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# Training Model of Basketball Offensive Route Based on Nonlinear Differential Equation

###### Accepté: 14 Apr 2022
Détails du magazine
Format
Magazine
eISSN
2444-8656
Première parution
01 Jan 2016
Périodicité
2 fois par an
Langues
Anglais
Introduction

Basketball attracts many people who love sports with its unique charm. Some scholars have studied the mechanical analysis of shooting aiming points. At the same time, they gave the trajectory of the basketball shot. Some scholars have studied the influence of moment of momentum on basketball shooting techniques [1]. They applied the Momentum Momentum Theorem and the Momentum Momentum Conservation Law to basketball shooting techniques and carried out an example analysis. Some scholars have studied the shooting percentage of basketball robots based on the normal cloud model [2]. At the same time, they propose a normal cloud model. They simulate the precision error of basketball robots in manufacturing design and conduct simulation experiments with MATLAB. Some scholars have studied the application of multiple regression in basketball shooting. Some scholars have studied the factors affecting the hit rate of the basketball spin ball [3]. Some scholars have studied the computer simulation problem of basketball's one-handed shooting technique. The author studies the influencing factors of two-point shooting and three-point shooting based on mathematics and mechanics.

In the text, l represents the horizontal distance between the pitching point and the front edge of the basket. d is the diameter of the basket. H represents the vertical height of the basket. h represents the shot height of the basketball. l1 is the horizontal distance between the two-point line and the front edge of the rim. l2 is the horizontal distance between the three-point line and the front edge of the basket. V0 represents the initial speed of the ball at the time of release. θ represents the angle between the shot speed and the horizontal direction. We assume Δh = Hh, d = 0.450m, H = 3.050m.

Main results
Shooting from the two-point line

We use the shot point as the coordinate origin to establish a plane rectangular coordinate system. l1 is the horizontal distance between the two-point shot and the front edge of the rim. The ball moves in a straight line at a constant speed in the horizontal direction, while in the vertical direction, it is in free fall due to the acceleration of gravity [4]. The trajectory equation of the center of the ball: ${l1=V0 cos θ tΔh=V0sinθ t+12gt2$ \left\{{\matrix{{{l_1} = {V_0}\,\cos \,\theta \,t} \hfill \cr {\Delta h = {V_0}\sin \theta \,t + {1 \over 2}g{t^2}} \hfill \cr}} \right.

Eliminating the parameter t we get $V0=gl122(l1 tan θ−Δh)cos2 θ=l1g2(l1 tan θ−Δh)cos2 θ$ {V_0} = \sqrt {{{gl_1^2} \over {2\left({{l_1}\,\tan \,\theta - \Delta h} \right){{\cos}^2}\,\theta}}} = {{{l_1}\sqrt g} \over {2\left({{l_1}\,\tan \,\theta - \Delta h} \right){{\cos}^2}\,\theta}}

Where g is the acceleration of gravity.

We treat θ as a variable. $V02$ V_0^2 is a continuous function over a finite interval of [θ1, θ2]. According to the properties of a continuous function on a closed interval, we find that its maximum value must exist [5]. The minimum value of $V02$ V_0^2 is more practical. The minimum initial velocity for a shot is V0 min. So we transform the above problem into finding the minimum value of $V02$ V_0^2 . At this point we $dV0(θ)dθ=(l1g2(l1tanθ−Δh)cos2θ)′=l1g[2(l1tanθ−Δh)cos2θ)−12]′get::=−12l1g[(2l1tan θ−Δh)cos2θ)−32](2(l1 tan θ−Δh)cos2 θ)′=−12l1g[(2(l1tan θ−Δh)cos2θ)−32]2[l1 sec2 θ cos2θ+2(Δh−l1 tan θ) sin θ cos θ]$ \matrix{{} \hfill & {{{d{V_0}\left(\theta \right)} \over {d\theta}} = {{\left({{{{l_1}\sqrt g} \over {2\left({{l_1}\tan \theta - \Delta h} \right){{\cos}^2}\theta}}} \right)}^{'}} = {l_1}\sqrt g {{\left[{^2\left({{l_1}\tan \theta - \Delta h} \right)\left. {{{\cos}^2}\theta} \right) - {1 \over 2}} \right]}^{'}}} \hfill \cr {{\rm{get:}}} \hfill & {= - {1 \over 2}{l_1}\sqrt g \left[{\left({^2{l_1}\tan \,\theta - \Delta h} \right)\left. {{{\cos}^2}\theta} \right) - {3 \over 2}} \right]{{\left({2\left({{l_1}\,\tan \,\theta - \Delta h} \right){{\cos}^2}\,\theta} \right)}^{'}} =} \hfill \cr {} \hfill & {- {1 \over 2}{l_1}\sqrt g \left[{\left({^2\left({{l_1}\tan \,\theta - \Delta h} \right){{\cos}^2}\theta} \right) - {3 \over 2}} \right]2\left[{{l_1}\,{{\sec}^2}\,\theta \,{{\cos}^2}\theta + 2\left({\Delta h - {l_1}\,\tan \,\theta} \right)\,\sin \,\theta \,\cos \,\theta} \right]} \hfill \cr}

We assume that $dV0(θ)dθ=0$ {{d{V_0}\left(\theta \right)} \over {d\theta}} = 0 , according to the method of finding the maximum value of the unary function, we get l1 sec2 θ cos2 θ + 2(Δhl1 tan θ)sin θ cos θ = 0.

We use sec $secθ=1cosθ$ \sec \theta = {1 \over {\cos \theta}} , 2 sin2 θ − 1 = cos 2θ, 2 sin θ cos θ = sin 2θ to get l1 + 2Δh sin θ cos θ − 2l1 sin2 θ = 0.

We can get l1 sec2 θ + 2Δh tan θ − 2l1 tan2 θ = 0 by dividing cos2 θ on both sides. $l1(tan2θ+1)−2l1tan2θ+2Δh tan θ=0−l1 tan2 θ+2Δh tan θ+l1=0tan θ=−2Δh±(2Δh)2+4l12−2l1=−Δh±(Δh)2+l12−l1$ \matrix{{{l_1}\left({{{\tan}^2}\theta + 1} \right) - 2{l_1}{{\tan}^2}\theta + 2\Delta h\,\tan \,\theta = 0} \hfill \cr {- {l_1}\,{{\tan}^2}\,\theta + 2\Delta h\,\tan \,\theta + {l_1} = 0} \hfill \cr {\tan \,\theta = {{- 2\Delta h \pm \sqrt {{{\left({2\Delta h} \right)}^2} + 4l_1^2}} \over {- 2{l_1}}} = {{- \Delta h \pm \sqrt {{{\left({\Delta h} \right)}^2} + l_1^2}} \over {- {l_1}}}} \hfill \cr}

Then the minimum throwing angle is $θ=arctanΔh+(Δh)2+l12l1$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + l_1^2}} \over {{l_1}}}

We bring Equation (3) into Equation (2) to get the best shot speed $Vmin=l1g2(l1 tan θ−Δh)cos2θ$ {V_{\min}} = {{{l_1}\sqrt g} \over {\sqrt {2\left({{l_1}\,\tan \,\theta - \Delta h} \right){{\cos}^2}\theta}}} .

Where $θ=arctanΔh+(Δh)2+l12l1$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + l_1^2}} \over {{l_1}}} .

If the ball is hit on the back edge of the basket, the trajectory equation of the basketball is: ${l1+d=V0 cos θ tΔh=V0sinθ t+12gt2$ \left\{{\matrix{{{l_1} + d = {V_0}\,\cos \,\theta \,t} \hfill \cr {\Delta h = {V_0}\sin \theta \,t + {1 \over 2}g{t^2}} \hfill \cr}} \right.

In the same way, we get the minimum ball angle and speed $Vmin=(l1+d)g2((l1+d) tan θ−Δh)cos2θ$ {V_{\min}} = {{\left({{l_1} + d} \right)\sqrt g} \over {\sqrt {2\left({\left({{l_1} + d} \right)\,\tan \,\theta - \Delta h} \right){{\cos}^2}\theta}}}

Where $θ=arctanΔh+(Δh)2+(l1+d)2(l1+d)$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {{\left({{l_1} + d} \right)}^2}}} \over {\left({{l_1} + d} \right)}} .

So = the best angle θ for the two-point shot is between $θ=arctanΔh+(Δh)2+(l1+d)2(l1+d)$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {{\left({{l_1} + d} \right)}^2}}} \over {\left({{l_1} + d} \right)}} and $θ=arctanΔh+(Δh)2+l12l1$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + l_1^2}} \over {{l_1}}} .

In the case of a two-point shot, the ball height h ∈ [1.95, 2.35], Δh ∈ [1.1,1.5]. We see from equation (3) that θ is a function of Δh. According to common sense, the two-point shot l1 = 4.375m, l1 + d = 4.825m, θh) also has a value range: θ(1.1) ≈ [0.863,0.867] ≈ [49.4°, 50.6°], θ(1.5) ≈ [0.89 9,0.923] ≈ [51.4°, 52.6].

Then we get the best shot speed $Vmin=(l1+d)g2((l1+d) tan θ−Δh)cos2θ$ {V_{\min}} = {{\left({{l_1} + d} \right)\sqrt g} \over {\sqrt {2\left({\left({{l_1} + d} \right)\,\tan \,\theta - \Delta h} \right){{\cos}^2}\theta}}} .

Δh ∈ [1.1.1.5], l1 = 4.375m, l1 + d = 4.825m. We can get the best two-point release speed V0 between [7.29,7.91]m / s. V0 ∈ [7.29, 7.91].

Since the shot is at the 2-point tee, the minimum 2-point shot release speed is required in this situation [6]. In fact, because of the influence of friction and any shooting point of the two-point shot is slightly farther than the free-throw line of the two-point shot, we can adjust the shooting speed appropriately according to the situation to improve the shooting rate.

Three-point line situation

l2 is the horizontal distance between the three-point line and the front edge of the rim [7]. If the ball hits the front edge of the basket, we also use the calculation method of Case 1 to obtain the minimum angle and speed $Vmin=(l1+d)g2((l1+d) tan θ−Δh)cos2θ$ {V_{\min}} = {{\left({{l_1} + d} \right)\sqrt g} \over {\sqrt {2\left({\left({{l_1} + d} \right)\,\tan \,\theta - \Delta h} \right){{\cos}^2}\theta}}} of the ball.

Where $θ=arctanΔh+(Δh)2+(l2+d)2(l2+d)$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {{\left({{l_2} + d} \right)}^2}}} \over {\left({{l_2} + d} \right)}} .

So the best angle θ for the three-point shot is between $θ=arctanΔh+(Δh)2+(l2+d)2(l2+d)$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {{\left({{l_2} + d} \right)}^2}}} \over {\left({{l_2} + d} \right)}} and $θ=arctanΔh+(Δh)2+l22l2$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + l_2^2}} \over {{l_2}}} .

The ball height h h ∈ [2.40.2.80], Δh ∈ [0.45, 0.85] in the case of a three-point shot [8]. We are based on basketball common sense three-pointer line l2 = 6. 525m, l2 + d = 6.975m, θ(0.45) ≈ [0.811, 0.814] ≈ [46.71°, 46.8 9°], θ(0.85) ≈ [0.844, 0.858] ≈ [48.61°, 48.92°].

According to the formula $Vmin=(l1+d)g2((l1+d) tan θ−Δh)cos2θ$ {V_{\min}} = {{\left({{l_1} + d} \right)\sqrt g} \over {\sqrt {2\left({\left({{l_1} + d} \right)\,\tan \,\theta - \Delta h} \right){{\cos}^2}\theta}}} , Δh ∈ [1.1.1.5], l2 = 4.375m, l2 + d = 4.825m, we can get the best shot speed V0 of the three-pointer between [7.98, 8.47]m / s. V0 ∈ [7.98, 8.47].

Shooting from anywhere

The calculation method of the two-point shot is easy to find that the best release angle θ is between $θ=arctanΔh+(Δh)2+l2l$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {l^2}}} \over l} and $θ=arctanΔh+(Δh)2+(l2+d)2(l+d)$ \theta = \arctan {{\Delta h + \sqrt {{{\left({\Delta h} \right)}^2} + {{\left({{l_2} + d} \right)}^2}}} \over {\left({l + d} \right)}} .

If the shooting point is within the two-point line, the best shot angle is θ < [51.4°, 52.6°]. If the shooting point is outside the two-point line and within the three-point line, the best shot angle is [46.71°, 46.89°] < θ < [49.4°, 50.6°]. If the shooting point is outside the three-point line, the best shot angle θ < [46.71°, 46.89°]. If the shooting point is within the two-point line, the best shot speed is V0 < [7.29, 7.91]. If the shooting point is outside the two-point line and within the three-pointer, the best shot speed is [7.29, 7.91] < V0 < [7.98, 8.47]. If the shooting point is outside the three-point line, the best release speed is V0 > [7.98,8.47].

Considering air resistance

If we consider air resistance, the trajectory of the two-pointer and three-pointer hitting the leading edge of the backboard is: ${l1=V0 cos θ t+12axt2,i=1,2Δh=V0sinθ t+12ayt2$ \left\{{\matrix{{{l_1} = {V_0}\,\cos \,\theta \,t + {1 \over 2}{a_x}{t^2},i = 1,2} \hfill \cr {\Delta h = {V_0}\sin \theta \,t + {1 \over 2}{a_y}{t^2}} \hfill \cr}} \right.

Where $ax=f cos θm$ {a_x} = {{f\,\cos \,\theta} \over m} , $ay=f sin θm−g$ {a_y} = {{f\,\sin \,\theta} \over m} - g , i = 1, 2. The trajectory of hitting the back edge of the backboard is: ${li+d=V0 cos θ t+12ax t2,i=1,2Δh=V0sinθ t+12ayt2$ \left\{{\matrix{{{l_i} + d = {V_0}\,\cos \,\theta \,t + {1 \over 2}{a_x}\,{t^2},i = 1,2} \hfill \cr {\Delta h = {V_0}\sin \theta \,t + {1 \over 2}{a_y}{t^2}} \hfill \cr}} \right.

Simplified ${li=V0 cos θ t+t2f cos θ2mΔh=V0sinθ t+t22(f sin θm−g)$ \left\{{\matrix{{{l_i} = {V_0}\,\cos \,\theta \,t + {{{t^2}f\,\cos \,\theta} \over {2m}}} \hfill \cr {\Delta h = {V_0}\sin \theta \,t + {{{t^2}} \over 2}\left({{{f\,\sin \,\theta} \over m} - g} \right)} \hfill \cr}} \right.

The second equation of equation (7) is multiplied by cosθ, and the first equation of equation (7) is multiplied by sinθ subtracting it. $Δh cos θ−li sin θ=−g cos θ t22$ \Delta h\,\cos \,\theta - {l_i}\,\sin \,\theta = {{- g\,\cos \,\theta \,{t^2}} \over 2}

From the square root of both sides of Equation (8), we can get $t=2(li sin θ−Δh cos θ)g cos θ$ t = \sqrt {{{2\left({{l_i}\,\sin \,\theta - \Delta h\,\cos \,\theta} \right)} \over {g\,\cos \,\theta}}}

We substitute equation (9) into the first equation of equation (7) to get $t=li−(li sin θ−Δh cos θ) fmgV0cosθ$ t = {{{l_i} - {{\left({{l_i}\,\sin \,\theta - \Delta h\,\cos \,\theta} \right)\,f} \over {mg}}} \over {{V_0}\cos \theta}}

We combine equations (9) and (10) to get $V0=li−(li sin θ−Δh cos θ) fmg2(li sin θ−Δh cos θ) cosθg$ {V_0} = {{{l_i} - {{\left({{l_i}\,\sin \,\theta - \Delta h\,\cos \,\theta} \right)\,f} \over {mg}}} \over {\sqrt {{{2\left({{l_i}\,\sin \,\theta - \Delta h\,\cos \,\theta} \right)\,\cos \theta} \over g}}}}

Take the derivative of the above formula concerning Δh $V′=g2(lisinθ−Δhcosθ)cosθ{fmgcos θ(lisinθ−Δhcosθ)cosθ−[li−f(lisinθ−Δh cos)mg]−cos2θ2(lisinθ−Δhcosθ)cosθ}=0fmgcosθ(lisinθ−Δhcosθ)cosθ−[li−f(lisinθ−Δhcosθ)mg]−cos2θ2(lisinθ−Δhcosθ)cosθ=0⇒fmg cos θ(lisinθ−Δhcosθ)cosθ=[li−f(li sin θ−Δh cos θ)mg]−cos2θ2(lisinθ−Δhcosθ)cosθ⇒2fmgcos2θ(lisinθ−Δh cos θ)=−cos2θ[li−f(lisinθ−Δh cos θ)mg]$ \matrix{{V' = {{\sqrt {{g \over 2}}} \over {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta}}\left\{{{f \over {mg}}\cos \,\theta \sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta} -} \right.} \hfill \cr {\left. {\left[{{l_i} - {{f\left({{l_i}\sin \theta - \Delta h\,\cos} \right)} \over {mg}}} \right]{{- {{\cos}^2}\theta} \over {2\sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta}}}} \right\}} \hfill \cr {= 0{f \over {mg}}\cos \theta \sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta} - \left[{{l_i} - {{f\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)} \over {mg}}} \right]} \hfill \cr {{{- {{\cos}^2}\theta} \over {2\sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta}}} = 0 \Rightarrow {f \over {mg}}\,\cos \,\theta \sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta} =} \hfill \cr {\left[{{l_i} - {{f\left({{l_i}\,\sin \,\theta - \Delta h\,\cos \,\theta} \right)} \over {mg}}} \right]{{- {{\cos}^2}\theta} \over {2\sqrt {\left({{l_i}\sin \theta - \Delta h\cos \theta} \right)\cos \theta}}} \Rightarrow {{2f} \over {mg}}{{\cos}^2}\theta \left({{l_i}\sin \theta - \Delta h\,\cos \,\theta} \right)} \hfill \cr {= - {{\cos}^2}\theta \left[{{l_i} - {{f\left({{l_i}\sin \theta - \Delta h\,\cos \,\theta} \right)} \over {mg}}} \right]} \hfill \cr}

Arranged to get: $Δh=lisinθ+mglifcosθ$ \Delta h = {{{l_i}\sin \theta + {{mg{l_i}} \over f}} \over {\cos \theta}}

Arranged to get $Δh=lisinθ(μcosθ−2+2μ sin θ)μcosθ(2 cos θ+1)$ \Delta h = {{{l_i}\sin \theta \left({\mu \cos \theta - 2 + 2\mu \,\sin \,\theta} \right)} \over {\mu \cos \theta \left({2\,\cos \,\theta + 1} \right)}} . $μ=fmg$ \mu = {f \over {mg}} is the coefficient of friction.

This allows us to find the best shot height for two-pointers and three-pointers [9]. Because of the presence of air friction, the optimal release angle for a two-point 3-pointer requires a smaller release angle and relatively higher release speed than a frictionless 2-point 3-pointer V0 to make a shot.

Simulation of the optimal solution for shooting

Examples of shooting angles and entry angles for different shooting speeds and shooting heights are shown in Table 1. We set the shooting height to 2.6 meters [10]. The horizontal distance is 7.3 meters. The shot speed is 8.8 m/s. The angle is 40° to 50°. At this time, the basketball trajectory we get is shown in Figure 1.

Shooting Angles and Entry Angles for Different Shooting Speeds and Shooting Heights

Shooting height The horizontal distance between the shooting point and the center of the basket Speed Angle range Hit rate
2.6m 7.4m 8.7m/s 40°–50° 60%
2.6m 7.4m 8.8m/s 40°–50° 90%
2.6m 7.4m 8.9m/s 40°–50° 20%
2.6m 7.4m 9.0m/s 40°–50° 90%
2.6m 7.3m 8.7m/s 40°–50° 80%
2.6m 7.3m 8.8m/s 40°–50° 100 %
2.6m 7.3m 8.9m/s 40°–50° 90%
2.6m 7.3m 9.0m/s 40°–50° 75%

We set the shooting height to 2.6 meters. The horizontal distance is 7.3 meters [11]. The shot speed is 8.7 m/s. The angle is 40° ~ 50°. At this point, the basketball trajectory we get is shown in Figure 2.

When the shooting height, horizontal distance, and shooting angle are the same, and the shooting speed is different, the shooting angle is between 40° and 50°. All shot trails end within the basket. The two-shot trajectories of 40° and 41° in Figure 2 end with no goal basket. The difference in shooting speed between the two pictures is only 0.1 m/s. The hit rate of Figure 2 is 20% lower than that of Figure 1. When the shooting height is 2.6 meters, the best shooting speed is 8.8 meters per second.

Conclusion

We looked at shooting percentages and the factors that affect them. The trajectory of a basketball depends on a certain release speed, release angle, and shot height. We have mastered the influencing factors of basketball shooting percentage, and we can improve the shooting quality by strengthening training and summing up some laws. The research conclusions of the article can provide a reference for basketball fans and basketball teaching and training.

#### Shooting Angles and Entry Angles for Different Shooting Speeds and Shooting Heights

Shooting height The horizontal distance between the shooting point and the center of the basket Speed Angle range Hit rate
2.6m 7.4m 8.7m/s 40°–50° 60%
2.6m 7.4m 8.8m/s 40°–50° 90%
2.6m 7.4m 8.9m/s 40°–50° 20%
2.6m 7.4m 9.0m/s 40°–50° 90%
2.6m 7.3m 8.7m/s 40°–50° 80%
2.6m 7.3m 8.8m/s 40°–50° 100 %
2.6m 7.3m 8.9m/s 40°–50° 90%
2.6m 7.3m 9.0m/s 40°–50° 75%

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