# Exploring the Mathematics of a 3-points Shot in Basketball

How it works

Since the age of 7 when I had my first encounter with the sport of basketball, it has been my passion ever since. Having my feet stamped on the wooden floor, the stinging sensation of leather in my palms and the net above my head, I have been devoting the past ten years for the “splashing” sound of the ball when it hits the basket. It was abnormal, yet satisfying. I was enthralled by how Stephen Curry just instinctively hit the ball in the exact spot at the basket every time that he scores. He was able to perform the shot with angle and speed that send the ball into the basket smoothly, without any bounce. Therefore, I decided to do an exploration on the movement of the ball, specifically ball trajectories, and factors that might influence its movement. Here, I choose to focus on a 3-points shot as it provides a clear process of ball trajectories.

To make a perfect 3-points shot, it is determined that the factors of speed and angle at which it is released are crucial effects upon the ball trajectories as well as the success of the performance. After the ball is released and in the air, there are four main forces that act on the ball, include gravitational force, buoyant force, drag force, and Magnus force when the ball is spinning (G. K. Hung).

For the purpose of comparison, I will first construct a “under vacuum condition”, where all of the forces are not taken into account for this basic model. Then, I will explore the aspects of Magnus effect or spin of the ball that could promote possible effects upon trajectory. Air resistance and drag force will be explored last for the construction of a theoretical experiment. As stated above, a comparison will be made between a theoretical experiment and experimental model in which I will construct a real-life sample of myself performing the shot and do an analysis of the record. As I am aware of how changes in angles and speed at the point of release could have of the success rate of my shots, an analysis and comparison would served the purpose of exploration of a perfect 3-point shot.

## BASIC MODEL

The basketball movement during a shot is defined as a projectile motion, which describes the movement of a particle travelling through air in time. Here, all of the variables are assumed using facts and examples.

Firstly, the point of release is assumed to be at approximately 2 meters (my own height is 1.65 meters, the height of the jump is approximately 20 centimeters and inflection of arm at the point of release and I measured half-arm, which is 44 centimeters).

The diagram illustrates the path of the basketball through the court into the hoop. The height of the hoop is 10 feet or approximately 3.048 meters (NBA, 2002) and the distance of the 3-points line to the hoop is approximately 8 meters (NBA, 2002). As the 3-points line is an arc, I will assume the distance from the top to the hoop.

These mathematical equations obtained above are essential in order to investigate the trajectory of a ball in motion as well as can be used to determine the perfect model which will ensure a scoring shot. However, it is quite complex and certainly cannot be calculated during a game in which a player is currently performing a shot. For real-life situations, it relies on the player’s intuition and practices to be able to perform with a suitable launch angle and speed of release. Also, there are other forces that will contribute to the trajectory of a ball which spinning is a crucial part for a scoring secret.

### TOPSPIN/BACKSPIN

A crucial factor for a perfect release condition is backspin, which deadens the motion of a basketball after it bounces off of a ring (CHAU M. TRAN & LARRY M. SILVERBERG, 2008). The effect ensures that the ball stay closer to the rim of the hoop after a bounce which results in a higher chance for a successful shot.

The graph above illustrates the motion of a ball bouncing of a surface in case of without spin and with backspin. As a ball bounces into a surface, there are two motions going on at the same time: one that acts straight into the surface whilst the other is parallel to the surface (Orzel, 2017). For a ball without spin, there is minimal effect of friction and the ball bounces off the same angle that it goes in, illustrated on the diagram. On the other hand, for a ball with a backspin, the effect of friction is opposite from the forward motion of the ball and therefore slow the ball down. The exit angle of a ball that is thrown with backspin is larger than without and if it is thrown with a significant amount of backspin, the ball will bounce back toward the angle of release.

As I begin to learn how to shoot the ball correctly, my coach always make note of how I need to put more spin on the ball. “Spin it! Flick your wrist!” He said. Back then, it appears to me that a fancy flick of the wrist will make the ball spin, with absolutely no idea why, only it will make the ball go into the hoop more often.

For a ball with backspin, the air pressure around the ball increase and experiences an upward force. Magnus force always acts perpendicular to the drag force and spin axis. The diagram on the left show an upward FL on the spinning ball.

### Formula of Magnus Force

- F=CL Ad(v22)

Where F is magnus force and CL is lift coefficient, A is cross-sectional area of the ball, d is density of air (d=1.21 kg/m3) and v is the ball speed (m/s).

- ? CL=1(2+vv(spin))this is formula for the lift coefficient as the ball experiences an upward lift force with backspin.
- ? v(spin)=Ras the peripheral speed of the ball.

=angular speed about the horizontal axis that is right-angle to the path of motion of the ball (Hz or radians/seconds).

R= radius of the ball. In this case, the standard diameter of NBA game ball is 9 inches, equivalent to 0.2286 m for radius.

Assume that =3 Hz for an optimum condition of a shot.

- 1 Hz=2rad/sec
- ? 1Hz ? 6.283 rad/sec
- ? 3 Hz ? 18.85 rad/sec
- ? v(spin)=0.2286 18.85=4.309 m/s

Here, I assume v=10.63 m/sas I took Stephen Curry’s release speed for a model as he is currently the best 3-point shooter in the NBA.

- ? CL=12+(10.634..309)=0.22

Drag force is a backward force applied on the ball as it travels through air as the air pressure on the front is larger than the back of the ball. The value of drag force depends on size of the ball, the speed it goes into motion and air density. The formula for drag force is generated below:

- F=CdAd(v22)

Where F is drag force and Cd is drag coefficient, A is the cross-sectional area of the ball (A=r2), d is air density(d=1.21 kg/m3), v is ball speed (m/s). From research, I assumed that Cd is 0.54. For a standard game ball, the radius is 0.2286 m with air density of approximately 1.225 kg/m3at sea level and 15oC.

The formula shows that there is a relationship between the drag force F and the ball speed v that drag force is proportional to the ball speed squared. Accordingly, for a large release speed, the ball slows down as it flies through air, resulting in a higher chance of bouncing into the hoop than hitting the rim. This equation seems to make sense as if taken the example of Stephen Curry’s shooting mechanic with a release speed of 10.63 m/s, a huge number and yet this gave him a perfect shooting form.

I was aware of the concept of air resistance, yet, I did not know that such a factor could propose a significant effect on the trajectory of a ball and the chances of a successful shot. During this investigation, I come to a point of realization for the difference in ball size and weight between sexes. Every time that we have a game, the coach always notify us and make sure that we use the girl’s ball which is smaller in size and lighter in weight compared to that of the boys. As there’s variation in strength between the two sexes, a lighter and smaller ball, therefore, allows the female player to shoot with higher speed as suitable strength can be put into the release action. I have tried to shoot with the boy’s ball myself and the result of made shots was much lower than when I am using the girl’s ball, which is explained through this equation.

From the two formulas for magnus force and drag force above, new equations for horizontal and vertical motions are generated:

mdvxdt= -Fd cos-FL sin (horizontal)

mdvydt= FL cos-Fd sin-mg(vertical)

- ? ax = -kv (Cd vx + CLvy) (horizontal)
- ? ay = kv (CL vx – Cdvy) -g (vertical)

Here, k=d r22mwith m is the mass of the ball.

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