So, when trying to measure the gravity constant $g$ at home through a pendulum I realized I wanted to try to model drag on the pendulum, to get a more accurate result.
I am asking how we can justify linear drag because that is the only easily solved model and that seems to be what people in literature and here on phys.stack do, but I don't understand why it should work.
Assuming $F_{drag}\propto v$ is equivalent to assuming a low Reynolds number right? They argue this in the linked paper, claiming $ R_e\approx 1200$ would be low enough.
From Wiki (horrifyingly they actually quote wiki in this paper but that's another discussion)
At low $R_{e}, C_{D} $is asymptotically proportional to ${\displaystyle R_{e}^{-1}} $, which means that the drag is linearly proportional to the speed, [...] given by the Stokes Law
But I have done several derivations of Stokes law myself and you really need to assume a very viscous environment, basically you can derive it, if almost all inertia can be neglected (like for a bacterium in water, Kundu cites mist drops, or molten plastic in this regime). Which certainly not the case for a pendulum.
To show a source on what low $R_e$ means, from the same wiki article:
[...] (i.e. low Reynolds number, ${\displaystyle R_{e}<1}).$ [...] The equation for viscous resistance is: $ \mathbf{F}_d = - b \mathbf{v}$
Kundu gives an upper limit of $R_e < 5$ for the validity of this approximation.
I am no expert but i am pretty sure that these Reynolds numbers are too low for my pendulum.
Concluding: Can anybody either help me justify linear drag in a more sensible way or put the notion of realistic linear drag for a pendulum in air to bed in a more rigorous way once and for all?
