Here is a function that describes the shape of a water balloon as it fills with water.

The beautiful teardrop shape that a water balloon takes on as you fill it with water has a shape defined by the following formula:

where \(v\) is the volume of liquid in the balloon, and \(d\) is the density of the liquid, and \(D\) is the maximum density for which the formula is valid.

That formula only looks daunting because it is a composition of several functions. Each of the functions defines a different aspect of the physics of the water balloon.

Let's look at an ideal water balloon. It is a sphere of water, surrounded by a film of latex rubber, whose shape is distorted by gravity.

If the liquid inside had a density of zero (or the same density as the medium outside the balloon), then the shape would remain a sphere, since gravity has no effect when the density is zero.

If the density is above zero, then the sphere would be stretched in the direction that gravity is pulling (down). It would become an ellipse.

The equation for the area of an ellipse is

$$ v = \pi a b $$

where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. When these are equal, they are simply the radius of a circle, and the area is the familiar \( \pi { r } ^2 \).

In our two dimensional graph of the water balloon, the volume of liquid is the area. The volume is one of our inputs, so it is a known quantity. This allows us to solve for \(b\) in terms of a, since everything else is a constant.

$$ b = { {v} \over {(\pi a)} } $$

This is important, since gravity is going to make \(a\) larger, and to keep the area (volume) constant, \(b\) has to shrink as \(a\) grows.

The value of \(a\) is a function of the volume and the density, our two inputs:

$$ a = { \sqrt { {( v + d )} \over {\pi} } } $$

You can see that as either the volume or the density get larger, so does the length \(a\) of the balloon. And the width \(b\) of the balloon gets bigger as the volume goes up, but it is also a function of \(a\), so the volume remains a function of \(a\) times \(b\).

The formula for the perimeter of an ellipse is

$$ {{x^2} \over {a^2}} + {{y^2} \over {b^2}} = 1 $$

Solving for \(y\) we have:

$$ y = { \sqrt { {b^2} {\left [ 1-{{{(x-a)}^2} \over { a^2 }} \right ] }} }$$

This gives us the shape of the ellipse as gravity stretches the circle. But it does not yet describe the effects of the pressure in the balloon. Pressure is a linear function of the depth, which is the \(x\) direction in our formula. We can multiply \(y\) by the relative depth of the liquid at each value of \(x\):

$$ p = { x \over a } $$

Now we need to handle the density. As the density goes to zero, so does the pressure. We need to add exactly enough to our pressure factor to make it equal to 1 when the density is zero, and yet leave it as \({ x \over a } \) when the density is the maximum we allow.

$$ p = { {x \over a} + \left [ 1-{x \over a} \right ] { \left [ {D-d} \right ] \over D } } $$

Now our formula is:

$$ y p $$

Which, when you expand all of those \(a\) values and \(b\) values to express them in terms of our inputs, \(v\) and \(d\), we get

$$ { \sqrt {  { \left [ { {v} \over {(\pi  { \sqrt { {( v + d )} \over {\pi} } })} } \right ]^2}  {\left [ 1-{{{(x-{ \sqrt { {( v + d )} \over {\pi} } })}^2} \over { { {( v + d )} \over {\pi} } }} \right ] }} }  \left [ { {x \over { \sqrt { {( v + d )} \over {\pi} } }} + \left [ 1-{x \over { \sqrt { {( v + d )} \over {\pi} } }} \right ] { \left [ {D-d} \right ] \over D } } \right ] $$