I have a (standard?) 2-torus $T^2$ with a non-standard Riemannian metric on it
\begin{array}{lcr} g(\theta_1,\theta_2) & = & \begin{bmatrix}Izz1 + Izz2 + (L1^2*m1)/4 + L1^2*m2 + (L2^2*m2)/4 + L1*L2*m2*cos(\theta_2) & (m2*L2^2)/4 + (L1*m2*cos(\theta_2)*L2)/2 + Izz2\\ (m2*L2^2)/4 + (L1*m2*cos(\theta_2)*L2)/2 + Izz2 & (m2*L2^2)/4 + Izz2 \end{bmatrix} \\ \end{array} (For definiteness, for this particular double-pendulum robot arm, the values of the parameters are $m_i = 23.142$ kg, $Izz_i = 2.668$ $\text{kg}\cdot\text{m}^2$, and $L_i = 1.000$ m, but these values are mutable.)
It is my understanding that for any Riemannian 2-torus $(T^2,g)$ with a "flat metric" on it, there is a lattice $\Lambda$ in $\mathbb{R}^2$ with $\mathbb{R}^2/\Lambda$ with its inherited quotient metric being isometric to $(T^2,g)$.
My question is, what is the lattice inducing this metric and how would I determine that for myself?
[EDIT: As some possibly helpful and possibly irrelevant information, (I) (a) the length of the path $\gamma_1(t) = (t,0)$ for $t \in [0, 2\pi]$ is Length1 = 49.947 $\text{kg}^{\frac{1}{2}}\cdot\text{m}$, (b) the length of the path $\gamma_2(t) = (0,t)$ for $t \in [0, 2\pi]$ is Length2 = 18.268 $\text{kg}^{\frac{1}{2}}\cdot\text{m}$, (c) the length of the path $\gamma_3(t) = (t,t)$ for $t \in [0, 2\pi]$ is Length3 = 48.980 $\text{kg}^{\frac{1}{2}}\cdot\text{m}$; (d) this leads to an "implied angle" of $\alpha = 76.394^{\circ}$ for The Law of Cosines, with $\text{Length3}^2 = \text{Length1}^2 + \text{Length2}^2 - 2\text{Length1}\text{Length2}\cos(\alpha)$. (II) (a) The path $\gamma_4(t) = (t,0.498t)$ for $t \in [0, 2\pi]$ has Length4 = 41.220 $\text{kg}^{\frac{1}{2}}\cdot\text{m}$ and appears to minimize the length of all paths going once around $\theta_1$ and some positive multiple $b$ around $\theta_2$ at the same time; (b) this has an implied angle of $\beta = 51.769^{\circ} = \arctan\left(\frac{0.66}{1}\right)$. See this Mathematica script.
EDIT 2: OK, so, This link I think I computes the sectional curvature correctly in MATLAB, and it is identically 0. Could someone help me find the lattice?
EDIT 3: This link I think confirms the computation of the sectional curvature of my Riemannian manifold in Mathematica (although it heavily uses the fact we are in dimension 2; I'll need to work on that).
EDIT 4: I hazard the guess that the lattice in $\mathbb{R}^2$ (with units for which being radians in each direction) with its standard Riemannian metric (with units for which being $\text{kg}\cdot\text{m}^2$) given by $\vec{v}_1 = \langle 49.947, 0\rangle \text{kg}^{\frac{1}{2}}\cdot\text{m}$ and $\vec{v}_2 = \langle 18.268\cos(\delta), 18.268\sin(\delta)\rangle \text{kg}^{\frac{1}{2}}\cdot\text{m}$, where $\delta = 180^{\circ} - 76.394^{\circ} = 103.606^{\circ}$ is the lattice I want (by properties of local isometries and covering maps); would anyone care to help me confirm or deny this? I'll post EDIT 4 as the answer if I don't get any feedback in a few days.]