Get 20M+ Full-Text Papers For Less Than $1.50/day. Start a 14-Day Trial for You or Your Team.

Learn More →

Fractional differential equations in National Sports Training in Colleges and Universities

Fractional differential equations in National Sports Training in Colleges and Universities 1IntroductionIn sports training in colleges and universities, people generally equate a team's offensive ability with the striker's level of ‘one-to-one’ [1]. In the fierce competition in football, the forward is an essential factor in winning, but the passing and receiving cooperation from the backcourt and midfielders is also essential.2Combination circuit for catching the ballNetwork topology has been widely used in communication networks, computer networks, and operations research. Here we use line graph analysis in network theory to discuss passing and receiving in football [2]. We first define the generalized path (1)L(ki)=∏i=1nkiL({k_i}) = \prod\limits_{i = 1}^n {k_i}Here, ki represents a quantity that affects the generalized path. In a football game, we mainly consider ‘the geometric distance between two points, the speed of the ball, the accuracy of passing, the ability to receive the ball, and the defense of the passing and receiving points.’ When multiple people pass and receive the ball multiple times, L(ki) forms a distance matrix, which follows the following rules (2)L2(ki)=L(ki)*L(ki)L3(ki)=L2(ki)*L(ki)⋯⋯{L^2}({k_i}) = L({k_i})*L({k_i}){L^3}({k_i}) = {L^2}({k_i})*L({k_i}) \cdots \cdots We define (3)C*D=E=[ejk]m×nC*D = E = [{e_{jk}}{]_{m \times n}}Among them C = [cjk]m×l, D = [djk]l×n, ejk=mincjl≠0,dlk≠0{cj1*d1k,cj2*d2k,⋯⋯cjl*dlk}{e_{jk}} = \mathop {\min }\limits_{{c_{jl}} \ne 0,{d_{lk}} \ne 0} \{ {c_{j1}}*{d_{1k}},{c_{j2}}*{d_{2k}}, \cdots \cdots {c_{jl}}*{d_{lk}}\} .For example, C = [1, 2, 3, 4, 0], D = [0, 2, 1, 2, 0]T, ejk=mincjl≠0,dlk≠0{0,2,4,6,0}{e_{jk}} = \mathop {\min }\limits_{{c_{jl}} \ne 0,{d_{lk}} \ne 0} \{ 0,2,4,6,0\} .We use Figure 1 to show the generalized path diagram of the pass when attacking. P1 represents the node formed by the defender of the possession side and the goalkeeper. P2, P3 corresponds to midfielders and full-backs. P4 stands for striker. P5 refers to the goal. Then (4)L=[0320000313000100000100000], L2=[1jk]=L*L=[0063600042000020000000000]L = \left[ {\matrix{ 0 & 3 & 2 & 0 & 0 \cr 0 & 0 & 3 & 1 & 3 \cr 0 & 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^2}{ = [1_{jk}}] = L*L = \left[ {\matrix{ 0 & 0 & 6 & 3 & 6 \cr 0 & 0 & 0 & 4 & 2 \cr 0 & 0 & 0 & 0 & 2 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right]Fig. 1Generalized path diagram of passing on offenseThe same (5)L3=L2*L=[0007400005000000000000000], L4=L3*L=[0000800000000000000000000], L5=L4*L=0{L^3} = {L^2}*L = \left[ {\matrix{ 0 & 0 & 0 & 7 & 4 \cr 0 & 0 & 0 & 0 & 5 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^4} = {L^3}*L = \left[ {\matrix{ 0 & 0 & 0 & 0 & 8 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^5} = {L^4}*L = 0To find the shortest distance from Pj to Pk, we define (6)C⊕D=[ejk]m×nC \oplus D = [{e_{jk}}{]_{m \times n}}among them (7)C=[ajk]m×nD=[djk]m×n,ejk={0,cjk=djk=0cjk,cjk>0,djk=0djk,cc8=0,djk>0min(cjk,djk),cjk≠0djk≠0Lmin=L⊕L2⊕L3⊕L4⊕L5=[0323400313000120000100000]=[1jk]min\matrix{ {C = [{a_{jk}}{]_{m \times n}}D = [{d_{jk}}{]_{m \times n}},{e_{jk}} = \left\{ {\matrix{ {0,{c_{jk}} = {d_{jk}} = 0} \hfill \cr {{c_{jk}},{c_{jk}} > 0,{d_{jk}} = 0} \hfill \cr {{d_{jk}},c{c^8} = 0,{d_{jk}} > 0} \hfill \cr {\min ({c_{jk}},{d_{jk}}),{c_{jk}} \ne 0{d_{jk}} \ne 0} \hfill \cr} } \right.} \hfill \cr {{L_{\min }} = L \oplus {L^2} \oplus {L^3} \oplus {L^4} \oplus {L^5} = \left[ {\matrix{ 0 & 3 & 2 & 3 & 4 \cr 0 & 0 & 3 & 1 & 3 \cr 0 & 0 & 0 & 1 & 2 \cr 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right] = \mathop {{{[1}_{jk}}]}\limits_{\min } } \hfill \cr } [1jk]min\mathop {{{[1}_{jk}}]}\limits_{\min } represents the shortest distance from Pj to Pk. The ultimate goal of passing and receiving is to get the (best) shot. There are many factors to consider when passing the ball. A node that is geometrically connected is not necessarily a valid point. From formula (1), it can be known that any factor other than the geometric distance is zero, and the generalized path is the point of leading and receiving the ball, which is the weak point [3]. From formula (2), we can know the importance of midfielders. Because P1, P5 is constant for the stadium. The value of [1jk]min\mathop {{{[1}_{jk}}]}\limits_{\min } mainly depends on the direction and size of P2, P3.3Best offensive intensityThe best-generalized path gives us the whereabouts of the ball in the pass. But how to form the best path is primarily affected by the defensive team's defensive lineup. Let us illustrate this point with a simple diagram [4]. The three U1, U2, U3 secant lines in Figure 2 below represent the three defense lines of the defender. And define (8)Ui=∑PkPk+1U1=5,U2=5,U3=4\matrix{ {{U_i} = \sum {P_k}{P_{k + 1}}} \hfill \cr {{U_1} = 5,{U_2} = 5,{U_3} = 4} \hfill \cr } We introduce the concept of generalized attack intensity and define the following formula. The attack intensity in football refers to the attack intensity, the reliability of passing and receiving the ball, and the degree of threat to the defense [5].(9)fjk↔Pjk{f_{jk}} \leftrightarrow {P_{jk}}A linear model can represent the generalized attack intensity secant (10)minfjkU=∑fjk\mathop {\min }\limits_{{f_{jk}}} U = \sum {f_{jk}}The minimum cut is equivalent to the maximum flow, so the generalized attack intensity cut is (11)minfjkU=∑fjk−∑fjk{U,j≠s0,j≠s,t−U,j=t\mathop {\min }\limits_{{f_{jk}}} U = \sum {f_{jk}} - \sum {f_{jk}}\left\{ {\matrix{ {U,j \ne s} \hfill\cr {0,j \ne s,t}\hfill \cr { - U,j = t} \hfill\cr } } \right.s, t represents a confident start and endpoint, respectively. Through the secant analysis, we know two ways to increase the attack intensity: one is to increase the attack line, and the other is to reduce the path to the attack point P5. This is the value that forms the minimum secant U3 to form the maximum attack intensity [6].Fig. 2The equivalent model of a football field pass4Unconstrained dynamic equationsWe can write the dynamic equations of the model by the Kane method (12)Fj+Fj*=0{F_j} + F_j^* = 0We replace the extra force acting on the system with an equal force system. Then the generalized active force Fj can be written as (13)Fj=∑K=115(∂Vk∂x˙jFk+∂ωk∂x˙jMk){F_j} = \sum\limits_{K = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}{F_k} + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}{M_k}} \right)Fk and Mk represent the ‘equivalent’ force and force matrix acting on the link Bk. The line of action of Fk passes through the center of the link. Gk ·Vk and ωk are respectively called the velocity and angular velocity through the point Gk in the inertial reference frame. ∂Vk∂x˙j{{\partial {V_k}} \over {\partial {{\dot x}_j}}}and ∂ωk∂x˙j{{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}are respectively referred to as ‘change rate of position deviation’ and ‘change rate of azimuth deviation.’ Mk is (14)Mk=Mkext+∑Mkl{M_k} = M_k^{ext} + \sum M_k^lMk is the moment produced by the external force acting on Bk. MklM_k^lis the force applied by Bl to Bk. It is generally caused by force Bl exerted on Bk. Therefore, rewriting (14) has (15)Mk=Fjext+Fjint{M_k} = F_j^{ext} + F_j^{\mathop {{\rm{int}}}}among them (16)Fjext=∑K=115(∂Vk∂x˙jFk+∂ωk∂x˙jMkext)F_j^{ext} = \sum\limits_{K = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}{F_k} + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}M_k^{ext}} \right)(17)Fjint=∑K=115(∑∂ωk∂x˙jMk)F_j^{\mathop {{\rm{int}}}} = \sum\limits_{K = 1}^{15} \left( {\sum {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}{M_k}} \right)Due to (18)∂ωk∂x˙j=ωijknoi, ∂Vk∂x˙j=υijknoi{{\partial {\omega _k}} \over {\partial {{\dot x}_j}}} = {\omega _{ijk}}{n_{oi}},\quad {{\partial {V_k}} \over {\partial {{\dot x}_j}}} = {\upsilon _{ijk}}{n_{oi}}Equation (17) can be rewritten as (19)Fjext=υkjkFki+ωkjkMkiextF_j^{ext} = {\upsilon _{kjk}}{F_{ki}} + {\omega _{kjk}}M_{ki}^{ext}Where Fki and MkiextM_{ki}^{ext}are the components of Fk and MkextM_k^{ext}in the noi direction, respectively [7]. Let's analyze the generalized inertial force. From Eq. (14), we can get:(20)Fj*=∑k=115(∂Vk∂x˙jFk*+∂ωk∂x˙jMk*)F_j^* = \sum\limits_{k = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}F_k^* + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}M_k^*} \right)Among them Fj*=−mkak=−mk(υijkx¨j+υ˙ijkx˙j)noi, Tk*=−Ik⋅ak−ωk×Ik⋅ωkF_j^* = - {m_k}{a_k} = - {m_k}({\upsilon _{ijk}}{\ddot x_j} + {\dot \upsilon _{ijk}}{\dot x_j}){n_{oi}},\quad T_k^* = - {I_k} \cdot {a_k} - {\omega _k} \times {I_k} \cdot {\omega _k}The inertial dyadic Ik is given by (21)Ik=Imnknomnon{I_k} = {I_{mnk}}{n_{om}}{n_{on}}Among them, the value of element Imnk related to nom and nmnk is related. It has the same value as nkr and nks related element Irskk(k=1,2,3;m,n,r,s,k)I_{rsk}^k(k = 1,2,3;m,n,r,s,k). O represents the origin of the reference coordinate of the inertial reference system [8]. The expression is as follows:(22)Imnk=nomnonnksnrskk{I_{mnk}} = {n_{om}}{n_{on}}{n_{ks}}n_{rsk}^k(23)S0Kmr=nomnkr, S0Kns=nonnks, Imnk=S0KmrS0KnsIrsk{S_0}{K_{mr}} = {n_{om}}{n_{kr}},\quad {S_0}{K_{ns}} = {n_{on}}{n_{ks}},\quad {I_{mnk}} = {S_0}{K_{mr}}{S_0}{K_{ns}}{I_{rsk}}Organized (24)Tk*=−(Imnknomnon)[(w˙ijkx¨j+ωijkx¨j)noi]−(ωijkx˙jnoi)×Imnknomnonωijkx˙jnoiTk*=−[Imnk(ω˙njkx¨j+ωnjkx¨j)+eiqmωipkωnjkx˙px˙jIqnk]nom\matrix{ {T_k^* = - ({I_{mnk}}{n_{om}}{n_{on}})\left[ {\left( {{{\dot w}_{ijk}}{{\ddot x}_j} + {\omega _{ijk}}{{\ddot x}_j}} \right){n_{oi}}} \right] - \left( {{\omega _{ijk}}{{\dot x}_j}{n_{oi}}} \right) \times {I_{mnk}}{n_{om}}{n_{on}}{\omega _{ijk}}{{\dot x}_j}{n_{oi}}} \hfill \cr {T_k^* = - [{I_{mnk}}({{\dot \omega }_{njk}}{{\ddot x}_j} + {\omega _{njk}}{{\ddot x}_j}) + {e_{iqm}}{\omega _{ipk}}{\omega _{njk}}{{\dot x}_p}{{\dot x}_j}{I_{qnk}}]{n_{om}}} \hfill \cr } Among them Tk*=∑k=115{mkυijk(υipkx¨j+υ˙ipkx˙j)+ωijk[Iink(ωnpkx¨p+ω˙npkx˙p)+emqiωmpkωnhkωnhkx˙px˙hIqnk]}T_k^* = \sum\limits_{k = 1}^{15} \left\{ {{m_k}{\upsilon _{ijk}}\left( {{\upsilon _{ipk}}{{\ddot x}_j} + {{\dot \upsilon }_{ipk}}{{\dot x}_j}} \right) + {\omega _{ijk}}\left[ {{I_{ink}}({\omega _{npk}}{{\ddot x}_p} + {{\dot \omega }_{npk}}{{\dot x}_p}) + {e_{mqi}}{\omega _{mpk}}{\omega _{nhk}}{\omega _{nhk}}{{\dot x}_p}{{\dot x}_h}{I_{qnk}}} \right]} \right\}i, m, n, q and p are summed from 1 to 3. Both j and h are summed from 1 to 40. The kinetic equation (14) can be expressed in the following concise form (25)Ajpx¨p=fj+Fjext+Fjint{A_{jp}}{\ddot x_p} = {f_j} + F_j^{ext} + F_j^{{\rm{int}}}Where p is summed from 1 to 40. Ajp and fj are given by (26)Ajp=∑k=115[mkυijkυipk+Iinkωijkωnpk]{A_{jp}} = \sum\limits_{k = 1}^{15} \left[ {{m_k}{\upsilon _{ijk}}{\upsilon _{ipk}} + {I_{ink}}{\omega _{ijk}}{\omega _{npk}}} \right](27)fj=∑k=115[mkυijkυ˙ipkx˙p+Iinkωijkω˙npkx˙p+emqiIqnkωmpkωnhkx˙px˙h]{f_j} = \sum\limits_{k = 1}^{15} \left[ {{m_k}{\upsilon _{ijk}}{{\dot \upsilon }_{ipk}}{{\dot x}_p} + {I_{ink}}{\omega _{ijk}}{{\dot \omega }_{npk}}{{\dot x}_p} + {e_{mqi}}{I_{qnk}}{\omega _{mpk}}{\omega _{nhk}}{{\dot x}_p}{{\dot x}_h}} \right]Equation (24) forms 40 second-order nonlinear differential equations. Use the Runge-Kutta method to find numerical solutions to these equations. We can find 40 generalized coordinates. If some equations of these generalized coordinates are known, a part of the Eq. (24) becomes an algebraic equation for finding the sum of unknown forces [9]. All the coefficients in the equation can be obtained by calculating the four coefficient matrices of ωijk, υijk, υ˙ijkx˙j{\dot \upsilon _{ijk}}{\dot x_j}and ω˙ijkx˙j{\dot \omega _{ijk}}{\dot x_j}. So we can solve the dynamic equation on the computer.5Constrained dynamic equationsTo obtain the dynamic equation of the constraint system, we only need to multiply the left side of the dynamic equation of the original system by transposing the coefficient matrix of the constraint equation [10]. Therefore, from Eq. (24), the dynamic equation of the multi-rigid body mechanical model of the human body under constraints can be obtained:(28)a(Fj+Fj*)=0      aAjpx¨p=a(fj+Fjext+Fjint)\matrix{ {a\left( {{F_j} + F_j^*} \right) = 0} \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,a{A_{jp}}{{\ddot x}_p} = a\left( {{f_j} + F_j^{ext} + F_j^{{\rm{int}}}} \right)} \hfill \cr } Where a is the transposition of the coefficient matrix of the constraint equation. In the formula, Ajp, fj, FjextF_j^{ext}and FjintF_j^{{\rm{int}}}are determined by formulas (18–21).6Application examplesThe election here is one of the ten classic World Cup events-a-goal matches between Argentina and the Netherlands. The picture is the eight sports characteristic pictures from the attack from the middle of the ball to the goal. Through image reconstruction, a panoramic view of the whole process is obtained [11]. From the reconstructed panorama, the combination line of passing and receiving for the offensive midfielder is obtained as shown in Figure 3:Fig. 3Midfielder pass and catch combination lineAccording to the offensive strength, the offensive midfielder's adequate passing directions, namely, to the right and pass back. The defense of the three forwards on the offensive side is successful for the defense. Still, the defensive guards are too densely positioned to create opportunities for the proper offensive back [12]. Although a ‘6-3 situation is formed in the local area, there is unnecessary defensive overlap, resulting in insufficient defensive space. At this time, the three strikers of the offensive side are effectively blocked, and the goalkeeper's proper position is not a fatal threat to the defensive side.Figures 4 and 5 are the characteristic pictures of the ‘back bow’ on the pole in the shooting action using the improved dynamic model and the original model [13]. It can be seen that the hip movement, which is the crucial part of the technical movement, more realistically describes the characteristics of the movement by increasing the degree of freedom.Fig. 4Improved dynamics modelFig. 5Original model7ConclusionThe essential physical fitness and technical level of a football player are essential. It is an integral part of the technical level of the entire team. However, the most considerable charm of football is the ‘acting’ of talented players and the unpredictable results. The team's tactical style and performance are also the most attractive. Research has found that establishing differential equation models in football can help us use mathematical ideas, methods, and knowledge to solve practical problems. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Applied Mathematics and Nonlinear Sciences de Gruyter

Fractional differential equations in National Sports Training in Colleges and Universities

Applied Mathematics and Nonlinear Sciences , Volume 7 (1): 8 – Jan 1, 2022

Loading next page...
 
/lp/de-gruyter/fractional-differential-equations-in-national-sports-training-in-u05QU1VXzS

References (10)

Publisher
de Gruyter
Copyright
© 2021 Yan Li, published by Sciendo
eISSN
2444-8656
DOI
10.2478/amns.2021.2.00158
Publisher site
See Article on Publisher Site

Abstract

1IntroductionIn sports training in colleges and universities, people generally equate a team's offensive ability with the striker's level of ‘one-to-one’ [1]. In the fierce competition in football, the forward is an essential factor in winning, but the passing and receiving cooperation from the backcourt and midfielders is also essential.2Combination circuit for catching the ballNetwork topology has been widely used in communication networks, computer networks, and operations research. Here we use line graph analysis in network theory to discuss passing and receiving in football [2]. We first define the generalized path (1)L(ki)=∏i=1nkiL({k_i}) = \prod\limits_{i = 1}^n {k_i}Here, ki represents a quantity that affects the generalized path. In a football game, we mainly consider ‘the geometric distance between two points, the speed of the ball, the accuracy of passing, the ability to receive the ball, and the defense of the passing and receiving points.’ When multiple people pass and receive the ball multiple times, L(ki) forms a distance matrix, which follows the following rules (2)L2(ki)=L(ki)*L(ki)L3(ki)=L2(ki)*L(ki)⋯⋯{L^2}({k_i}) = L({k_i})*L({k_i}){L^3}({k_i}) = {L^2}({k_i})*L({k_i}) \cdots \cdots We define (3)C*D=E=[ejk]m×nC*D = E = [{e_{jk}}{]_{m \times n}}Among them C = [cjk]m×l, D = [djk]l×n, ejk=mincjl≠0,dlk≠0{cj1*d1k,cj2*d2k,⋯⋯cjl*dlk}{e_{jk}} = \mathop {\min }\limits_{{c_{jl}} \ne 0,{d_{lk}} \ne 0} \{ {c_{j1}}*{d_{1k}},{c_{j2}}*{d_{2k}}, \cdots \cdots {c_{jl}}*{d_{lk}}\} .For example, C = [1, 2, 3, 4, 0], D = [0, 2, 1, 2, 0]T, ejk=mincjl≠0,dlk≠0{0,2,4,6,0}{e_{jk}} = \mathop {\min }\limits_{{c_{jl}} \ne 0,{d_{lk}} \ne 0} \{ 0,2,4,6,0\} .We use Figure 1 to show the generalized path diagram of the pass when attacking. P1 represents the node formed by the defender of the possession side and the goalkeeper. P2, P3 corresponds to midfielders and full-backs. P4 stands for striker. P5 refers to the goal. Then (4)L=[0320000313000100000100000], L2=[1jk]=L*L=[0063600042000020000000000]L = \left[ {\matrix{ 0 & 3 & 2 & 0 & 0 \cr 0 & 0 & 3 & 1 & 3 \cr 0 & 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^2}{ = [1_{jk}}] = L*L = \left[ {\matrix{ 0 & 0 & 6 & 3 & 6 \cr 0 & 0 & 0 & 4 & 2 \cr 0 & 0 & 0 & 0 & 2 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right]Fig. 1Generalized path diagram of passing on offenseThe same (5)L3=L2*L=[0007400005000000000000000], L4=L3*L=[0000800000000000000000000], L5=L4*L=0{L^3} = {L^2}*L = \left[ {\matrix{ 0 & 0 & 0 & 7 & 4 \cr 0 & 0 & 0 & 0 & 5 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^4} = {L^3}*L = \left[ {\matrix{ 0 & 0 & 0 & 0 & 8 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right],\quad {L^5} = {L^4}*L = 0To find the shortest distance from Pj to Pk, we define (6)C⊕D=[ejk]m×nC \oplus D = [{e_{jk}}{]_{m \times n}}among them (7)C=[ajk]m×nD=[djk]m×n,ejk={0,cjk=djk=0cjk,cjk>0,djk=0djk,cc8=0,djk>0min(cjk,djk),cjk≠0djk≠0Lmin=L⊕L2⊕L3⊕L4⊕L5=[0323400313000120000100000]=[1jk]min\matrix{ {C = [{a_{jk}}{]_{m \times n}}D = [{d_{jk}}{]_{m \times n}},{e_{jk}} = \left\{ {\matrix{ {0,{c_{jk}} = {d_{jk}} = 0} \hfill \cr {{c_{jk}},{c_{jk}} > 0,{d_{jk}} = 0} \hfill \cr {{d_{jk}},c{c^8} = 0,{d_{jk}} > 0} \hfill \cr {\min ({c_{jk}},{d_{jk}}),{c_{jk}} \ne 0{d_{jk}} \ne 0} \hfill \cr} } \right.} \hfill \cr {{L_{\min }} = L \oplus {L^2} \oplus {L^3} \oplus {L^4} \oplus {L^5} = \left[ {\matrix{ 0 & 3 & 2 & 3 & 4 \cr 0 & 0 & 3 & 1 & 3 \cr 0 & 0 & 0 & 1 & 2 \cr 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 & 0 \cr } } \right] = \mathop {{{[1}_{jk}}]}\limits_{\min } } \hfill \cr } [1jk]min\mathop {{{[1}_{jk}}]}\limits_{\min } represents the shortest distance from Pj to Pk. The ultimate goal of passing and receiving is to get the (best) shot. There are many factors to consider when passing the ball. A node that is geometrically connected is not necessarily a valid point. From formula (1), it can be known that any factor other than the geometric distance is zero, and the generalized path is the point of leading and receiving the ball, which is the weak point [3]. From formula (2), we can know the importance of midfielders. Because P1, P5 is constant for the stadium. The value of [1jk]min\mathop {{{[1}_{jk}}]}\limits_{\min } mainly depends on the direction and size of P2, P3.3Best offensive intensityThe best-generalized path gives us the whereabouts of the ball in the pass. But how to form the best path is primarily affected by the defensive team's defensive lineup. Let us illustrate this point with a simple diagram [4]. The three U1, U2, U3 secant lines in Figure 2 below represent the three defense lines of the defender. And define (8)Ui=∑PkPk+1U1=5,U2=5,U3=4\matrix{ {{U_i} = \sum {P_k}{P_{k + 1}}} \hfill \cr {{U_1} = 5,{U_2} = 5,{U_3} = 4} \hfill \cr } We introduce the concept of generalized attack intensity and define the following formula. The attack intensity in football refers to the attack intensity, the reliability of passing and receiving the ball, and the degree of threat to the defense [5].(9)fjk↔Pjk{f_{jk}} \leftrightarrow {P_{jk}}A linear model can represent the generalized attack intensity secant (10)minfjkU=∑fjk\mathop {\min }\limits_{{f_{jk}}} U = \sum {f_{jk}}The minimum cut is equivalent to the maximum flow, so the generalized attack intensity cut is (11)minfjkU=∑fjk−∑fjk{U,j≠s0,j≠s,t−U,j=t\mathop {\min }\limits_{{f_{jk}}} U = \sum {f_{jk}} - \sum {f_{jk}}\left\{ {\matrix{ {U,j \ne s} \hfill\cr {0,j \ne s,t}\hfill \cr { - U,j = t} \hfill\cr } } \right.s, t represents a confident start and endpoint, respectively. Through the secant analysis, we know two ways to increase the attack intensity: one is to increase the attack line, and the other is to reduce the path to the attack point P5. This is the value that forms the minimum secant U3 to form the maximum attack intensity [6].Fig. 2The equivalent model of a football field pass4Unconstrained dynamic equationsWe can write the dynamic equations of the model by the Kane method (12)Fj+Fj*=0{F_j} + F_j^* = 0We replace the extra force acting on the system with an equal force system. Then the generalized active force Fj can be written as (13)Fj=∑K=115(∂Vk∂x˙jFk+∂ωk∂x˙jMk){F_j} = \sum\limits_{K = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}{F_k} + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}{M_k}} \right)Fk and Mk represent the ‘equivalent’ force and force matrix acting on the link Bk. The line of action of Fk passes through the center of the link. Gk ·Vk and ωk are respectively called the velocity and angular velocity through the point Gk in the inertial reference frame. ∂Vk∂x˙j{{\partial {V_k}} \over {\partial {{\dot x}_j}}}and ∂ωk∂x˙j{{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}are respectively referred to as ‘change rate of position deviation’ and ‘change rate of azimuth deviation.’ Mk is (14)Mk=Mkext+∑Mkl{M_k} = M_k^{ext} + \sum M_k^lMk is the moment produced by the external force acting on Bk. MklM_k^lis the force applied by Bl to Bk. It is generally caused by force Bl exerted on Bk. Therefore, rewriting (14) has (15)Mk=Fjext+Fjint{M_k} = F_j^{ext} + F_j^{\mathop {{\rm{int}}}}among them (16)Fjext=∑K=115(∂Vk∂x˙jFk+∂ωk∂x˙jMkext)F_j^{ext} = \sum\limits_{K = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}{F_k} + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}M_k^{ext}} \right)(17)Fjint=∑K=115(∑∂ωk∂x˙jMk)F_j^{\mathop {{\rm{int}}}} = \sum\limits_{K = 1}^{15} \left( {\sum {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}{M_k}} \right)Due to (18)∂ωk∂x˙j=ωijknoi, ∂Vk∂x˙j=υijknoi{{\partial {\omega _k}} \over {\partial {{\dot x}_j}}} = {\omega _{ijk}}{n_{oi}},\quad {{\partial {V_k}} \over {\partial {{\dot x}_j}}} = {\upsilon _{ijk}}{n_{oi}}Equation (17) can be rewritten as (19)Fjext=υkjkFki+ωkjkMkiextF_j^{ext} = {\upsilon _{kjk}}{F_{ki}} + {\omega _{kjk}}M_{ki}^{ext}Where Fki and MkiextM_{ki}^{ext}are the components of Fk and MkextM_k^{ext}in the noi direction, respectively [7]. Let's analyze the generalized inertial force. From Eq. (14), we can get:(20)Fj*=∑k=115(∂Vk∂x˙jFk*+∂ωk∂x˙jMk*)F_j^* = \sum\limits_{k = 1}^{15} \left( {{{\partial {V_k}} \over {\partial {{\dot x}_j}}}F_k^* + {{\partial {\omega _k}} \over {\partial {{\dot x}_j}}}M_k^*} \right)Among them Fj*=−mkak=−mk(υijkx¨j+υ˙ijkx˙j)noi, Tk*=−Ik⋅ak−ωk×Ik⋅ωkF_j^* = - {m_k}{a_k} = - {m_k}({\upsilon _{ijk}}{\ddot x_j} + {\dot \upsilon _{ijk}}{\dot x_j}){n_{oi}},\quad T_k^* = - {I_k} \cdot {a_k} - {\omega _k} \times {I_k} \cdot {\omega _k}The inertial dyadic Ik is given by (21)Ik=Imnknomnon{I_k} = {I_{mnk}}{n_{om}}{n_{on}}Among them, the value of element Imnk related to nom and nmnk is related. It has the same value as nkr and nks related element Irskk(k=1,2,3;m,n,r,s,k)I_{rsk}^k(k = 1,2,3;m,n,r,s,k). O represents the origin of the reference coordinate of the inertial reference system [8]. The expression is as follows:(22)Imnk=nomnonnksnrskk{I_{mnk}} = {n_{om}}{n_{on}}{n_{ks}}n_{rsk}^k(23)S0Kmr=nomnkr, S0Kns=nonnks, Imnk=S0KmrS0KnsIrsk{S_0}{K_{mr}} = {n_{om}}{n_{kr}},\quad {S_0}{K_{ns}} = {n_{on}}{n_{ks}},\quad {I_{mnk}} = {S_0}{K_{mr}}{S_0}{K_{ns}}{I_{rsk}}Organized (24)Tk*=−(Imnknomnon)[(w˙ijkx¨j+ωijkx¨j)noi]−(ωijkx˙jnoi)×Imnknomnonωijkx˙jnoiTk*=−[Imnk(ω˙njkx¨j+ωnjkx¨j)+eiqmωipkωnjkx˙px˙jIqnk]nom\matrix{ {T_k^* = - ({I_{mnk}}{n_{om}}{n_{on}})\left[ {\left( {{{\dot w}_{ijk}}{{\ddot x}_j} + {\omega _{ijk}}{{\ddot x}_j}} \right){n_{oi}}} \right] - \left( {{\omega _{ijk}}{{\dot x}_j}{n_{oi}}} \right) \times {I_{mnk}}{n_{om}}{n_{on}}{\omega _{ijk}}{{\dot x}_j}{n_{oi}}} \hfill \cr {T_k^* = - [{I_{mnk}}({{\dot \omega }_{njk}}{{\ddot x}_j} + {\omega _{njk}}{{\ddot x}_j}) + {e_{iqm}}{\omega _{ipk}}{\omega _{njk}}{{\dot x}_p}{{\dot x}_j}{I_{qnk}}]{n_{om}}} \hfill \cr } Among them Tk*=∑k=115{mkυijk(υipkx¨j+υ˙ipkx˙j)+ωijk[Iink(ωnpkx¨p+ω˙npkx˙p)+emqiωmpkωnhkωnhkx˙px˙hIqnk]}T_k^* = \sum\limits_{k = 1}^{15} \left\{ {{m_k}{\upsilon _{ijk}}\left( {{\upsilon _{ipk}}{{\ddot x}_j} + {{\dot \upsilon }_{ipk}}{{\dot x}_j}} \right) + {\omega _{ijk}}\left[ {{I_{ink}}({\omega _{npk}}{{\ddot x}_p} + {{\dot \omega }_{npk}}{{\dot x}_p}) + {e_{mqi}}{\omega _{mpk}}{\omega _{nhk}}{\omega _{nhk}}{{\dot x}_p}{{\dot x}_h}{I_{qnk}}} \right]} \right\}i, m, n, q and p are summed from 1 to 3. Both j and h are summed from 1 to 40. The kinetic equation (14) can be expressed in the following concise form (25)Ajpx¨p=fj+Fjext+Fjint{A_{jp}}{\ddot x_p} = {f_j} + F_j^{ext} + F_j^{{\rm{int}}}Where p is summed from 1 to 40. Ajp and fj are given by (26)Ajp=∑k=115[mkυijkυipk+Iinkωijkωnpk]{A_{jp}} = \sum\limits_{k = 1}^{15} \left[ {{m_k}{\upsilon _{ijk}}{\upsilon _{ipk}} + {I_{ink}}{\omega _{ijk}}{\omega _{npk}}} \right](27)fj=∑k=115[mkυijkυ˙ipkx˙p+Iinkωijkω˙npkx˙p+emqiIqnkωmpkωnhkx˙px˙h]{f_j} = \sum\limits_{k = 1}^{15} \left[ {{m_k}{\upsilon _{ijk}}{{\dot \upsilon }_{ipk}}{{\dot x}_p} + {I_{ink}}{\omega _{ijk}}{{\dot \omega }_{npk}}{{\dot x}_p} + {e_{mqi}}{I_{qnk}}{\omega _{mpk}}{\omega _{nhk}}{{\dot x}_p}{{\dot x}_h}} \right]Equation (24) forms 40 second-order nonlinear differential equations. Use the Runge-Kutta method to find numerical solutions to these equations. We can find 40 generalized coordinates. If some equations of these generalized coordinates are known, a part of the Eq. (24) becomes an algebraic equation for finding the sum of unknown forces [9]. All the coefficients in the equation can be obtained by calculating the four coefficient matrices of ωijk, υijk, υ˙ijkx˙j{\dot \upsilon _{ijk}}{\dot x_j}and ω˙ijkx˙j{\dot \omega _{ijk}}{\dot x_j}. So we can solve the dynamic equation on the computer.5Constrained dynamic equationsTo obtain the dynamic equation of the constraint system, we only need to multiply the left side of the dynamic equation of the original system by transposing the coefficient matrix of the constraint equation [10]. Therefore, from Eq. (24), the dynamic equation of the multi-rigid body mechanical model of the human body under constraints can be obtained:(28)a(Fj+Fj*)=0      aAjpx¨p=a(fj+Fjext+Fjint)\matrix{ {a\left( {{F_j} + F_j^*} \right) = 0} \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,a{A_{jp}}{{\ddot x}_p} = a\left( {{f_j} + F_j^{ext} + F_j^{{\rm{int}}}} \right)} \hfill \cr } Where a is the transposition of the coefficient matrix of the constraint equation. In the formula, Ajp, fj, FjextF_j^{ext}and FjintF_j^{{\rm{int}}}are determined by formulas (18–21).6Application examplesThe election here is one of the ten classic World Cup events-a-goal matches between Argentina and the Netherlands. The picture is the eight sports characteristic pictures from the attack from the middle of the ball to the goal. Through image reconstruction, a panoramic view of the whole process is obtained [11]. From the reconstructed panorama, the combination line of passing and receiving for the offensive midfielder is obtained as shown in Figure 3:Fig. 3Midfielder pass and catch combination lineAccording to the offensive strength, the offensive midfielder's adequate passing directions, namely, to the right and pass back. The defense of the three forwards on the offensive side is successful for the defense. Still, the defensive guards are too densely positioned to create opportunities for the proper offensive back [12]. Although a ‘6-3 situation is formed in the local area, there is unnecessary defensive overlap, resulting in insufficient defensive space. At this time, the three strikers of the offensive side are effectively blocked, and the goalkeeper's proper position is not a fatal threat to the defensive side.Figures 4 and 5 are the characteristic pictures of the ‘back bow’ on the pole in the shooting action using the improved dynamic model and the original model [13]. It can be seen that the hip movement, which is the crucial part of the technical movement, more realistically describes the characteristics of the movement by increasing the degree of freedom.Fig. 4Improved dynamics modelFig. 5Original model7ConclusionThe essential physical fitness and technical level of a football player are essential. It is an integral part of the technical level of the entire team. However, the most considerable charm of football is the ‘acting’ of talented players and the unpredictable results. The team's tactical style and performance are also the most attractive. Research has found that establishing differential equation models in football can help us use mathematical ideas, methods, and knowledge to solve practical problems.

Journal

Applied Mathematics and Nonlinear Sciencesde Gruyter

Published: Jan 1, 2022

Keywords: fractional differential equations; colleges and universities; physical training; generalized path; line graph analysis; transfer coordination; 35A24

There are no references for this article.