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Modeling Of Volleyball

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Modeling of Volleyball Bumps with Vectors and Planes
Introduction
In my search for a Mathematics Internal Assessment topic, I have decided to model one of my favorite sports, volleyball. Over the past few years of playing volleyball, the one action that I have the most difficult time executing properly is the bump. Bumps in volleyball are when you extend both your arms straight out in front of you and hit the ball up by contacting it with the underside of your forearms.

The main aspect of the bump that is most difficult for me is the angle of the arm extensions, which is the most important component for aiming the bumps. After years of trial and error, I have become better at bumping semi-accurately, but I still have trouble keeping my bumps …show more content…

This is because vectors can be used to describe things with magnitude and direction. A volleyball's displacement has both of magnitude (distance) and direction. The arms extended out can be modeled as planes, and by looking at how different angles affect the rebound trajectory of the volleyball, I can manipulate the angle of the plane in relation to the bumper to get the right direction of reflection towards the setter.

The Problem
How do I angle myself in relation to the setter when bumping a spike?

Volleyball Basics
My exploration will start with investigating the basics of volleyball.
Volleyball Court Dimensions: 18m x 9m
The center line divides the court into two halves (9m x 9m), with each team occupying one.
The attack line is 3m away from the center line and 6m away from the back (service) line.

The top of the net in men’s volleyball is about 2.43m high, which is the number I will be using in my exploration.
All other dimensions are irrelevant.

Positions:
There are six positions the team’s half of the court, each position held by one player, totaling up to six players in each team. The main receivers are positioned in the two back corners, left and right. I personally do most of my receiving at the back left corner when the other team spikes from the right diagonal. Therefore, in my modeling, I will only focus on the back left corner …show more content…

This was one of the trickier parts that I was not prepared for. It took me some time to figure out, but I managed to use trigonometry to solve it. In order to use the trigonometry, I had to first make a right triangle. This I found using Points F, G, and a third, Point H: (5, -4.35, 0.825). This third point forms the right angle between ("FH" ) ̅ "and" ("GH" ) ̅, modeled below:

Now I need to use this angle to find the two elbow points.
The blue angle was calculated by: tan^(-1)⁡〖0.015/0.1071=〖7.97〗^° 〗 cos⁡〖〖(25.77〗^°)*0.1081=〗 0.0973 cos⁡〖〖(9.83〗^°)*0.1081=〗 0.1065 sin⁡(⁡〖〖25.77〗^°)*0.1081=〗 0.0470 sin⁡(⁡〖〖9.83〗^°)*0.1081=0.0185〗 These orange and blue numbers represent the corresponding lengths of the figure, which defines the relative position of the two elbow points. Also, the elbows are 0.105m above Point F. So, using these numbers:
Point I =(3.925i-4.35j+0.825k)+(-0.0470i-0.0973j+0.105k)
= (3.878, -4.4473, 0.93)
Point J =(3.925i-4.35j+0.825k)+(-0.0185i-0.1065j+0.105k)
= (3.9065, -4.4565, 0.93)
Finally we can create the forearms' plane:
("FI" ) ⃑=(-0.0470i-0.0973j+0.105k)
("FJ" ) ⃑=(-0.0185i-0.1065j+0.105k)
Plane Y =(3.925i-4.35j+0.825k)+λ(-0.0470i-0.0973j+0.105k)+μ(-0.0185i-0.1065j+0.105k)
The following depicts the model so far (Plane Y is

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