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Shape of meniscus

Hello All,

I am trying to find the governing equation to the shape of a meniscus outside of a small cylinder. I know that by solving the youngs-laplace equation this equation can be derrived, but that is beyond me. I need the equation to be height of meniscus as a function of radius. Any insight to this would be appriciated.

Thanks

I suspect that this problem cannot be solved analytically because of the interplay between stress and mean curvature.  There is also the matter of variation in surface tension between fluids (e.g. mercury vs. water).  That said,

1. By Laplace-Young, P=-2*sig*h*H
   P=pressure, sig=stress, h=thickness, H=mean curvature

2. Surface Tension + Self-Weight -> Height-Dependent Downward Pressure
   P=rho*g*z, rho=density, g=accel due to gravity, z=height

Hence, rho*g*z=-2*sig*h*H.  If the bounding film can be treated as an elastic membrane with nu=1/2, then the uncertainty is as follows:

a: sig*h ~ linear function of z, H ~ constant
   -> see surfaces of constant mean curvature

b: sig*h ~ constant, H ~ linear function of z
   -> refer to Theory of Plates & Shells by Timoshenko & Krieger, 1969
      under weight optimization (of storage tanks), sessile waterdrop problem

c: both are non-linear, but their product is a linear function of z
   -> most likely scenario
       solution requires numerical treatment (i.e. computer software)

I've investigated a fourth possibility: sig~linear, h*H~uniform.  The following satisfies the former (through constitutive relations) but not the latter.

X={(R-p)*cos(theta),(R-p)*sin(theta),-log(cos(2c*p))/(2c)}
0<=p<=r, 0<=theta<=2Pi, R>r, c~rho*g/(-2*h*H) (approx.)

or as an explicit cylindrical equation,

z=-log(cos(2c*(r-R)))/(2c)

The curve might not have a vertical slope at the point of contact, but I'll bet it's close.

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