Lesson 7 · 30 min
Radius of Curvature
\(a_n = v^2/\rho\) is only useful if you can find \(\rho\). For a circle it is the radius, but roads, cams, wires and trajectories are rarely circles. This lesson gives the formulas that work for any smooth path.
Learning objectives
- Compute \(\rho\) for a path \(y = f(x)\) from its first and second derivatives.
- Compute \(\rho\) from a motion given as \(x(t)\), \(y(t)\), or from \(\vvec\) and \(\avec\).
- Find \(\rho\) quickly at crests, sags and inflection points, and at the top of a projectile's flight.
- Decide which side of the path \(\en\) (and the center of curvature) is on.
Radius of curvature of \(y = f(x)\)
Lesson 6 used \(d\psi = ds/\rho\): the radius of curvature is how much path length it takes to turn the tangent through one radian, \(\rho = ds/d\psi\). For a graph \(y = f(x)\), the tangent angle satisfies \(\tan\psi = y'\) and the arc length satisfies \(ds = \sqrt{1 + y'^2}\,dx\). Putting these together (details below) gives
Radius of curvature of a path \(y = f(x)\)
\[ \rho = \frac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\left|\dfrac{d^2y}{dx^2}\right|} \]The center of curvature is on the side where the path is concave: above the path when \(y'' \gt 0\), below it when \(y'' \lt 0\).
Optional: where the formula comes from
Differentiate \(\tan\psi = y'\) with respect to \(x\): \(\sec^2\psi\,\dfrac{d\psi}{dx} = y''\), and \(\sec^2\psi = 1 + \tan^2\psi = 1 + y'^2\). So
\[ \frac{d\psi}{dx} = \frac{y''}{1 + y'^2}, \qquad \frac{ds}{dx} = \sqrt{1 + y'^2} \] \[ \rho = \left|\frac{ds}{d\psi}\right| = \left|\frac{ds/dx}{d\psi/dx}\right| = \frac{(1 + y'^2)^{3/2}}{|y''|} \]Two quick consequences:
- Crests and sags. Where the path is horizontal (\(y' = 0\)), the numerator is 1, so \(\rho = 1/|y''|\). For the parabola \(y = kx^2\), the vertex has \(\rho = 1/(2|k|)\).
- Inflection points. Where \(y'' = 0\), \(\rho = \infty\): the path is momentarily straight, and \(a_n = 0\) there however fast the particle moves.
Example 7.1 — A sag curve on a highway
A highway dips through a valley along the vertical curve \(y = x^2/400\) (metres, \(x\) horizontal). (a) Find \(\rho\) at the lowest point, and the normal acceleration of a car passing through it at \(25\ \text{m/s}\). (b) Find \(\rho\) at \(x = 20\ \text{m}\).
Show solution
Derivatives. \(y' = x/200\) and \(y'' = 1/200\ \text{m}^{-1}\) (constant).
(a) Lowest point, \(x = 0\). \(y' = 0\), so \(\rho = 1/y'' = 200\ \text{m}\). Then
\[ a_n = \frac{v^2}{\rho} = \frac{25^2}{200} = 3.125\ \text{m/s}^2 \ \text{(upward, toward the center above the road)} \](b) At \(x = 20\ \text{m}\). \(y' = 0.1\):
\[ \rho = \frac{(1 + 0.1^2)^{3/2}}{1/200} = 200(1.01)^{3/2} = 203.0\ \text{m} \]Interpret. Road slopes are small, so \(y'^2\) barely matters: engineers often use \(\rho \approx 1/|y''|\) along the whole of a gentle vertical curve.
When the motion is given as \(x(t)\) and \(y(t)\)
You do not need to eliminate \(t\) to get \(y = f(x)\). Lesson 6 said \(a_n = v^2/\rho\), so if you can find \(a_n\) from the rectangular components, you have \(\rho\). The normal acceleration is the part of \(\avec\) perpendicular to \(\vvec\), which is \(|\vvec \times \avec|/v\) (Lesson 8 shows why). That gives
Radius of curvature from the motion
\[ \rho = \frac{v^2}{a_n} = \frac{v^3}{|\vvec \times \avec|} = \frac{\left(\dot x^2 + \dot y^2\right)^{3/2}}{\left|\dot x\,\ddot y - \dot y\,\ddot x\right|} \]In the plane, the "cross product" \(\vvec \times \avec\) reduces to the single number \(\dot x\,\ddot y - \dot y\,\ddot x\).
Example 7.2 — A parametric path
A particle moves with \(x = 2t^2\) and \(y = t^3\) (metres, seconds). Find the radius of curvature of its path at \(t = 1\ \text{s}\).
Show solution
Derivatives at \(t = 1\). \(\dot x = 4t = 4\), \(\dot y = 3t^2 = 3\), \(\ddot x = 4\), \(\ddot y = 6t = 6\).
\[ v = \sqrt{4^2 + 3^2} = 5\ \text{m/s}, \qquad \dot x\,\ddot y - \dot y\,\ddot x = 4(6) - 3(4) = 12 \] \[ \rho = \frac{5^3}{|12|} = \frac{125}{12} = 10.42\ \text{m} \]Check with \(y = f(x)\). Eliminating \(t\): \(y = (x/2)^{3/2}\). At \(x = 2\): \(y' = \tfrac34\), \(y'' = \tfrac{3}{16}\), and \(\rho = (1 + \tfrac{9}{16})^{3/2}/\tfrac{3}{16} = 10.42\ \text{m}\). ✓ Same path, same answer.
Example 7.3 — The top of a projectile's flight
A ball is launched at \(20\ \text{m/s}\), \(30^\circ\) above the horizontal. What is the radius of curvature of its path at the highest point?
Show solution
At the top the velocity is horizontal, \(v = v_0\cos\theta_0 = 17.32\ \text{m/s}\), and the acceleration \(g\) points straight down, perpendicular to \(\vvec\). So all of \(g\) is normal acceleration: \(a_n = g\).
\[ \rho = \frac{v^2}{a_n} = \frac{(v_0\cos\theta_0)^2}{g} = \frac{17.32^2}{9.81} = 30.58\ \text{m} \]This trick (find \(v\) and \(a_n\) from the physics, then \(\rho = v^2/a_n\)) is often the fastest route to a radius of curvature.
Check your understanding
Key takeaways
- \(\rho = (1 + y'^2)^{3/2}/|y''|\) for a path \(y = f(x)\); at a crest or sag this is simply \(1/|y''|\).
- \(\rho = v^3/|\dot x\ddot y - \dot y\ddot x|\) for a motion \(x(t), y(t)\), or \(\rho = v^2/a_n\) whenever \(a_n\) is known.
- At an inflection point (\(y'' = 0\)), \(\rho = \infty\) and \(a_n = 0\).
- The center of curvature, and \(\en\), are on the concave side: above the path if \(y'' \gt 0\), below if \(y'' \lt 0\).
- Next: Lesson 8 converts freely between \(x\)–\(y\) and \(n\)–\(t\) components.