Human Body Movement Coupling Model in Physical Education Class in the Educational Mathematical Equation of Reasonable Exercise Course

Analyzing human movement from biomechanical characteristics is a multi-link chain system characteristic. Different joints play different roles in this multi-link limb chain. At present, there is not much research on the multi-joint motion chain of the lower limbs, and the mechanism of motion control is not precise. The kinematic characteristics of the knee joint when the human body falls from a height are important factors that affect the ground reaction force. The impact at the moment of landing is a complex physical process, and the human body is viscoelastic [1]. Therefore, the human body has a series of dynamic responses to the impact at the moment of landing. This article attempts to use the dominant joint decomposition theory to establish a collaborative mathematical model of the lower limb system at the moment of landing. This article hopes that the conclusions drawn in the thesis will provide a theoretical basis for analyzing the complex motion mechanism of lower limbs.

The thesis defines the moment when the foot and ankle joints are in two states during the moment when the human body lands from a height to land (using a single foot as the research object). The first state is when the toe touches the ground, and the second state is when the foot is flat [2]. The moment the toe touches the environment comes from the muscle Achilles tendon force. For the ankle system, it is the longitudinal internal force F, and on the vertical plane F₁ + F_G = mg + F_G and F₁ = mg. (1){F1=mgF1l1sinα1=Fl1cosα1FGl2sinα2=Fl2cosα2\begin{equation}\left\{\begin{array}{l}F_{1}=m g \\F_{1} l_{1} \sin \alpha_{1}=F l_{1} \cos \alpha_{1} \\F_{G} l_{2} \sin \alpha_{2}=F l_{2} \cos \alpha_{2}\end{array}\right.\end{equation}

We put together formula (1) to get a mathematical model of the moment the toe touches the ground (2){F1=mgF=F1tanα1FG=F/tanα2\begin{equation}\left\{\begin{array}{l}F_{1}=m g \\F=F_{1} \tan \alpha_{1} \\F_{G}=F / \tan \alpha_{2}\end{array}\right.\end{equation}

Fig. 1

For the second state, the foot is always flat, assuming that the human body maintains a balanced and static condition F₁ + F₂ = mg. Decomposing from the moment of the ankle joint we get: (3){F1l1sinα1=Fl1cosα1F2l2sinα2=Fl2cosα2\begin{equation}\left\{\begin{array}{l}F_{1} l_{1} \sin \alpha_{1}=F l_{1} \cos \alpha_{1} \\F_{2} l_{2} \sin \alpha_{2}=F l_{2} \cos \alpha_{2}\end{array}\right.\end{equation}

Organize formula (2) to get (4){F1=F/tanα1F2=F/tanα2\begin{equation}\left\{\begin{array}{l}F_{1}=F / \tan \alpha_{1} \\F_{2}=F / \tan \alpha_{2}\end{array}\right.\end{equation}

From this we get the mathematical model of the human foot and ankle joint (5){F1+F2=mgF1=F/tanα1F2=F/tanα2\begin{equation}\left\{\begin{array}{l}F_{1}+F_{2}=m g \\F_{1}=F / \tan \alpha_{1} \\F_{2}=F / \tan \alpha_{2}\end{array}\right.\end{equation}

From model (2) and formula (5), the joint mathematical model of the ankle joint of the human body at the moment of landing impact is obtained. F₁ is the ground reaction force at the forefoot (forefoot). F₂ is the ground reaction force at the back of the foot (heel). m is the weight of the human body. g is the acceleration due to gravity. F is the resultant force of the longitudinal structure. l₁ The distance from the sole to the center of the ankle joint. l₁ The length from the sole of the back of the foot to the center of the ankle joint. α₁ is the angle between the line between the sole of the forefoot and the center of the ankle common and the vertical (y) direction. α₂ is the angle between the line from the sole of the back of the foot to the center of the ankle joint and the straight movement (y).

From this model, the longitudinal structural force of the foot can be calculated [3]. For the soles of the same size, the lower the arch (flat feet), the greater the angle between α₁ and α₂. The greater the longitudinal structural stress acting on the sole that maintains the integrity of the turn. This causes excessive pressure on the connective tissue of the plantar, and the clinical response is to cause plantar fascia disease. The moment the toe touches the ground, the force on the longitudinal arch structure increases significantly. At the same time, a significant internal pressure occurs at the Achilles tendon inside the ankle system. At this time, people with plantar fascia pain will adapt to the pathological condition.

According to the leading joint decomposition theory, we disassemble the common knee axis. We assume that both the hip joint and the knee joint rotate shaft structures around the center [4]. The direction of the axis of rotation is perpendicular to the forward direction. The mathematical model of the knee joint is as follows: (6){FCcosθC=mCaCcosφCFCsinθC=mCaCsinφC+mCg\begin{equation}\left\{\begin{array}{l}F_{C} \cos \theta_{C}=m_{C} a_{C} \cos \varphi_{C} \\F_{C} \sin \theta_{C}=m_{C} a_{C} \sin \varphi_{C}+m_{C} g\end{array}\right.\end{equation}

Fig. 2

The mathematical model of the hip joint obtained by inverse dynamics is as follows: (7){−FCcosθC+FDcosθD=mDαDcosφD−FCsinθC+FDsinθD=mDαDsinφD+mDg\begin{equation}\left\{\begin{array}{l}-F_{C} \cos \theta_{C}+F_{D} \cos \theta_{D}=m_{D} \alpha_{D} \cos \varphi_{D} \\-F_{C} \sin \theta_{C}+F_{D} \sin \theta_{D}=m_{D} \alpha_{D} \sin \varphi_{D}+m_{D} g\end{array}\right.\end{equation}

We can obtain the combined mathematical model of the knee joint and hip joint torque from the angular movement of the calf around the knee joint (8){MC−mCglCcosβC=(IC+mClC2)αCMD−MC−mDglDcosβD=(ID+mDlD2)αD\begin{equation}\left\{\begin{array}{l}M_{C}-m_{C} g l_{C} \cos \beta_{C}=\left(I_{C}+m_{C} l_{C}^{2}\right) \alpha_{C} \\M_{D}-M_{C}-m_{D} g l_{D} \cos \beta_{D}=\left(I_{D}+m_{D} l_{D}^{2}\right) \alpha_{D}\end{array}\right.\end{equation}

F_C, F_D is the resultant force on the knee joint and hip joint, respectively. θ_C, θ_D is the angle between the knee joint and hip joint force and the horizontal direction (x), respectively. M_C, M_D is the resultant moment on the knee joint and hip joint, respectively. m_C, m_D is the mass of calf and thigh, respectively. a_C, a_D is the acceleration on the mass center of the calf and thigh, respectively. ϕ_C, ϕ_D is the angle between the force generated by the mass center of the calf and the thigh and the horizontal direction (x). l_C is the distance from the center of the knee joint to the center of mass of the calf. l_D is the distance from the center of the hip joint to the center of mass of the thigh. β_C is the angle between the calf and the horizontal direction (x), β_D and is the angle between the thigh and the horizontal direction (x). α_C, α_D is the angular acceleration of the knee joint and hip joint, respectively. I_C, I_D is the moment of inertia of the calf and thigh, respectively. g is the acceleration of gravity. The combined torque represents the movement of the joint flexor muscles [5]. Since the human body is a complex structure, muscle contraction causes it to produce a resultant moment.

We assume f˙=∂θf\dot{f} = \partial_\theta f and f^′ = ∂_x f, considering the dynamic coupling problem, that is, the Lagrange function is: (9)L(ε)=F(R−r)2r2ε2+G2[(ε″)2+(ε¨)2+2μβε¨+2(1−μ)(ε˙')2]+Nx2(ε)2−Nxθε'ε˙\begin{equation}L(\varepsilon)=\frac{F(R-r)}{2 r^{2}} \varepsilon^{2}+\frac{G}{2}\left[\left(\varepsilon^{\prime \prime}\right)^{2}+(\ddot{\varepsilon})^{2}+2 \mu \beta \ddot{\varepsilon}+2(1-\mu)\left(\dot{\varepsilon}^{\prime}\right)^{2}\right]+\frac{N_{x}}{2}(\varepsilon)^{2}-N_{x \theta} \varepsilon^{\prime} \dot{\varepsilon}\end{equation}

We use the space coordinate to simulate the time coordinate t, according to the Hamilton variational principle: (10)δ∬{F(R−r)2r2ε2+G2[(ε″)2+(ε¨)2+2με″ω¨ε¨+2(1−μ)(ε˙')2+Nx2(ε')2−Nxθε'ε}rdθdx=0\begin{equation}\delta \iint\left\{\frac{F(R-r)}{2 r^{2}} \varepsilon^{2}+\frac{G}{2}\left[\left(\varepsilon^{\prime \prime}\right)^{2}+(\ddot{\varepsilon})^{2}+2 \mu \varepsilon^{\prime \prime} \ddot{\omega} \ddot{\varepsilon}+2(1-\mu)\left(\dot{\varepsilon}^{\prime}\right)^{2}+\frac{N_{x}}{2}\left(\varepsilon^{\prime}\right)^{2}-N_{x \theta} \varepsilon^{\prime} \varepsilon\right\} r d \theta d x=0\right.\end{equation}

We double-integrate the formula (2) to get: (11)δ∬{F(R−r)2r2ε2+G2[(ε″)2+(ε¨)2+2(ε″)2]+Nx2(ε')2−Nxθε'ε˙}rdθdx=0\begin{equation}\delta \iint\left\{\frac{F(R-r)}{2 r^{2}} \varepsilon^{2}+\frac{G}{2}\left[\left(\varepsilon^{\prime \prime}\right)^{2}+(\ddot{\varepsilon})^{2}+2\left(\varepsilon^{\prime \prime}\right)^{2}\right]+\frac{N_{x}}{2}\left(\varepsilon^{\prime}\right)^{2}-N_{x \theta} \varepsilon^{\prime} \dot{\varepsilon}\right\} r d \theta d x=0\end{equation}

We set the initial variable q={ε,−ε˙}q = \{\varepsilon,-\dot{\varepsilon}\} and the dual variable p = {p₁, p₂}. p1=Qθ=−D(ε˙″+ε)p_{1}=Q_{\theta}=-D\left(\dot{\varepsilon}^{\prime \prime}+\varepsilon\right) , p2=(Mx+Mθ)/[1+(R−r)]=−D(ε″+ε¨)p_{2}=\left(M_{x}+M_{\theta}\right) /[1+(R-r)]=-D\left(\varepsilon^{\prime \prime}+\ddot{\varepsilon}\right) , p₁ is the equivalent shear force in the θ direction and p₂ is the comparable torsional moment in the θ order [6]. The Hamilton function that we use the initial variable and the dual variable to express is: (12)M(q,p)=pTq−L(q,p)=−F(R−r)2r2ε2+12Gp22+p1ε+p2ε″−Nx2(ε')2+Nxθε'ε\begin{equation}M(q, p)=p^{T} q-L(q, p)=-\frac{F(R-r)}{2 r^{2}} \varepsilon^{2}+\frac{1}{2 G} p_{2}^{2}+p_{1} \varepsilon+p_{2} \varepsilon^{\prime \prime}-\frac{N_{x}}{2}\left(\varepsilon^{\prime}\right)^{2}+N_{x \theta} \varepsilon^{\prime} \varepsilon\end{equation}

The Hammon equation is described as follows: (13){ε˙−ε˙p˙1p˙2}={δM/δP1δM/δP1−δM/δεδM/δε˙}=[0−100∂x2ε001/DF(R−r)r2+Nx∂x2ε−Nxθ∂xε0−∂x2εNxθ∂xη010]{ε−p2p1p2}\begin{equation}\left\{\begin{array}{l}\dot{\varepsilon} \\-\dot{\varepsilon} \\\dot{p}_{1} \\\dot{p}_{2}\end{array}\right\}=\left\{\begin{array}{l}\delta M / \delta P_{1} \\\delta M / \delta P_{1} \\-\delta M / \delta \varepsilon \\\delta M / \delta \dot{\varepsilon}\end{array}\right\}=\left[\begin{array}{cccc}0 & -1 & 0 & 0 \\\partial_{x}^{2} \varepsilon & 0 & 0 & 1 / D \\\frac{F(R-r)}{r^{2}}+N_{x} \partial_{x}^{2} \varepsilon & -N_{x \theta} \partial_{x} \varepsilon & 0 & -\partial_{x}^{2} \varepsilon \\N_{x \theta} \partial_{x} \eta & 0 & 1 & 0\end{array}\right]\left\{\begin{array}{l}\varepsilon \\-p_{2} \\p_{1} \\p_{2}\end{array}\right\}\end{equation}

We abbreviate formula (13) as: (14)ψ˙=Hψ\begin{equation}\dot{\psi} = H\psi\end{equation}

The entire state vector: (15)ψ={qT,pT}={ε,−ε˙,p1,p2}\begin{equation}\psi=\left\{q^{T}, p^{T}\right\}=\left\{\varepsilon,-\dot{\varepsilon}, p_{1}, p_{2}\right\}\end{equation}

The operator matrix is: (16)M=[0−100∂x2ε001/GE(R−r)r2+Nx∂x2ε−Nxθ∂xε0−∂x2εNxθ∂xε010]\begin{equation}M=\left[\begin{array}{cccc}0 & -1 & 0 & 0 \\\partial_{x}^{2} \varepsilon & 0 & 0 & 1 / G \\\frac{E(R-r)}{r^{2}}+N_{x} \partial_{x}^{2} \varepsilon & -N_{x \theta} \partial_{x} \varepsilon & 0 & -\partial_{x}^{2} \varepsilon \\N_{x \theta} \partial_{x} \varepsilon & 0 & 1 & 0\end{array}\right]\end{equation}

The identity matrix is expressed as (17)J=[0I2−I20]\begin{equation}J = \left[\begin{array}{cc}0 & I_2 \\-I_2 & 0 \end{array}\right]\end{equation}

We have (18)〈ψ1,ψ2〉=∫0xeψ1Jψ2dx\begin{equation}\left\langle\psi_{1}, \psi_{2}\right\rangle=\int_{0}^{x_{e}} \psi_{1} J \psi_{2} d x\end{equation}

ψ = {ɛ, − ɛ, p₁, p₂} forms a symplectic space vector. In the linear Hamilton system, we use the generalized variable separation method to make (19)ψ(x,θ)=ϕ(x)eλθ\begin{equation}\psi(x, \theta)=\phi(x) e^{\lambda \theta}\end{equation}

φ(x) is the intrinsic solution, which is a function of x. The Eigen equation is expressed as follows: (20)Hϕ=λϕ\begin{equation}H\phi = \lambda\phi\end{equation}(21)ϕ(x)=ϕ(x)e2λπ\begin{equation}\phi(x)=\phi(x) e^{2 \lambda \pi}\end{equation}

Among them, when λ = in(n = 0,±1,…),n = 0 is the eigenvalue λ = 0, the formula (12) becomes Hφ = 0. At this time, the characteristic equation is (22)η4−2n2r2η2+n4r4+E(R−r)Dr3=0\begin{equation}\eta^{4}-\frac{2 n^{2}}{r^{2}} \eta^{2}+\frac{n^{4}}{r^{4}}+\frac{E(R-r)}{D r^{3}}=0\end{equation}

φ(x) has nothing to do with N_xθ. When n ≠ 0, the eigenvalue λ_n ≠ 0,φ_n(x) is a non-zero eigensolution. If λ_i + λ_j ≠ 0, then (23)〈ϕi,ϕj〉=∫0xeϕiTJϕjdx=0\begin{equation}\left\langle\phi_{i}, \phi_{j}\right\rangle=\int_{0}^{x_{e}} \phi_{i}^{T} J \phi_{j} d x=0\end{equation}

Solutions φ_i and φ_j have a symplectic orthogonal relationship. (24)〈ϕi,ϕj〉=∫0xeϕiTJϕjdx≠0\begin{equation}\left\langle\phi_{i}, \phi_{j}\right\rangle=\int_{0}^{x_{e}} \phi_{i}^{T} J \phi_{j} d x \neq 0\end{equation}

The relationship between φ_j and φ_j is symplectic conjugate [7]. The vector space is complete, and any eigenstate vector can be represented by a combination of function eigenvectors. (25)ψ(x,θ)=∑n=0∞[anϕn(x)einθ+bnϕ−n(x)e−inθ]\begin{equation}\psi(x, \theta)=\sum_{n=0}^{\infty}\left[a_{n} \phi_{n}(x) e^{i n \theta}+b_{n} \phi-n(x) e^{-i n \theta}\right]\end{equation}

We can get this by solving the Eigen equation: (26)ϕn(x)=∑k=14CkxknAkn(x)=∑k=14Ckxkneηkx∑k=14Ck{1−in/rD(in3/r3+ηkn2/r2)D(n2/r2−ηk2)}eηkx\begin{equation}\phi_{n}(x)=\sum_{k=1}^{4} C_{k} x_{k}^{n} A_{k}^{n}(x)=\sum_{k=1}^{4} C_{k} x_{k}^{n} e^{\eta k x} \sum_{k=1}^{4} C_{k}\left\{\begin{array}{l}1 \\-i n / r \\D\left(i n^{3} / r^{3}+\eta k n^{2} / r^{2}\right) \\D\left(n^{2} / r^{2}-\eta_{k}^{2}\right)\end{array}\right\} e^{\eta k x}\end{equation}

C_k,(k = 1,2,3,4) is the undetermined Eigen coefficient [8]. The eigenvalue equation of symplectic space is shown below. η_k,(k = 1,2,3,4) is the solution of the Eigen-characteristic equation in symplectic space. (27)η4+(NxDr2−2n2r2)η2+2niNxyDr2η+n4r4+E(R−r)Dr3=0\begin{equation}\eta^{4}+\left(\frac{N_{x}}{D r^{2}}-\frac{2 n^{2}}{r^{2}}\right) \eta^{2}+\frac{2 n i N_{x y}}{D r^{2}} \eta+\frac{n^{4}}{r^{4}}+\frac{E(R-r)}{D r^{3}}=0\end{equation}

We simulate the action of the human body jumping from a height of 66 cm. The load is the mechanical parameter that occurs in the dynamic experiment [9]. When the subject is landing from a height, the angle between the extension line of the calf.

Fig. 3

From the simulation results, it can be seen that the knee joint has an excellent cushioning effect on the vertical ground reaction force, and the angular velocity of the hip joint has an advantage in cushioning the horizontal backward reaction force.

We set the unit properties of the material of the parts and the contact between the components. The model setting boundary conditions and loads have been developed. During the simulation, a rotational load was applied to the femur during buckling [10]. We simulate the internal and external rotation of the femur, and the angle of internal and external rotation is gradually increased from 5° to 30°, as shown in Figure 4. We set the measurement parameters and then solve them. The result is shown in Figure 5 [11].

Fig. 4

Fig. 5

The landing method that reduces the maximum vertical reaction force has a greater degree of joint flexion than the proper landing method [12]. Therefore, the human body delays the cushioning of the landing process by increasing the degree of flexion of the joints and reduces the reaction force of the ground under the state of falling motion-fixed.

The skeletal muscle system of the lower limb system of the human body immediately reaches a large load during the landing. Therefore, the method used to control the reaction force must be activated before touching the ground. Thus, the muscles of the lower limbs of the human body enter the contraction state before the end of the link chain touches the ground, and the joint chain should be in a flexion state before touching the ground [13].

According to the decomposition theory of dominant joints, a joint mathematical model of the lower limb system of the human body is established at the moment of landing. Based on the Hamiltonian system, the mathematical model of the coupling problem of the lower limbs of the human body is analyzed. From the simulation results, it can be known that since the skeletal muscle system reaches a large load immediately when it touches the ground, the method used to control the reaction force must be activated before the ground touches.