Anyone who has ever been to the gym has most likely seen or used an exercise ball. They’re big, inflatable balls that many people use to help them facilitate exercise or to stretch. They’re good for strengthening your core and improving your balance. Incorporating an exercise ball into your daily workout will add a challenge to your routine. A lot of people may also use them at home due to the fact that they are very affordable and versatile. While they may seem harmless, there can be serious risks when using one. It is important to use the correct size ball based on your height in order to get the most effective workout. It has to be inflated properly in order to work correctly. An over or under inflated exercise ball will put you at a greater risk for injury. The biggest danger of an exercise ball is having it burst. Over-inflation, improper use, and excessive user weight are some common ways that the ball may burst. The first step in preventing an exercise ball from bursting is choosing the right material. The material needs to be able to support the user and keep the integrity of the ball. Some common materials are vinyl and plastic. It’s important to figure out the force that is being applied to the ball, in order to figure out the strength of the material needed. The material needs to have a high tensile strength to support the user and prevent the ball from bursting and injuring the user.
The first step is figuring out the appropriate size of the ball based on height. When sitting on the ball, your knees should be at a right angle and your thighs should be parallel to the ground. The average woman in the United States according to the CDC is 64 inches tall and 166 lbs (75.3 Kg). According to anthropometric dimensional data, the ground to her knees is .285 (Anthropometric data-197t3u3) of her height.
.285 * 64 inches = 18.24 inches
This means that the proper exercise ball should be 18 inches in diameter or 46 centimeters.
To find the best material, we can calculate Young’s modulus based on Hooke’s Law.
?= stress = Force/Area
E= Young’s modulus
ɛ= strain= Δl/l
Surface Area of Sphere = 4 * π * r^2
- Exercise ball undergoes elastic and linear response to force
- All of the users weight is being applied over the top half of the ball
- Assume the exercise ball displaces 1 inch when force is applied
Force = 75.3 Kg * 9.8 m/s^2 = 738 Newtons
Half the area of sphere = ½ * 4 * π * 9.12^2= 523 in^2 = .337 m^2
? = 738 N / .337 m^2 = 2190 N/m^2
ɛ= strain= Δl/l
Ɛ = 1 inch/18.24 inch = .0548
2190 N/m^2 = E * .0548
Young’s modulus: E = 4.00 x 10^4 Pa
From this calculation, an exercise ball would need to be made of a plastic or vinyl with a Young’s modulus of around 4.00 x 10^4 Pa.
The assumption that the user’s total weight is distributed on the entire top half of the exercise ball led to the Young’s modulus to be underestimated. In real life, the user’s full weight wouldn’t be on the ball and it would also be distributed over a smaller surface area. The actual material would have a higher Young’s modulus than the calculated value.
Learn more about the things you can do with an exercise ball!