I am currently building a computational model for a simulation of a rover landing on the surface of mars. As part of the assignment I need to model airbags being used to cushion the fall. (Assuming the airbags are spherical). I don't know how I would go about modelling the forces in this, when the rover gets squished between the rover and the planet? I am using euler integration (simulating some delta time step) for the simulation.

Can anyone please offer some help ~ Please bear in mind I am only in yr 12.


  • $\begingroup$ This would be better fit for matter modelling/ engineering $\endgroup$ Apr 21, 2021 at 21:46

1 Answer 1


As a first pass, you could assume that the balloon material is inextensible (i.e., that it can't stretch) and has no bending stiffness. Then the resistance to squashing a balloon arises solely from the pressurization of the internal gas, which we'll assume to be ideal.

Assume that the compression makes the balloon look like a sphere that's lost a spherical cap:

The remaining volume $V(h)$ is then

$$V(h)=\frac{4}{3}\pi r^3-\frac{\pi h^2}{3}(3r-h). $$

The resistance force $F$ when you compress a system arises from the increase in energy $U$ per unit compression $h$:

$$F=\left(\frac{\partial U}{\partial h}\right)_S.$$

Here, I've written this statement in partial differential form; don't worry if you haven't covered this material yet. The $S$ at the lower right stands for constant entropy, which simply means that the squashing occurs quickly (as opposed to very slowly, where we might assume that instead it's the gas temperature that stays constant).

I can apply the chain rule of differentiation to expand this term into things I know:

$$F=\left(\frac{\partial U}{\partial h}\right)_S=\left(\frac{\partial U}{\partial V}\right)_S\left(\frac{\partial V}{\partial h}\right)_S=-P\left(\frac{\partial V}{\partial h}\right)_S,$$

where $P$ is the pressure. (We know from the fundamental relation that $(\partial U/\partial V)_S=-P$ for simple closed systems.) It turns out that for an ideal gas being compressed quickly, $$PV^\gamma=P_0V_0^\gamma,$$ where $P_0$ and $V_0$ are the initial pressure and volume before compression and $\gamma$ is the heat capacity ratio, which can be approximated by 1.4 for air, for example. Thus,

$$F=-P_0\left(\frac{V_0}{V}\right)^\gamma\left(\frac{\partial V}{\partial h}\right)_S.$$

Note that $V_0$ is simply $\frac{4}{3}\pi r^3$. We can determine $\left(\frac{\partial V}{\partial h}\right)_S=\frac{dV(h)}{dh}$ from differentiating the volume expression above:

$$\frac{dV(h)}{dh}=\pi h^2-2\pi hr,$$

if I differentiated correctly. The final unsimplified expression is

$$F=-P_0\left(\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi r^3-\frac{\pi h^2}{3}(3r-h)}\right)^\gamma\left(2\pi hr-\pi h^2\right).$$

Try simplifying this and plugging in some expected values. Do the results make sense?

  • $\begingroup$ Thank you very much for this in depth analysis, I am very grateful for your time. I will try and implement this in code and let you know if I have any questions / concerns. Thanks. $\endgroup$
    – Tim
    Apr 22, 2021 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.