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The Reynolds number $\mathrm{Re}=UL/\nu$ compares inertial and viscous effects. Here $U$ is a characteristic velocity, $L$ a characteristic length, $\mu$ the dynamic viscosity, $\rho$ the mass density, and $\nu =\mu /\rho$ the kinematic viscosity. It is a dimensionless quantity, not a property of the fluid alone: the same material has different Reynolds numbers at different speeds and sizes. Specify the chosen scales when reporting it.

The comparison comes from the Navier-Stokes equation. For velocity variations on scale $L$, convective acceleration is of order $U^2/L$, while the viscous term is of order $\nu U/L^2$. Their ratio is $UL/\nu$. Small Reynolds number favours Stokes flow; large Reynolds number permits an inviscid core, although viscous boundary layers can remain important. Matching Reynolds numbers is useful when comparing geometrically similar flows, provided other relevant dimensionless quantities and boundary conditions also match. ^[ref]

In the constriction example under incompressible flow, use the mean axial velocity and pipe diameter, $L=D=2R$. Taking $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and a stipulated dynamic viscosity $\mu =0.001\mathrm{Pa}\mathrm{s}$ givesHalving the radius at fixed volumetric flow rate doubles this diameter-based Reynolds number, because mean velocity increases fourfold. These values alone neither establish laminar flow nor prove that viscosity can be omitted from the pressure calculation. Increasing $\mu$ to $0.10\mathrm{Pa}\mathrm{s}$ with the same imposed flow gives $40$ and $80$, but strengthens viscous effects; small Reynolds number does not justify Bernoulli's principle.

The name recalls Osborne Reynolds's 1883 experimental investigation of water motion and resistance in pipes. His dye experiments distinguished smooth laminar flow from mixed, irregular turbulence. ^{[ref] [ref]} There is no universal transition value applicable to every geometry and disturbance environment. ^[ref]

In his 1883 paper, Osborne Reynolds investigated how the character of pipe flow depends jointly on velocity, pipe diameter and viscosity. His final dye apparatus drew water from a large glass tank through a horizontal glass tube with a smooth trumpet-shaped entrance. A separate reservoir supplied strongly colored water just upstream of the entrance; a valve below the tank controlled the outflow. The glass tube was immersed in the tank, so its internal dye band could be watched. Reynolds repeated the trials with different tube diameters and water temperatures. ^[ref]

The volumetric flow rate was measured independently of the appearance of the dye. Reynolds timed a one-inch fall of the tank's water level, corresponding to a calibrated discharged volume. In modern notation, if this volume is $V$ and the elapsed time is $\Delta t$, then $Q=V/\Delta t$ and $U=4V/(\pi D^2\Delta t)$. He determined the glass tubes' internal diameters by filling them with water. For a uniform bore, a known filled length $\ell$ and measured water volume $V_{\mathrm{tube}}$ determine $D=\sqrt{4V_{\mathrm{tube}}/(\pi \ell )}$; checking bore uniformity matters because the same volume would only give an average cross-section in a tapered tube. Thermometers at different tank levels checked both the temperature and its uniformity. Reynolds compared the viscosity at different temperatures using Poiseuille's results and checked his own water by separate experiments of that type. Thus the Reynolds number estimate did not come from judging the dye band or from assuming that transition occurred at a preset value. ^[ref]

A practical repetition can instead collect a known outflow volume in a measuring vessel and time its collection, or weigh the collected water and divide its mass by mass density. Measure the internal diameter, rather than the external diameter, and obtain kinematic viscosity at the measured water temperature from independently established properties or a separate viscosity measurement. Record the dye appearance and observation position alongside $Q$, $D$, temperature and the resulting Reynolds number. These are alternative measurement procedures following $Q=V/\Delta t$, not additional historical claims about Reynolds' dye apparatus.

Begin with settled water and a small dye supply, then open the outlet valve in small steps. Reynolds saw a thin straight band at sufficiently low speeds. With a tank that had not fully settled, the band could wander while retaining its coherence. At higher discharge it could abruptly spread into the surrounding water downstream of the entrance. Short illumination by an electric spark resolved curls within what otherwise looked like a colored cloud. This connected the visible breakup with internal irregular motion rather than merely the appearance of an opaque pipe. ^[ref]

The controls are part of the result. Reynolds used a smooth entrance and let the tank rest to reduce inlet disturbances. He found that a heated tank could retain slow circulation as it cooled, so measuring one temperature was insufficient: thermal uniformity and residual motion also mattered. His wire-coil trial broke up the band below the previous onset speed, but he explicitly cautioned that once the dye was mixed he could not tell from that dye alone whether the motion became orderly again farther along the tube. In a repetition, keep the dye supply small enough to act approximately as a passive scalar, avoid a strong injection jet, and record the inlet condition and downstream observation distance. If also testing the Hagen-Poiseuille equation, allow the profile to develop and measure pressure drop between positions within the developed region; do not treat the entrance and outlet losses as part of its uniform pressure gradient. The dye experiment by itself does not establish fully developed flow. ^[ref]

The experiment's usefulness is that it compares a visible change in the motion with independently measured scales. Repeating it with several diameters and temperatures, while controlling disturbances, tests the organizing role of the Reynolds number. Its limitations are equally instructive: dye is a tracer of transport, diffusion still acts in laminar flow, and a local dye observation cannot by itself prove that turbulence persists throughout an arbitrarily long pipe.

For a sinusoidal small-amplitude bed, constant current $U>0$ and buoyancy frequency $N$, put $m=\sqrt{N^2/U^2-k^2}>0$. The linear vertical velocity amplitude is $W(z)=ikU{\eta }_0\sin [m(H-z)]/\sin (mH)$. The upward-wave coefficient is $A_{+}=-kU{\eta }_0{e}^{-imH}/[2\sin (mH)]$. At $mH=n\pi$, the prescribed nonzero bed motion and homogeneous rigid-lid condition are incompatible with a bounded stationary inviscid solution. With angle $\theta$ between intrinsic group velocity and vertical, $\tan \theta =m/k$, giving $kH\tan \theta =n\pi$.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

The rocket exhaust speed is the magnitude of the emitted material's velocity relative to the instantaneous rocket. In the ideal rearward-ejection convention of the rocket equation, a rocket with inertial velocity $v$ emits material with inertial velocity $v-u$ to first order. A constant $u$ does not make $v-u$ constant.

The rocket mass ratio compares the initial and final total rocket masses for a burn. It is dimensionless, and the ideal rocket equation becomes $\Delta v=u\ln R$. For the same fixed rocket exhaust speed, two successive burns with rocket mass ratios $R_1$ and $R_2$ have combined rocket mass ratio $R_1{R}_2$ and total velocity gain $u\ln R_1+u\ln R_2$.

For stress-free impermeable fixed-temperature plates, thermal-time growth rate $s$, horizontal wavenumber $k$ and vertical index $n$, put $a=k^2+n^2{\pi }^2$ and $P=\mathrm{Pr}$. Vertical velocity, temperature and vertical vorticity amplitudes satisfy $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, $(s+a)\Theta =W$. Taking their determinant givesThis is a cubic and remains valid without dividing by a potentially zero factor. Setting $s=0$ or $s=i\omega$ determines the stationary or admissible oscillatory neutral curves.