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# Modelling Workshop Biomechanics 1: Solution

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The Modelling Workshop Biomechanics 1 Solution of the Modelling Course of Industrial Design of the TU Delft

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### Modelling Workshop Biomechanics 1: Solution

1. 1. Workshop H-W-1: The gym Machine Analysis
2. 2. Modelling - Using sum(M) = 0 > > Ileg d mThigh\$L12 CmShank\$L32 CmFoot\$L52; 2 2 2 (3.1) Ileg := mThigh L1 CmShank L3 CmFoot L5 > equ dK Ileg\$diff theta t , t\$2 CmThigh\$g\$L1\$sin theta t CmShank\$g\$L3\$sin theta t CmFoot\$g\$L5\$sin theta t K mb\$g Cmb\$diff x t , t\$2 \$GL5 CMLeg = 0; d2 equ := K mThigh L12 CmShank L32 CmFoot L52 θ t CmThigh g L1 sin θ t (3.2) dt2 CmShank g L3 sin θ t CmFoot g L5 sin θ t K mb g Cmb CMLeg = 0 Modelling - Explore geometric relations d2 dt2 x t GL5
3. 3. x(t) & thets(a) relations > x d t/sqrt 2 GL1 CL4\$sin theta t GL1 CL4 sin θ t x := t/ 2 C GL2KL4\$cos theta t 2 ; 2 C GL2 KL4 cos θ t (4.1) Explore the length of GL5 > GL5 d sqrt GL12 CGL22 \$sin alpha t ; GL5 := GL12 CGL22 sin α t (4.2) Explore alpha(t) arctan(GL4/GL3) GL1 CL4\$sin theta t > beta d t/arctan ; GL2KL4\$cos theta t GL1 CL4 sin θ t β := t/arctan GL2 KL4 cos θ t > delta d t/arctan GL1 GL2 GL1 GL2 (4.4) α := t/β t Kδ t (4.5) δ := t/arctan > alpha d t/beta t Kdelta t ; > GL5 GL1 CL4 sin θ t GL2 KL4 cos θ t K GL12 CGL22 sin K arctan > equ d2 K mThigh L12 CmShank L32 CmFoot L52 CmShank g L3 sin θ t CL4 sin θ t KL4 cos θ t L4 cos θ t d θ t dt L4 sin θ t d θ t dt C GL2 KL4 cos θ t 1 C 2 GL1 GL2 (4.6) CmThigh g L1 sin θ t θ t dt2 CmFoot g L5 sin θ t 2 3/2 (4.3) (4.7) Carctan 1 4 C mb g Cmb K 2 GL1 C2 GL2 2 2 GL1 CL4 sin θ t 2 L4 cos θ t 2 d θ t dt 2 2 K2 GL1
4. 4. L4 sin θ t d θ t dt CL4 sin θ t L4 cos θ t KL4 cos θ t L4 cos θ t d θ t dt KL4 cos θ t L4 sin θ t 2 d2 θ t dt2 d2 θ t dt2 CL4 sin θ t GL1 CL4 sin θ t K arctan 2 C2 GL1 2 C2 L4 sin θ t Carctan 2 C2 GL2 C2 GL2 GL1 GL2 > Parameters Body segment & Antropometric data mass & g value > m d 60; g d 9.81; d θ t dt 2 C GL2 KL4 cos θ t GL1 CL4 sin θ t GL2 KL4 cos θ t 2 2 GL12 CGL22 sin CMLeg = 0
5. 5. m := 60 g := 9.81 mass of each parts of the leg > mFoot d m\$0.0145; > mShank d m\$0.0465 > mThigh d m\$0.1 (5.1.1) mFoot := 0.8700 (5.1.2) mShank := 2.7900 (5.1.3) mThigh := 6.0 (5.1.4) L1 := 0.1781362 (5.1.5) L2 := 0.4114 (5.1.6) L3 := 0.5851196 (5.1.7) L4 := 0.8126 (5.1.8) L5 := 0.8701 (5.1.9) GL1 := 1.2 (5.2.1) GL2 := 1.5 (5.2.2) mb := 5 (5.2.3) MLeg := 35 (5.3.1) > Length of different parts of the leg > > L1 d 0.4114\$0.433 > L2 d 0.4114; > L3 d L2 C0.4012\$0.433 > L4 d L2 C0.4012 > L5 d L4 C0.115\$0.5 > Gym machine parameters Length of the machine > GL1 d 1.2; > GL2 d 1.5; > mb d 5; Input The torque applied on hip hoints > MLeg d 35;
6. 6. Question 1: Simulation Review equation > equ; K 1.804247624 K 5 4 d2 θ t dt2 C33.92580414 sin θ t 1.6252 1.2 C0.8126 sin θ t K0.8126 cos θ t sin θ t C 1.5 K0.8126 cos θ t 2 3/2 K1.6252 1.2 C0.8126 sin θ t C0.8126 sin θ t cos θ t C1.6252 1.5 K0.8126 cos θ t K0.8126 cos θ t sin θ t 1.2 C0.8126 sin θ t sin arctan 2 d θ t dt cos θ t 2 d θ t dt 5 2 C K1.920937271 49.05 2 d θ t dt2 d θ t dt 1.5 K0.8126 cos θ t 2 cos θ t 2 d θ t dt d θ t dt 2 2 C1.6252 1.2 C1.32063752 sin θ t 2 d θ t dt 2 2 C1.6252 1.5 d2 θ t dt2 C 1.5 K0.8126 cos θ t 1.2 C0.8126 sin θ t C1.6252 1.5 1.2 C0.8126 sin θ t 1.32063752 cos θ t sin θ t (6.1) 2 K0.6747409422 C35 = 0 > ics d theta 0 = 0, D theta 0 = 0; ics := θ 0 = 0, D θ 0 =0 equ, ics , theta t , numeric, output = listprocedure, maxfun = 100000 d res := t = proc t ... end proc, θ t = proc t ... end proc, θ t = proc t dt ... end proc (6.2) > res d dsolve > angleTheta d rhs res 2 ; angularVelocity d rhs res 3 ; angularAcceleration d t1 /fdiff angularVelocity t , t = t1 ; (6.3)
7. 7. angleTheta := proc t ... end proc angularVelocity := proc t ... end proc angularAcceleration := t1/fdiff angularVelocity t , t = t1 3.1415926 > timeTo90Degree d fsolve angleTheta t = , t = 1 ..3 ; 2 timeTo90Degree := 2.602802873 > p00 d plot 90, t = 0 ..timeTo90Degree, color = yellow ; p00 := PLOT ... > angleTheta t \$180 > p01 d plot , t = 0 ..timeTo90Degree, color = blue ; 3.14 p01 := PLOT ... angularVelocity t \$180 > p02 d plot , t = 0 ..timeTo90Degree, color = green ; 3.14 p02 := PLOT ... angularAcceleration\$180 , 0 ..timeTo90Degree, color = red ; 3.14 p03 := PLOT ... > display p00, p01, p02, p03 (6.4) (6.5) (6.6) (6.7) (6.8) > p03 d plot (6.9)
8. 8. 1200 1000 800 600 400 200 0 0 0.5 1 1.5 2 2.5 t After 90 degrees, the model is not valid anymore since angle conventions Green line congratulation, you finished the workshop, howeve, question 2 will be very helpful Question 2: Model > consider equ 3.7 If we know angular acceleration, angular velocity and angle, then the moment generated by human is (copy the equation, put and -1 infront)
9. 9. 2 2 d2 θ t dt2 2 > MomentLeg1 dK K mShank L1 CmThigh L3 CmFoot L5 1\$ CmShank g L1 sin θ t 1 4 C mb g Cmb K CmThigh g L3 sin θ t 2 GL1 CL4 sin θ t C2 GL2 KL4 cos θ t CL4 sin θ t C 1 2 2 L4 sin θ t d θ t dt CL4 sin θ t L4 cos θ t d2 θ t dt2 KL4 cos θ t GL1 CL4 sin θ t K arctan L4 cos θ t L4 sin θ t 2 2 3/2 d2 dt2 2 C2 GL1 C2 L42 sin θ t d θ t dt 2 2 2 d θ t dt C2 GL2 θ t C GL2 KL4 cos θ t GL1 CL4 sin θ t GL2 KL4 cos θ t GL1 K2 GL1 CL4 sin θ t C2 GL2 KL4 cos θ t 2 2 d θ t dt 2 d θ t dt L4 cos θ t d θ t dt L4 sin θ t C GL2 KL4 cos θ t 2 L42 cos θ t CmFoot g L5 sin θ t Carctan GL1 GL2 2 2 2 GL1 CGL2 sin : > Introduce angular acceleration, angular velocity and angle as parameters, to avoid Maple confusion, we do it from angular acceletation First, we replace all diff(theta(t), t\$2)) as a parameter angularAcceleration > MomentLeg1 d subs diff theta t , t\$2 = angularAcceleration, MomentLeg1 : Then we replace all diff(theta(t), t) as a parameter angularVelocity
10. 10. > MomentLeg1 d subs diff theta t , t = angularVelocity, MomentLeg1 : Finally, we replace all theta(t) as angle > MomentLeg1 d subs theta t = angle, MomentLeg1 : Check > MomentLeg1 2.801377758 angularAcceleration K46.74175211 sin angle C1.920937271 49.05 K 5 4 1.6252 1.2 C0.8126 sin angle C1.6252 1.5 K0.8126 cos angle 2 1.2 C0.8126 sin angle C 5 2 cos angle angularVelocity sin angle angularVelocity 2 2 2 3/2 C 1.5 K0.8126 cos angle 1.32063752 cos angle (8.1.1) angularVelocity2 K1.6252 1.2 C0.8126 sin angle sin angle angularVelocity2 C1.6252 1.2 C0.8126 sin angle cos angle angularAcceleration C1.32063752 sin angle 2 angularVelocity2 C1.6252 1.5 K0.8126 cos angle cos angle angularVelocity2 C1.6252 1.5 K0.8126 cos angle sin angle angularAcceleration 1.2 C0.8126 sin angle 2 C 1.5 K0.8126 cos angle 1.2 C0.8126 sin angle sin arctan K0.6747409422 1.5 K0.8126 cos angle 2 Now we create a function of MomentLeg1, which has three parameters (angularAcceleration, angularVelocity, angle) > MomentLeg1 d unapply MomentLeg1, angularAcceleration, angularVelocity, angle : Input parameters number of points > n d 25; Re-conformed data, not necessary n := 25 (8.2.1)
11. 11. Time > Time d MeasurementTime : Angle > an d MeasurementAngle : Velocity > aV d MeasurementAngularVelocity : Acceleration > aA d MeasurementAngularAcceleration : Solution Model > for i from 1 to n do M i d MomentLeg1 aA i , aV i , an i : end: > In the plot, we display angle in degree > p11 d plot seq Time i , M i , i = 1 ..n , color = red, style = line : an i \$180 > p12 d plot seq Time i , , i = 1 ..n , color = cyan, style = line : 3.14 aV i \$180 > p13 d plot seq Time i , , i = 1 ..n , color = green, style = line : 3.14 aA i \$180 > p14 d plot seq Time i , , i = 1 ..n , color = blue, style = line : 3.14 > display p11, p12, p13, p14 ;
12. 12. 150 100 50 0 K 50 K 100 K 150 > 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0