A study of shuttlecock's trajectory in badminton.

Introduction

Most studies in the past explored the speed of shuttlecocks. Virtually no research has been done on the flight trajectory of the badminton shuttlecock. By understanding the flight trajectory of the badminton shuttlecock we can predict the speed, time, direction and path, and this would provide helpful information for training of players.

Essentially, it is easy to understand that the force and angle of stroke can affect the speed and trajectory of the shuttlecock. Moreover, air resistance force is an important factor that influences performance. It is generally recognized that the Reynolds number R play a crucial role in determining whether the law of linear or quadratic air resistance force is used (The Engineering Tool Box, 2005). Moreover, an experiment on vertical fall shows that the best model is the quadratic air resistance force in the instantaneous speed of the falling shuttlecock (Peastrel, Lynch and Angelo, 1980). Therefore, a motion equation of the badminton shuttlecock's flying trajectory could be constructed by determining the terminal velocity in aerodynamics. It is meaningful that we try to construct an aerodynamics equation of the badminton shuttlecock trajectory.

The main purpose of this study is to construct and confirm the motion equation of a shuttlecock's flying trajectory. Applying the equation to badminton, the speed, time, direction and path of a shuttlecock will be able to be predicted, thus contributing to sports training in badminton.

Methods

When a shuttlecock is in flight, according to Newton's Second Law:

W + [F.sub.v] + B = ma (1)

In equation (1): W is gravitational force, [F.sub.v] is aerodynamic drag force, and B is buoyancy. As a matter of fact, the measured air buoyancy of standard shuttlecock, for example which has a volume of 19 [cm.sup.3] in the air (density~1.205kg*[m.sup.3]), to be about 0.02 gw. We may neglect its very slight influence, when compared with the force of gravity and aerodynamic drag. The magnitude of aerodynamic drag force depends on the relative speed of the shuttlecock through the air, and its direction is always opposite to the direction of the shuttlecock. Generally, the magnitude of resistance force can be expressed as:

[F.sub.V] = [bv.sup.n] = (Thornton and Marion, 2003) (2)

Where v is the speed of the shuttlecock relative to air, parameter n is a real number and b is a constant that depends on the properties of the air and the shape and dimension of the shuttlecock. The two parameters b and n in general were determined by experiments.

As a shuttlecock is falling down vertically, the speed and resistance force will increase (Eq. 2). The rate of acceleration becomes zero when the increased resistance force eventually balances the weight. At this point, the shuttlecock reaches its terminal velocity [v.sub.T] and from then on it continues to move with zero acceleration. After this point, the motion of a shuttlecock is under constant velocity. The terminal velocity can be obtained from Eq. (1) and (2) by neglecting buoyancy and setting a = dv/dt = 0. This gives

mg - [bv.sup.n.sub.t] = 0 [right arrow] [v.sub.T] = [(mg/b).sup.1/n] (3)

Here, measuring the parameters b and n could be replaced measuring its terminal velocity [v.sub.T]. This helps us to find the eligible trajectory of the shuttlecock.

Resistance force could be modeled in two ways, either proportional to object speed or to the speed squared. Assume the shuttlecock is hit with initial velocity [v.sub.i], then, the horizontal and vertical velocities are expressed by

[v.sub.xi] = [v.sub.i] cos [[theta].sub.i, [v.sub.yi] = [v.sub.i] sin [[theta].sub.i] (4)

where angle [[theta].sub.i] is the initial angle.

n = 1, we have [F.sub.v] = [F.sub.vx]i + [F.sub.vy] j (5)

[F.sub.vx] = [bv.sub.x] and [F.sub.vy] = [bv.sub.y] are the x - component and y - component of air resistance.

We consider vertical and horizontal directions respectively.

(a) Vertical component:

-mg - [bv.sub.y] = m [dv.sub.y]/dt (6)

The value of [vy is "+" for upward and "-" for downward. Then we may integrate Eq. (6) to obtain vertical velocity.

[v.sub.y(t) = (v.sub.t + v.sub.yi) [e.sup.-gt/[v.sub.t] - [v.sub.t] (7)

and height

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

Where m is the mass of shuttlecock, [v.sub.yi] is the initial vertical velocity and we have taken y = 0 at t = 0, g is the gravitational acceleration, and [v.sub.t] = mg/b is the terminal velocity.

From Eq. (7), at the top of trajectory, that is, the point of maximum height H, we have [v.sub.y] = 0 , this gives the time of flight

t = [v.sub.t]/g 1n [v.sub.t] + [v.sub.yi]/[v.sub.t] (9)

(b) Horizontal component:

[-bv.sub.x] = m [dv.sub.x]/dt (10)

We may integrate Eq. (10) to obtain horizontal velocity

[v.sub.x] = [v.sub.xi][e.sup.-gt/[v.sub.t]] (11)

and horizontal distance

x = [v.sub.t.sub.xi/g (1 - [e.sup.-gt/[v.sub.t]] (12)

Combine Eq. (8) and (12), then we have the equations of the trajectory as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)

n = 2, consider the air resistance force of the shuttlecock as [F.sub.v] = [F.sub.vx]i + [F.sub.vy]j, here [F.sub.vx] = cos [theta] [bv.sup.2] cos [theta] and [F.sub.vy] = [bv.sup.2] sin 2 [theta] are the x - component and y - component of air resistance force, respectively. Obviously, the factor v in each of the force component expressions is the essential coupling between the x and y equations that prevents any analytic solutions from being found. These equations can only be solved satisfactorily accurately using analytical method. But in order to describe the role of terminal velocity and approach the practical trajectory of a shuttlecock, let us consider motion in the vertical and horizontal directions separately.

A. Vertical directional motion: in this case

[F.sub.] = [F.sub.vy] = [bv.sup.2.sub.y] - mg - [bv.sup.2.sub.y] = m [dv.sub.y]/dt (14)

After integration, we have the solution of vertical velocity as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (15)

Where [v.sub.t'] = [(mg/b).sup.1/2] is the terminal velocity. When the shuttlecock reaches the highest point where we have [v.sub.y] = 0, , the time of flight 't' is expressed as follows:

t = [v.sub.t']/g [tan.sup-1]([v.sub.yt]/[v.sub.t']) (16)

[FIGURE 1 OMITTED]

At this moment, the height of apex is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (17)

B. Horizontal direction: here we have

[F.sub.v] = [F.sub.vx] = [bv.sup.2.sub.x] and - [bv.sup.2.sub.x] = m [dv.sub.x]/dt (18)

and could find the horizontal speed as

[v.sub.x] = [v'.sup.2.sub.t]/g 1n ([v.sub.xi]gt + v'.sup.2.sub.t]/[v'.sup.2.sub.t]) (19)

and horizontal distance as

x = [v'.sup.2.sub.t]/g 1n ([v.sub.xi]gt + [v'.sup.2.sub.t]/[v'.sup.2.sub.t]) (20)

Combine Eq.(17) and (20), we have the equation of the trajectory as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (21)

Theoretically, the relationship between air resistance and speed was revealed in n and b. At first sight, it seems that to find the trajectory of the shuttlecock, we need to measure n and b in the beginning. However, from the derived results of Eq. (13) or Eq. (21), we have found that the coefficients n and b could be determined by the terminal velocity [v.sub.t]. In other words, if the terminal velocity was measured, the shuttlecock's trajectory might be found.

Results

Figure 1 is the experimental schematic diagram, where X is the video camera, [Q.sub.A] is the initial shuttlecock position, [Q.sub.Adj] is the position after parallax correction, and d is the distance between the shuttlecock and scale. We found that the terminal velocity of some shuttlecocks ranged from 6.51 to 6.87 m*[sup.-1]. One of the experimental shuttlecocks has the mass of 5.19 g, a head diameter of 2.70 cm and neck diameter of 4.20 cm, a tail diameter of 6.50 cm and a head-to-tail length of 9.00 cm. The shuttlecock was released from a height of 18 m, and experimental y(t) was measured. In Figure 2a and 2b, it was found that the best equation for measuring and calculating a shuttlecock trajectory was that of the quadratic air resistance force. From the measured data, we could plot the v(t)-t diagram in Figure 2c and the terminal velocity 6.86 m*[sup.-1] was found. This is very close to the value 6.80 m*[sup.-1] that Peastrel, Lynch and Angelo (1980) had measured. Figure 3 and 4 are the theoretical trajectories from Eq. (13) and (21). Where Y (n = 1) and Y1(n = 2) in Figure 3 both have same initial velocity [v.sub.i] = 30 m*[sup.-1] [[theta].sub.i]= 30[degrees] but consider the different condition tha [F.sub.V] = [b.sub.v] and [F.sub.v] = [bv.sup.2] respectively. We found the shuttlecock landed with a horizontal distance of 17.48 m and 9.65 m respectively. Therefore, the shuttlecock in the first case will land out side of the standard court, which has a length of 13.41 m (44 ft). Generally, in practice, the initial speed of a shuttlecock in overhead stroke hit is below 56 m*[sup.-1]. In other words, at [[theta].sub.i] 30[degrees] = the initial speed of most players is often greater than [v.sub.i] = 30[degrees] m*[sup.-1]. Very often, the flying shuttlecock lands inside the court. This also helps us to judge intuitionally. What is the relationship between air drag resistance force and the speed of a shuttlecock that have been described above. Apparently, the possibility of the first condition, which states that air drag resistance force is proportional to the speed of a shuttlecock (i.e. [F.sub.v] = bv), is very low. Nevertheless, we must point out, in particular, the relationship between air drag resistance force and the speed of a shuttlecock. The trajectory of a shuttlecock could be described by using terminal velocity. However, we can test again with fast overhead stroke that has an initial velocity of vi = 56 m*[sup.-1] [[theta].sub.i] 30[degrees] and 60[degrees]. Their trajectories Y2 (n = 2) and Y3 (n = 2) are also shown in Figure 3. From Eq. (16) and (17), we also found that the shuttlecock needs 0.88s and 0.94s to reach the top of 6.81m and 9.31m high respectively, and totally 2.33s and 2.63s to alight with a horizontal distance of 12.52m and 10.65m. From Eq. (15) and (19), we could also found that the vertical and horizontal velocities. For example, if the initial angle is adjusted to 60 degree, their vertical and horizontal landing velocities are 6.4 m/s and 1.5 m*[sup.-1]. Their ratio is about 4.3 : 1, which interplays the almost vertical landing phenomenon observed. Next in Fig. 4, the trajectories of clear, smash, and drop have their initial velocity of [v.sub.i] = 45m/s, [[theta].sub.i] = 0[degrees], [v.sub.i] = 70m/s, [[theta].sub.i] = -10[degrees] and [v.sub.t] = 25m/s, [[theta].sub.i] = -20[degrees] were shown. Y4 (n=2), Y5 (n=2), and Y6 (n=2) respectively. Their reaction time for received player also could be found if necessary.

[FIGURE 2 OMITTED]

[FIGURE 3 OMITTED]

[FIGURE 4 OMITTED]

Discussion

The special structure of a shuttlecock makes its trajectory perform unsymmetrical motion when playing. From the terminal velocity of a shuttlecock, we could find that the air drag force is proportional to the square of the speed of the shuttlecock. The result is consistent with Peastrel, Lynch and Angelo (1980), who performed an experiment on vertical fall to measure the terminal velocity of a shuttlecock. They found that the best model was the quadratic air resistance force. Tong (2004) performed an identical experiment that also came up with the same result. Furthermore, the angle and strength of a stroke could influence its trajectory. And the equation we found can predict the trajectory of the shuttlecock. Some known values were substituted to check the experimental data. This equation also works adequately on the motion of a shuttlecock under smashing. In this paper, we have considered both the linear and the quadratic air resistance force laws respectively. We found that the trajectory of a shuttlecock could be expressed in terms of its terminal velocity, which means it is unnecessary to find the other parameters, like air drag force. The results revealed that the motion equation of a shuttlecock's flying trajectory could fit experimental data. It also revealed that the angle and force of a stroke could influence trajectory, therefore playing an important part in making strategic plans. This should be helpful for badminton player training. These findings were the same as what Tsai et al. (1995; 1997) had come up with. From the motion equations, it is easy to understand why the order of a shuttlecock speed is: jumpsmash, smash, clear, and drop among badminton forehand overhead strokes--because stroke force and angle can affect the trajectory of a shuttlecock significantly.

Moreover, testing a shuttlecock for speed, according to the "Laws of Badminton" by the Badminton World Federation (2006), was specified as law (3.1): to test a shuttlecock, using full underhand stroke which makes contact with the shuttlecock over the back boundary line. The shuttlecock shall be hit at an upward angle and in a direction parallel to the side lines. As law (3.2) suggests, a shuttlecock will land no less than 530 mm and no more than 990 mm short of the other back boundary line as illustrated in Diagram B (not shown). This subjective test is worthy of consideration since the stroke force affects the initial speed, and the initial angle affects the landing distance. For example, if the initial speed is 40 m*[sup.-1], with the release angle of 30 and 60 degrees, their landing distances are 12.50 m and 10.82 m, respectively. Therefore, an objective and revised test is necessary. As an illustration, if we let a shuttlecock fall from a height of 5 m, it will land on the ground after ~1.20 s.

Conclusion

From the terminal velocity of a shuttlecock, the conclusion reveals that the equation of this study could predict the trajectory of a shuttlecock, and it shows that air drag force is proportional to the square of a shuttlecock velocity. Additionally, the angle and strength of a stroke could also influence trajectory. Finally, this study presented a suggestion that we could use a scientific approach to test a shuttlecock's velocity objectively and to replace the traditional and subjective method made by the Badminton World Federation. Finally, research findings can be applied to improving the professional knowledge of badminton player training.

Key points

* The motion equation of a shuttlecock's flying trajectory could be constructed by determining the terminal velocity in aerodynamics.

* Air drag force is proportional to the square of a shuttlecock velocity. Furthermore, the angle and strength of a stroke could influence trajectory.

Received: 17 November 2008 / Accepted: 14 October 2009 / Published (online): 01 December 2009

References

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Lung-Ming Chen (1), Yi-Hsiang Pan (2) * and Yung-Jen Chen (3)

(1) Department of Computer and Communication, Sue-Te University, Taiwan ROC,

(2) Graduate Institute of Physical Education, National Taiwan Sport University, Taiwan ROC,

(3) Taipei County Ell-Chorng Elementary School, Taipei County, Taiwan ROC

* Yi-Hsiang Pan

Graduate Institute of Physical Education, National Taiwan Sport University, 250, Wenhua 1st Rd., Kueishan, Taoyuan 333, Taiwan ROC.

AUTHORS BIOGRAPHY

Lung-Ming CHEN

Employment Professor, Department of Computer and Communication, Sue-Te University, Taiwan ROC.

Degree PhD

Research interests Sport biomechanics

E-mail: chenlm@mail.stu.edu.tw

Yi-Hsiang PAN

Employment Assistant Professor, Graduate Institute of Physical Education, National Taiwan Sport University, Taiwan ROC.

Degree PhD

Research interests Physical Education teaching and training

E-mail: poterpan@seed.net.tw

Yung-Jen CHEN

Employment Taipei County Ell-Chorng Elementary School, Taiwan ROC.

Degree BSc

Research interests Physical education teaching and training

E-mail: chicken1008@yahoo.com.tw
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