The network below is formed by hexagons, octagons, and squares, with its unit cell highlighted in the red rectangle.
My goal is to obtain unit cells from replications of this rectangle in different directions from the conventional ($x$ and $y$) ones. I started with the chiral vector, which can be easily obtained. Assuming that there are $m$ replications in the $x$ direction and $n$ replications in the $y$ direction, this chiral vector is $\vec C_h = m\vec a_1 + n\vec a_2$, where $\vec a_1$ and $\vec a_2$ are the lattice vectors in the $x$ and $y$ directions, respectively. To obtain the translational vector, $\vec T = t_1\vec a_1 + t_2 \vec a_2$, perpendicular to $\vec C_h$, I have $\vec C_h\cdot \vec T = 0$, but it needs to be the smallest possible values for $t_1$ and $t_2$ to replicate the system, so $\gcd(t_1, t_2) = 1$.
This is the point where I got stuck – obtaining the values of $t_1$ and $t_2$ to make the system periodic.
I managed to solve the problem computationally with the following strategy: i) I create a large system, cut it according to $\vec C_h$ to ensure the correct width, randomly select three points from this lattice that resemble a ribbon, and create perpendicular lines to the $\vec C_h$ vector. By checking all points on the lattice, I measure the distance between the points. If they are equal for the three randomly selected points, I have the magnitude of $\vec T$. However, I need the analytical form since I would like to calculate the magnitude of $\vec T$ (and also the number of points) as a function of replications $m$ and $n$, without having to run this script for every case I want to analyze.
If you could help me, I would especially like to obtain the values of $t_1$ and $t_2$ to calculate $|\vec T|$.
Any two non-collinear vectors (which are integer linear combinations of $\vec{a}_1$ and $\vec{a}_2$) specify a doubly-periodic system. They don't have to be perpendicular.
If you still want the two vectors to be perpendicular, how to achieve that depends on the ratio $a_1/a_2$.
From the image, it seems that the unit cell is a square, so the ratio is equal to $1$. If so, take $t_1=-n$ and $t_2=m$. If they are not coprime and you want them to be coprime, divide them both by their gcd.
In general, we want $0=\vec{C}_h\cdot\vec{T}=mt_1a_1^2+nt_2a_2^2$, so $t_1/t_2=-na_2^2/(ma_1^2)$. The r.h.s. is only known within some error bounds, I suppose, so choose the simplest fraction within that interval or an approximation to it (e.g. using continued fractions).
"$a_1=3.88$" actually means something like $3.875<a_1<3.885$, and same for $a_2$. This leads to $0.5998 < -t_1/t_2 < 0.6057$. The ratio $3/5$ falls in this range.