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Basketball Shooting Rate Based on Multiple Regression Logical-Mathematical Algorithm

Pubblicato online: 15 Jul 2022
Volume & Edizione: AHEAD OF PRINT
Pagine: -
Ricevuto: 15 Apr 2022
Accettato: 12 Jun 2022
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License
Formato
Rivista
eISSN
2444-8656
Prima pubblicazione
01 Jan 2016
Frequenza di pubblicazione
2 volte all'anno
Lingue
Inglese
Introduction

Basketball is one of the most popular sports in our lives. Improving shooting accuracy is something we all look forward to. We found that shot accuracy was related to the angle and speed at which the basketball entered the hoop. In this paper, a dynamic model is established for the fixed-point shooting process. This paper theoretically explores the conditions for the hollow of basketball into the frame to improve the shooting accuracy in basketball. The process of basketball from the shot to the hollow into the frame can be divided into two movement processes [1]. There is no need to consider the basketball shape during the previous exercise. We can think of basketball as a particle. At this time, the movement process of basketball can be regarded as oblique throwing movement. During the last movement, the movement process is very short. This process can be approximated as a uniform linear motion. We need to consider the impact of the actual shape of the basketball on the frame. Therefore, we first regard the basketball as a mass point and establish the trajectory model of the basketball center. We get the condition that the basketball center passes through the center of the basket. On this basis, we analyze the conditions that the incident angle should satisfy when the basketball is about to enter the frame. Finally, the analysis of the article gives the angle of the shot (lateral declination) and the allowable deviation of the speed.

Conditions for the center of the ball to pass through the center of the basket
Conditions for the basketball trajectory to pass through the center of the basket

Consider first the case where the center of the basketball ball hits the center of the basket. At this time, we can regard the basketball as a particle without considering the blocking of the basket. From the oblique throw motion knowledge, we can decompose the shot release speed along with the horizontal and vertical directions [2]. At this time, we obtain the kinematic equations of the basketball in the horizontal and vertical directions and then establish the trajectory model of the basketball center. Since the center of the basket is on the trajectory of the basketball, we substitute the coordinates of the center of the basket into the trajectory equation of the basketball. At this point, we can get the functional relationship between the shot speed and the angle.

Assume that the release velocity is in the same plane as the center of the basket when shooting. There is no lateral deflection in the model at this time. Basketball only moves in the plane determined by the center of the ball, the center of the basket, and the shot’s speed. Without considering the rotation of the basketball, we regard the basketball’s motion as a two-dimensional oblique throwing motion [3]. So we set the origin of the coordinates at the center of the ball when the basketball is shot. The horizontal direction is set to the x-axis direction, and the vertical direction is set to the y-axis. At this point, we establish a plane rectangular coordinate system, as shown in Figure 1.

Figure 1

Establishment of shooting parabola and coordinate system

The x-axis is the horizontal direction, and the y-axis is the vertical direction [4]. The basketball is thrown at t = 0 with release speed v0 and release angle θ. According to Newton’s second law, the dynamic equations in the horizontal and vertical directions can be expressed as follows: { Md2x(t)dt2=0Md2y(t)dt2=Mg

Where g is the acceleration of gravity and M is the mass of the basketball. Substitute the initial condition { dx(t)dt|t=0=υ0cosθx(0)=0{ dy(t)dt|t=0=υ0sinθy(0)=0 . In this way, we solve the equations of motion of the basketball in the horizontal and vertical directions: { x(t)=υ0tcosθy(t)=υ0tsinθgt22

From equation (2), we can get the shooting trajectory model: y=xtanθx2g2υ02cos2θ

We substitute the coordinate x = L, y = H - h of the center of the rim into the shooting trajectory model. This way we can get Hh=LtanθL2g2υ02cos2θ . After tidying up, we can get L2g2υ02tan2θLtanθ+L2g2υ02+Hh=0 . This is a quadratic equation in one variable for tanθ. The preconditions for its solution are: Δ=L2(12gυ02(Hh+gL22υ02))0 : υ02g[ Hh+L2+(Hh)2 ]

From the root formula we get: tanθ=υ02gL[ 1±12gυ02(Hh+gL22υ02) ]

Under a certain shooting height h, we make the initial velocity of equation (4) equal to be Vmin. Obviously Vmin is a decreasing function of h. The higher the shot height, the smaller the minimum shot speed. This is consistent with everyday life experience. There are two shot angles determined by the formula (5). The unified standard for the height of the basket is H=3.05m, the horizontal distance between the three-point line and the center of the basket is L=6.75m, and the acceleration of gravity g is 9.8m/s2. The shot height is h=2m. We use MATLAB for simulation [5]. The simulation results are shown in Figure 2.

Figure 2

The relationship between release speed and release angle

We think of basketball as a particle. The relationship between the release angle θ and the release speed v0 when the basketball center passes through the center of the basket is tanθ=υ02gL[ 1±12gυ02(Hh+gL22υ02) ] . Among them, the shot speed v0 needs to meet: υ0g[ Hh+L2+(Hh)2 ] .

Conditions for basketball into the basket

The basketball cannot be considered a particle when it is about to enter the basket due to its blocking. When we look at the disk on the table, the closer our eyes are to the plane of the disk, the more “flat” we see. A disk is only round when viewed directly above it. The area is also the largest at this time [6]. The basketball’s trajectory was flying diagonally when shooting is a parabola. Still, when the basketball is close to the basket, we can think that the basketball enters the basket in a straight line at a uniform speed. The diameter of the basket is D. According to the projection of the side view of FIG. 3, the major semi-axis of the ellipse is a=D/2, and the minor semi-axis is b = D sin β / 2 .

Figure 3

Projection of side view with basketball in the hoop

So we make the center of the basketball pass through the center of the hoop. We must make the short axis of the incident section not less than the diameter of the basketball sphere: 2b ≥ 2r. Arranged: sinβrR

The basketball diameter is 0.246m and the diameter of the basket is 0.45m. After sorting out, we can get the basketball incident angle β ≥ 33.1°. When the basketball is just put into the basket (β = 33.1°), we see the trajectory from the direction of the basketball movement, as shown in Figure 4.

Figure 4

The schematic diagram of the basketball just hitting the hoop

The negative value of the derivative of the trajectory of the basketball center at x = L is equal to the tangent of the angle of incidence, i.e.: β=dydx|x=L . We substitute it into the motion trajectory equation (3) to solve the relationship between the incident angle and the shooting angle as follows: tanβ=tanβ+gLυ02cos2θ

Since the range of the shot angle and the incident angle is (0°, 90°), we can get sin[ arctan(tanθ+gLυ02cos2θ) ]rR by substituting it into (6).

If the basketball is regarded as a solid sphere, the release angle and release speed should satisfy { tanθυ02gL[ 1±12gυ02(Hh+gL22υ02) ]sin[ arctan(tanθ+gLυ02cos2θ) ]rRυ02g[ Hh+L2+(Hh)2 ] when the basketball center passes through the center of the basket.

The maximum lateral declination of a basketball shot

We project the basket to a perpendicular plane to the incident direction and pass through the front of the basket. The basket is an ellipse, as shown in figure [7]. The basketball is tangent to this projected ellipse at maximum offset. We only need to consider what happens when the center of the basketball hits the major axis AB of the ellipse and establish the equation of the ellipse and the great circle of the basketball. At this point, we calculate the maximum deviation distance between the center of the ellipse and the center of the great circle. We divide the maximum offset distance by the distance OO’ from the release point to the center of the basket to obtain the tangent of the lateral deflection angle. Since the maximum lateral deflection angle is small, this angle is approximately equal to the tangent of this angle. From this we can find the maximum lateral deflection angle [8]. The established coordinate system is shown in Figure 5.

Figure 5

Coordinate system for lateral declination

The parameter of the ellipse is a=D2,b=D2sinβ,c=D2cosβ . Assume that the center coordinate of the circle is (m, 0) when it is tangent to the ellipse. At this point, we assume m > 0. Due to symmetry, we only consider the case on the right [9]. According to the properties of the inscribed circle of the ellipse, we can discuss it in the following two cases:

When the circle and the ellipse are tangent to the ellipse vertex (a,0), it should satisfy: c2a|m|<a . The radius of the circle is now r = a- | m |. Because of r = d/2, there is m=D2d2 . And because of c2a|m|<a , we can get m=D2d2>c2a .

Substituting in the ellipse parameters we can get D2d2>Dacos2β . D2sin2β>d2 . So at this time sin2β>d2 .

There is a formula since the basketball incident angle is in the range of (0, 90°) sinβ>dD .

When the circle and the ellipse are not tangent to the ellipse vertex (a,0), there should be | m |<c2a . The radius of the circle is now r=bc(c2m2) .

Because of r = d/2, we get m=D24cos2βd24cot2β . And because of | m |<c2a , we get D24cos2βd24cot2β<c2a . We can get D24cos2βd24cot2β<D2acos4β . D24cos2β(1cos2β)>d2cos2β4sin2β by substituting the parameters of the ellipse. Finished to get sin4β<d2D2 . Since the basketball incident angle β has a value range of (0, 90°), sinβ<dD .

The distance m between the center of the great circle and the center of the ellipse of the basketball satisfies: m={ D24cos2βd24cot2β,dD<sinβ<dDm=D2d2,sinβ>dD

We use MATLAB simulation to obtain the relationship between m and β as shown in Figure 6.

Figure 6

Relationship between m and β

From the above conclusions, it can be seen that the maximum offset distance of left and right is m=D2d2=0.102m . The basketball incident angle β is at least 47.67°. From Figure 6, it can be seen that the lateral declination angle is very small, so: α=tanα=mL2+(Hh)2 . When m takes the maximum value of 0.102, the lateral deflection angle α is the largest [10]. At this point, we solve for the maximum lateral deflection angle α 0.8594°.

Analysis of allowable deviation when basketball is released
Tolerance of release speed and angle

First, find the relationship between the release angle deviation and the landing deviation [11].

The basketball trajectory equation is as follows: y=xtanθx2g2υ02cos2θ

The above formula can be transformed into ycos2θ=xsinθcosθx2g2υ02=12xsin2θx2g2υ02 . The partial derivative of the above formula can be obtained: 2ycosθsinθ=12xθsin2θ+xcos2θx2g2υ02xθ . 2ycosθsinθ=xcos2θ=(12sin2θx2g2υ02)xθ .

From this, it can be deduced that the influence formula of the deviation of the release angle on the deviation of the landing point is as follows: xθ=2ycosθsinθ+xcos2θx2g2υ0212sin2θ .

We look for the relationship between the release speed deviation ∂υ0 and the resulting drop deviation ∂x0. Due to the convenience of finding partial derivatives, we let w=υ02 . At this time, equation (9) can be transformed into 2ycos2θ=xsin2θx2gω . Thus ω=gx2xsin2θ2ycos2θ .

Taking the partial derivative of w concerning x we can obtain ωx=gx2sin2θ4xycos2θ(xsin2θ2ycos2θ)2 . xω=(xsin2θ2ycos2θ)2gx2sin2θ4xycos2θ=(2xsinθ2ycosθ)22gxtanθ4gxy .

Suppose k = y/x we can get xω=(sinθyxcosθ)2g2tanθgyx=2g(sinθkcosθ)2tanθ2k . Hence xυ0=xωωυ0=4υ0g(sinθkcosθ)2tanθ2k . We have obtained the relationship between the landing point deviation caused by different ideal projection angles and projection speeds through the above process: { xθ=ysin2θ+xcos2θx2gυ0212sin2θxυ0=4υ0g(sinθkcosθ)2tanθ2k .

We can analyze the size of the landing deviation caused by different ideal projection angles and projection speeds [12]. At this time, we need to find the size of xθ,xυ0 and then analyze the relationship of hits when there is a deviation.

Since the deviation of angle and speed is very low when shooting with superior release speed and angle, the following approximate relationship { ΔxΔθxθΔxΔυ0xυ0 can be obtained [13].

Therefore, the allowable deviations of different shot angles and speeds corresponding to Δx are: { Δυ0=ΔxΔxΔυ0=Δx4υ0g(sinθkcosθ)2tanθ2kΔθ=ΔxΔxΔθΔxysin2θ+xcos2θx2gυ212sin2θ

We can calculate the allowable deviation of Δx corresponding to different shot angles and speeds [14].

Example analysis

When the shot angle is 60°, we set the shot speed size v0 to be 8.7m/s. From the formula (7), it can be obtained that the basketball incident angle β = 68.75°. Therefore, it can be seen from (8) that the maximum allowable deviation of the drop point Δx = 0.102m, and at this time x = 6.25m, y = 0.85m. We can substitute it into (10) to calculate Δυ0=0.1024×8.79.8cos260°(tan60°0.136)2tan60°2×0.136=0.1562m/s .

When the shot speed is 9m/s, we can take out the hand angle of 55°. According to formula (7), it can be obtained that the basketball incident angle β = 55.9894° at this time, so it can be known from (8) that the maximum allowable deviation of the landing point is Δx = 0.102m. At this time, x=6.25m, y=0.85m. We substitute it into (10) to calculate: ΔθΔxxθ=0.1020.85×sin100°+6.25×cos110°6.252×9.89212sin110°=0.3242° .

After the above solution, it can be obtained that the shot angle is 60° when x and y are fixed. When the shot speed size v0 is 8.6m/s, we allow the speed deviation to be 0.1562m/s. At this time, the deviation of the shot speed can be controlled between -0.1562m/s ~ 0.1562m/s. When the given shot speed v0 = 9m/s, the shot angle is 55°. At this time, the shooting angle deviation can be controlled between -0.3242° and 0.3242°.

Conclusion

The results of this paper have theoretical and practical application significance. We can generalize the results to a basketball player shooting drills. We generalize the theory to practical training. The basketball shooting angle and shooting force can be controlled within a certain range for training using some scientific methods. Athletes can respond quickly by flexibly applying the scientific principles of this article when preparing to shoot, depending on their position and situation. Athletes save physical strength to improve the hit rate and win the game with a more relaxed attitude.

Figure 1

Establishment of shooting parabola and coordinate system
Establishment of shooting parabola and coordinate system

Figure 2

The relationship between release speed and release angle
The relationship between release speed and release angle

Figure 3

Projection of side view with basketball in the hoop
Projection of side view with basketball in the hoop

Figure 4

The schematic diagram of the basketball just hitting the hoop
The schematic diagram of the basketball just hitting the hoop

Figure 5

Coordinate system for lateral declination
Coordinate system for lateral declination

Figure 6

Relationship between m and β
Relationship between m and β

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