I can't find the formula, I can see some formulas for circular windows are given here [1] (thanks CERN!). For example with a circular (flat/dome?) window, buckling pressure ($p_\text{CL}$) and max deflection ($w_\text{max}$) are:
$$
\begin{equation}
\begin{aligned}
p_\text{CL}&=\frac{2E}{\sqrt{3\left(1-v^2\right)}}\left(\frac{t}{R}\right)^2 \\
Where \\
E &= \text{material's Modulus of Elasticity} \\
v &= \text{material's Poisson’s Ratio} \\
t &= \text{thickness of the window} \\
R &= \text{radius of the curvature of the spherical dome window}
\end{aligned}
\end{equation}
$$
And
$$
\begin{equation}
\begin{aligned}
w_\text{max}&=\frac{3}{16}\frac{p}{E\left(1-v^2\right)}\frac{R^4}{t^3} \\
Where \\
R &= \text{the radius of the circular window} \\
p &= \text{pressure}
\end{aligned}
\end{equation}
$$
And
Non-circular windows exhibit high loads concentrated in the corners somewhat difficult to treat in case of a weld and a source of nightmares when clamped inside flanges.
I'm looking to use a square/rectangular window, not a circular one. I know acrylic would work as I've seen that used in a DIY all-acrylic glove box (rectangular 1/4"-3/8" thickness) at around 32mbar. It had very noticeable deflection at that pressure and thickness.
I'm still trying to make sense of those equations, the paper lacked details there. I am going though the bibliography on this paper. Meanwhile I was hoping for any advice if I am misguided, or an outright answer 🙂
More details:
Pressure inside cube vessel (with windows) can reach at theoretical lowest pressure of $1$ to $5$ mbar (low vacuum).
I'm at $1403.6$ meters altitude, so atmospheric pressure[2] in pascals is:
$$
\begin{equation}
\begin{aligned}
p&=101325*(1 – 2.25577*10^{-5}*h)^5.25588 \\
&=85561.0486383\text{Pa} \\
Where \\
p &= \text{air pressure (Pa)} \\
h &= \text{altitude above sea level (m)}
\end{aligned}
\end{equation}
$$
This is about $0.85$ atm outside pressure of the container.
So $0.0145037738$ psi inside, and $12.4095809$ psi outside. So the difference $12.3950771262$ psi is roughly the maximal pressure the window would be exposed to.
[1] https://cds.cern.ch/record/1046848/files/p31.pdf
[2] http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html
Best Answer
Found the formula for the use I was looking for [1].
$$ \begin{equation} \begin{aligned} T&=L\sqrt{SF\frac{K}{2}}\times\sqrt{\frac{P}{F_a\left(1+\left(\frac{L}{W}\right)^2\right)}} \\ Where \\ T&=\text{thickness} \\ P&=\text{pressure} \\ L&=\text{length} \\ W&=\text{width} \\ F_a&=\text{apparent elastic limit or rupture modulus} \\ K&=\text{empirical constant: 0.75 clamped, 1.125 unclamped} \\ SF&=\text{safety factor: 4 seems suggested in both sources I've read} \\ \end{aligned} \end{equation} $$
Additional details are in the link.
[1] http://www.crystran.co.uk/userfiles/files/design-of-pressure-windows.pdf