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The flow is steady and irrotational, so the Bernoulli equation has the same constant throughout the connected exterior. Far from the sphere it equals $v^2/2+p_{\infty }/\rho$. On the surface the speed is $3v|\sin \theta |/2$, givingThe maximum stagnation pressure is $p_{\infty }+\rho v^2/2$ at both poles. At the equator, where the flow is fastest, the pressure is $p_{\infty }-5\rho v^2/8$. The fore-aft symmetry gives zero integrated pressure drag in this inviscid model, the content of D'Alembert's paradox; it does not predict zero drag for a real viscous separated flow.

Locally, zero curl permits $\mathbf{v}=\nabla \Phi$; for a global single-valued potential the circulation around every noncontractible loop must also vanish. incompressibility gives $\nabla \cdot \mathbf{v}={\nabla }^2\Phi =0$. The exterior of a cylinder is multiply connected, so local irrotationality alone does not exclude a circulating solution.

For inviscid constant-density flow under conservative gravity, Euler's equation is ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (p/\rho +{\varphi }_g)$. The identity $(\mathbf{v}\cdot \nabla )\mathbf{v}=\nabla (v^2/2)-\mathbf{v}\times (\nabla \times \mathbf{v})$ gives, for irrotational flow,For steady flow, absorb the potential's time-only gauge term and obtain $p+\rho v^2/2+\rho {\varphi }_g=\mathrm{constant}$ throughout the connected fluid region. Without irrotationality the steady constant need only be common along each streamline.

Use $U$ for the uniform upstream speed and take the flow along $+x$. The boundary conditions are $v_r(a,\theta )=0$ and $\mathbf{v}\to U{\mathbf{e}}_x$ at infinity. A perfect inviscid model imposes no penetration, not no slip. For the intended no-circulation solution, require a globally single-valued potential, so the coefficient of $\theta$ is zero. The logarithmic term would be a net source and violates no penetration, so its coefficient is also zero. Growing terms with $n\ge 2$ violate the uniform far field. For each remaining $n\ge 2$, the surface normal condition then forces its decaying coefficient to vanish. The $n=1$ terms satisfy $C_1=U$ and $D_1=U{a}^2$:An irrelevant additive potential constant is understood. At the same elevation, Bernoulli balance with the upstream stream givesThe force per unit height ispressure symmetry gives zero drag and zero lift in this zero-circulation idealization. D'Alembert's paradox is the failure of this inviscid model to represent real separated viscous wakes and their drag, not a claim that a real pole feels no force.

The zero-circulation assumption matters: $\Phi \to \Phi +\Gamma \theta /(2\pi )$ leaves the stated velocity boundary conditions intact but adds $\Gamma /(2\pi a)$ to the surface tangential speed. It produces $F_y^{\prime }=-\rho U\Gamma$ for positive counterclockwise circulation. Thus the boundary conditions alone do not justify excluding the angular term; an initially noncirculating inviscid start, or an explicitly single-valued potential, supplies the missing condition.^[ref]

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Take a fixed sphere of radius $R$ in an unbounded incompressible inviscid fluid, with uniform velocity $U{\mathbf{e}}_z$ at infinity. Assume irrotational flow, so $\mathbf{v}=\nabla \varphi$ and incompressibility gives ${\nabla }^2\varphi =0$. Axisymmetry and the dipolar angular dependence of the far field suggestMatching the far field gives $A=U$. The no-penetration condition $v_r(R)=0$ then gives $B=U{R}^3/2$. ThusThe surface has stagnation points at its front and rear poles, while the equatorial speed is $3U/2$. Fluid slips tangentially past the sphere; imposing a no-slip condition would contradict this inviscid potential-flow model.

With gravity omitted or the hydrostatic pressure subtracted, steady Bernoulli givesAt the surface,The front–rear symmetry makes the integral of pressure force along the flow vanish: terms proportional to $\cos \theta$ integrate to zero over the sphere. This is D'Alembert's paradox, not a prediction that real spheres have no drag. Viscosity creates boundary layers; separation and a wake break the fore–aft symmetry. Potential flow remains valuable for the outer velocity field when viscous regions are thin, but it cannot supply their drag by itself.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.