Innovative Systems Design and Engineering ISSN 2222-1727 (Paper) ISSN 2222-2871 (Online) Vol.4, No.10, 2013

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Kinematics of Lower Limb Segments during Cycling Session *

AJAYI, A. B., DAMISA, O., AJIBOYE, J. S.

l1 [cosθ1 + j sin θ1 ] + l 2 [cosθ 2 − j sin θ 2 ] + l3 [cosθ 3 + j sin θ 3 ] − l 4 [cosθ 4 − j sin θ 4 ] + l 5 [cos π / 2 + j sin π / 2] + l 6 [cos π + j sin π ] = 0

(1)

The vector loop equation 1 can be written for the real and imaginary axes as

l1 cosθ1 + l 2 cosθ 2 + l3 cosθ 3 − l 4 cosθ 4 + l 5 cos π / 2 + l 6 cos π = 0

(2a)

l1 sin θ1 − l 2 sin θ 2 + l 3 sin θ 3 + l 4 sin θ 4 + l 5 sin π / 2 + l 6 sin π = 0

(2b)

59

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b2 O

A

b1 l1

θ1

θ2

B

Knee

l6

l2

Ankle

l3 θ3 Crank Spindle

D

l5

l4 C

θ4

Pedal Spindle

Figure 1: A model of a lower limb segments, pedal crank and bicycle frame 2.2 Derivation of Thigh and Knee Angles Fig 2: shows the resolutions of the leg segments in the real and imaginary axis as well as the position analysis of the leg segments and the space analysis of the thigh and knee angles. From triangle ABF in fig 2, it can be found that

tan θ 2 =

l 4 Sinθ 4 − l 3 Sinθ 3 − l1 Sinθ1 + l5 l 4 Cosθ 4 − l 3Cosθ 3 − l1Cosθ1 + l 6

(3)

 l 4 Sinθ 4 − l3 Sinθ 3 − l1 Sinθ1 + l 5    l 4 Cosθ 4 − l 3Cosθ 3 − l1Cosθ1 + l 6 

(4)

θ 2 = tan −1 

The expression for the thigh angle θ1 can also be gotten from triangle OAH in fig 2 as shown below: Using Pythagoras’ theorem,

l 22 = [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) − l1Cosθ 1 ] 2 + [(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) − l1 Sinθ 1 ] 2 (5)

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b2 O

Hip

l1Sinθ1

b1 θ1

l1 θ2

l1Cosθ1

A

H Knee

l6

(l 6 + l 4 cosθ 4 − l 3 cosθ 3 − l1 cosθ1 )

l2 Ankle

(l 5 + l 4 sin θ 4 − l 3 sin θ 3 − l1 sin θ1 ) B

G

F

l3Cosθ3

l3

E

l3Cosθ3 θ3 C Crank l4 Spindle l5 D

Pedal Spindle

l4Sinθ4

θ4

l4Cosθ4

Figure 2: Space Analysis of thigh, θ1 and knee, θ2 angles

(l + l Cosθ 4 − l3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12  l 22 =  6 4  − 2[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )(l1Cosθ1 )] − 2[(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )(l1 Sinθ1 )]

{2[(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )(l1 Sinθ1 )] + 2[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )(l1Cosθ1 )]}

{

= (l 6 + l 4 Cosθ 4 − l3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l3 Sinθ 3 ) 2 + l12 − l 22

}

(6)

(7)

If we let

A = (l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 − l 22

(8)

B = 2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1

(9)

C = 2(l 6 + l 4 Cosθ 4 − l3 Cosθ 3 )l1

(10)

Then, equation (7) becomes

CSinθ + BCosθ = A

(11)

Using trigonometry identity and let tan θ / 2 = t

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Sin

θ

Cos

2

θ 2

=

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t

(12)

1+ t2

=

1

(13)

1+ t2

θ θ t 1 2t Sinθ = 2Sin Cos = 2 ⋅ ⋅ = 2 2 2 1+ t2 1+ t2 1+ t

(14)

2

θ

 1  2 2 −1− t 2 1− t 2 Cosθ = 2Cos − 1 = 2 −1 = =  −1 = 2 2 1+ t2 1+ t2 1+ t2  1+ t  2

(15)

Recall equation (11)

CSinθ + BCosθ = A

(11)

Substituting equations (14) and (15) into equation (11), we have

1 − t 2   2t  C + B =A  2  2  1 + t  1 + t 

(12)

Multiplying through by 1+t2, we have

C (2t ) + B(1 − t 2 ) = A(1 + t 2 )

(13)

2Ct + B − Bt 2 = A + At 2

(14)

A + At 2 − 2Ct − B + Bt 2 = 0

(15)

( A + B)t 2 − 2Ct + ( A − B) = 0

(16)

Recall equations (8), (9) and (10)

A = (l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 − l 22

(8)

B = 2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1

(9)

C = 2(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )l1

(10)

Then equation (16) becomes

[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2  2   2 2 t    + l1 − l 2 ] + [2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1 ]   − 2[2(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )l1 ]t =0  2 2  + [(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 ) + (l5 + l 4 Sinθ 4 − l 3 Sinθ 3 )     + l 2 − l 2 ] − [2(l + l Sinθ − l Sinθ )l ]   2 5 4 4 3 3 1   1 Equation (17) is a quadratic equation of the form

62

(17)

ax + bx + c = 0 where 2

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[(l + l Cosθ 4 − l 3Cosθ 3 ) 2 + (l5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 − l 22 ] a= 6 4  + [2(l 5 + l 4 Sinθ 4 − l3 Sinθ 3 )l1 ] 

(18)

b = −2[2(l 6 + l 4 Cosθ 4 − l3 Cosθ 3 )l1 ] ,

(19)

[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 ) 2 + (l5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 − l 22 ] c=  − [2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1 ] 

(20)

and

The solution of equation (17) is then,

         2(2(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )l1 )  0.5     (− 2[ 2(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )l1 ])2      [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) 2     2      [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )  + (l + l Sinθ − l Sinθ ) 2   t = ±   2 2 2 5 4 4 3 3    − 4 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) + l1 − l 2 ]  2 2  + l1 − l 2 ] −      + [2(l + l Sinθ − l Sinθ )l ]  5 4 4 3 3 1  [2(l + l Sinθ − l Sinθ )l ]      5 4 4 3 3 1      −1   2     [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )    2 + (l + l Sinθ − l Sinθ ) 2 + l 2 − l 2 ]  5 4 4 3 3 1 2         + [2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1 ]    Recall

t = tan

(21)

θ1 2

2(2(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )l1 )   0.5     (− 2[2(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )l1 ])2     2   θ θ [( ) l l Cos l Cos + −     2 4 3 3     6 4 ±   [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) 2   − 4 + (l + l Sinθ − l Sinθ ) 2 + l 2 − l 2 ]  + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )       5 4 4 3 3 1 2    + l12 − l 22 ] − θ1 = 2 tan −1    + [2(l + l Sinθ − l Sinθ )l ]     5 4 4 3 3 1  [2(l + l Sinθ − l Sinθ )l ]      5 4 4 3 3 1      − 1   2    [(l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 )    2 + (l + l Sinθ − l Sinθ ) 2 + l 2 − l 2 ]   5 4 4 3 3 1 2        + [ 2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1 ]     (23)

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2.3 Derivation of the Foot Angle The foot angle is θ3 in figure 2 above, looking at triangle BCD, if a perpendicular is drawn from point C to line BD, and called h, as shown in figure 3 below. BC is assumed to be the foot segment while CD is the crank arm,

B

θ3 h C

θ4

D

Figure 3: Analysis of foot angle

h = l 3 sin θ 3 = −l 4 sin θ 4

(24)

l 3 sin θ 3 = −l 4 sin θ 4

(25)

 − l 4 sin θ 4 l3 

θ 3 = sin −1 

  

(26)

2.4 Velocity Analysis The velocity derivatives of the thigh and shank segments can be gotten by taking the first time-derivatives of (1) above we obtain the angular velocities

l1 [cos θ1 + j sin θ 1 ] + l 2 [cos θ 2 − j sin θ 2 ]  d   + l 3 [cos θ 3 + j sin θ 3 ] − l 4 [cos θ 4 − j sin θ 4 ]  = 0 dt   + l 5 [cos π / 2 + j sin π / 2] + l 6 [cos π + j sin π ]

(27)

Equation (27) is

l1 ⋅ θ& [ j cos θ1 − sin θ1 ] + l 2 ⋅ θ&2 [− j cos θ 2 − sin θ 2 ]   =0 + l3 ⋅ θ&3 [ j cos θ 3 − sin θ 3 ] − l 4 ⋅ θ&4 [ j cos θ 4 − sin θ 4 ] Separating equation (28) into real and imaginary parts and equating to zero Real Part:

64

(28)

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l1 [ − sin θ 1 ]θ&1 + l 2 [ − sin θ 2 ]θ&2 + l 3 [ − sin θ 3 ]θ&3 − l 4 [ − sin θ 4 ]θ&4 = 0

(29)

{

}

(30)

i l1 [cos θ ]θ& + l 2 [ − cos θ1 ]θ&1 + l 3 [cos θ 2 ]θ&2 − l 4 [ − cos θ 3 ]θ&3 = 0

]

(31)

l1 [cos θ 1 ]θ&1 + l 2 [ − cos θ 2 ]θ&2 + l3 [cos θ 3 ]θ&3 − l 4 [ − cos θ 4 ]θ&4 = 0

(32)

{

(33)

l1 [ − sin θ 1 ]θ&1 + l 2 [ − sin θ 2 ]θ&2 + − l 3 [sin θ 3 ]θ&3 + l 4 [sin θ 4 ]θ&4 = 0 Imaginary part:

[

Since i ≠ 0 , then

Therefore,

}

l1 [cos θ 1 ]θ&1 + l 2 [ − cos θ 2 ]θ&2 + l 4 [cos θ 4 ]θ&4 + l3 [cos θ 3 ]θ&3 = 0 Equations (31) and (33) can be set as simultaneous equations, and then solved for

θ&1 , andθ&2

using second order determinant and simplify,

θ&1 =

l 4 l 2θ&4 sin (θ 2 + θ 4 ) + l 3l 2θ&3 sin (θ 2 − θ 3 ) l1l 2 sin (θ1 + θ 2 )

(34)

θ&2 =

l1l 4θ&4 sin (θ1 + θ 4 ) + l1l 3θ&3 sin (θ1 − θ 3 ) l1l 2 sin (θ1 + θ 2 )

(35)

The velocity of the foot angle is gotten by taken the time derivatives of the foot angle thus

 − l sin θ 4 d d  (θ 3 ) = sin −1  4 dt dt  l3 

  

− l 4θ&4 cos θ 4

θ&3 = l3

2  l  1 −  4 sin θ   4   l 3   

(36)

(37)

2.5. Acceleration Analysis Taking the second time-derivates of loop equation (1), we have:

d2 dt 2

l1 [cos θ1 + j sin θ1 ] + l 2 [cosθ 2 − j sin θ 2 ] + l 3 [cos θ 3 + j sin θ 3 ]    − l 4 [cos θ 4 − j sin θ 4 ] + l 5 [cos π / 2 + j sin π / 2] + l 6 [cos π + j sin π ]

l1 [−(θ&&1 sin θ1 + θ&12 cos θ1 ) + j (θ&&1 cos θ1 − θ&12 sin θ1 )]    2 2 − l 2 [(θ&&2 sin θ 2 + θ&2 cos θ 2 ) + j (−θ&2 sin θ 2 + θ&&2 cos θ 2 )]  =  && &2 &2 && + l3 [−(θ 3 sin θ 3 + θ 3 cos θ 3 ) + j (−θ 3 sin θ 3 + θ 3 cos θ 3 )] + l [(θ&& sin θ + θ& 2 cos θ ) − j (−θ& 2 sin θ + θ&& cos θ )]  4 4 4 4 4 4 4  4 4  65

(38)

(39)

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Separating equation (39) into real and imaginary parts, we have Real part:

l1 (− sin θ1 )θ&&1 − l1 (θ&12 cos θ1 ) + l 2 (− sin θ 2 )θ&&2 − l 2θ&22 cos θ 2 ]  =0 + l3 (− sin θ 3 )θ&&3 − l 3θ&32 cos θ 3 + l 4θ&&4 sin θ 4 + l 4θ&42 cos θ 4 )]  − l1θ&12 cos θ1 − l 2 (θ&22 cos θ 2 ) − l3θ&&3 sin θ 3  l1 (− sin θ1 )θ&&1 + l 2 (− sin θ 2 )θ&&2 +  =0 − l 3θ&32 cos θ 3 + l 4θ&&4 sin θ 4 + l 4θ&42 cos θ 4 

(40)

(41)

Imaginary part:

l1[(θ&&1 cos θ1 − θ&12 sin θ1 )] − l 2 [(−θ&22 sin θ 2 + θ&&2 cos θ 2 )]  i =0 + l 3 [(−θ&32 sin θ 3 + θ&&3 cos θ 3 )] − l 4 [(−θ&42 sin θ 4 + θ&&4 cos θ 4 )]

(42)

Since i ≠ 0 , then

l1 (cos θ1 )θ&&1 − l1θ&12 sin θ1 − l 2 (cos θ 2 )θ&&2 + l 2θ&22 sin θ 2   =0 − l 3θ&32 sin θ 3 + l3θ&&3 cos θ 3 + l 4θ&42 sin θ 4 − l 4θ&&4 cos θ 4 

(43)

− l1θ&12 sin θ1 + l 2θ&22 sin θ 2 − l 3θ&32 sin θ 3  & & & & l1 (cos θ1 )θ1 − l 2 (cos θ 2 )θ 2 +  =0 + l 3θ&&3 cos θ 3 + l 4θ&42 sin θ 4 − l 4θ&&4 cos θ 4 

(44)

The angular accelerations of the thigh and shank can be determined from the simultaneous equation (32) and equation (33), using second order determinant and simplify,

θ&&1 =

θ&&2 =

2l 22θ&22 cos 2 θ 2 + l1l 2θ&12 cos(θ1 − θ 2 ) + l 3l 2θ&32 cos(θ 3 − θ 2 ) −    l 4 l 2θ&42 cos(θ 2 − θ 4 ) + l 3l 2θ&&3 sin (θ 3 − θ 2 ) + l 4 l 2θ&&4 sin (θ 2 − θ 4 ) l1l 2 sin (θ1 + θ 2 )

− l12θ&22 − l1l 2θ&22 cos(θ1 + θ 2 ) − l3 l1θ&32 cos(θ1 − θ 3 ) +   &2  l1l 4θ 4 cos(θ1 − θ 4 ) + l 3l1θ&&3 sin (θ1 − θ 3 ) − l 4 l1θ&&4 sin (θ1 + θ 4 ) l1l 2 sin (θ1 + θ 2 )

(45)

(46)

The acceleration of the foot is gotten by taking the time derivatives of the angular velocity expression equation 38 above, assuming the differentiation of a quotient of two functions:

   d & d  θ3 =  dt dt   l3  

   − l 4θ&4 cos θ 4   2  l   1 −  4 sin θ    4   l 3    

( )

If

[

u = −l 4θ&4 cosθ 4 , u& = l 4 θ&4 sin θ 4 − θ&&4 cosθ 4

] 66

(47)

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and

v = l3

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2 2  l    1 −  4 sin θ   , v& = l θ& cos θ 1 −  l 4 sin θ   4 4 4 4 4    l 3   l 3     

− 12

( )

d & u vu& − uv& θ 3 = θ&&3 = = dt v v2   l 3  & & θ3 = 

(48)

 2  l   4 1 −  sin θ   l θ& sin θ − θ&& cos θ +  l θ& cos θ  4 4 4 4 4 4 4   l3  4 4      2  l   2 4 l 3 1 −  sin θ 4    l3   

([

]) (

2  l   2 4   1 −  sin θ 4     l3  

)

−1

2

    

(50) 3. Results and Discussions For the studies of these segments, the values in Table 1, below is used. The profile of crank angles and the leg segments angles is shown in fig 4, below. These plots were obtained from our derivations for thigh, shank and foot angles in equations 4, 23 and 26 respectively.

Table 1: Parameters used for simulation S/N DESCRIPTION

SYMBOL

VALUES USED

1

Length of thigh bone

l1

0.396m

2

Length of shank bone

l2

0.435m

3

Length of foot bone

l3

0.213m

4

Crank arm length

l4

0.170m

5

Hip to crank axis (horizontal)

l5

0.212m

6

Hip to crank axis (vertical)

l6

0.673m

7

Angle between the vertical and thigh bone

θ1

870

8

Angle between the vertical and the shank bone

θ2

1570

9

Angle between the vertical axis and the foot bone

θ3

390

10

Angle of the crank arm

θ4

450

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Figs. 4 -6 are the profile of the segments angles versus the crank angles: The foot angle profile in Fig. 4 is sinusoidal which agrees with experimental observation of Hull and Jorge [6]. The profile of the thigh and shank angles versus crank angles also agrees with standard kinematics of four bar linkage mechanism [13], if the foot angle is fixed. The velocity profile of foot segment versus crank angles in figure 5 is cosine shape which agrees in theory with differentiation of the foot segment angle (a sine curve). The profile of the angular velocity of the shank and thigh segments peak at around 450 and also at 3150 with lesser peaks at 1350. The acceleration profiles in Fig. 6 reveal that the magnitude of the shank segment profile peaks at 450 and again at 3300 but with lesser magnitude. It can be inferred from the graphs that the cycling activities will be felt more in the shank segment of the lower limb at around 450 and 3150 which suggests likely stress and strain in the region of the knee since it is the knee angle that is associated with the shank segment in the analysis. Segment Angles 7.0000

6.0000

5.0000

4.0000

3.0000

Knee hip ankle

2.0000

1.0000

5

0

5

0

5

0

5

0 36

34

33

31

30

28

27

25

5

0 24

0

22

21

0

5

5 19

18

5

0

5

0

16

15

13

12

90

10

75

60

45

30

0

15

0.0000

-1.0000

-2.0000 Crank angle (degrees)

Figure 4: The profile of segments angles and crank angles

The profile of velocity and crank angles for thigh, shank and foot are shown in Fig. 5 and it was obtained from equations 34, 35 and 37 respectively.

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Segments angular velocity 300

200

100

34 5 36 0

28 5 30 0 31 5 33 0

24 0 25 5 27 0

18 0 19 5 21 0 22 5

13 5 15 0 16 5

90 10 5 12 0

75

60

45

30

0

15

0 foot shank thigh

-100

-200

-300

-400 crank angles (degrees)

Figure 5: The profile of segments angular velocity and crank angles The profile of angular acceleration and crank angles for thigh, shank and foot are shown in Fig. 6 as obtained from equations 45, 46 and 50 respectively. Segments angular acceleration 5

34 5 36 0

33 0

31 5

30 0

28 5

27 0

25 5

24 0

22 5

19 5 21 0

18 0

16 5

15 0

13 5

12 0

10 5

90

75

60

45

30

0

15

0

-5

-10 ankle shank thigh

-15

-20

-25

-30

-35 Crank angle (degrees)

Figure 6: The profile of segments angular acceleration and crank angles

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Innovative Systems Design and Engineering ISSN 2222-1727 (Paper) ISSN 2222-2871 (Online) Vol.4, No.10, 2013

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4. Conclusion Kinematics analysis of the lower limb segments during cycling session has been presented. The position, velocity and acceleration analysis of the lower limb segments during cycling which will afford theoretical validation of experiments that can be carried out in this field of study. The equations for the position, velocity and acceleration of the foot, shank and thigh segments were derived. The profiles of position, velocity and acceleration and crank angles plotted. The plots agree with experiment results. References 1. Cerone, V., Andreo, D., Larsson, M., Regruto, D.: Stabilization of a riderless bicycle: A linear-parametervarying approach. IEEE Control Systems Magazine. October, 2010, 23 – 32. 2. Hoes, J. J. A. J. M., Binkhorst, R. A., Smeeks-Kuyl, A. E. M. C. Vissers, A. C. A.: Measurement of forces exerted on pedal and crank during work on a bicycle ergometer at different loads. Int. Z. Angew. Physiol. 26, 1968, 33 - 42. 3. Gregor, R. J.: A Biomedical Analysis of Lower Limb Action during Cycling at four different loads. University Microfilms International, Ann Arbor, MI. (1976). 4. Gregor, R. J., Green, D. Garhammer, J. J.: An electromyographic analysis of selected muscle activity in elite competitive cyclists. Biomechanics VII, 1982, 537 - 541. 5. Hull, M. L. Davies, R. R.: Measurement of pedal loading during bicycling – I. Instrumentation". J. Biomech. 14, 1981, 843 - 855. 6. Hull, M. L Jorge, M.: A method for biomechanical analysis of bicycle pedaling. J. Biomech. 18, 1985, 631 644. 7. Brown, N. A. T., Jensen, J. L.: The development of contact force construction in the dynamic-contact task of cycling. J. Biomech. 36, 2003, 1 - 8 . 8. Haapala, S. A., Faghri, P., Adams, D. J.: Leg joint power output during progressive resistance FES-LCE cycling in SCI subjects: developing an index of fatigue. Journal of Neuro-Engineering and rehabilitation. 5,(14) 2008. 9. Yoshihuku, Y, Herzog, W.: Maximal muscle power output in cycling: a modeling approach. J. Sport Science. 14, 1990, 139 - 157. 10. Yoshihuku, Y, Herzog, W.: Optimal Design parameters of the bicycle-rider system for maximal muscle power output. J. Biomech. 23, 1996, 1069-1079. 11. Chang, Z.: Nonlinear Dynamics and Analysis of a four bar linkage with clearance. In the proceedings of IFToMM World Congress, Besancon (France). 2007. 12. Chavdarov, I.: Kinematics and forces of a five-link mechanism by the four spaces Jacoby Matrix. Problems of Engineering Cybernetics and robotics, 55, 2005, 53 – 63. 13. Doughty, S.: Mechanics of Machines. (Wiley, New York. 1988).

Nomenclature

b1 , b2 : real and imaginary axes respectively ; l1 , l 2 , l3 : respectively lengths of thigh, shank and ankle to pedal spindle;

l 4 : crank arm length – from pedal to crank spindle; l5 , l 6 : respectively horizontal and vertical distances of the seat from the crank arm spindle;

θ1 , θ&1 , θ&&1 : positional angle, angular velocity and angular acceleration for thigh with the vertical respectively;

θ 2 , θ&2 , θ&&2 : positional angle, angular velocity and angular acceleration for thigh with the vertical respectively;

θ 3 , θ&3 , θ&&3 : positional angle, angular velocity and angular acceleration for thigh with the respectively;

θ 4 , : crank angle measured from the vertical. 70

vertical