A few months ago, I tried to tackle the problem of finding the shape, height and other properties of a puddle (whenever a few mL of some liquid are dropped in a "uniform way" so that the puddle remains approximately circular). Trying to find the height, dimensional analysis with density, gravity and superficial tension gives the "famous" (appears in most of the books about this problem) formula:
$$ H = \sqrt{\frac{\sigma}{\rho g}} $$
But as soon as I found this, I discarded it (at this point I hadn't researched anything about the problem) because it seemed to contradict the experiments I'd done so far! I had dropped a few mL of water and honey, and, as I expected, the honey puddle was higher than the water one (3 mm vs 1.5 mm approx). But this is contradictory! The surface tension of honey is similar to that of water, but the density is significantly higher, so the height should be lower.
What does this mean? Is it because the viscosity of the honey makes it "stop" from spilling on the ground and makes it stick on top of the honey that's already on the ground? I've tried a different approach which leads me to a formula that agrees much more with the experiments, but it's completely different from this one and doesn't have the theoretical justification that this one does.
Any suggestions or ideas? Thanks in advance.
EDIT: Here's what I obtained so far and some of the experiments I did (keep in mind that I didn't have much precision due to the lack of any sophisticated instruments). (All measurements were taken at about 20-25 degrees Celsius)
$$ H = \kappa \sqrt[\leftroot{-1}\uproot{2}\scriptstyle 15]{\frac{\eta^2 \sigma^6}{g^7 \rho^8}} $$ Where $\eta$ is the viscosity, $\sigma$ the superficial tension, $g$ is gravity and $\rho$ is density. The value of the constant is $\kappa = 0.89 \pm 0.24 $. With this, I can "predict" the height of a puddle of water, honey, glycerin and alcohol. $$ H_{water}^{theory} = (1.1 \pm 0.4) mm \leftrightarrow H_{water}^{experiment} = (1.5 \pm 0.5) mm$$ $$ H_{alcohol}^{theory} = (0.8 \pm 0.3) mm \leftrightarrow H_{alcohol}^{experiment} = (0.8 \pm 0.5) mm$$ $$ H_{honey}^{theory} = (3.0 \pm 1.1) mm \leftrightarrow H_{honey}^{experiment} = (3.0 \pm 0.5) mm $$ $$ H_{glycerin}^{theory} = (2.4 \pm 0.9) mm \leftrightarrow H_{glycerin}^{experiment} = (2.6 \pm 0.2) mm$$
Also, the formula I've obtained for the shape of the drop is (in two dimensions):
$$ H(r) = H_m(1-e^{\frac{tan{\theta_C}}{H_m}(\left|r \right| - R)}) $$
With $R$ the radius of the puddle, $\theta_C$ the contact angle and $H_m$ the maximum height of the puddle (can be obtained by the formula before) and the radius is approximately $R = \sqrt{\frac{V}{\pi H_m}} $.
I have a formula for the contact angle but requires extreme precision, so it's not of much use itself.
$$\theta_C = \arctan \left(\frac{\pi R H_m^2}{\pi R^2 H_m - V}\left(1+ \sqrt{\frac{2V}{\pi R^2 H_m} -1} \right) \right) $$
