How to estimate the difference in viscosity between 2 liquids without lab instruments? Having 2 liquids I wonder which of the two is more viscous.
I'm not looking at precise values, but something like roughly "twice as viscous".
Without lab instruments, only using kitchen elements, is there a way to estimate roughly the difference in viscosity of 2 liquids?
 A: Use two identical cups with identical circular holes in their bottoms, both stoppered. Fill both with identical amounts of liquids $1$ and $2$, at identical temperature.
Then remove remove the stoppers and measure the time for the cups to empty.
These times are proportional to the liquids' viscosities.
A variant of this method is to insert a paper straw in the bottom of the cup, as shown below (use bathroom silicone sealant or similar to stem any leaking):

Use a yard stick (shown in green) to make sure $h_1$ and $h_3$ are always the same.
This may provide better resolution,
A simple model can show that the time $\Delta t$ to empty the cup is proportional to the viscosity $\mu$.
According to the Hagen-Poisuelle equation (laminar) flow through a pipe obeys:
$$\Delta p=\frac{8 \mu h_2 Q}{\pi R^4}$$
And with Pascal:
$$\Delta p=\rho g h_1=\rho g y$$
So that:
$$\rho g y=\frac{8 \mu h_2 Q}{\pi R^4}$$
We also know that:
$$Q=\frac{\text{d}V}{\text{d}t}$$
Assume the cup to be cylindrical (not shown that way) with radius $R_{cup}$, then:
$$\text{d}V=-\pi R_{cup}^2 \text{d}y$$
So that:
$$\text{d}t=-\frac{8\mu R_{cup}^2h_2}{\rho g R^4}\frac{\text{d}y}{y}$$
Integrated between $(0, h_1)$ and $(\Delta t, h_1')$ we obtain:
$$\boxed{\Delta t=\frac{8 R_{cup}^2 h_2}{\rho g R^4}\mu\ln{\Big(\frac{h_1}{h_3}\Big)}}$$
So this suggests strongly that, all other things being equal:
$$\boxed{\Delta t \propto \mu}$$

If the cup isn't cylindrical but a truncated cone (as is often the case and as pictured) then $R_{cup}$ is a function of $y$, i.e. $R_{cup}(y)$.
Thus $\text{d} t$ becomes:
$$\text{d} t=-\frac{8 h_2}{\rho g R^4}\mu \frac{R_{cup}(y)^2 \text{d}y}{y}=-\frac{8 h_2}{\rho g R^4}\mu I$$
$$R_{cup}(y)=ay+b$$
where:
$$a=\frac{R_1-R_0}{h_1}\text{ and }b=R_0$$
with $R_0=R_{cup}$ at $y=0$ and $R_1=R_{cup}$ at $y=h_1$
So:
$$I=\frac{(ay+b)^2}{y}\text{d}y=\Big(a^2 y+2ab+\frac{b^2}{y}\Big)\text{d}y$$
$$I_{det}=\int_{h_1}^{h_3}I\text{d}y=\frac{a^2}{2}(h_3^2-h_1^2)+2ab(h_3-h_1)+b^2 \ln\Big(\frac{h_3}{h_1}\Big)$$
$$\boxed{\Delta t=\frac{8 h_2}{\rho g R^4}\mu \Big(\frac{a^2}{2}(h_1^2-h_3^2)+2ab(h_1-h_3)+b^2 \ln\Big(\frac{h_1}{h_3}\Big)\Big)}$$
