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Measure pressure relative to its hydrostatic value and let $\Delta P$ be the lower-side excess pressure minus the upper-side excess pressure. The long unobstructed parts of the tube act as Poiseuille resistances in series. Their lengths add to $L$ at leading order because $l\ll L$; entrance and exterior resistances are smaller for $L\gg a$. ConsequentlyHere $Q$ is downward laboratory-frame volume flux.

Take $z$ upwards in the particle frame and let $s$ measure distance from the particle surface to the tube wall. In this frame the particle is stationary and the wall moves upwards at speed $U$. In the thin, slowly varying gap, pressure is uniform across $s$ and the leading axial Stokes equation is $\mu w_{ss}=p_z$. No slip givesThe gap's upward relative flux is constant, while conservation of displaced volume gives $2\pi aq=\pi a^{2}U-Q$. Thus $p_z=6\mu U/h^2-12\mu q/h^3$, and integration from bottom to top yieldsCurvature corrections to the gap circumference are relative order $h/a$. Rapid end regions do not contribute to the leading singular neck resistance.

For the force use a cylindrical fluid control volume containing the particle, with full-area end caps just beyond it and the particle surface as an inner boundary. The upward pressure force on the caps is $\pi a^2\Delta P$. The outer wall exerts axial traction $\mu w_s(h)$ on the fluid, whereas the particle's traction on the fluid integrates to minus its upward hydrodynamic resistance $F$. The zero-inertia momentum balance is thereforeThe velocity profile gives $w_s(h)=4U/h-6q/h^2$, provingUsing the outer-wall traction is essential; directly using the particle-side shear and a flat-end pressure force would omit the pressure on the sloping particle surface. This is control-volume drag formula for a sphere in a tube.

The four dimensionless relations areSolving them givesWithin the preceding lubrication and tube-resistance approximations, the force isSince $h/a\ll 1$, $I_3^{\ast }\gg I_2^{\ast }\gg I_1^{\ast }$. Dropping the smaller additive $-2I_2^{\ast }+4I_1^{\ast }/3$ terms in the numerator yields the requested leading formulaThe Cauchy-Schwarz inequality gives $(I_2^{\ast }{)}^2\le I_1^{\ast }{I}_3^{\ast }$, so the retained $\delta$ term is positive. The term multiplied by $\delta$ must be retained: it can compete with $I_3^{\ast }$ even though the unmultiplied lower integrals cannot.

For a sphere of radius $a(1-ϵ)$, the neck has $h=aϵ+z^2/(2a)$ at leading order. With $z=a\sqrt{2ϵ}t$, the integrals reduce to $\sqrt{2}{ϵ}^{1/2-n}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-n}dt$. For $n=1,2,3$ these last integrals are $\pi ,\pi /2,3\pi /8$, giving the full two-sided integrals. PutThe printed sphere integrals are half these values: they correspond to integrating over one side of the neck rather than the stated interval spanning both sides. The substitution above establishes the radius-deficit normalization of sphere-tube lubrication integrals. Using the two-sided values, the force becomesso the three answers areThese are the corrected sphere-tube drag regimes with a radius-deficit gap. If the supplied half-integrals are treated as stipulated data instead, the same algebra gives $3\pi {ϵ}^{-5/2}/(8\sqrt{2})$, $1/\delta$ and $\pi \sqrt{2}{ϵ}^{-1/2}/3$ in the three regimes. Thus the first and third numerical drag coefficients double when the geometrically correct integrals are used; the regime boundaries and dominant flow patterns are unchanged.

In the first regime, $Q^{\ast }\sim \delta c_3{ϵ}^{-5/2}\ll 1$, $q^{\ast }\sim 1/4$, and $\Delta P^{\ast }\sim c_3{ϵ}^{-5/2}\sim F^{\ast }$. Almost no displaced fluid escapes through the long tube; it bypasses upwards through the thin annulus. The large pressure-driven bypass resistance sets the drag, with shear smaller.

In the second regime, $Q^{\ast }\sim 1$, $\Delta P^{\ast }\sim F^{\ast }\sim 1/\delta$, andAlmost all displaced volume travels through the unobstructed tube, whose pressure resistance dominates the force. Both terms in this small gap flux are worth displaying: their crossover at $\delta \sim {ϵ}^{3/2}$ changes the balance between pressure-driven and wall-driven gap transport but does not create a fourth leading force regime.

In the third regime, $1-Q^{\ast }\sim 4ϵ/3$, $q^{\ast }\sim ϵ/3$, and $\Delta P^{\ast }\sim 1/\delta \ll F^{\ast }$. Tube throughflow still carries almost all the displaced volume, but the dominant resistance is local annular lubrication rather than the net tube pressure drop. The relative gap flux is now of order $Uh$, so wall-driven and pressure-driven components both matter locally. The formulas describe hydrodynamic resistance after subtracting hydrostatic buoyancy; terminal settling would balance it against the particle's excess weight.

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence

For the force, enclose the fluid between sections just above and below the sphere, including the annular layer. The net upward pressure force is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the sphere exerts a downward force equal to the upward hydrodynamic force $F$ on itself. From the gap solution,The axial control volume force balance therefore gives the control-volume drag formula for a sphere in a tube:Using a control volume avoids having to integrate the curved-sphere normal pressure and tangential shear separately.

With $Q^{\ast }=Q/(\pi a^{2}U)$, $\Delta P^{\ast }=a\Delta P/(6\mu U)$, $q^{\ast }=2q/(aU)$, $F^{\ast }=F/(6\pi \mu aU)$, $I_n^{\ast }=a^{n-1}{I}_n$ and $L^{\ast }=4L/(3a)$, the three flux/pressure equations becomeSolving these linear relations givesThe force relation is $F^{\ast }=\Delta P^{\ast }+\frac{4}{3}{I}_1^{\ast }-q^{\ast }{I}_2^{\ast }$. Substitution first yieldsBecause $I_2^{\ast }/I_3^{\ast }=2ϵ/3$ and $I_1^{\ast }/I_3^{\ast }=2{ϵ}^2/3$, the terms $-2L^{\ast }{I}_2^{\ast }$ and $\frac{4}{3}{L}^{\ast }{I}_1^{\ast }$ are lower order than $L^{\ast }{I}_3^{\ast }$. Keeping the uniformly relevant leading terms givesThe squared $I_2^{\ast }$ is essential; replacing it by $I_2^{\ast }{I}_3^{\ast }$ would give the wrong sign and scaling. This establishes the three drag regimes for a nearly occluding sphere.

For $L^{\ast }\ll {ϵ}^{-1/2}$,The sphere nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant force comes from the local annular shear term. The pressure force associated with the overall tube resistance is smaller.

For ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$,The flow remains approximately piston-like, but the dominant force is now the pressure needed to drive Hagen-Poiseuille flow through the long clear tube. Both displayed terms in $q^{\ast }$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.

For $L^{\ast }\gg {ϵ}^{-5/2}$,The large tube resistance makes its net through-flow negligible. The sphere's displaced fluid instead passes upwards relative to it through the thin annular gap, with gap velocity of order $U/ϵ$. Its pressure drop and resulting pressure force dominate.

Finally consider two identical co-moving spheres, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^{2}U$, so the same total tube pressure drop is shared equally between the two identical spheres. The load sharing between co-moving spheres in a tube gives half the single-sphere force on each sphere in regime (ii). In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each sphere. In regime (iii), each sphere must pass essentially its own displacement flux through its annular gap, giving the same local pressure drag as before; adding another pressure jump changes the already small tube flux but not either leading drag. The force is unchanged in regimes (i) and (iii).

Use radius-deficit $ϵ$, $F^{\ast }=F/(6\pi \mu aU)$ and $\delta =3a/(4L)$. The control-volume drag formula for a sphere in a tube and the radius-deficit normalization of sphere-tube lubrication integrals give three leading resistances: $c_3{ϵ}^{-5/2}$ for $\delta \ll {ϵ}^{5/2}$, $1/\delta$ for ${ϵ}^{5/2}\ll \delta \ll {ϵ}^{1/2}$, and $2\pi \sqrt{2}/(3\sqrt{ϵ})$ for $\delta \gg {ϵ}^{1/2}$. They correspond to gap bypass with almost no tube throughflow, pressure resistance of tube throughflow, and local annular resistance, respectively.

Use $F^{\ast }=F/(6\pi \mu aU)$ and $L^{\ast }=4L/(3a)$. The control-volume drag formula for a sphere in a tube gives the displayed uniform leading drag. If $L^{\ast }\ll {ϵ}^{-1/2}$, $F^{\ast }\sim (4\pi /3){ϵ}^{-1/2}$ and local annular shear dominates. If ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$, $F^{\ast }\sim L^{\ast }$ and the Hagen-Poiseuille flow resistance of the long tube dominates. If $L^{\ast }\gg {ϵ}^{-5/2}$, $F^{\ast }\sim 3\pi {ϵ}^{-5/2}$ and pressure-driven bypass through the narrow gap dominates. The laboratory flow is piston-like in the first two limits and negligible in the third.