Correlation between badminton shot and height

After being able to put together all the results from the calculations to find the angle and plot it against the maximum height for each service on the graph, it becomes clearer visually the relationship between the two variables. We see as the angle of the service increases, so does the maximum height. This then proves my hypothesis to be correct. I’ve realized a trend where the angle increases, so does the b variable in each of the equations of the parabola. Serve #3, which had the highest maximum height and angle, has the highest b variable of a 4. While some of the serves with lower maximum heights and angles have a number below 0 as their b variable. The trend line is roughly placed in equally between each of the points, and it’s obvious to see the curve that the points make.

I would have to say that technically, there aren’t any outliers within the data. Both #2 and #3 are farther in maximum height compared to the rest but that data is what supports my hypothesis. As seen from the diagrams of the slope that is then used to create the right triangle and find the angle, when the angle increases, the line becomes slowly more into a straight horizontal line. I’ve also realized that as the maximum decrease, the vertical distance of the shuttlecock increases.

Now applying this into a real-life situation, it’s much easier for your opponent to hit back the shuttlecock when it's at a higher angle because it causes it increases in height, giving the opponent actually the time to possibly even smash the shuttlecock back to you, earning them a point. It’s better to keep the angle towards a lower range, allowing the shuttlecock to move faster and closer towards the net, making it difficult for the opponent to hit it back. It also allows the player to have more control as to where they want the shuttlecock to go.

We can see the range of outcomes of an upper hand serve from this. The difference between the three parabolas, f(x), g(x), and j(x), is how far the shuttlecock could travel. However, compared to the rest of the serves, parabolas h(x) and i(x) were able to rise higher.

To find the different parabolas of the trajectory of the shuttlecocks, I’ll be having a volunteer who’ll repeatedly hit underhand serves five times in a row, which I’ll video from the side view. For it to be reliable, I’ve chosen a distance that will be constant throughout the recording of the video. This player will also be standing from the same point, right behind the front server line, and using an underhand serve through all of the trials.

In terms of then graphing the curves on Desmos, I’ve taken screenshots from the video at the point where the player is about to hit the shuttlecock and another point where the shuttlecock has touched the ground. These images are then imported into Desmos, which will help me find the equation for the parabolas.

I’ll put the point where the shuttlecock is hit at (0,0).

As can be seen from the tangent function calculation for service number 3, the slope was found to be the same as the equation's constant. I can see from this that the term x is solely represented by its derivatives, or the outcomes of differentiation. being used to determine the slope when x = 0 makes it significant.

By exploring this topic, its application can be used in sports widely throughout the world for different purposes, from those who professionally play to beginners and even just those who enjoy watching the sport. Personally as well, I can understand math better through a real-life situation that I take an interest in.

Comparing various skill manoeuvres, such as a serve, a smash, and a drop, and examining how their differences in parabolas, including their slope and angle, could be another relevant topic that could be studied to increase the knowledge. The relationship between probability and statistics in terms of the likelihood that someone would win or receive a point based on their abilities may also be interesting to observe.

The results that I expect to come out of this research are that as the angle of the service increases, so will the maximum height. As an angle increases, it turns closer to a straight horizontal line; therefore, it’s as if the shuttlecock is aimed more upwards instead of in a forward direction. Because of this, then, I would also assume that the distance of how far the shuttlecock would go would decrease as the angle increases. Following the physics law, seeing that the shuttlecock is increasing in height, its fall would be much sharper at a similar angle downwards.

We must first determine the slope of each parabola at a point to comprehend the relationship between the angle at which a cork is launched from the ground and its greatest height (0,0). Calculus plays a key role in this. I can obtain the derivative of each equation using differentiation and use that information to determine the angle.

The equation for the first serve was \(f(x)\ =\ \frac{-3x^2\ +\ 27x}{20.025}\) then plug into the differentiation formula to find the slope at only one point.

Slope = \(\frac{change\ in\ Y}{change\ in\ X}\ =\ \frac{dy}{dx}\)

\(\frac{dy}{dx},\ \frac{-3x^2\ +\ 27x}{20.25}\)

\(f'(x)\ =\ \frac{20.25(-6x\ +\ 27)}{410.0625}\)

If 𝑥 = 0, then by plugging that into the equation, we get 𝑓'𝑥 = 1.33.

Now that we have the derivative, I can then find the angle. If 𝑥 = 1, then 𝑦 = 1.33, we can now create a right-angled triangle to find the missing angle, which is the angle the shuttlecock was hit at point (0,0). Using trigonometry functions, in this case using the tangent function since we know the opposite and adjacent side of the angle, we find the angle:

\(\theta\ =\ tan^{-1}\ \frac{1.33}{1}\)

= 53.06^o

Badminton has been a sport that’s been a part of my life in various ways. It’s a sport that my father has always loved to play since he was young, and now I have also followed in the same footsteps. But also being Indonesian, our country has won several medals in badminton through the years at the Olympics and other highly accredited tournaments, which my family sometimes likes to watch together. It’s definitely one of our many proud achievements and reasons to be patriotic, and we are hoping to win more this year at the 2016 Olympics in Rio. Being a player myself, I’ve understood the struggles and challenges when trying to hit the shuttlecock as to where I want to but having it then go another direction. Being both frustrated and curious, I wanted to figure out exactly why it moved the way it did. I’m not a physics student as. Personally, it’s not my strongest subject. However, I can still try to understand it through the maths that I’ve learned so far.

Badminton is a sport for the fast-thinkers, having to be able to move swiftly in split-second decisions in order to keep the shuttlecock off the ground and back to the opponent. However, as important as having fast reflexes, being able to understand the projectile of a shuttlecock as well allows a player to play even better. Unlike other sports, the projectile of a shuttlecock is asymmetrical due to its characteristic of having a denser cork end compared to its feathered ‘skirt’. With my curiosity of this fact, I’ve decided to explore the curves formed when serving, specifically the angles of the shot and how it affects its maximum height. I’ll find the parabola for numerous serves and then apply areas of algebra and calculus to find its mathematical functions to establish an answer.

Finding mathematical correlations and calculating numbers in areas of arithmetic that I thought to be fairly difficult have been some of my personal struggles and challenges during the exploration of this topic. But now that I've challenged myself to discover answers to numerous puzzles and comprehend even better in both the mathematical or badminton-related areas I've learned and in areas I wasn't even aware of before this assessment, I'm really proud of my work. I did my best to maintain my resolve and attention as I searched for more information to learn more about calculus and its applications. My interest in completing this research was maintained by my strong passion for sports, whether I was participating or just observing. It provided me the motivation to keep looking for an answer to satisfy my curiosity.

This exam also allowed me to identify my strengths, which I had doubted would be possible. As I personally believe that I am better with language than I am with actual arithmetic, I enjoyed writing about the steps I took to compute things in order to reach an answer. As I mentioned before, I personally find that visual learning helps me understand concepts better, so I also like utilising graphs, tables, and other visual representations to present data. Of course, I still have areas in which I can work to strengthen. If I had used exact numbers—some of which I had to estimate—or even values with more decimal points, the results might have been more dependable or accurate, but I opted to keep the computations as straightforward as possible. Additionally, I can do a lot better in math, especially in calculus as I have never taken it before. Although I have a fundamental understanding of it, there are some ideas or details that I haven't yet been able to grasp.

Once I’ve established which points of the screenshots from the video that I will use to help me find my parabolas, I can now find each curve’s equation. This could be found through the quadratic equation:

𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0

But since the starting point of where the shuttlecock will be hit is at the (0,0) point for each service, thus then there is no need for the variable c. The maximum height will also be estimated for all serves as it shows calculations from other areas of maths related to the quadratic equation. I’ve also decided that as a way to keep the calculations simple that I should round up the numbers to only one decimal place. Now taking what I can see from the graph, I can then substitute it into this formula to find b:

\(\frac{-b}{2a}=4.5\)

−𝑏 = 4.5 × 2𝑎

b = -9a

This can then be plugged into the quadratic equation as,

𝑎𝑥² − 9𝑎𝑥 = 0

From here, we can substitute more numbers:

3 = 𝑎 4.9² − 9𝑎 (4.9)

= 𝑎[ (4.9)² − 9 (4.9) ]

= 𝑎 (24.01 - 44.1)

\(a=\frac{3}{-20.25}\)

Therefore the equation for the first parabola would be\(f(x)=\frac{-{3x}^2+27x}{20.25}\) Plugging the equation into Desmos, I could then check if the equation were roughly correct.

What is the correlation between the angle of a shuttlecock when hit and its maximum height?

I made sure that all of the screenshots were roughly the same size when inserted onto Desmos in order for the results to be as accurate as possible. I had to combine two separate screenshots in order to view the entire result because there were some issues where the shuttlecock had fallen outside of the frame. Additionally, I've discovered around halfway through solving each equation that it would be lot simpler to perform calculus if there were no denominator because there would be fewer rules to adhere to. It becomes more simpler to compute and comprehend, making it far more accurate.