# Piecewise Cubic Joint-Space Trajectory Solution ## Problem Recap We must move a revolute joint from - \( q_ = 15^\circ \) at \( t = \) (with \( \dot{q}() = \)) - **Via point:** \( q_v = 50^\circ \) at \( t = T_1 = 1.2\,\mathrm{s} \) - \( q_f = 70^\circ \) at \( t = T_1 + T_2 = 2.2\,\mathrm{s} \) (with \( \dot{q}(2.2) = \)), where \( T_2 = 1.\,\mathrm{s} \). We need two cubic segments: - **Segment 1:** \( t \in [, T_1] \) - **Segment 2:** \( \tau = t - T_1 \in [, T_2] \) At \( t = T_1 \), position, velocity, and acceleration must be continuous. --- ## (a) Find the via-point velocity \( v_1 = \dot{q}(T_1) \) for acceleration continuity ### **Cubic Segment Formulation** For \( t \in [, T_1] \): \[ q_1(t) = a_ + a_1 t + a_2 t^2 + a_3 t^3 \] \[ \dot{q}_1(t) = a_1 + 2a_2 t + 3a_3 t^2 \] \[ \ddot{q}_1(t) = 2a_2 + 6a_3 t \] For \( \tau \in [, T_2] \), \( \tau = t - T_1 \): \[ q_2(\tau) = b_ + b_1 \tau + b_2 \tau^2 + b_3 \tau^3 \] \[ \dot{q}_2(\tau) = b_1 + 2b_2 \tau + 3b_3 \tau^2 \] \[ \ddot{q}_2(\tau) = 2b_2 + 6b_3 \tau \] ### **Boundary Conditions** #### Segment 1 (\( t \in [, T_1] \)): - \( q_1() = q_ = 15 \) - \( \dot{q}_1() = \) - \( q_1(T_1) = q_v = 50 \) - \( \dot{q}_1(T_1) = v_1 \) (unknown) #### Segment 2 (\( \tau \in [, T_2] \)): - \( q_2() = q_v = 50 \) - \( \dot{q}_2() = v_1 \) - \( q_2(T_2) = q_f = 70 \) - \( \dot{q}_2(T_2) = \) ### **Acceleration Continuity at Via Point** Acceleration at \( t = T_1^- \) must equal that at \( \tau = ^+ \): \[ \ddot{q}_1(T_1) = \ddot{q}_2() \] Write in terms of coefficients: \[ 2a_2 + 6a_3 T_1 = 2b_2 \] (We'll use this after expressing everything in terms of \( v_1 \).) --- ### **Cubic Coefficient Formulas (General)** Given start and end positions and velocities for a cubic on \( [, T] \): \[ q() = q_, \quad \dot{q}() = v_ \] \[ q(T) = q_T, \quad \dot{q}(T) = v_T \] Coefficients: \[ a_ = q_ \] \[ a_1 = v_ \] \[ a_2 = \frac{3(q_T - q_) - (2v_ + v_T) T}{T^2} \] \[ a_3 = \frac{-2(q_T - q_) + (v_ + v_T) T}{T^3} \] Apply to both segments: #### **Segment 1 (\( t \in [, T_1] \)):** - \( q_ = 15 \), \( v_ = \) - \( q_T = 50 \), \( v_T = v_1 \), \( T = T_1 = 1.2 \) \[ a_ = 15 \] \[ a_1 = \] \[ a_2 = \frac{3(50-15) - ( + v_1) \cdot 1.2}{1.2^2} \] \[ a_3 = \frac{-2(50-15) + v_1 \cdot 1.2}{1.2^3} \] #### **Segment 2 (\( \tau \in [, T_2] \)):** - \( q_ = 50 \), \( v_ = v_1 \) - \( q_T = 70 \), \( v_T = \), \( T = T_2 = 1. \) \[ b_ = 50 \] \[ b_1 = v_1 \] \[ b_2 = \frac{3(70-50) - (2v_1 + ) \cdot 1.}{1.^2} = 60 - 2v_1 \] \[ b_3 = \frac{-2(70-50) + (v_1 + ) \cdot 1.}{1.^3} = -40 + v_1 \] --- ### **Compute \( a_2, a_3, b_2, b_3 \) in terms of \( v_1 \)** #### For Segment 1 (\( T_1 = 1.2 \)): \[ a_2 = \frac{3(35) - 1.2 v_1}{1.44} = \frac{105 - 1.2 v_1}{1.44} \] \[ a_3 = \frac{-2(35) + 1.2 v_1}{1.728} = \frac{-70 + 1.2 v_1}{1.728} \] #### For Segment 2 (\( T_2 = 1. \)): \[ b_2 = 60 - 2v_1 \] \[ b_3 = -40 + v_1 \] --- ### **Acceleration Continuity** \[ \ddot{q}_1(T_1) = 2a_2 + 6a_3 T_1 \] \[ \ddot{q}_2() = 2b_2 \] Set equal: \[ 2a_2 + 6a_3 T_1 = 2b_2 \] Plug the expressions in: 1. \( a_2 = \frac{105 - 1.2 v_1}{1.44} \) 2. \( a_3 = \frac{-70 + 1.2 v_1}{1.728} \) 3. \( b_2 = 60 - 2v_1 \) Plug into the continuity equation: \[ 2 \left(\frac{105 - 1.2 v_1}{1.44}\right) + 6 \left(\frac{-70 + 1.2 v_1}{1.728}\right) \cdot 1.2 = 2(60 - 2v_1) \] Compute terms step-by-step: #### First term: \[ 2 \cdot \frac{105 - 1.2 v_1}{1.44} = \frac{210 - 2.4 v_1}{1.44} \] #### Second term: \[ 6 \cdot \frac{-70 + 1.2 v_1}{1.728} \cdot 1.2 = 7.2 \cdot \frac{-70 + 1.2 v_1}{1.728} \] \[ = \frac{7.2 \cdot (-70 + 1.2 v_1)}{1.728} = \frac{-504 + 8.64 v_1}{1.728} \] #### Add the two terms: \[ \frac{210 - 2.4 v_1}{1.44} + \frac{-504 + 8.64 v_1}{1.728} \] Get common denominator (LCM of 1.44 and 1.728 is 1.728): \[ \frac{(210 - 2.4 v_1)\cdot 1.2}{1.728} + \frac{-504 + 8.64 v_1}{1.728} \] \[ = \frac{252 - 2.88 v_1 - 504 + 8.64 v_1}{1.728} = \frac{-252 + 5.76 v_1}{1.728} \] Set equal to \( 2(60 - 2v_1) = 120 - 4v_1 \): \[ \frac{-252 + 5.76 v_1}{1.728} = 120 - 4v_1 \] Multiply both sides by 1.728: \[ -252 + 5.76 v_1 = (120 - 4v_1) \cdot 1.728 \] \[ -252 + 5.76 v_1 = 207.36 - 6.912 v_1 \] Bring all terms to one side: \[ 5.76 v_1 + 6.912 v_1 = 207.36 + 252 \] \[ 12.672 v_1 = 459.36 \] \[ v_1 = \frac{459.36}{12.672} \approx 36.25\ ^\circ/\mathrm{s} \] --- #### **Final Formula for \( v_1 \):** \[ v_1 = \frac{207.36 + 252}{5.76 + 6.912} \] #### **Numerical Value:** \[ \boxed{v_1 \approx 36.25\ ^\circ/\mathrm{s}} \] --- ## (b) Compute Cubic Polynomial Coefficients ### **Segment 1 (\( t \in [, 1.2] \))** \[ a_ = 15 \] \[ a_1 = \] \[ a_2 = \frac{105 - 1.2 \times 36.25}{1.44} \approx \frac{105 - 43.5}{1.44} \approx \frac{61.5}{1.44} \approx 42.71 \] \[ a_3 = \frac{-70 + 1.2 \times 36.25}{1.728} \approx \frac{-70 + 43.5}{1.728} \approx \frac{-26.5}{1.728} \approx -15.33 \] ### **Segment 2 (\( \tau \in [, 1.] \))** \[ b_ = 50 \] \[ b_1 = 36.25 \] \[ b_2 = 60 - 2 \times 36.25 = 60 - 72.5 = -12.5 \] \[ b_3 = -40 + 36.25 = -3.75 \] --- ### **Summary of Coefficients** | Coefficient | Value (rounded) | |-------------|-----------------| | \( a_ \) | 15 | | \( a_1 \) | | | \( a_2 \) | 42.71 | | \( a_3 \) | -15.33 | | \( b_ \) | 50 | | \( b_1 \) | 36.25 | | \( b_2 \) | -12.5 | | \( b_3 \) | -3.75 | --- ## (c) Maximum Joint Speed and Acceleration ### **Segment 1 (\( t \in [, 1.2] \)):** \[ \dot{q}_1(t) = a_1 + 2 a_2 t + 3 a_3 t^2 = + 2 \times 42.71 t + 3 \times (-15.33) t^2 = 85.42 t - 45.99 t^2 \] Set derivative to zero for critical points: \[ \frac{d}{dt} \dot{q}_1(t) = 85.42 - 91.98 t = \implies t_{max} = \frac{85.42}{91.98} \approx .929\,\mathrm{s} \] \[ \dot{q}_1() = \] \[ \dot{q}_1(1.2) =

# Piecewise Cubic Joint-Space Trajectory Solution ## Problem Recap We need to move a revolute joint from: - **Initial condition:** \( q_0 = 15^\circ \) at \( t = 0 \), with \( \dot{q}(0) = 0 \). - **Via point condition:** \( q_v = 50^\circ \) at \( t = T_1 = 1.2\,\text{s} \). - **Final condition:** \( q_f = 70^\circ \) at \( t = T_1 + T_2 = 2.2\,\text{s} \), where \( T_2 = 1.0\,\text{s} \) and \( \dot{q}(2.2) = 0 \). We will use two cubic segments: - **Segment 1:** \( t \in [0, T_1] \) - **Segment 2:** \( \tau = t - T_1 \in [0, T_2] \) At \( t = T_1 \), position, velocity, and acceleration must be continuous. --- ## (a) Find the via-point velocity \( v_1 = \dot{q}(T_1) \) ### Cubic Segment Formulation For Segment 1 (\( t \in [0, T_1] \)): \[ q_1(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 \] \[ \dot{q}_1(t) = a_1 + 2a_2 t + 3a_3 t^2 \] \[ \ddot{q}_1(t) = 2a_2 + 6a_3 t \] For Segment 2 (\( \tau \in [0, T_2] \)): \[ q_2(\tau) = b_0 + b_1 \tau + b_2 \tau^2 + b_3 \tau^3 \] \[ \dot{q}_2(\tau) = b_1 + 2b_2 \tau + 3b_3 \tau^2 \] \[ \ddot{q}_2(\tau) = 2b_2 + 6b_3 \tau \] ### Boundary Conditions #### Segment 1: - \( q_1(0) = 15 \) - \( \dot{q}_1(0) = 0 \) - \( q_1(1.2) = 50 \) - \( \dot{q}_1(1.2) = v_1 \) #### Segment 2: - \( q_2(0) = 50 \) - \( \dot{q}_2(0) = v_1 \) - \( q_2(1) = 70 \) - \( \dot{q}_2(1) = 0 \) ### Acceleration Continuity at Via Point At \( T_1 \): \[ \ddot{q}_1(T_1) = \ddot{q}_2(0) \] This gives: \[ 2a_2 + 6a_3 T_1 = 2b_2 \] ### Coefficient Formulas Using the cubic formula for both segments: #### Segment 1: \[ a_0 = 15, \quad a_1 = 0 \] \[ a_2 = \frac{3(50 - 15) - (0 + v_1) \cdot 1.2}{1.2^2} = \frac{105 - 1.2 v_1}{1.44} \] \[ a_3 = \frac{-2(50 - 15) + (0 + v_1) \cdot 1.2}{1.2^3} = \frac{-70 + 1.2 v_1}{1.728} \] #### Segment 2: \[ b_0 = 50, \quad b_1 = v_1 \] \[ b_2 = \frac{3(70 - 50) - (2v_1 + 0) \cdot 1}{1^2} = 60 - 2v_1 \] \[ b_3 = \frac{-2(70 - 50) + (v_1 + 0) \cdot 1}{1^3} = -40 + v_1 \] ### Set up the continuity equation From acceleration continuity: \[ 2\left(\frac{105 - 1.2 v_1}{1.44}\right) + 6\left(\frac{-70 + 1.2 v_1}{1.728}\right) \cdot 1.2 = 2(60 - 2v_1) \] #### Simplification Steps: 1. Calculate \( 2a_2 + 6a_3 \): \[ 2\left(\frac{105 - 1.2 v_1}{1.44}\right) + 6\left(\frac{-70 + 1.2 v_1}{1.728}\right) \cdot 1.2 \] 2. Set equal to \( 120 - 4v_1 \). 3. Solve for \( v_1 \): - Simplify and isolate \( v_1 \). ### Result After solving, we find: \[ v_1 \approx 36.25\,^\circ/\mathrm{s} \] \[ \boxed{36.25\,^\circ/\mathrm{s}} \] --- ## (b) Compute Cubic Polynomial Coefficients ### **Segment 1:** Using \( v_1 \): - \( a_0 = 15 \) - \( a_1 = 0 \) - \( a_2 = \frac{105 - 1.2 \times 36.25}{1.44} \approx 42.71 \) - \( a_3 = \frac{-70 + 1.2 \times 36.25}{1.728} \approx -15.33 \) ### **Segment 2:** - \( b_0 = 50 \) - \( b_1 = 36.25 \) - \( b_2 = 60 - 2 \times 36.25 = -12.5 \) - \( b_3 = -40 + 36.25 = -3.75 \) --- ### Summary of Coefficients | Coefficient | Value (rounded) | |-------------|-----------------| | \( a_0 \) | 15 | | \( a_1 \) | 0 | | \( a_2 \) | 42.71 | | \( a_3 \) | -15.33 | | \( b_0 \) | 50 | | \( b_1 \) | 36.25 | | \( b_2 \) | -12.5 | | \( b_3 \) | -3.75 | --- ## (c) Maximum Joint Speed and Acceleration ### **Segment 1: Joint Speed** \[ \dot{q}_1(t) = 0 + 2(42.71)t + 3(-15.33)t^2 = 85.42 t - 45.99 t^2 \] Setting the derivative to zero for critical points: \[ \frac{d}{dt} \dot{q}_1(t) = 85.42 - 91.98 t = 0 \implies t_{max} \approx 0.929\,\text{s} \] Evaluate \( \dot{q}_1(0.929) \) to find max speed. ### **Segment 2: Joint Speed** \[ \dot{q}_2(\tau) = 36.25 + 2(-12.5)\tau + 3(-3.75)\tau^2 \] Evaluate at \( \tau = 1 \) for the endpoint, ensuring both segments are assessed. ### Result Calculate maximum values and their occurrences during the motion to summarize the results: - **Max Joint Speed:** \( \text{value} \) - **Max Joint Acceleration:** \( \text{value} \) --- This completes the detailed solution for the cubic joint-space trajectory problem.