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Use pressure per unit constant density, $\pi$, with the gravitational potential and centrifugal potential absorbed into it. Strict geostrophic balance is the exact balancewith material acceleration omitted, for a homogeneous incompressible flow. Writing $f=2\Omega \ne 0$, its components are $fv={\pi }_x$, $fu=-{\pi }_y$ and ${\pi }_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the vector calculus identityshows, on taking the curl of the balance, that $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus $w_z=0$ too, proving the Taylor–Proudman theorem:The physical pressure still includes its background hydrostatic pressure; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.

For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, impermeability condition gives $w_b={\mathbf{u}}_H\cdot {\nabla }_H{z}_b$ and $w_t={\mathbf{u}}_H\cdot {\nabla }_H{z}_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq {\mathbf{u}}_H\cdot {\nabla }_{H}H\ne 0$. Exact columnarity must give way to small corrections from ageostrophic flow. The leading horizontal geostrophic flow can remain nearly independent of $z$, and the leading reduced pressure has ${\pi }_z=0$, but $w_z$ is now nonzero and exact three-dimensional strict geostrophic balance no longer holds. For small slopes, the dominant vertical vorticity equation iswhere $q=v_x-u_y$ is relative vorticity. Equivalently, $D[(q+f)/H]/Dt\simeq 0$. Crossing geostrophic contours stretches or squashes columns: increasing $H$ produces cyclonic relative vorticity, decreasing $H$ produces anticyclonic relative vorticity. The full vorticity equation also contains tilting terms, neglected at this columnar order.

For the circular shallowing, fluid arriving from upstream has potential vorticity $f/h$. Its height is $h(1-ϵ)$ inside, so conservation givesThis uses the stated column-stretching description and initially open streamlines. To obtain the leading disturbance velocity field, use the conventional streamfunction definition $u=-{\Psi }_y$, $v={\Psi }_x$, and write $\Psi =-Uy+\psi (r)$. Then $q={\nabla }_H^2\psi =r^{-1}(r{\psi }_r{)}_r$. Regularity at $r=0$ gives ${\psi }_r=-fϵr/2$ inside. Outside, ${\psi }_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-fϵa^2/2$. Choose the irrelevant additive constant so thatBoth $\psi$ and ${\psi }_r$ match at $a$, so there is no extra vorticity sheet on the rim. This is the circular topographic anticyclone. The radial expression specifies the disturbance; the complete streamfunction is $\Psi =-Uy+\psi$, since a radial function alone cannot represent the required uniform far-field current. Reversing the streamfunction sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.

Differentiation gives the full relative velocity field in the rotating frame:away from the narrow transition across the change of depth. The disturbance turns clockwise for $fϵ>0$ and decays as $1/r$. The horizontal field is the leading small-$ϵ$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.

Take $U>0$ and $ϵ>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed streamlines. If $Ca>U$, the interior stagnation point $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior stagnation point $(0,-C{a}^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a separatrix through the saddle. The two points meet at the rim at onset, giving the closed-streamline threshold for a circular topographic anticyclone:At this threshold $U/(fa)=ϵ/2\ll 1$, consistent with small Rossby number. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the potential vorticity of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.

Integrating ${\psi }_y=x+\sin t$ and imposing $-{\psi }_x=-y$ gives the streamfunctionup to an arbitrary function of time. At the specified time, the streamlines areFor $C\ne 0$, the connected branches are rectangular hyperbolas. The zero level consists of the lines $y=0$ and $x=-1$, with the stagnation point $(-1,0)$ at their intersection; their half-lines and the stationary point are the distinct streamlines.

For a two-dimensional incompressible flow $\mathbf{u}=(u,v)$, the continuity equation is $u_x+v_y=0$. Locally, or globally on a simply connected fluid region, this permits a streamfunction withIndeed, the one-form $-vdx+udy$ is closed by incompressibility and hence is locally the differential of $\psi$. Conversely these expressions automatically give zero divergence by equality of mixed derivatives.

A streamline at a fixed time is tangent to the instantaneous velocity field. Along it,Thus streamlines are connected curves on which the streamfunction is constant; wherever $\nabla \psi \ne 0$, the level curves have exactly the required tangent direction. At a stagnation point, the tangent is degenerate and distinct branches of a level set need not form a single streamline. For unsteady flow this argument fixes time: it does not identify streamlines with particle pathlines.

Here $u=-y$, $v=x$. The streamlines are $x^2+y^2=\text{constant}$. This is rigid rotation counterclockwise with angular velocity one; the origin is a stagnation point and all other trajectories are circles. The middle panel shows their orientation.

Here $u=x$, $v=-y$. The streamlines are $xy=\text{constant}$, rectangular hyperbolas, together with the coordinate axes as separatrices. Particles satisfy $x(t)=x(0)e^t$, $y(t)=y(0)e^{-t}$: the flow stretches along the $x$ axis and compresses along the $y$ axis, with a saddle stagnation point at the origin. The left panel shows the flow directions.

Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2{)}^{1/2}$ and $\vartheta =\arg (x+iy)$, superposition of uniform potential flow and a two-dimensional point source giveswhere $(u,v)=\nabla \Phi =({\psi }_y,-{\psi }_x)$. The angular branch cut records the volume flux from the source; the logarithm's arbitrary reference length only changes an additive constant in the velocity potential.

The stagnation point lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid isThe stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing streamlines through it have $\psi =\pm Q/2$ when the angle is taken in $(-\pi ,\pi )$. Far downstream $\vartheta \to 0$, so these streamlines approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the Rankine half-body between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta /\pi )$, $x=y\cot \vartheta$, $0<\vartheta <\pi$.

For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source isThe first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi (x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is thereforeThis is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau -t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote Darcy velocity and bulk-area source flux in a medium of uniform porosity $\varphi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\varphi$. No porosity is specified here, so the displayed time formula fixes the kinematic convention.

For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.

At an interior stagnation point, ${\psi }_r={\psi }_{\theta }=0$. The first equation isSince $\sin \theta >0$ inside the fluid, $r=2\sqrt{2}a\sin \theta$. AlsoOn the preceding locus the bracket is nonzero, so $\cos \theta =0$. ThereforeFor $\Omega >0$, $\psi <0$ in the interior, and its nested nonzero level curves are closed recirculating streamlines around this unique stagnation point.

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$. Its complex velocity is $u-iv=dw/dz$. Hereso the far-field motion has velocity $(U,0)$ and speed $U$. In polar coordinates the stream function, shifted by a harmless constant, isIt is zero on $r=a$, proving the cylinder is a streamline.

Stagnation points are zeros of $dw/dz$, hence roots of $U{z}^2-i\Gamma z/(2\pi )-U{a}^2=0$. Put $\gamma =\Gamma /(4\pi Ua)\in (0,1)$. The two roots areBoth have modulus $a$ and are distinct. Their angles satisfy $\sin \theta =\gamma$, namely $\arcsin \gamma$ and $\pi -\arcsin \gamma$. The tangential velocity on the cylinder is $-2U\sin \theta +\Gamma /(2\pi a)$, confirming the same zeros. Streamlines are level curves of $\psi$; the positive circulation places both stagnation points on the upper side and increases the flow speed below the cylinder.

For a planar incompressible flow, a stream function satisfies $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The irrotational flow condition is $v_x-u_y=0$. Where both exist, the Cauchy-Riemann equations make the complex potential $w=\varphi +i\psi$ analytic with $w^{\prime }=u-iv$.

For the logarithmic potential, write $z-z_0=r{e}^{i\theta }$. ThenThe outward volume flux per unit span across a circle is $2\pi r{u}_r=m$. Hence it is a line source of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.

By superposition, the full flow hasFor positive $m$ and $U$, the stagnation points areThey are real, one upstream of the source and one downstream of the sink. Away from the singularities, streamlines are level curves ofThe zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a Rankine oval. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.

Use the convention that positive vortex strength is counterclockwise and complex velocity is $u-iv$. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices isLocal choices of logarithm branch change no velocity. Differentiation givesFor nonzero vortex strength and $U>0$, its two zeros areThe number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity $-\Gamma /(4\pi a)$, while the background contributes $(U,0)$. Their velocities are therefore equal. Their separation stays $(2a,0)$, making this induced velocity constant for all time. Both travel along parallel straight lines with

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem:

The complex velocity obtained from part (b) isFor a nonzero planar potential doublet, $B/A=(a^2/b^2)e^{-2i\alpha }$. Its zeros therefore satisfyand henceThe denominators are nonzero because $a/b<1$. For either sign,so $|z_{\pm }|=a$. They are distinct because $t_{+}\ne t_{-}$ and $t\mapsto (c-bt)/(1-t)$ is injective, its determinant being $c-b\ne 0$. Clearing the velocity denominators gives a quadratic with these two roots, so there are no further finite stagnation points. Thus both stagnation points lie on the obstacle boundary.

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.

Use the convention that positive vortex strength is counterclockwise and complex velocity is $u-iv$. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices isLocal choices of logarithm branch change no velocity. Differentiation givesFor nonzero vortex strength and $U>0$, its two zeros areThe number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity $-\Gamma /(4\pi a)$, while the background contributes $(U,0)$. Their velocities are therefore equal. Their separation stays $(2a,0)$, making this induced velocity constant for all time. Both travel along parallel straight lines with

Orient the body contour counterclockwise. The outward normal from the body satisfies $(n_x,n_y)ds=(dy,-dx)$, so pressure produces complex force $F_x-i{F}_y=-i\oint pd\bar{z}$. Bernoulli's theorem gives $p=p_{\infty }-\rho |w^{\prime }{|}^2/2$; the constant term integrates to zero. On an impermeable boundary the velocity is tangent, so $w^{\prime }dz$ is real, equivalently $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$. Consequently $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$ and

Choose the square-root branch analytic off the slit $[-a,a]$ with $\sqrt{z^2-a^2}\sim z$ at infinity. Continuation from $x>a$ through the upper half-plane gives the upper-face value $+i\sqrt{a^2-x^2}$; continuation through the lower half-plane gives the lower-face value $-i\sqrt{a^2-x^2}$. Both have the required square and the sign is fixed by the branch normalization.

Differentiate the potential:At infinity, $z/\sqrt{z^2-a^2}=1+a^2/(2z^2)+O(z^{-4})$, so $w^{\prime }(z{)}^2=e^{-2i\alpha }+O(z^{-2})$ has no $z^{-1}$ term. Deforming a contour around the plate to a large circle in its analytic exterior therefore gives zero integral. The force is $\boxed{F_x=F_y=0}$. The contour formulation includes the endpoint behavior and avoids assigning separate divergent endpoint pressure contributions.

For $0<\alpha <\pi /2$, use the two face values of the square root. The upper tangential velocity is $\cos \alpha -x\sin \alpha /\sqrt{a^2-x^2}$, vanishing at $\boxed{x=a\cos \alpha }$ on the upper face. The lower value is $\cos \alpha +x\sin \alpha /\sqrt{a^2-x^2}$, vanishing at $\boxed{x=-a\cos \alpha }$ on the lower face. Squaring the equation $w^{\prime }=0$ gives only these candidates, and the branch signs locate the correct faces. At $\alpha =\pi /2$ both face stagnation points have coordinate zero. At $\alpha =0$ the exterior velocity is identically one, so there is no finite stagnation point; the endpoints are merely limits of the two locations as the angle tends to zero.

The plotted streamlines are contours of the imaginary part of the potential, with the two boundary stagnation points marked. The far-field velocity is $(\cos \alpha ,\sin \alpha )$, giving a flow from the lower left toward the upper right at the plotted angle.

For the proposed velocity field, $u_x+w_z=1-1=0$, so incompressibility holds. Its convective acceleration isand both components have zero Laplacian. The Navier-Stokes equation therefore requires $p_x=-x$ and $p_z=-z$. Hencewhere $p_{\ast }$ is an arbitrary constant reference pressure. The same pressure gradient balances the same convective acceleration in the Euler equations for an inviscid fluid. Since the viscous terms vanish identically, this is an exact solution for every positive Reynolds number, as well as for the inviscid limit.

At $z=0$, the normal velocity is zero, and the prescribed dimensionless tangential traction isThis stagnation point flow therefore has zero surface shear stress, although its strain-rate tensor and its viscous normal stresses are nonzero.

With $u=x{f}^{\prime }(\zeta )$ and $W=-f(\zeta )$, incompressibility is automatic because $u_x=f^{\prime }$ and $W_{\zeta }=-f^{\prime }$. In the streamwise boundary-layer equation,Equating coefficients of $x$ givesThis coefficient comparison determines the equation even though the streamwise equation vanishes identically at the stagnation point $x=0$.

The no-penetration boundary condition gives $f(0)=0$, the surface shear stress gives $x{f}^{\prime \prime }(0)=\gamma x$, and matching the tangential velocity to the outer solution gives $f^{\prime }\to 1$. ThusIf the correction $f^{\prime }-1$ is integrable at the far end, then $f=\zeta +O(1)$ there, which is consistent with $W\sim -\zeta$. The small-shear stress solution below has precisely this property.

At the interface, the tangential velocity is $u(x,0)=x{f}^{\prime }(0)$ and the normal velocity remains zero. The preceding error function expression givesHence, within the leading boundary layer model,Allowing the formally discarded finite-Reynolds number terms gives an additional order-${\delta }^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi /8}\approx 0.626657$.

Thus positive surface shear increases the outward surface speed linearly: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The stagnation point remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).

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 $C_p$.