Day: March 22, 2017

Attitude matrix and quaternion path for m>1

Posted on by arkajad

I noticed that somehow I did not finish with the case m>1. So today, without further ado, I am posting the algorithm.
We have the body with I_1<I_2<I_3, and we are considering the case with d>1/I_2, where d is, as always, the ratio of the doubled kinetic energy to angular momentum squared.
Then we define

(1)   \begin{eqnarray*} A_1&=&\sqrt{\frac{I_1(dI_3-1)}{I_3-I_1}},\\ A_2&=&\sqrt{\frac{I_2 (1-dI_1)}{I_2-I_1}},\\ A_3&=&\sqrt{\frac{I_3 (1-dI_1)}{I_3-I_1}},\\ B&=&\sqrt{\frac{(dI_3-1)(I_2-I_1)}{I_1I_2I_3}},\\ \mu&=&\frac{1}{m}=\frac{(1-dI_1)(I_3-I_2)}{(dI_3-1)(I_2-I_1)}. \end{eqnarray*}

(2)   \begin{eqnarray*} L_1(t)&=&A_1\,\dn(Bt,\mu),\\ L_2(t)&=&A_2\,\sn(Bt,\mu),\\ L_3(t)&=&A_3\,\cn(Bt,\mu). \end{eqnarray*}

(3)   \begin{equation*} \alpha=\frac{I_3-I_1}{\sqrt{\frac{I_1(dI_3-1)(I_2-I_1)I_3}{I_2}}}, \end{equation*}

(4)   \begin{equation*} \nu=\frac{I_3-dI_1I_3}{I_1-dI_1I_3}. \end{equation*}

(5)   \begin{equation*} \psi(t)=\frac{t}I_3-\arctan\left((A_2/A_1)\mathrm{sd}(Bt,\mu) \right)+\alpha \Pi(\nu,\am(Bt,\mu),\mu),\ \end{equation*}

(6)   \begin{equation*} Q_1(t)=\begin{bmatrix}1-\frac{L_1(t)^2}{1+L_3(t)}&-\frac{L_1(t)L_2(t)}{1+L_3(t)}&-L_1(t)\\ -\frac{L_1(t)L_2(t)}{1+L_3(t)}&1-\frac{L_2(t)^2}{1+L_3(t)}&-L_2(t)\\L_1(t)&L_2(t)&L_3(t)\end{bmatrix}. \end{equation*}

(7)   \begin{equation*} Q_2(t)=\begin{bmatrix}\cos\psi(t)&-\sin\psi(t)&0\\\sin\psi(t)&\cos\psi(t)&0\\0&0&1 \end{bmatrix}. \end{equation*}

(8)   \begin{equation*} Q(t)=Q_2(t)Q_1(t). \end{equation*}

(9)   \begin{eqnarray*} q_0(t)&=&\sqrt{\frac{1+L_3(t)}{2}}\cos\frac{\psi(t)}{2},\\ q_1(t)&=&\frac{1}{\sqrt{2(1+L_3(t))}}\left(L_2(t)\cos\frac{\psi(t)}{2}+L_1(t)\sin\frac{\psi(t)}{2}\right),\\ q_2(t)&=&\frac{1}{\sqrt{2(1+L_3(t))}}\left(L_2(t)\sin\frac{\psi(t)}{2}-L_1(t)\cos\frac{\psi(t)}{2}\right),\\ q_3(t)&=&\sqrt{\frac{1+L_3(t)}{2}}\,\sin\frac{\psi(t)}{2},\\ q(t)&=&(q_0(t),q_1(t),q_2(t),q_3(t))=q_0(t)+\mathbf{i}\,q_1(t)+\mathbf{j}\,q_1(t)+\mathbf{k}\,q_3(t). \end{eqnarray*}

I use the above formulas to draw a stereographic projection of one particular path. So, I take I_1=1,I_2=2,I_3=3,d=0.5000001, and do the parametric plot of the curve \mathbf{r}(t) in \mathbf{R}^3,
with

    \[\mathbf{r}(t)=\left(\frac{q_1(t)}{1-q_0(t)},\frac{q_2(t)}{1-q_0(t)},\frac{q_3(t)}{1-q_0(t)}\right).\]

I show below two plots. One with t\in(-1000,1000), and one with t\in(-10000,10000). For this selected value of d, the time between consecutive flips, given by the formula

(10)   \begin{equation*}\tau =4\mathrm{EllipticK}(\mu)/B= 116.472.\end{equation*}

So, for t\in(-10000,10000) we have somewhat less than 200 flips, and the lines are getting rather densely packed in certain regions.
Notice that is just one geodesic line, geometrically speaking the straightest possible line in the geometry determined by the inertial properties of the body.

It is this “geometry”that will become the main subject of the future notes.

Geodesic line for t between -1000 and 1000

Geodesic line for t between -10000 and 10000

How can such a line be “straight”? Well, it is….