# The Official Equation for the relationship between the Volume of a balloon and time

I would like to ask a question on whether there is a proportionality between volume of a balloon, and the time it takes to deflate.

I have conducted several balloon hovercraft experiments. I need to find the relationship between the amount of air pumped into the balloon and how long the hovercraft hovers. The balloon is tied on a sport bottle cap, with an acrylic disc as its base. The cap is glued onto the disc.

I have plotted such a graph, with volume of each of the balloons on the x axis and time of flight that resulted from them on the other. I have achievement a seemingly exponential trend. I would like to know if there is a formula that can find my proportionality of the relationship between volume and how long the it takes for the balloon to deflate (the hovercraft to float).

Thank you.

Edit: Here is the picture of one of the graphs,

• Can you provide the raw data of your experiments and/or your graph? Commented Apr 23, 2019 at 8:16
• Time normally goes on the horizontal axis, but I can understand why you want time on the vertical axis for this problem. Also, for the title on your plot, the correct form is "y" vs. "x", so there is an inconsistency between your plot and its title. And, as has been pointed out, the raw data would be very helpful. Note - in my opinion, your function will be quadratic, not exponential. Commented Apr 23, 2019 at 15:16
• Heh, I almost want to +1 this just for "Official Equation". But in all seriousness, what did you mean by that?
– Nat
Commented Apr 25, 2019 at 23:25

The pressure inside a balloon inflated to radius $$R$$ from its equilibrium radius $$R_0$$ is $$P=2Et_0 \frac{(R-R_0)R_0}{R^3},$$ where $$E$$ is the elasticity modulus of the material and $$t_0$$ the initial thickness. If we use $$R=(3V/4\pi)^{1/3}$$ to re-express it in terms of volume we get $$P=(8/3)\pi^{2/3} (3/4)^{1/3} E t_0 \frac{(V^{1/3}-V_0^{1/3})V_0^{1/3}}{V},$$

Using the Hagen-Poiseuille equation we get $$Q=\frac{\pi r^4 \Delta P}{8\mu L}$$ where $$r$$ is the radius of the "pipe" of the balloon, $$L$$ the length, $$\mu$$ the dynamic viscosity of air and $$Q$$ the volumetric flow rate. This assumes an incompressible fluid, which might be a bit dodgy in this case, but I assume balloons can't reach that many atmospheres of pressure and hence the approximation is OK.

We get the differential equation $$V'(t) = -Q(V) = \left[\frac{ (8/3)\pi^{5/3} (3/4)^{1/3} E t_0 r^4 V_0^{1/3}}{8\mu L}\right ] \frac{(V^{1/3}-V_0^{1/3})}{V}$$ or, if we represent the bracketed mess as $$\alpha$$: $$V'=-\alpha \left(\frac{1}{V^{2/3}} - \frac{V_0^{1/3}}{V}\right )$$ This equation can be integrated into the implicit solution $$3V_0^{5/3}\log(V^{1/3}-V_0^{1/3})+3V_0^{4/3}V^{1/3}+(3/2)V_0^{2/3}V+(3/4)V_0^{1/3}V^{4/3}+(3/5)V^{5/3}=C-\alpha t$$ where $$C$$ is a constant of integration we can set so that $$V(0)$$ has the right value. While clear as mud, one can plot it.

If we set the scale so that $$V_0=1$$ and $$\alpha=1$$, we get the following curve if we start from $$V=3$$: The initial deflation is nearly linear in terms of volume, and then it slows down into an exponential decline (due to the log-term in the equation).

Based on general rules of geometrical similarity, the time is proportional to $$size^{1.5}$$
as the volume is proportional to $$size^{3}$$
while the rates of spontaneous or forced loses (via cooling, opening, fabric leaking)
are all proportional to $$size^{2}$$

Therefore 4 times bigger ( by diameter) balloon takes 8 times more time to deflate.