[latexpage]

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

Then we define

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}

\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}

\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}

\begin{equation}

\nu=\frac{I_3-dI_1I_3}{I_1-dI_1I_3}.

\end{equation}

\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}

\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}

\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}

\begin{equation}

Q(t)=Q_2(t)Q_1(t).

\end{equation}

\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

\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.**

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