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On this page
  • Inverted pendulum model
  • Human pendulum model
  • Dynamic human pendulum model
  • Analytic solution
  • Simulation solution in MATLab
  • Passive torque (assume no time delay)
  • Active torque (assume time delay)
  • Center of Pressure (CoP)
  • References

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  1. Models

Inverted Pendulum Model

For postural sway modelling - control theory

Inverted pendulum model

Newton’s second law of motion states that the force acting on an object is proportional to its mass and the rate of change of its velocity. In the case of rotational motion, the corresponding law is given by the equation:

τ=Iα\tau =I\alphaτ=Iα

where τ is the torque applied to the object, I is the moment of inertia of the object about the axis of rotation, and α is the angular acceleration of the object.

For a single-link pendulum which pivot point is fixed, we have:

τ=Iθ¨ \tau =I\ddot{\theta}τ=Iθ¨

where θ is is the angle of the pendulum from the earth vertical.

For a point mass, the moment of inertia is

I=mr2I =mr^2 I=mr2

Thus, we have

τ=ml2θ¨\tau =ml^2\ddot{\theta}τ=ml2θ¨

where lll is the length of the pendulum.

Given Newton’s second law of motion, we have

F=maF=maF=ma

We also know that the acceleration of an inverted pendulum is due to gravity:

F=mgsin⁡θF =mg \sin \thetaF=mgsinθ

Thus, the torque can also be expressed as

τ=mglsin⁡θ\tau =mgl \sin \thetaτ=mglsinθ

Thus we have

ml2θ¨=mglsin⁡θml^2\ddot{\theta}=mgl \sin \thetaml2θ¨=mglsinθ

Human pendulum model

From the inverted pendulum model we have

mh2θ¨=mghsin⁡θ+ηmh^2\ddot{\theta}=mgh \sin \theta +\eta mh2θ¨=mghsinθ+η

where h is the height of the human, and η is noise of this system.

(Ibody+mh2)θ¨=Mankle+mghsin⁡θ+η(I_{body}+mh^2)\ddot{\theta}=M_{ankle} + mgh \sin \theta +\eta (Ibody​+mh2)θ¨=Mankle​+mghsinθ+η

where IbodyI_{body}Ibody​represents the moment of inertia of body and MankleM_{ankle}Mankle​ represents a corrective ankle torque that resists postural sway.

Dynamic human pendulum model

Because postural sway angle and ankle torque change in time, we have:

(Ibody+mh2)θ¨(t)=Mankle(t)+mghsin⁡(θ(t))+η(I_{body}+mh^2)\ddot{\theta}(t)=M_{ankle}(t) + mgh \sin (\theta(t)) +\eta(Ibody​+mh2)θ¨(t)=Mankle​(t)+mghsin(θ(t))+η

Ankle torque MankleM_{ankle}Mankle​ can be divided into passive torque MpassiveM_{passive}Mpassive​ and active torqueMactiveM_{active}Mactive​ (2):

Mankle=Mpassive+MactiveM_{ankle}=M_{passive} + M_{active}Mankle​=Mpassive​+Mactive​

where MpassiveM_{passive}Mpassive​ is caused by the stiffness and damping of the muscles, whereas MactiveM_{active}Mactive​ is based on “active” time-delayed neuromuscular feedback from proprioceptive, vestibular, and visual sensory inputs.

This paper (1) reports that people with profound bilateral vestibular loss had a higher level of muscle stiffness, possibly as a strategy for compensating for their vestibular deficit.

Assume η = 0, and that when θ is small sinθ≈θsin \theta ≈ \thetasinθ≈θ. We have

Mankle=Aθ(t)+Bθ˙+Cθ¨M_{ankle} = A \theta(t) + B \dot{\theta} + C\ddot{\theta}Mankle​=Aθ(t)+Bθ˙+Cθ¨

Assume there is no delay in feedback

(mh2−C)θ¨−Bθ˙−(mgh+A)θ=0(mh^2-C)\ddot{\theta} - B \dot{\theta} - (mgh+A)\theta = 0(mh2−C)θ¨−Bθ˙−(mgh+A)θ=0

where θ\thetaθ represents angular distance from heading, θ˙\dot\thetaθ˙ represents angular velocity from optic flows, and θ¨\ddot\thetaθ¨ represents angular acceleration from vestibular inputs.

Let x=mh2−Cx = mh^2-Cx=mh2−C and y=−By = -By=−B and z=−(mgh+A)z = - (mgh+A)z=−(mgh+A), we have

xθ¨+yθ˙+zθ=0x\ddot{\theta} + y \dot{\theta} + z\theta = 0xθ¨+yθ˙+zθ=0

Analytic solution

Assume answer is

θ=αeβt\theta = \alpha e^{\beta t}θ=αeβt
θ˙=βαeβt=βθ\dot{\theta} = \beta \alpha e^{\beta t} = \beta \thetaθ˙=βαeβt=βθ
θ¨=β2αeβt=β2θ\ddot{\theta} = \beta^2 \alpha e^{\beta t} = \beta^2 \thetaθ¨=β2αeβt=β2θ

Thus we have

xβ2θ+yβθ+zθ=0x \beta^2 \theta + y \beta \theta + z \theta = 0xβ2θ+yβθ+zθ=0

which is xβ2+yβ+z=0x \beta^2 + y \beta + z = 0xβ2+yβ+z=0. Thus we have

β±=−y±y2−4xz2x\beta_{\pm} = \frac{-y \pm\sqrt{y^2-4xz}}{2x}β±​=2x−y±y2−4xz​​

Condition 1

If y2−4xz>0y^2 - 4xz > 0 y2−4xz>0

Simulation solution in MATLab

For xθ¨+yθ˙+zθ=0x\ddot{\theta} + y \dot{\theta} + z\theta = 0xθ¨+yθ˙+zθ=0, we have xθ¨=−yθ˙−zθx\ddot{\theta} = - y \dot{\theta} - z\thetaxθ¨=−yθ˙−zθ. Let θ˙=ϕ\dot{\theta} =\phiθ˙=ϕ, we have xϕ˙=−yϕ−zθx\dot{\phi} = -y\phi - z \thetaxϕ˙​=−yϕ−zθ. Thus

ϕ˙=−yxϕ−zxθ\dot{\phi} = -\frac{y}{x}\phi - \frac{z}{x} \thetaϕ˙​=−xy​ϕ−xz​θ
θ˙=ϕ\dot{\theta} =\phiθ˙=ϕ

Combine we have V˙=MV\dot{V} = MVV˙=MV:

(ϕ˙θ˙)=(−yx−zx10)\begin{pmatrix} \dot{\phi}\\ \dot{\theta} \end{pmatrix} = \begin{pmatrix} -\frac{y}{x} & - \frac{z}{x} \\ 1 & 0 \end{pmatrix}(ϕ˙​θ˙​)=(−xy​1​−xz​0​)

and V0˙=(θ0˙θ)\dot{V_0} = \begin{pmatrix} \dot{\theta_0}\\ \theta \end{pmatrix}V0​˙​=(θ0​˙​θ​).

From V˙=MV\dot{V} = MVV˙=MV we have

Vi+1−Viδti=MVi\frac{V_{i+1}-V_i}{\delta t_i} = M V_iδti​Vi+1​−Vi​​=MVi​
Vi+1=Vi+δtiMViV_{i+1} = V_i +\delta t_iMV_iVi+1​=Vi​+δti​MVi​

and because t=δtit = \delta t_it=δti​.

Passive torque (assume no time delay)

Mpassive(t)=K(t)[θ(t)+βθ3(t)]+C(t)θ˙(t)M_{passive}(t)=K(t)[\theta(t)+\beta\theta^3(t)] + C(t)\dot{\theta}(t)Mpassive​(t)=K(t)[θ(t)+βθ3(t)]+C(t)θ˙(t)

Active torque (assume time delay)

Mactive(t)=M_{active}(t) =Mactive​(t)=

Center of Pressure (CoP)

CoP(−mhθ¨sinθ−mhθ˙2cosθ+mg+mfg)CoP (-mh\ddot{\theta}sin\theta-mh\dot{\theta}^2cos\theta+mg+m_fg)CoP(−mhθ¨sinθ−mhθ˙2cosθ+mg+mf​g)

where −mhθ¨sinθ−mhθ˙2cosθ-mh\ddot{\theta}sin\theta-mh\dot{\theta}^2cos\theta−mhθ¨sinθ−mhθ˙2cosθ is the second derivative of −mhcosθ-mhcos\theta−mhcosθ and hfmh(θ¨cosθ−θ˙2sinθ)h_fm_h(\ddot{\theta}cos\theta-\dot{\theta}^2sin\theta)hf​mh​(θ¨cosθ−θ˙2sinθ) is the second derivative of −hfmhsinθ-h_fm_hsin\theta−hf​mh​sinθ

References

(1) Peterka, R. J. Sensorimotor integration in human postural control. Journal of Neurophysiology 88, 1097– 1118 (2002). URL https://www.physiology. org/doi/10.1152/jn.2002.88.3.1097.

(2) Chagdes, J. R. et al. Limit cycle oscilla- tions in standing human posture. Journal of Biomechanics 49, 1170–1179 (2016). URL https://www.sciencedirect.com/science/ article/pii/S002192901630269X.

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Because human legs can be considered a series of interconnected segments separated by the joints, we will include ankle joint (1). With , we have:

parallel-axis theorem