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The relevant comparison is between gravity and surface tension. With the mass density difference $\Delta \rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, the Bond number and capillary length areThus the drop is small compared with the capillary length; hydrostatic pressure variations only weakly perturb its shape. The Young–Laplace equation then gives approximately constant interfacial mean curvature, and an axisymmetric drop on a uniform flat substrate is a spherical cap. The substrate selects the contact angle, measured through the water.

Write $R$ for the radius of the generating sphere, $s$ for the contact-circle radius, and $h$ for the apex height. Elementary trigonometric identities giveTaking the stated droplet radius to mean its radius of curvature, $R=r$, givesThere is no diagram or explicit radius convention in the original PDF. If $r$ instead denotes the contact-circle radius, the answer is $h=r\tan 27.5^{\circ }=0.260\mathrm{mm}$. If it denotes the radius of an equal-volume free sphere, use the volume of a spherical cap, $\pi h^2(3R-h)/3=4\pi r^3/3$, to obtain $h=r[4(1-\cos \theta )/(2+\cos \theta ){]}^{1/3}=0.436\mathrm{mm}$. The spherical cap approximation is justified in all three conventions; a unique numerical height requires specifying which radius is meant.

At a point of a smooth surface, a normal section of a surface can bend by different amounts depending on its tangent vector. The principal curvatures are the largest and smallest signed normal curvatures among unit tangent vectors. Equivalently they are the two eigenvalues $k_1,k_2$ of the shape operator, with an orthonormal basis $e_1,e_2$ of corresponding principal directions. Here the labels follow the selected directions, without requiring $k_1\le k_2$. ^[ref]

For a unit tangent vector making angle $\theta$ with $e_1$, Euler formula for normal curvature givesThe normal curvature is a weighted average of the principal curvatures; the extremes occur in the principal directions. If the principal curvatures coincide, every direction in the tangent plane is a principal direction: this is an umbilical point. The Gaussian curvature and mean curvature summarize the pair by

For an outward-oriented sphere of radius $R$, both principal curvatures are $-1/R$, so the normal curvature is $-1/R$ in every direction. For an outward-oriented circular cylinder, choose $e_1$ around the circumference and $e_2$ along the axis. The principal curvatures are $-1/R,0$, giving $k_n(\theta )=-{\cos }^2\theta /R$; halfway between the principal directions, $k_n(\pi /4)=-1/(2R)$.

At the origin of the upward-oriented hyperbolic paraboloid $z=(x^2-y^2)/2$, the principal directions are the coordinate axes and the principal curvatures are $1,-1$ in the chosen length unit. Thus $k_n(\theta )=\cos (2\theta )$, with zero normal curvature at $\theta =\pi /4,3\pi /4$. Zero normal curvature in these directions coexists with Gaussian curvature $-1$: the shape operator is not zero. The mean curvature is zero at this point; this does not say that the whole hyperbolic paraboloid is a minimal surface.

Each graph uses angle $\theta$ in the tangent plane, measured from the indicated first principal direction. The sphere and circular cylinder have $R=1$; the saddle graph refers only to the origin. Reversing the unit normal reflects each normal curvature graph through zero, negates both principal curvatures and the mean curvature, and preserves their product, the Gaussian curvature.

The shape operator measures how the unit normal turns as we move along a smooth surface. A plane has a constant unit normal, whereas the outward unit normal of a sphere turns in every direction in the tangent plane. Keeping the direction of this change, rather than just its magnitude, lets one recover the principal curvatures and their principal directions.

Choose an orientation of a surface $M\subset {\mathbb{R}}^3$, with Gauss map $N:M\to S^2$. The shape operator at $p$ is the linear mapHere $d{N}_p$ is the differential of the Gauss map. Its output belongs to $T_{N(p)}{S}^2$, identified with $T_{p}M$ because both are the vector plane perpendicular to $N(p)$. The minus sign is our convention throughout this branch. ^[ref]

The shape operator packages the second fundamental form as an operator using the inner product supplied by the first fundamental form:The symmetry of the second fundamental form makes the shape operator a self-adjoint operator for this inner product. Thus an orthonormal basis of principal directions diagonalizes it. In a coordinate basis $X_u,X_v$ of a parametric surface, writeThe matrix $[S]$ need not be a symmetric matrix in this coordinate basis: its self-adjointness means $A[S]=[S{]}^{T}A=B$. Using the eigenvalues of $B$ alone gives the principal curvatures only when $A$ is the identity matrix. ^[ref]

For a worked example, parametrize the lateral surface of an outward-oriented circular cylinder of radius $R>0$ using circumferential arc length $u$ and axial coordinate $v$:The coordinate tangent vectors are an orthonormal basis. Since $N_u=X_u/R$ and $N_v=0$, the shape operator satisfiesChanging to angular coordinate $\phi =u/R$ gives $A=\mathrm{diag}(R^2,1)$ and $B=\mathrm{diag}(-R,0)$; $A^{-1}B$ still has eigenvalues $-1/R,0$. This is why the first fundamental form must participate in the calculation.

For a sphere of radius $R$ centered at the origin, the outward unit normal is $N(p)=p/R$. Hence $S_{p}v=-v/R$, so both principal curvatures are $-1/R$. For a plane, $dN=0$ and the shape operator vanishes. Reversing the unit normal negates the shape operator and both principal curvatures; the Gaussian curvature remains unchanged while the mean curvature changes sign, as in orientation reversal of surface curvature.

Two equal circular rings of radius $R$ in parallel planes at $z=\pm L$ provide a boundary for an open film. In the ideal axisymmetric model, a stationary connected film has the catenoid profileThe catenoid existence threshold for equal rings explains why this equation can have two solutions, one solution, or none. The cancellation of principal curvatures holds for both solutions whenever they exist; the mean curvature equation alone does not select a stable minimal surface.

This ideal geometry uses $R=1$ and $L/R=0.60$. The two profiles have different waist radii $a$, although both have $H=0$. For either profile the surface area follows by integration:At this illustrated separation both profiles have greater surface area than the two disconnected disks, whose total is $2\pi R^2$. This comparison disproves a claim of global area minimization in a class allowing disconnected competitors; it does not determine the sign of the second variation for small variations of a connected film.

A reported experiment used coaxial wire rings of radius $4\mathrm{cm}$, drawn from a surfactant solution and slowly separated until collapse. The solution contained tetradecyltrimethylammonium bromide in deionized water, with glycerol added to vary viscosity. Fluorescein enabled fluorescence imaging under cyan LED arrays; a Phantom V641 camera recorded 4,000 or 5,600 frames per second. Extracted contours gave neck radius, attachment angle, and inferred surface area. They showed neck narrowing and eventual breakup, rather than indefinite persistence of a stationary catenoid. Comparing solutions across viscosities helped distinguish liquid dissipation from the surrounding air's inertia, but the ring displacement beyond the last stable state was not precisely controlled. These are collapse observations, not direct measurements of equilibrium $H$. ^[ref]

A proposed classroom measurement would instead hold two equal rings coaxial on a translation stage, leave both sides open to the same gas, and photograph the settled silhouette beside a scale. Fit $r(z)=a\cosh (z/a)$ and inspect fit residuals before inferring cancellation of the principal curvatures. Repeat at several separations and with the camera rotated about the axis to test rotational symmetry. Keep solution composition, ring wetting, and elapsed drainage time comparable, and shield the film from drafts. Film weight, drainage, evaporation, surfactant gradients, and rupture can invalidate the ideal soap film pressure balance. A silhouette fit alone neither measures the pressure jump nor tests all possible perturbations of a stable minimal surface.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.