Today, while contemplating on the absurdity of life (or, in other words, procrastinating as usual), I stumbled across a strange inflated balloon, the first of its kind that I have ever seen as per my memory.

When I squeezed the inflated balloon on any of the ends, it changed shape (as expected). But, intriguingly, it retained the new shape even after I stopped squeezing it! I could do this as many times as I wanted.

The balloon was made of rubber, filled with air and was absolutely normal in all other respects.

The photos below show one such cycle of making the balloon change shape by squeezing from both ends one after the other.

Now I can imagine why this would happen with a balloon filled with viscous fluid, but why is it happening with air which has almost 0 viscosity? Shouldn't the air always move in the balloon to attain the state of maximum stability?

In the pictures below, one can clearly see (from the shade of balloon colour) that sometimes the top side of the balloon is more pressurized than the bottom side and vice versa. How and why is this possible? Why doesn't the air pressure distribute itself evenly throughout the balloon?

enter image description here

  • $\begingroup$ Can you comment on the temperature (variation) within the room while you were doing that? Also - how long did those new shapes remain stable? $\endgroup$
    – Sanya
    Commented Nov 11, 2016 at 16:38
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Two-balloon_experiment $\endgroup$ Commented Nov 11, 2016 at 17:00
  • $\begingroup$ @Sanya I carried out each shape change cycle within a minute. So, the temperature variation would have been negligible. $\endgroup$ Commented Nov 13, 2016 at 10:59
  • $\begingroup$ @KeithMcClary Thank you for the VERY interesting link. And yet a part of my question remains. The pressure at any one end of my balloon should have been more than the other. So, the balloon should have changed shape from one to another instead of changing shape both ways as per my desire. $\endgroup$ Commented Nov 13, 2016 at 11:04
  • $\begingroup$ @Sanya The shapes remained stable for as long as I didn't change them by squeezing one of the ends (even hours). $\endgroup$ Commented Nov 13, 2016 at 11:32

3 Answers 3


The resistance to movement comes from hysteresis in the stress strain curve of the rubber. The shape of the balloon magnifies the effect of this hysteresis.

If you consider a cylindrical balloon, the equations to balance tensional stress and pressure are as follows:


The thickness of the rubber will typically vary approximately with $\frac1{r^2}$ so:



The strain will vary as $\epsilon=\frac{r}{r_0}-1$ so we can put pressure in terms of strain:


The fraction portion is the balloon's shape magnifying the natural hysteresis in the stress strain curve.

The stress strain curve of the rubber might look like this:

Stress strain curve with hysteresis

Note the stress is slightly lower when relaxing than when stretching.

When we multiply by the balloon effect:

Balloon effect plot.

We get something proportional to pressure vs strain when expanding vs contracting:

Pressure vs strain for balloon inflating vs deflating.

Here I've added a constant pressure line that goes between the high pressure (expanding) and low pressure (contracting) lines. If the pressure in the balloon was at the pressure corresponding to that line, then different parts of the balloon could happily maintain a stain anywhere from 1 to 6.

When the portion of the balloon that is stretched to 6 is squeezed, it reduces the volume of the balloon, increasing the pressure, causing the portion of the balloon that was down near 1 to stretch and move towards 6. Once the extra squeeze is removed the pressure drops back down and the balloon keeps its new shape.


It depends on the strength of the latex compound used in the manufacture of the balloon.

Here is another example:

enter image description here

So, using the correct manufacturing techniques and latex compounds, it can overcome the pressure of the air inside, and subsequent tendency to assume a spherical shape.

  • $\begingroup$ This is totally correct. But OP has changed the shape of his baloon after blowing it up if I understood the post correctly. $\endgroup$
    – Sanya
    Commented Nov 11, 2016 at 16:36
  • $\begingroup$ @Sanya fair enough, but I think it might just be a variation of the compound used, if it can allow the above shape, possibly there is another mix than allows temporary shapes. Actually, sorry, you are asking the same question. $\endgroup$
    – user108787
    Commented Nov 11, 2016 at 16:42
  • 1
    $\begingroup$ There's no apologies necessary here - I hope? I didn't want to appear rude by my comment. $\endgroup$
    – Sanya
    Commented Nov 11, 2016 at 16:55

The balloon is in a state of equilibrium for all the above cases. So, the forces are balanced. Consider taking a small cross-section of the balloon. The force on either side of this section balances out to keep it in steady state. For larger section, pressure is less and for smaller section pressure is more but together they compliment and balance out to make the force same. Now think the same along all cross sections of the balloon. In the event of squeezing the shape changes, so does cross sectional area and by Bernoulli's Principle maintains equilibrium and remains in steady state.

  • $\begingroup$ Are you saying that the pressure is different in different parts of the balloon? $\endgroup$ Commented Nov 17, 2016 at 16:00

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