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| {{Refimprove|article|date=March 2012}}
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| [[File:Terminal velocity.gif|thumb|Terminal velocity where the downward force of gravity (''F<sub>g</sub>'') equals the air resistance/force of drag (''F<sub>d</sub>'') and buoyancy, velocity becomes asymptotic to a constant value with net acceleration = 0]][[File:Terminal velocity.svg|thumb|200px|right|The downward force of gravity (''F<sub>g</sub>'') equals the restraining force of drag (''F<sub>d</sub>''). The net force on the object is then zero, and the result is that the velocity of the object remains constant.]]
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| The '''terminal velocity''' of a falling object is the velocity of the object when the sum of the [[Drag (physics)|drag]] force (''F<sub>d</sub>'') and [[buoyancy]] equals the downward force of [[gravity]] (''F<sub>G</sub>'') acting on the object. Since the [[net force]] on the object is zero, the object has zero [[acceleration]].<ref>
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| {{cite web
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| | publisher = NASA Glenn Research Center
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| | title = Terminal Velocity
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| | url = http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html
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| | accessdate =2009-03-04
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| }}</ref>
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| In [[fluid dynamics]], an object is moving at its '''terminal velocity''' if its speed is constant due to the restraining force exerted by the fluid through which it is moving.
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| As the [[speed]] of an object increases, the drag force acting on the object, resultant of the substance (e.g., air or water) it is passing through, increases. At some speed, the drag or force of resistance will equal the gravitational pull on the object (buoyancy is considered below). At this point the object ceases to accelerate and continues falling at a constant speed called terminal velocity (also called settling velocity). An object moving downward with greater than terminal velocity (for example because it was thrown downwards or it fell from a thinner part of the atmosphere or it changed shape) will slow down until it reaches terminal velocity. Drag depends on the projected area, and this is why objects with a large projected area relative to mass, such as parachutes, have a lower terminal velocity than objects with a small projected area relative to mass, such as bullets.
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| ==Examples==
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| Based on wind resistance, for example, the terminal velocity of a [[skydiving|skydiver]] in a belly-to-earth (i.e., face down) [[free-fall]] position is about 195 [[km/h]] (122 [[Miles per hour|mph]] or 54 [[m/s]]).<ref name="Huang 1999">{{cite web |url=http://hypertextbook.com/facts/JianHuang.shtml |last=Huang |first=Jian |work=The Physics Factbook |publisher=Glenn Elert, Midwood High School, Brooklyn College |title=Speed of a Skydiver (Terminal Velocity)|year=1999}}</ref> This velocity is the [[asymptote|asymptotic]] limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.
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| Higher speeds can be attained if the skydiver pulls in his or her limbs (see also [[freeflying]]). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s),<ref name="Huang 1999" /> which is almost the terminal velocity of the [[Peregrine Falcon]] diving down on its prey.<ref>{{cite web |url=http://web.archive.org/web/20090512003331/http://www.fws.gov/endangered/recovery/peregrine/QandA.html |title=All About the Peregrine Falcon (archived)|publisher=U.S. Fish and Wildlife Service |date=2007-12-20}}</ref> The same terminal velocity is reached for a typical [[.30-06 Springfield|.30-06]] bullet dropping downwards—when it is returning to earth having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study.<ref>{{cite web |url=http://www.loadammo.com/Topics/March01.htm |title=Bullets in the Sky |author= The Ballistician |publisher=W. Square Enterprises, 9826 Sagedale, Houston,Texas 77089 |date=March 2001}}</ref>
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| Competition speed skydivers fly in the head down position and reach even higher speeds. The current world record is 1,357.6km/h (843.6mph/Mach 1.25) by [[Felix Baumgartner]] who skydived from 38,969.4m (127,852.4ft) above earth on 14 October 2012. The record was set due to the high altitude where the lesser density of the atmosphere decreased drag.<ref>{{cite web|title=Red Bull Stratos: Mission Accomplished|url=https://www.redbullcontentpool.com/content/stratos/products/red_bull_stratos_mission_accomplished|work=Red Bull Stratos Mission Press Releases|publisher=Red Bull Stratos Mission|accessdate=14 October 2012}}</ref>
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| ==Physics==
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| Mathematically, terminal velocity—without considering [[buoyancy]] effects—is given by
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| : <math>V_t= \sqrt{\frac{2mg}{\rho A C_d }}</math>
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| where
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| *<math>V_t</math> is terminal velocity,
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| *<math>m</math> is the [[mass]] of the falling object,
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| *<math>g</math> is the [[Earth's gravity|acceleration due to gravity]],
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| *<math>C_d</math> is the [[drag coefficient]],
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| *<math>\rho</math> is the [[density]] of the fluid through which the object is falling, and
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| *<math>A</math> is the [[projected area]] of the object.
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| Mathematically, an object approaches its terminal velocity [[asymptote|asymptotically]].
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| Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using [[buoyancy|Archimedes' principle]]: the mass <math>m</math> has to be reduced by the displaced fluid mass <math>\rho\mathcal{V}</math>, with <math>\mathcal{V}</math> the [[volume]] of the object. So instead of <math>m</math> use the reduced mass <math>m_r=m-\rho\mathcal{V}</math> in this and subsequent formulas.
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| On Earth, the terminal velocity of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional [[surface area]].
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| Air density increases with decreasing altitude, ca. 1% per {{convert|80|m|ft}} (see [[barometric formula]]). For objects falling through the atmosphere, for every {{convert|160|m|ft}} of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed ''decreases'' to change with the local terminal velocity.
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| ===Derivation for terminal velocity===
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| Mathematically, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the [[drag equation]]):
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| :<math>F_{net} = m a = m g - {1 \over 2} \rho v^2 A C_\mathrm{d}</math>
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| At [[equilibrium]], the [[net force]] is zero (F = 0);
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| :<math>m g - {1 \over 2} \rho v^2 A C_\mathrm{d} = 0</math>
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| Solving for ''v'' yields
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| :<math>v = \sqrt\frac{2mg}{\rho A C_\mathrm{d}}</math>
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| {| class="toccolours collapsible collapsed" width="90%" style="text-align:left"
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| !Derivation of the solution for the velocity ''v'' as a function of time ''t''
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| |-
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| The drag equation is
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| :<math> m a = m \frac{\mathrm{d}v}{\mathrm{d}t} = m g - \frac{1}{2} \rho v^2 A C_\mathrm{d}</math>
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| A more practical form of this equation can be obtained by making the substitution ''k'' = {{frac|1|2}}''ρAC''<sub>d</sub>.
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| Dividing both sides by ''m'' gives
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| :<math>\frac{\mathrm{d}v}{\mathrm{d}t}=g-\frac{kv^2}{m}</math>
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| The equation can be re-arranged into
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| :<math>dt = \frac{\mathrm{d}v}{g - \frac{kv^2}{m}}</math>
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| Taking the integral of both sides yields
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| :<math>\int_0^t {\mathrm{d}t^\prime} = \int_0^v \frac{\mathrm{d}v^\prime}{g-\frac{kv^{\prime 2}}{m}} = {1 \over g}\int_0^v \frac{\mathrm{d}v^\prime}{1-\alpha^2 v^{\prime 2}}</math>
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| where ''α'' = ( {{frac|''k''|''mg''}} )<sup>{{frac|1|2}}</sup>.
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| After integration, this becomes
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| :<math>t - 0 = {1 \over g}\left[{\ln(1 + \alpha v^\prime) \over 2\alpha} - \frac{\ln(1 - \alpha v^\prime)}{2\alpha} + C \right]_{v^\prime=0}^{v^\prime=v}={1 \over g} \left[{\ln \frac{1 + \alpha v^\prime}{1 - \alpha v^\prime} \over 2\alpha} + C \right]_{v^\prime=0}^{v^\prime=v}</math>
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| or in simpler a form
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| :<math>t = {1 \over 2\alpha g} \ln \frac{1 + \alpha v}{1 - \alpha v}</math>
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| The [[inverse hyperbolic tangent]] is defined as:
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| :<math>\frac{1}{2} \ln \frac{1+\alpha v}{1-\alpha v}=\mathrm{arctanh}(\alpha v)</math>
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| So the solution of the integral is
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| :<math>t = \frac{\mathrm{arctanh}(\alpha v)}{\alpha g}</math>
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| or alternatively,
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| :<math>\frac{1}{\alpha}\tanh(\alpha g t) = v</math>
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| with tanh the [[hyperbolic tangent]] function. Assuming that ''g'' is positive (which it was defined to be), and substituting ''α'' back in, the velocity ''v'' becomes
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| :<math> v=\sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{k}{mg}} g t\right)</math>
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| Next, after ''k'' = {{frac|1|2}}''ρAC''<sub>d</sub> has been substituted, the velocity ''v'' is in the desired form:
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| :<math>v = \sqrt\frac{2mg}{\rho A C_d} \tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right)</math>
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| As time tends to infinity ( ''t'' → ∞ ), the hyperbolic tangent tends to 1, resulting in the terminal velocity
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| :<math> v = \sqrt\frac{2mg}{\rho A C_d}</math>
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| |}
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| === Terminal velocity in creeping flow ===
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| [[File:Stokes sphere.svg|thumb|right|200px|Creeping flow past a sphere: [[Streamlines, streaklines, and pathlines|streamlines]], drag force ''F''<sub>d</sub> and force by gravity ''F''<sub>g</sub>]]
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| For very slow motion of the fluid, the inertia forces of the fluid are negligible (assumption of massless fluid) in comparison to other forces. Such flows are called [[Stokes flow|creeping flows]] and the condition to be satisfied for the flows to be creeping flows is the [[Reynolds number]], <math>Re \ll 1</math>. The equation of motion for creeping flow (simplified [[Navier-Stokes equation]]) is given by
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| :<math>\nabla p = \mu \nabla^2 {\mathbf v} </math>
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| where
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| *<math>{\mathbf v}</math> is the fluid velocity vector field
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| *<math>p</math> is the fluid pressure field
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| *<math>\mu</math> is the fluid [[viscosity]]
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| The analytical solution for the creeping flow around a sphere was first given by [[George Gabriel Stokes|Stokes]] in 1851. From Stokes' solution, the drag force acting on the sphere can be obtained as
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| :<math>\quad (6) \qquad D = 3\pi \mu d V \qquad \qquad \text{or} \qquad \qquad C_d = \frac{24}{Re} </math>
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| where the Reynolds number, <math>Re = \frac{1}{\mu} \rho d V</math>. The expression for the drag force given by equation (6) is called [[Stokes' law]].
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| When the value of <math>C_d</math> is substituted in the equation (5), we obtain the expression for terminal velocity of a spherical object moving under creeping flow conditions:
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| :<math>V_t = \frac{g d^2}{18 \mu} \left(\rho_s - \rho \right)</math>
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| ==== Applications ====
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| The creeping flow results can be applied in order to study the settling of sediment particles near the ocean bottom and the fall of moisture drops in the atmosphere. The principle is also applied in the [[Viscometer#Falling sphere viscometers|falling sphere viscometer]], an experimental device used to measure the viscosity of high viscous fluids.
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| ===Finding the terminal velocity when the drag coefficient is not known===
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| In principle one doesn't know beforehand whether to apply the creeping flow solution, or what coefficient of drag to use, because the coefficient depends on the speed. What one can do in this situation is to calculate the product of the coefficient of drag and the square of the [[Reynolds number]]:
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| :<math>C_d \mathrm{Re}^2 = \frac{mgD^2}{A\rho\nu^2}</math>
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| where ν is the [[kinematic viscosity]], equal to μ/ρ. This product is a function of Reynolds number, and one can consult a graph of C<sub>d</sub> versus Re to find where along the curve the product attains the correct value (a qualitative example of such a graph for spheres is found at this NASA site: [http://www.grc.nasa.gov/WWW/k-12/airplane/dragsphere.html]) From this one know'''s the coefficient of drag and one can then use the formula given higher up to find the terminal velocity.'''
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| '''For a spherical object, the above-mentioned product can be simplified:'''
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| :<math>C_d \mathrm{Re}^2 = \frac{4mg}{\pi\rho\nu^2}</math>
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| We can see from this that the regime and the drag coefficient depend only on the sphere's weight and the fluid properties. There are three regimes: creeping flow, intermediate-Reynolds number Newton's Law (almost constant drag coefficient), and a high-Reynolds number regime.<ref>''Chemical Engineer's Handbook'', edited by [[Robert H. Perry]] and [[Cecil Chilton]], fifth edition, ISBN 978-0070494787, pp. 5-61 and 6-62.</ref> In the latter regime the boundary layer is everywhere turbulent (see [[Golf ball#Aerodynamics]]). These regimes are given in the following table. The weight range for each regime is given for water and air at 1 [[atm (unit)|atm]] pressure and 25 °C. Note though that the weight (given in terms of mass in [[standard gravity]]) is the weight in the fluid, which is less than the mass times the local gravity because of buoyancy.
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| {| class="wikitable"
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| |-
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| ! Regime
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| ! Range of Reynolds number
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| ! Range of C<sub>d</sub>Re<sup>2</sup>
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| ! Range of weight in water
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| ! Range of weight in air
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| |-
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| | Creeping flow
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| | Quite accurate up to 0.3
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| | Up to 7.2
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| | Up to {{convert|0.00058|mg-f|nN|abbr=on}}
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| | Up to {{convert|0.00017|mg-f|nN|abbr=on}}
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| |-
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| | C<sub>d</sub> between 0.4 and 0.5
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| | {{gaps|1|000}} to {{gaps|200|000}}
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| | {{gaps|500|000}} to 2{{e|10}}
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| | {{convert|40|mg-f|mN|abbr=on}} to {{convert|1.6|kg-f|N|abbr=on}}
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| | {{convert|11|mg-f|mN|abbr=on}} to {{convert|470|g-f|N|abbr=on}}
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| |-
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| | C<sub>d</sub> between 0.1 and 0.2
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| | Over {{gaps|400|000}}
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| | Over 1.6{{e|10}}
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| | Over {{convert|1.3|kg-f|N|abbr=on}}
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| | Over {{convert|375|g-f|N|abbr=on}}
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| |}
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| Between the first two regimes there is a smooth transition. But notice that there is overlap between the ranges of C<sub>d</sub>Re<sup>2</sup> for the last two regimes. Spheres in this weight range have two stable terminal velocities. A rough surface, such as of a dimpled [[golf ball]], allows transition to the lower drag coefficient at a lower Reynolds number.
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| == Terminal velocity in the presence of buoyancy force ==
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| [[File:Settling velocity quartz.png|thumb|right|200px|Settling velocity Ws of a sand grain (diameter d, density 2650 kg/m<sup>3</sup>) in water at 20 °C, computed with the formula of Soulsby (1997).]]
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| When the buoyancy effects are taken into account, an object falling through a fluid under its own weight can reach a terminal velocity (settling velocity) if the net force acting on the object becomes zero. When the terminal velocity is reached the weight of the object is exactly balanced by the upward [[buoyancy force]] and drag force. That is
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| :<math> \quad (1) \qquad W = F_b + D </math>
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| where
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| *<math>W</math> = weight of the object,
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| *<math>F_b</math> = buoyancy force acting on the object, and
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| *<math>D</math> = drag force acting on the object.
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| If the falling object is spherical in shape, the expression for the three forces are given below:
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| :<math>\begin{align}
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| \quad &(2) \qquad& W &= \frac{\pi}{6} d^3 \rho_s g \\
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| \quad &(3) \qquad& F_b &= \frac{\pi}{6} d^3 \rho g \\
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| \quad &(4) \qquad& D &= C_d \frac{1}{2} \rho V^2 A
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| \end{align}</math>
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| where
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| *<math>d</math> is the diameter of the spherical object
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| *<math>g</math> is the gravitational acceleration,
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| *<math>\rho</math> is the density of the fluid,
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| *<math>\rho_s</math> is the density of the object,
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| *<math>A = \frac{1}{4} \pi d^2</math> is the projected area of the sphere,
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| *<math>C_d</math> is the drag coefficient, and
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| *<math>V</math> is the characteristic velocity (taken as terminal velocity, <math>V_t </math>).
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| Substitution of equations (2–4) in equation (1) and solving for terminal velocity, <math>V_t</math> to yield the following expression
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| :<math> \quad (5) \qquad V_t = \sqrt{\frac{4 g d}{3 C_d} \left( \frac{\rho_s - \rho}{\rho} \right)} </math> | |
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| ==See also==
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| *[[Stokes' law]]
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| *[[Free fall]]
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| *[[Terminal ballistics]]
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| ==References==
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| {{reflist}}
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| ==External links==
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| *[http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html Terminal Velocity] - NASA site
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| *[http://io9.com/5893615/absolutely-mindblowing-video-shot-from-the-space-shuttle-during-launch Onboard video of Space Shuttle Solid Rocket Boosters rapidly decelerating to terminal velocity on entry to the thicker atmosphere], from {{convert|2900|mph|Mach}} at 5:15 in the video, to 220 mph at 6:45 when the parachutes are deployed 90 seconds later—NASA video and sound, @ io9.com.
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| [[Category:Fluid dynamics]]
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