Workshop H-W-1: The gym Machine

Analysis

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(3.2)(3.2)

Modelling - Using sum(M) = 0

IlegdmThigh$L12CmShank$L32CmFoot$L52;Ileg := mThigh L12CmShank L32CmFoot L52

equdKIleg$diff theta t , t$2 CmThigh$g$L1$sin theta t CmShank$g$L3$sin theta tCmFoot$g$L5$sin theta t Kmb$gCmb$diff x t , t$2 $GL5CMLeg= 0;

equ := K mThigh L12CmShank L32CmFoot L52 d2

dt2 θ t CmThigh g L1 sin θ t

CmShank g L3 sin θ t CmFoot g L5 sin θ t K mb gCmb d2

dt2 x t GL5

CMLeg= 0

Modelling - Explore geometric relations

(4.7)(4.7)

(4.2)(4.2)

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(4.5)(4.5)

(4.3)(4.3)

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(4.4)(4.4)

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(4.6)(4.6)

(4.1)(4.1)

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x(t) & thets(a) relations

xd t/sqrt GL1CL4$sin theta t 2C GL2KL4$cos theta t 2 ;

x := t/ GL1CL4 sin θ t2C GL2KL4 cos θ t

2

Explore the length of GL5

GL5d sqrt GL12CGL22 $sin alpha t ;

GL5 := GL12CGL22 sin α t

Explore alpha(t) arctan(GL4/GL3)

betad t/arctanGL1CL4$sin theta tGL2KL4$cos theta t

;

β := t/arctanGL1CL4 sin θ tGL2KL4 cos θ t

delta d t/arctanGL1GL2

δ := t/arctanGL1GL2

alphad t/beta t Kdelta t ;α := t/β t Kδ t

GL5

K GL12CGL22 sin KarctanGL1CL4 sin θ tGL2KL4 cos θ t

CarctanGL1GL2

equ

K mThigh L12CmShank L32CmFoot L52 d2

dt2 θ t CmThigh g L1 sin θ t

CmShank g L3 sin θ t CmFoot g L5 sin θ t C mb gCmb K14

2 GL1

CL4 sin θ t L4 cos θ t ddt

θ t C2 GL2

KL4 cos θ t L4 sin θ t ddt

θ t2

GL1CL4 sin θ t2

C GL2KL4 cos θ t2 3/2

C12

2 L42 cos θ t2

ddt

θ t2

K2 GL1

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CL4 sin θ t L4 sin θ t ddt

θ t2

C2 GL1

CL4 sin θ t L4 cos θ t d2

dt2 θ t C2 L42 sin θ t

2

ddt

θ t2

C2 GL2

KL4 cos θ t L4 cos θ t ddt

θ t2

C2 GL2

KL4 cos θ t L4 sin θ t d2

dt2 θ t

GL1CL4 sin θ t2C GL2KL4 cos θ t

2 GL12CGL22 sin

KarctanGL1CL4 sin θ tGL2KL4 cos θ t

CarctanGL1GL2

CMLeg= 0

Parameters

Body segment & Antropometric data

mass & g value md 60; gd 9.81;

(5.2.3)(5.2.3)

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(5.3.1)(5.3.1)

(5.1.6)(5.1.6)

(5.1.4)(5.1.4)

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(5.1.9)(5.1.9)

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(5.1.2)(5.1.2)

(5.2.2)(5.2.2)

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(5.1.1)(5.1.1)

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(5.1.5)(5.1.5)

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(5.1.3)(5.1.3)

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(5.1.8)(5.1.8)

(5.1.7)(5.1.7)

m := 60g := 9.81

mass of each parts of the legmFootdm$0.0145;

mFoot := 0.8700

mShankdm$0.0465mShank := 2.7900

mThighdm$0.1mThigh := 6.0

Length of different parts of the leg

L1d 0.4114$0.433L1 := 0.1781362

L2d 0.4114;L2 := 0.4114

L3d L2C0.4012$0.433L3 := 0.5851196

L4d L2C0.4012L4 := 0.8126

L5d L4C0.115$0.5L5 := 0.8701

Gym machine parametersLength of the machine

GL1d 1.2;GL1 := 1.2

GL2d 1.5;GL2 := 1.5

mbd 5;mb := 5

InputThe torque applied on hip hoints

MLeg d 35;MLeg := 35

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(6.3)(6.3)

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(6.1)(6.1)

Question 1: Simulation

Review equation

equ;

K1.804247624 d2

dt2 θ t C33.92580414 sin θ t K1.920937271 49.05

K54

1.6252 1.2C0.8126 sin θ t cos θ t ddt

θ t C1.6252 1.5

K0.8126 cos θ t sin θ t ddt

θ t2

1.2C0.8126 sin θ t2

C 1.5K0.8126 cos θ t2 3/2

C52

1.32063752 cos θ t2

ddt

θ t2

K1.6252 1.2C0.8126 sin θ t sin θ t ddt

θ t2

C1.6252 1.2

C0.8126 sin θ t cos θ t d2

dt2 θ t C1.32063752 sin θ t

2

ddt

θ t2

C1.6252 1.5K0.8126 cos θ t cos θ t ddt

θ t2

C1.6252 1.5

K0.8126 cos θ t sin θ t d2

dt2 θ t

1.2C0.8126 sin θ t2C 1.5K0.8126 cos θ t

2

sin arctan1.2C0.8126 sin θ t1.5K0.8126 cos θ t

K0.6747409422 C35 = 0

icsd theta 0 = 0, D theta 0 = 0;ics := θ 0 = 0, D θ 0 = 0

resd dsolve equ, ics , theta t , numeric, output = listprocedure, maxfun = 100000

res := t = proc t ... end proc, θ t = proc t ... end proc,ddt

θ t = proc t

...

end proc

angleThetad rhs res 2 ; angularVelocityd rhs res 3 ; angularAccelerationd t1/fdiff angularVelocity t , t = t1 ;

(6.8)(6.8)

(6.4)(6.4)

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(6.7)(6.7)

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(6.9)(6.9)

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(6.5)(6.5)

angleTheta := proc t ... end proc

angularVelocity := proc t ... end proc

angularAcceleration := t1/fdiff angularVelocity t , t = t1

timeTo90Degreed fsolve angleTheta t =3.1415926

2, t = 1 ..3 ;

timeTo90Degree := 2.602802873

p00d plot 90, t = 0 ..timeTo90Degree, color = yellow ;p00 := PLOT ...

p01d plotangleTheta t $180

3.14, t = 0 ..timeTo90Degree, color = blue ;

p01 := PLOT ...

p02d plotangularVelocity t $180

3.14, t = 0 ..timeTo90Degree, color = green ;

p02 := PLOT ...

p03d plotangularAcceleration$180

3.14, 0 ..timeTo90Degree, color = red ;

p03 := PLOT ...

display p00, p01, p02, p03

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t0 0.5 1 1.5 2 2.5

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After 90 degrees, the model is not valid anymore since angle conventions

Green linecongratulation, 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)

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MomentLeg1dK1$ K mShank L12CmThigh L32CmFoot L52 d2

dt2 θ t

CmShank g L1 sin θ t CmThigh g L3 sin θ t CmFoot g L5 sin θ t

C mb gCmb K14

2 GL1CL4 sin θ t L4 cos θ t ddt

θ t

C2 GL2KL4 cos θ t L4 sin θ t ddt

θ t2

GL1

CL4 sin θ t2C GL2KL4 cos θ t

2 3/2

C12

2 L42 cos θ t2

ddt

θ t2

K2 GL1

CL4 sin θ t L4 sin θ t ddt

θ t2

C2 GL1

CL4 sin θ t L4 cos θ t d2

dt2 θ t C2 L42 sin θ t

2

ddt

θ t2

C2 GL2KL4 cos θ t L4 cos θ t ddt

θ t2

C2 GL2

KL4 cos θ t L4 sin θ t d2

dt2 θ t

GL1CL4 sin θ t2C GL2KL4 cos θ t

2 GL12CGL22 sin

KarctanGL1CL4 sin θ tGL2KL4 cos θ t

CarctanGL1GL2

:

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

MomentLeg1d subs diff theta t , t$2 = angularAcceleration, MomentLeg1 :

Then we replace all diff(theta(t), t) as a parameter angularVelocity

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(8.1.1)(8.1.1)

MomentLeg1d subs diff theta t , t = angularVelocity, MomentLeg1 :

Finally, we replace all theta(t) as angle

MomentLeg1d subs theta t = angle, MomentLeg1 :

CheckMomentLeg1

2.801377758 angularAccelerationK46.74175211 sin angle C1.920937271 49.05

K54

1.6252 1.2C0.8126 sin angle cos angle angularVelocity

C1.6252 1.5K0.8126 cos angle sin angle angularVelocity 2

1.2C0.8126 sin angle 2C 1.5K0.8126 cos angle 2 3/2

C52

1.32063752 cos angle 2 angularVelocity2K1.6252 1.2

C0.8126 sin angle sin angle angularVelocity2C1.6252 1.2

C0.8126 sin angle cos angle angularAcceleration

C1.32063752 sin angle 2 angularVelocity2C1.6252 1.5

K0.8126 cos angle cos angle angularVelocity2C1.6252 1.5

K0.8126 cos angle sin angle angularAcceleration

1.2C0.8126 sin angle 2C 1.5K0.8126 cos angle 2

sin arctan1.2C0.8126 sin angle1.5K0.8126 cos angle

K0.6747409422

Now we create a function of MomentLeg1, which has three parameters (angularAcceleration,angularVelocity, angle)

MomentLeg1d unapply MomentLeg1, angularAcceleration, angularVelocity, angle :

Input parametersnumber of points

nd 25;n := 25

Re-conformed data, not necessary

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TimeTimedMeasurementTime :

AngleandMeasurementAngle :

VelocityaVdMeasurementAngularVelocity :

AccelerationaAd MeasurementAngularAcceleration :

SolutionModel

for i from 1 to n do M i dMomentLeg1 aA i , aV i , an i : end:

In the plot, we display angle in degreep11d plot seq Time i , M i , i = 1 ..n , color = red, style= line :

p12d plot seq Time i , an i $180

3.14, i = 1 ..n , color = cyan, style = line :

p13d plot seq Time i ,aV i $180

3.14, i = 1 ..n , color = green, style= line :

p14d plot seq Time i ,aA i $180

3.14, i = 1 ..n , color = blue, style= line :

display p11, p12, p13, p14 ;

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1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

K150

K100

K50

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