Chestermiller said:
I think I'd still like to see the formulation of the differential equations and boundary conditions.
Before I continue, I forgot a very important scale: ##w##. If we assume pressure balances viscosity, then dimensional constant density NS yields in ##z## component
$$\partial_zP = \mu \nabla^2 w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2}+\frac{1}{H^2\tan^2\alpha}+\frac{1}{L^2}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\tan^2\alpha + 1}{H^2\tan^2\alpha}\right)w\implies\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{\sec^2\alpha}{H^2\tan^2\alpha}\right)w\implies\\
\frac{\sigma}{H f L} \sim \mu\left(\frac{1}{H^2\sin^2\alpha}\right)w\implies\\
w \sim \frac{\sigma H\sin^2\alpha}{ f L \mu}\implies\\
w \sim \frac{\epsilon \sigma \sin^2\alpha}{ f \mu}.
$$
Notice I neglected ##1/L^2## term since ##\epsilon = H^2/L^2\ll 1## (slender column). Also, I would guess time scales as ##t \sim L/w##, though I'm not sure why I'm inclined to use ##w## rather than ##u,v##. Most of the flow is ##z## directed; is that good enough reason? Anyways, the formulation of the differential equations, assuming we start with dimensional NS and constant density, would be (for ##z## component, though I could do the other two if you'd like as well)
$$
\rho\frac{D w}{Dt} = \partial_zP+\mu \nabla^2w\implies\\
\frac{\rho W}{L/W}\frac{D w}{Dt} = \frac{\sigma}{HfL}\partial_zP+\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w\implies\\
\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\frac{HfL}{\sigma}\frac{\mu W}{H^2\sin^2\alpha} \nabla^2w
$$
but we already balanced pressure and viscosity, so we know both terms on the RHS are now ##O(1)##, thus we have
$$\frac{\rho W^2}{L}\frac{HfL}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon \sigma \sin^2\alpha}{ f \mu}\right)^2\frac{\rho }{1}\frac{Hf}{\sigma}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{\sigma \rho H}{\mu^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\
\left(\frac{\epsilon^2 \sin^4\alpha}{ f }\right) \frac{1}{Oh^2}\frac{D w}{Dt} = \partial_zP+\nabla^2w\implies\\$$
and ##Oh \equiv \mu/\sqrt{\rho \sigma H}##. How does this look?
.
But only when I'm in a bad mood, of course. And my roommate is a mathematician.
Ok then, maybe it's simply from scaling the equations! Thanks for your guys' help!
