--- title: "Loop closure equation and Newton-Raphson method" subject: "Theory of Mechanism and Machine I" chapter: 2 key: "semester_6/theory_of_mechanism_and_machine_i:loop-closure-equation-newton-raphson" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/loop-closure-equation-newton-raphson site: BCT Engineering --- # Loop closure equation and Newton-Raphson method Theory of Mechanism and Machine I · Chapter 2: Linkages and Mechanisms - Asked in: year not given, 6 marks - Video: [Mechanisms: Vector Loop Analytical Position Analysis Intro (S21 ME401 Class 8)](https://www.youtube.com/watch?v=14smhv_tepU) — Professor Ted Diehl - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/loop-closure-equation-newton-raphson - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-2 ## Question Write the vector loop-closure equation of a four-bar linkage and explain how the unknown coupler and follower angles are found by Freudenstein's equation and by the Newton-Raphson iterative method. ## Short answer Vector loop equation: r2cosθ2 + r3cosθ3 - r4cosθ4 - r1 = 0 and r2sinθ2 + r3sinθ3 - r4sinθ4 = 0. Freudenstein's equation eliminates θ3 giving K1cosθ4 - K2cosθ2 + K3 = cos(θ2-θ4), solved via quadratic for both assembly modes. Newton-Raphson iteratively solves f1=0, f2=0 using Jacobian matrix, requiring initial guess but converging quickly for computer implementation. ## Answer #### Loop-closure equation Represent the links as vectors $\vec r_1$ (frame), $\vec r_2$ (crank), $\vec r_3$ (coupler), $\vec r_4$ (follower) with angles $\theta_1 = 0, \theta_2, \theta_3, \theta_4$ measured from the frame. The closed loop gives $$ \vec r_2 + \vec r_3 - \vec r_4 - \vec r_1 = 0 $$ Splitting into components: $$ \begin{aligned} r_2\cos\theta_2 + r_3\cos\theta_3 - r_4\cos\theta_4 - r_1 &= 0 \\ r_2\sin\theta_2 + r_3\sin\theta_3 - r_4\sin\theta_4 &= 0 \end{aligned} $$ These are two equations in the two unknowns $\theta_3$ and $\theta_4$ for a given input $\theta_2$. #### Freudenstein's equation (closed form) Isolate the $r_3$ terms, square both equations and add them to eliminate $\theta_3$. Dividing by $2r_2r_4$ gives $$ K_1\cos\theta_4 - K_2\cos\theta_2 + K_3 = \cos(\theta_2 - \theta_4) $$ with $K_1 = r_1/r_2$, $K_2 = r_1/r_4$, $K_3 = \dfrac{r_1^2 + r_2^2 + r_4^2 - r_3^2}{2r_2r_4}$. (Check: for $r_1 = 300$, $r_2 = 100$, $r_3 = 250$, $r_4 = 200$ mm at $\theta_2 = 60^\circ$, $\theta_4 = 97.56^\circ$: $K_1 = 3$, $K_2 = 1.5$, $K_3 = 1.9375$, and the left side is $0.793 = \cos(-37.56^\circ)$.) Using the half-angle substitution $t = \tan(\theta_4/2)$ gives a quadratic $A t^2 + Bt + C = 0$, whose two roots are the **open** and **crossed** configurations. Then $\theta_3$ follows from the component equations. #### Newton-Raphson iteration Write the two component equations as $f_1(\theta_3,\theta_4) = 0$ and $f_2(\theta_3,\theta_4) = 0$. 1. Guess $\theta_3^{(0)}$ and $\theta_4^{(0)}$ (from a sketch or the previous crank position). 2. Form the Jacobian $$ J = \begin{bmatrix} -r_3\sin\theta_3 & r_4\sin\theta_4 \\ r_3\cos\theta_3 & -r_4\cos\theta_4 \end{bmatrix} $$ 3. Solve $J\,\Delta = -\begin{bmatrix} f_1 \\ f_2 \end{bmatrix}$ for the corrections $\Delta = [\Delta\theta_3,\ \Delta\theta_4]^T$. 4. Update $\theta_3 \leftarrow \theta_3 + \Delta\theta_3$, $\theta_4 \leftarrow \theta_4 + \Delta\theta_4$. 5. Repeat until $|f_1|$ and $|f_2|$ are below a tolerance (about $10^{-6}$); it usually converges in 3 to 5 steps. #### Comparison | Point | Freudenstein | Newton-Raphson | |---|---|---| | Nature | Closed form | Iterative | | Initial guess | Not needed | Needed | | Assembly mode | Both roots found | Depends on guess | | Use | Four-bar only | Any loop, complex mechanisms | The iterative method is preferred in computer programs because the previous solution gives an excellent starting guess for the next crank angle, and the same Jacobian is reused for velocity analysis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The vector loop-closure equation of the four-bar linkage is: A. r2 + r3 + r4 + r1 = 0 B. r2 + r3 - r4 - r1 = 0 C. r2 - r3 + r4 - r1 = 0 D. r2 x r3 = r4 x r1 Answer: B. The closed loop gives r2 + r3 - r4 - r1 = 0, which splits into two scalar equations in theta3 and theta4. (Key point: Loop-closure equation) 2. How is Freudenstein's equation obtained and what does its quadratic give? A. Eliminate theta4; the two roots are crank angles B. Differentiate the loop equation; the roots are velocities C. Eliminate theta3 by squaring and adding; the two roots are the open and crossed configurations D. Linearise the loop equation; the roots are Jacobian entries Answer: C. Squaring and adding eliminates theta3. The half-angle substitution gives a quadratic whose roots are the open and crossed assembly modes. (Key point: Freudenstein's equation) 3. What does the Newton-Raphson method need that Freudenstein's closed form does not? A. A sketch of cam profile B. A fixed link of length zero C. Only the crank length D. An initial guess for theta3 and theta4 Answer: D. Newton-Raphson is iterative and starts from a guess, and the assembly mode found depends on that guess. (Key point: Newton-Raphson requirement) 4. In each Newton-Raphson iteration, how are the corrections found? A. Solve J*Delta = -[f1, f2] using the Jacobian J B. Set f1 and f2 equal to the Jacobian C. Average the last two guesses D. Divide f1 by theta3 Answer: A. The Jacobian system J*Delta = -[f1 f2]^T gives the corrections that update theta3 and theta4 until f1 and f2 are below the tolerance. 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