Attitude Dynamics and Cartoons

Introduction

A couple of years ago, maybe more, I had to estimate the maximum torque over a small Cubesat. This was quite an interesting engineering problem and, in the process, I came across a pure mathematical problem I would like to share here. Let's review some concepts first.

As you may know, if you're not here just for the cartoons, the Earth-orbiting bodies can be perturbed in their free-fall trajectory by interactions (that may alter also their attitude) with things like the upper atmosphere, the solar radiation pressure, the magnetic field, and so on. If we fix our attention on "drag-like" perturbations (interaction with the atmosphere or the solar radiation pressure), we may end asking ourselves, what attitude maximizes the frontal area of the orbiting body facing the incident "wind"? In other terms, what would be the maximum area of the hole that a cartoon like poor Coyote will make through a wall? (This is the maximum cross-sectional area).

You may want to review some bibliography here or just go on and enjoy the following picture.

Poor Coyote Maximum cross-suffering Area

The model

Cubesats are geometrically diverse, the may have deployables antennas or solar arrays. But in an estimation phase we can assume they are, well, cuboids (be aware of the limitation of this model with long deployables in your torque estimations). So, we have a cuboid, and we would like to know which orientation will give the maximum cross-sectional area (which will be related somehow with maximum forces and torques we will estimate at a later stage).

The first observation here is that, at the very least, only one face of the cuboid is seen by the perturbation (important if you're interested in finding the minimum cross-section area) and at most three faces are seen (directly) by the (solar or atmospheric) "wind". Let's assign to these three faces vectors with their area magnitude ( \(a_i\) are the dimensions of each edge),

\(\overrightarrow{S_{x}} = a_2 a_3 \overrightarrow{i}\) , \(\overrightarrow{S_{y}} = a_1 a_3 \overrightarrow{j}\) , \(\overrightarrow{S_{z}} = a_1 a_2 \overrightarrow{k}\)

And also suppose that a unit vector with the direction of the "wind" (or the wall) of the perturbation.

\[\overrightarrow{n} = n_1 \overrightarrow{i} + n_2 \overrightarrow{j} + n_3 \overrightarrow{k}\]

The cross-sectional area will be then, \(n_1 a_2 a_3 + n_2 a_1 a_3 + n_3 a_1 a_2\) .

Long story short, let's maximize \(A = n_1 a_2 a_3 + n_2 a_1 a_3 + n_3 a_1 a_2\) given \(n_1^{2} + n_2^{2} + n_3^{3} = 1\)

How do we do that? Lagrange multipliers method to the rescue!

\[\nabla (n_1 a_2 a_3 + n_2 a_1 a_3 + n_3 a_1 a_2 + \lambda (n_1^{2} + n_2^{2} + n_3^{3} - 1)) = 0\]

That's it!

The solution

After some manipulation we end with the expression for the unit vector that maximizes the cross-sectional area of the cuboid.

\[n_{i} = \sqrt{ \frac{1}{1 + a_i^{2} ( 1/a_j^{2} + 1/a_k^{2})}}\]

It is interesting to obtain this vector and the cross-sectional area for two common cases in cubesats, assuming that one unit cubesat is a square cube with edge \(a_1 = a_2 = a_3 = a\)

One-unit cubesat

\(n_{i} = \sqrt{1/3}\) and the resulting area is \(A = \sqrt{3} a^{2}\) .

Two-unit cubesat

We assume \(a = a_1 = a_2 = \frac{a_3}{2}\) and \(n_{1} = n_{2} = \frac{2}{3}, n_{3} = \frac{1}{3}\) . The resulting area is \(A = 3 a^{2}\) .

Numerical comparison

Yes, of course, you can put your computer to work on numerical methods to cross-validate the former analytical solutions. In the general case I remember I followed an original approach to projecting all vertices of the geometry over a plane and then finding the convex hull polygon over a parametrization. But here we can follow another interesting approach: How about brute-force with a ray tracer? Yes, I've made my custom ray tracer for this (and if you ask, yes, I've made it in Fortran: it's amazing fast, and I've learn a few tricks along the way).

Final words

I'm somehow in love with this problem and the solution, and I'm sure it contains more interesting approaches to solving it for general shapes, taking into account shadows of the geometry with itself. Maybe you know them; I would love to know. Finally, maybe you're here because of the clickbait, may be you are here because you're considering hiring me, or you're just curious. The message I would like to send is this: what ever you do (in engineering), do it well. And maths will help you with that (isn't it obvious?).

Original entry first published here.

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