Cam-Follower Root Selection and Newton Stability
A cam-follower clearance check requires solving f(θ)=0.8θ−sinθ−1.2=0 in radians, where θ is the cam angle at which the follower just touches the housing. The design tolerance is ±0.001 rad, and the manufacturing group needs a defensible root before cutting metal. A junior engineer proposes starting Newton–Raphson at θ0=0.5 rad because 'it is close to zero and converges fast.' At θ=0.5, f≈−1.279 and f′≈−0.078, so the first Newton step is about −16 rad; the spreadsheet returns a nonsense angle and a warning. A second engineer suggests guessing θ0=2.5 rad. The lead wants a bracket, two iterations from 2.5 rad, and a rule for when to abandon Newton because the derivative is too small. The cam profile cannot tolerate a root that is only accidentally close; the check must be repeatable for the production drawing.
1. Explain why θ0=0.5 rad is unsafe, establish a valid bracket for the root, and state the sign change you would document.
Read the worked answer
Compute f(2.0)=0.8(2.0)−sin2.0−1.2=1.6−0.909−1.2=−0.509. f(2.6)=0.8(2.6)−sin2.6−1.2=2.08−0.516−1.2=0.364. Since f is continuous and changes sign on [2.0, 2.6], a root exists there. At 0.5, f′=−0.0776 is tiny; the Newton correction is f/f′≈16.5, throwing the iterate far away. A small derivative means the linear model is a poor approximation, so this is a stability failure, not just slow convergence. Document the bracket endpoints, function values, units (rad), and that the root near 2.37 rad is the one required, not some other root.
2. Carry out two Newton–Raphson iterations from θ0=2.5 rad. Report θ1, θ2, and whether the ±0.001 rad tolerance is met; if not, state what is needed.
Read the worked answer
f(2.5)=0.8(2.5)−sin2.5−1.2=2.0−0.5985−1.2=0.2015. f′(2.5)=0.8−cos2.5=0.8−(−0.8011)=1.6011. θ1=2.5−0.2015/1.6011=2.3741 rad. f(2.3741)≈0.0047; f′(2.3741)≈0.8−(−0.7195)=1.5195. θ2=2.3741−0.0047/1.5195≈2.3710 rad. The change |θ2−θ1|=0.0031 rad is larger than 0.001 rad, so the tolerance is not met after two iterations. A third iteration gives θ3≈2.3709 rad and a residual near zero. Report the stopping criterion with the answer rather than simply saying 'converged.'
3. Recommend a solution strategy when |f′| falls below 0.2 during iteration, and explain why a false convergence check such as |f|<1e−6 alone is not enough.
Read the worked answer
If |f′|<0.2, Newton is poorly conditioned because the correction divides by a small number. Switch to a bracketed method such as bisection or Brent, or use a damped/safeguarded Newton that never leaves the bracket. A small residual |f| can occur on a flat region far from the root, so it is not a convergence proof. Require both a bracket-width or step-size criterion and a residual criterion, and verify that the sign change is still enclosed. For the cam check, keep the root inside [2.0, 2.6] and confirm f′ magnitude before accepting the angle.