solution
Tangent. The slope is \(dy/dx = x/2 = 1\) at \(x = 2\), so the tangent direction is \(\ihat + \jhat\) (angle \(45^\circ\)). Moving in \(+x\):
\[ \et = \frac{\ihat + \jhat}{\sqrt2} = 0.7071\,\ihat + 0.7071\,\jhat \]Normal. The two perpendicular unit vectors are \(\pm(-\ihat + \jhat)/\sqrt2\). The parabola opens upward (\(d^2y/dx^2 = \tfrac12 \gt 0\)), so its concave side is above it and \(\en\) must have a positive \(y\) component:
\[ \en = \frac{-\ihat + \jhat}{\sqrt2} = -0.7071\,\ihat + 0.7071\,\jhat \]Moving in \(-x\). \(\et\) reverses: \(\et = -0.7071\,\ihat - 0.7071\,\jhat\). The parabola still bends upward, so \(\en\) is unchanged.
Example 5.2 — A car on a highway curve
A car travels at \(90\ \text{km/h}\) round a curve of radius \(300\ \text{m}\). Write its velocity in path coordinates.
Show solution
Convert: \(90\ \text{km/h} = 90/3.6 = 25\ \text{m/s}\). In path coordinates \(\vvec = v\,\et\), so \(\vvec = 25\,\et\ \text{m/s}\). That is the whole answer: the radius of the curve does not affect the velocity (it will affect the acceleration), and there is no \(\en\) part.
Check your understanding
Key takeaways
- Path coordinates ride with the particle: \(\et\) points along the path in the direction of motion, \(\en\) points toward the center of curvature.
- The osculating circle fits the path best at a point; its radius is \(\rho\). A straight line has \(\rho = \infty\); an inflection point flips \(\en\).
- \(\vvec = v\,\et\): the velocity has no normal component.
- \(\et = \cos\psi\,\ihat + \sin\psi\,\jhat = \vvec/v\); \(\en\) is perpendicular to it, on the concave side.
- Next: Lesson 6 differentiates \(\vvec = v\,\et\) and finds the famous \(v^2/\rho\).