# Difficult Toy Ball Analysis

Note, as I have my degree in chemistry, I denote "=>" as the next step in the equation. Also, sorry for any obvious mistakes that I made - as I mentioned, my degree is not in engineering. Thanks!

I am considering a design for a PET spherical toy ball. The ball will have a radius of 0.07 m that will have pressurized air in it at a pressure of 5.81 atm (588698.3 Pa). Also, the ball will include a 35 nm coating of Reduced Graphene Oxide (RGO) as a durable coating on the outside of the ball.

To get the thickness of the walls of the ball, I know that the ultimate tensile strength of PET is 55 MPa. Because I want a 15% safety factor for the ball, I do the following calculation: $$(1-15\%)\cdot55 = 46.75\text{ MPa} = 46750000\text{ Pa}$$

I then use the reduced ultimate tensile strength to find the thickness of the wall using the formula for a thin-walled sphere where $p$ is pressure, $t$ is thickness, $r$ is radius, and $\sigma$ is stress:

$$\begin{gather} \sigma = \dfrac{pr}{2t} \\ 46750000 = \dfrac{588698.3 \cdot 0.07}{2t} \\ \therefore t = 4.41\times 10^{-4}\text{ m} \end{gather}$$

I can calculate the burst pressure of the sphere $P$ for pressure (gauge) inside sphere, $FS$ for factor of safety, $\sigma$ for allowable stress and additionally $R_i$ for inner radius, $R_o$ for outer radius:

$$P = \dfrac{(R_o^2-R_i^2)\sigma}{(R_i^2)FS}$$

For calculation of bursting pressure, we take $\sigma$ as ultimate stress for a given material and put $FS=1$.

Note here that I chose the express the ultimate tensile strength of PET as $5.5 \times 10^7$:

$$P = \dfrac{((0.07 + 4.41 \times 10^{-4})^2 - 0.07^2)\cdot 5.5 \times 10^7}{0.07^2 \times 1} = 695183.67\text{ Pa} = 6.86\text{ atm}$$

I now want to calculate the volume of PET needed to create a sphere of the desired dimensions:

\begin{align} V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3 - \left(\dfrac{0.140}{2}\right)^3\right) \\ &= 1.36 \times 10^{-5}\text{ m}^3 \end{align}

I now want to calculate the volume of Reduced Graphene Oxide (RGO) to apply to the surface of the sphere:

\begin{align} V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4} + 35 \times 10^{-9}}{2}\right)^3 - \left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3\right) \\ &= 1.084366 \times 10^{-9}\text{ m}^3 \end{align}

Can you see any mistakes in my calculations?

Please do not ask why I am using PET as a material or other design decisions like that. There are good reasons for them.

• What is the question here? Jun 6, 2018 at 20:19
• The question is whether the calculations are done correctly for the intended project and whether the margins allocated (such as safety margin) are adequate for the task. I don't think that a 15% safety margin is enough, but I don't know what safety margin is good enough for this application. Jun 6, 2018 at 22:16
• I wonder who will be able to afford a graphene-coated ball toy.
– SF.
Jun 7, 2018 at 13:22

I don't understand what you're trying to achieve, whether a 15% factor of safety is correct, whether having a factor of safety only on the material strength (instead of a combination of strength-minimizing and load-maximizing factors) or any such thing is adequate.

I also didn't bother checking your calculations (as in, punching numbers into a calculator). But just looking at your reasoning, it seems fine enough.

The only mistake I spotted was in the volume calculations, where you calculate the volume using diameters, but then add the wall thicknesses only once. If you're using diameters, you should add the thicknesses twice, once for each side of the ball.

Or just simplify your life and use radii instead (as you did for the rest of the question):

\begin{align} V &= \dfrac{4}{3}\pi \left(r_o^3 - r_i^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4}\right)^3 - 0.07^3\right) \\ &= 2.73\times 10^{-5}\text{ m}^3\text{ of PET} \\ V &= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4} + 35\times 10^{−9} \right)^3 - \left(0.07 + 4.41 \times 10^{-4}\right)^3\right) \\ &= 2.18\times 10^{-9}\text{ m}^3\text{ of RGO} \end{align}

Note that I did not check that your calculations for the values needed for the above (such as 4.41x10-4 m needed for the wall thickness) were correct.

• That is a fascinating response. I cannot understand why I would add the thickness twice when using diameters. Could you explain? Jun 10, 2018 at 17:25
• Also, is having a high safety factor good enough to make the object safe (I revised the safety factor to 450% instead of a mere 15% since it is common practice to have higher pressure applications with a safety factor of 3.5-4.5? Jun 10, 2018 at 17:38
• @hyp You should include the thickness twice because the sphere's external diameter is equal to its internal diameter plus the thickness at each end. And a high enough safety factor makes anything safe. It's just a matter of what's "high enough", and I have no experience with such cases and therefore can offer no judgement.
– Wasabi
Jun 10, 2018 at 17:53