# Rooster Tail @ 10 kph

Note in this video the speed of the Rover is 10kph. Also note that Charlie Duke (not driving) mentions the ‘rooster tails’ made on sharp turns and going over bumps. Now look at the images below of car racing (on Earth) on dirt tracks.

Big Rooster Tails there!

More Rooster Tails. The two Earth photographs of cars on dirt roads clearly show the same type of Rooster Tail as seen on the Moon made by the Rover. Now the Rover was moving at 10 kph and the Earth cars at various speeds much higher than 10 kph. The question here is: How high and long would the rooster tail go in 1/6 Earth’s gravity in a vacuum?

Newton’s Laws of Motion states:[Credit: Wikipedia]

 First law: In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3] Second law: In an inertial reference frame, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

So how can we tell the shape of the rooster tail or arch when driving on the Moon as opposed to driving on Earth?

We need to know; the speed of the particles being pushed out from behind the rear wheels, the force of gravity, atmosphere or vacuum. They start on the ground and are pushed upward to a particular height and then slowly, or quickly, gravity wins and pulls the particles back to the surface.

The escape velocity for the Moon is: 2375.89 meters per second as seen below.

The formula for escape velocity is:

Vesc=((2GM)/r)Vesc=((2∗G∗M)/r)

Where:

G = 6.67384E-11 N*m^2*kg^-2 (Gravitational Constant)
M = 7.34767E22 kg (mass of the moon)
r = 1737400 m (radius of the moon)

Plug those in and we get an escape velocity of 2375.89 m/s [Quora]

The estimated speed of the pushed regolith particles is 2.77 m/s.  The altitude of the particles should be ~3 meters before they begin to lose speed because of the Moon’s gravity. The rooster tails on the video only had a height of about 1-2 meters as they do on Earth. Also, the roosters tails on the Moon are not a parabola as they should be. With NO atmosphere to stop the the particles moving in an arch there should be a distinct parabola that has a ground width of 6 foot (~1.8 meters) and an apex of ~3 meters (as can be seen in the images below).
That is far different to what we see on the Moon Rover video. Their rooster tails stop abruptly on the down side. They only have half a parabola.

# Trajectory

trajectory or flight path is the path that a moving object follows through space as a function of time.[1] The object might be a projectile or a satellite. For example, it can be an orbit—the path of a planet, an asteroid, or a comet as it travels around a central mass. A trajectory can be described mathematically either by the geometry of the path or as the position of the object over time.

In control theory a trajectory is a time-ordered set of states of a dynamical system (see e.g. Poincaré map). In discrete mathematics, a trajectory is a sequence {\displaystyle (f^{k}(x))_{k\in \mathbb {N} }} of values calculated by the iterated application of a mapping {\displaystyle f} to an element {\displaystyle x} of its source.

Illustration showing the trajectory of a bullet fired at an uphill target.

## Contents

[hide]

• 1Physics of trajectories
• 2Examples
• 2.1Uniform gravity, neither drag nor wind
• 2.1.1Derivation of the equation of motion
• 2.1.2Range and height
• 2.1.3Angle of elevation
• 2.2Uphill/downhill in uniform gravity in a vacuum
• 2.2.1Derivation based on equations of a parabola
• 2.3Orbiting objects
• 3Catching balls
• 4Notes
• 6References

## Physics of trajectories

A familiar example of a trajectory is the path of a projectile, such as a thrown ball or rock. In a significantly simplified model, the object moves only under the influence of a uniform gravitational force field. This can be a good approximation for a rock that is thrown for short distances, for example at the surface of the moon. In this simple approximation, the trajectory takes the shape of a parabola. Generally when determining trajectories, it may be necessary to account for nonuniform gravitational forces and air resistance (drag and aerodynamics). This is the focus of the discipline of ballistics.

One of the remarkable achievements of Newtonian mechanics was the derivation of the laws of Kepler. In the gravitational field of a point mass or a spherically-symmetrical extended mass (such as the Sun), the trajectory of a moving object is a conic section, usually an ellipse or a hyperbola.[a] This agrees with the observed orbits of planets, comets, and artificial spacecraft to a reasonably good approximation, although if a comet passes close to the Sun, then it is also influenced by other forces such as the solar wind and radiation pressure, which modify the orbit and cause the comet to eject material into space.

Newton’s theory later developed into the branch of theoretical physics known as classical mechanics. It employs the mathematics of differential calculus (which was also initiated by Newton in his youth). Over the centuries, countless scientists have contributed to the development of these two disciplines. Classical mechanics became a most prominent demonstration of the power of rational thought, i.e. reason, in science as well as technology. It helps to understand and predict an enormous range of phenomena; trajectories are but one example.

Consider a particle of mass{\displaystyle m}, moving in a potential field{\displaystyle V}. Physically speaking, mass represents inertia, and the field {\displaystyle V} represents external forces of a particular kind known as “conservative”. Given {\displaystyle V} at every relevant position, there is a way to infer the associated force that would act at that position, say from gravity. Not all forces can be expressed in this way, however.

The motion of the particle is described by the second-order differential equation

{\displaystyle m{\frac {\mathrm {d} ^{2}{\vec {x}}(t)}{\mathrm {d} t^{2}}}=-\nabla V({\vec {x}}(t)){\text{ with }}{\vec {x}}=(x,y,z).}

On the right-hand side, the force is given in terms of {\displaystyle \nabla V}, the gradient of the potential, taken at positions along the trajectory. This is the mathematical form of Newton’s second law of motion: force equals mass times acceleration, for such situations.

## Examples

### Uniform gravity, neither drag nor wind

Trajectories of a mass thrown at an angle of 70°:
without drag
with Stokes drag
with Newton drag

The ideal case of motion of a projectile in a uniform gravitational field in the absence of other forces (such as air drag) was first investigated by Galileo Galilei. To neglect the action of the atmosphere in shaping a trajectory would have been considered a futile hypothesis by practical-minded investigators all through the Middle Ages in Europe. Nevertheless, by anticipating the existence of the vacuum, later to be demonstrated on Earth by his collaborator Evangelista Torricelli[citation needed], Galileo was able to initiate the future science of mechanics.[citation needed] In a near vacuum, as it turns out for instance on the Moon, his simplified parabolic trajectory proves essentially correct.

In the analysis that follows, we derive the equation of motion of a projectile as measured from an inertial frame at rest with respect to the ground. Associated with the frame is a right-hand coordinate system with its origin at the point of launch of the projectile. The {\displaystyle x}-axis is parallel to the ground, and the {\displaystyle y}axis is perpendicular to it ( parallel to the gravitational field lines ). Let {\displaystyle g} be the acceleration of gravity. Relative to the flat terrain, let the initial horizontal speed be {\displaystyle v_{h}=v\cos(\theta )} and the initial vertical speed be {\displaystyle v_{v}=v\sin(\theta )}. It will also be shown that the range is {\displaystyle 2v_{h}v_{v}/g}, and the maximum altitude is {\displaystyle v_{v}^{2}/2g}. The maximum range for a given initial speed {\displaystyle v} is obtained when {\displaystyle v_{h}=v_{v}}, i.e. the initial angle is 45{\displaystyle ^{\circ }}. This range is {\displaystyle v^{2}/g}, and the maximum altitude at the maximum range is {\displaystyle v^{2}/(4g)}.

#### Derivation of the equation of motion

Assume the motion of the projectile is being measured from a free fall frame which happens to be at (x,y) = (0,0) at t = 0. The equation of motion of the projectile in this frame (by the principle of equivalence) would be {\displaystyle y=x\tan(\theta )}. The co-ordinates of this free-fall frame, with respect to our inertial frame would be {\displaystyle y=-gt^{2}/2}. That is, {\displaystyle y=-g(x/v_{h})^{2}/2}.

Now translating back to the inertial frame the co-ordinates of the projectile becomes {\displaystyle y=x\tan(\theta )-g(x/v_{h})^{2}/2} That is:

{\displaystyle y=-{g\sec ^{2}\theta \over 2v_{0}^{2}}x^{2}+x\tan \theta ,}

(where v0 is the initial velocity, {\displaystyle \theta } is the angle of elevation, and g is the acceleration due to gravity).

#### Range and height

Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed. t= time from launch, T = time of flight, R = range and H = highest point of trajectory (indicated with arrows).

The rangeR, is the greatest distance the object travels along the x-axis in the I sector. The initial velocityvi, is the speed at which said object is launched from the point of origin. The initial angleθi, is the angle at which said object is released. The g is the respective gravitational pull on the object within a null-medium.

{\displaystyle R={v_{i}^{2}\sin 2\theta _{i} \over g}}

The heighth, is the greatest parabolic height said object reaches within its trajectory

{\displaystyle h={v_{i}^{2}\sin ^{2}\theta _{i} \over 2g}}

#### Angle of elevation

In terms of angle of elevation {\displaystyle \theta } and initial speed {\displaystyle v}:

{\displaystyle v_{h}=v\cos \theta ,\quad v_{v}=v\sin \theta \;}

giving the range as

{\displaystyle R=2v^{2}\cos(\theta )\sin(\theta )/g=v^{2}\sin(2\theta )/g\,.}

This equation can be rearranged to find the angle for a required range

{\displaystyle \theta ={\frac {1}{2}}\sin ^{-1}\left({\frac {gR}{v^{2}}}\right)} (Equation II: angle of projectile launch)

Note that the sine function is such that there are two solutions for {\displaystyle \theta } for a given range {\displaystyle d_{h}}. The angle {\displaystyle \theta } giving the maximum range can be found by considering the derivative or {\displaystyle R} with respect to {\displaystyle \theta } and setting it to zero.

{\displaystyle {\mathrm {d} R \over \mathrm {d} \theta }={2v^{2} \over g}\cos(2\theta )=0}

which has a nontrivial solution at {\displaystyle 2\theta =\pi /2=90^{\circ }}, or {\displaystyle \theta =45^{\circ }}. The maximum range is then {\displaystyle R_{\max }=v^{2}/g\,}. At this angle {\displaystyle \sin(\pi /2)=1}, so the maximum height obtained is {\displaystyle {v^{2} \over 4g}}.

To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height {\displaystyle H=v^{2}\sin ^{2}(\theta )/(2g)} with respect to {\displaystyle \theta }, that is {\displaystyle {\mathrm {d} H \over \mathrm {d} \theta }=v^{2}2\cos(\theta )\sin(\theta )/(2g)} which is zero when {\displaystyle \theta =\pi /2=90^{\circ }}. So the maximum height {\displaystyle H_{\mathrm {max} }={v^{2} \over 2g}} is obtained when the projectile is fired straight up.

### Uphill/downhill in uniform gravity in a vacuum

Given a hill angle {\displaystyle \alpha } and launch angle {\displaystyle \theta } as before, it can be shown that the range along the hill {\displaystyle R_{s}} forms a ratio with the original range {\displaystyle R} along the imaginary horizontal, such that:

{\displaystyle {\frac {R_{s}}{R}}=(1-\cot \theta \tan \alpha )\sec \alpha } (Equation 11)

In this equation, downhill occurs when {\displaystyle \alpha } is between 0 and -90 degrees. For this range of {\displaystyle \alpha } we know: {\displaystyle \tan(-\alpha )=-\tan \alpha } and {\displaystyle \sec(-\alpha )=\sec \alpha }. Thus for this range of {\displaystyle \alpha }{\displaystyle R_{s}/R=(1+\tan \theta \tan \alpha )\sec \alpha }. Thus {\displaystyle R_{s}/R} is a positive value meaning the range downhill is always further than along level terrain. The lower level of terrain causes the projectile to remain in the air longer, allowing it to travel further horizontally before hitting the ground.

While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to {\displaystyle R_{s}/R=1} (i.e. the slant range is equal to the level terrain range) and solving for the “critical angle” {\displaystyle \theta _{\text{cr}}}:

{\displaystyle 1=(1-\tan \theta \tan \alpha )\sec \alpha \quad \;}
{\displaystyle \theta _{\text{cr}}=\arctan((1-\csc \alpha )\cot \alpha )\quad \;}

Equation 11 may also be used to develop the “rifleman’s rule” for small values of {\displaystyle \alpha } and {\displaystyle \theta } (i.e. close to horizontal firing, which is the case for many firearm situations). For small values, both {\displaystyle \tan \alpha } and {\displaystyle \tan \theta }have a small value and thus when multiplied together (as in equation 11), the result is almost zero. Thus equation 11 may be approximated as:

{\displaystyle {\frac {R_{s}}{R}}=(1-0)\sec \alpha }

And solving for level terrain range, {\displaystyle R}

{\displaystyle R=R_{s}\cos \alpha \ } “Rifleman’s rule”

Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. “In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position.”[1]

#### Derivation based on equations of a parabola

The intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope {\displaystyle m} in standard linear form at coordinates {\displaystyle (x,y)}:

{\displaystyle y=mx+b\;} (Equation 12) where in this case, {\displaystyle y=d_{v}}{\displaystyle x=d_{h}} and {\displaystyle b=0}

Substituting the value of {\displaystyle d_{v}=md_{h}} into Equation 10:

{\displaystyle mx=-{\frac {g}{2v^{2}\cos ^{2}\theta }}x^{2}+{\frac {\sin \theta }{\cos \theta }}x}
{\displaystyle x={\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-m\right)} (Solving above for x)

This value of x may be substituted back into the linear equation 12 to get the corresponding y coordinate at the intercept:

{\displaystyle y=mx=m{\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-m\right)}

Now the slant range {\displaystyle R_{s}} is the distance of the intercept from the origin, which is just the hypotenuse of x and y:

{\displaystyle {\begin{aligned}R_{s}&={\sqrt {x^{2}+y^{2}}}={\sqrt {\left({\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-m\right)\right)^{2}+\left(m{\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-m\right)\right)^{2}}}\\[6pt]&={\frac {2v^{2}\cos ^{2}\theta }{g}}{\sqrt {\left({\frac {\sin \theta }{\cos \theta }}-m\right)^{2}+m^{2}\left({\frac {\sin \theta }{\cos \theta }}-m\right)^{2}}}\\[6pt]&={\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-m\right){\sqrt {1+m^{2}}}\end{aligned}}}

Now {\displaystyle \alpha } is defined as the angle of the hill, so by definition of tangent, {\displaystyle m=\tan \alpha }. This can be substituted into the equation for {\displaystyle R_{s}}:

{\displaystyle R_{s}={\frac {2v^{2}\cos ^{2}\theta }{g}}\left({\frac {\sin \theta }{\cos \theta }}-\tan \alpha \right){\sqrt {1+\tan ^{2}\alpha }}}

Now this can be refactored and the trigonometric identity for {\displaystyle \sec \alpha ={\sqrt {1+\tan ^{2}\alpha }}} may be used:

{\displaystyle R_{s}={\frac {2v^{2}\cos \theta \sin \theta }{g}}\left(1-{\frac {\cos \theta }{\sin \theta }}\tan \alpha \right)\sec \alpha }

Now the flat range {\displaystyle R=v^{2}\sin 2\theta /g=2v^{2}\sin \theta \cos \theta /g} by the previously used trigonometric identity and {\displaystyle \cos \theta /\sin \theta =\cot \theta } so:

{\displaystyle R_{s}=R(1-\cot \theta \tan \alpha )\sec \alpha \;}
{\displaystyle {\frac {R_{s}}{R}}=(1-\cot \theta \tan \alpha )\sec \alpha }

### Orbiting objects

If instead of a uniform downwards gravitational force we consider two bodies orbiting with the mutual gravitation between them, we obtain Kepler’s laws of planetary motion. The derivation of these was one of the major works of Isaac Newton and provided much of the motivation for the development of differential calculus.

## Catching balls

If a projectile, such as a baseball or cricket ball, travels in a parabolic path, with negligible air resistance, and if a player is positioned so as to catch it as it descends, he sees its angle of elevation increasing continuously throughout its flight. The tangent of the angle of elevation is proportional to the time since the ball was sent into the air, usually by being struck with a bat. Even when the ball is really descending, near the end of its flight, its angle of elevation seen by the player continues to increase. The player therefore sees it as if it were ascending vertically at constant speed. Finding the place from which the ball appears to rise steadily helps the player to position himself correctly to make the catch. If he is too close to the batsman who has hit the ball, it will appear to rise at an accelerating rate. If he is too far from the batsman, it will appear to slow rapidly, and then to descend.

# Drag (physics)

Shape and flow Form
Drag
Skin
friction
0% 100%
~10% ~90%
~90% ~10%
100% 0%

In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid.[1] This can exist between two fluid layers (or surfaces) or a fluid and a solid surface. Unlike other resistive forces, such as dry friction, which are nearly independent of velocity, drag forces depend on velocity.[2][3] Drag force is proportional to the velocity for a laminar flowand the squared velocity for a turbulent flow. Even though the ultimate cause of a drag is viscous friction, the turbulent drag is independent of viscosity.[4]

Drag forces always decrease fluid velocity relative to the solid object in the fluid’s path.

## Contents

[hide]

• 1Examples of drag
• 2Types of drag
• 3Drag at high velocity
• 3.1Power
• 3.2Velocity of a falling object
• 4Very low Reynolds numbers: Stokes’ drag
• 5Drag in aerodynamics
• 5.1Lift-induced drag
• 5.2Parasitic drag
• 5.3Power curve in aviation
• 5.4Wave drag in transonic and supersonic flow
• 8References
• 9Bibliography

## Examples of drag

Examples of drag include the component of the net aerodynamic or hydrodynamic force acting opposite to the direction of movement of a solid object such as cars, aircraft[3] and boat hulls; or acting in the same geographical direction of motion as the solid, as for sails attached to a down wind sail boat, or in intermediate directions on a sail depending on points of sail.[5][6][7] In the case of viscous drag of fluid in a pipe, drag force on the immobile pipe decreases fluid velocity relative to the pipe.[8][9]

In physics of sports, the drag force is necessary to explain the performance of runners, particularly of sprinters.[10]

## Types of drag

Types of drag are generally divided into the following categories:

• parasitic drag, consisting of
• form drag,
• skin friction,
• interference drag,
• lift-induced drag, and
• wave drag (aerodynamics) or wave resistance (ship hydrodynamics).

The phrase parasitic drag is mainly used in aerodynamics, since for lifting wings drag it is in general small compared to lift. For flow around bluff bodies, form and interference drags often dominate, and then the qualifier “parasitic” is meaningless.[citation needed]

Further, lift-induced drag is only relevant when wings or a lifting body are present, and is therefore usually discussed either in aviation or in the design of semi-planing or planing hulls. Wave drag occurs either when a solid object is moving through a fluid at or near the speed of sound or when a solid object is moving along a fluid boundary, as in surface waves.

Drag coefficient Cd for a sphere as a function of Reynolds number Re, as obtained from laboratory experiments. The dark line is for a sphere with a smooth surface, while the lighter line is for the case of a rough surface.

Drag depends on the properties of the fluid and on the size, shape, and speed of the object. One way to express this is by means of the drag equation:

{\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A}

where

{\displaystyle F_{D}} is the drag force,
{\displaystyle \rho } is the density of the fluid,[11]
{\displaystyle v} is the speed of the object relative to the fluid,
{\displaystyle A} is the cross sectional area, and
{\displaystyle C_{D}} is the drag coefficient – a dimensionless number.

The drag coefficient depends on the shape of the object and on the Reynolds number

{\displaystyle R_{e}={\frac {vD}{\nu }}},

where {\displaystyle D} is some characteristic diameter or linear dimension and {\displaystyle {\nu }} is the kinematic viscosity of the fluid (equal to the viscosity {\displaystyle {\mu }} divided by the density). At low {\displaystyle R_{e}}{\displaystyle C_{D}} is asymptotically proportional to {\displaystyle R_{e}^{-1}}, which means that the drag is linearly proportional to the speed. At high {\displaystyle R_{e}}{\displaystyle C_{D}} is more or less constant and drag will vary as the square of the speed. The graph to the right shows how {\displaystyle C_{D}} varies with {\displaystyle R_{e}} for the case of a sphere. Since the power needed to overcome the drag force is the product of the force times speed, the power needed to overcome drag will vary as the square of the speed at low Reynolds numbers and as the cube of the speed at high numbers.

## Drag at high velocity

Explanation of drag by NASA.

As mentioned, the drag equation with a constant drag coefficient gives the force experienced by an object moving through a fluid at relatively large velocity (i.e. high Reynolds number, Re > ~1000). This is also called quadratic drag. The equation is attributed to Lord Rayleigh, who originally used L2 in place of A (L being some length).

{\displaystyle F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{d}\,A,}

see derivation

The reference area A is often orthographic projection of the object (frontal area)—on a plane perpendicular to the direction of motion—e.g. for objects with a simple shape, such as a sphere, this is the cross sectional area. Sometimes a body is a composite of different parts, each with a different reference areas, in which case a drag coefficient corresponding to each of those different areas must be determined.

In the case of a wing the reference areas are the same and the drag force is in the same ratio to the lift force as the ratio of drag coefficient to lift coefficient.[12]Therefore, the reference for a wing is often the lifting area (“wing area”) rather than the frontal area.[13]

For an object with a smooth surface, and non-fixed separation points—like a sphere or circular cylinder—the drag coefficient may vary with Reynolds number Re, even up to very high values (Re of the order107). [14] [15] For an object with well-defined fixed separation points, like a circular disk with its plane normal to the flow direction, the drag coefficient is constant for Re > 3,500.[15] Further the drag coefficient Cdis, in general, a function of the orientation of the flow with respect to the object (apart from symmetrical objects like a sphere).

### Power

The power required to overcome the aerodynamic drag is given by:

{\displaystyle P_{d}=\mathbf {F} _{d}\cdot \mathbf {v} ={\tfrac {1}{2}}\rho v^{3}AC_{d}}

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW).[16] With a doubling of speed the drag (force) quadruples per the formula. Exerting 4 times the force over a fixed distance produces 4 times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, 4 times the work done in half the time requires 8 times the power.

### Velocity of a falling object

An object falling through viscous medium accelerates quickly towards its terminal speed, approaching gradually as the speed gets nearer to the terminal speed. Whether the object experiences turbulent or laminar drag changes the characteristic shape of the graph with turbulent flow resulting in a constant acceleration for a larger fraction of its accelerating time.

The velocity as a function of time for an object falling through a non-dense medium, and released at zero relative-velocity v = 0 at time t = 0, is roughly given by a function involving a hyperbolic tangent(tanh):

{\displaystyle v(t)={\sqrt {\frac {2mg}{\rho AC_{d}}}}\tanh \left(t{\sqrt {\frac {g\rho C_{d}A}{2m}}}\right).\,}

The hyperbolic tangent has a limit value of one, for large time t. In other words, velocity asymptoticallyapproaches a maximum value called the terminal velocity vt:

{\displaystyle v_{t}={\sqrt {\frac {2mg}{\rho AC_{d}}}}.\,}

For a potato-shaped object of average diameter d and of density ρobj, terminal velocity is about

{\displaystyle v_{t}={\sqrt {gd{\frac {\rho _{obj}}{\rho }}}}.\,}

For objects of water-like density (raindrops, hail, live objects—mammals, birds, insects, etc.) falling in air near Earth’s surface at sea level, the terminal velocity is roughly equal to

{\displaystyle v_{t}=90{\sqrt {d}},\,}

with d in metre and vt in m/s. For example, for a human body ({\displaystyle \mathbf {} d} ~ 0.6 m) {\displaystyle \mathbf {} v_{t}} ~ 70 m/s, for a small animal like a cat ({\displaystyle \mathbf {} d} ~ 0.2 m) {\displaystyle \mathbf {} v_{t}} ~ 40 m/s, for a small bird ({\displaystyle \mathbf {} d} ~ 0.05 m) {\displaystyle \mathbf {} v_{t}} ~ 20 m/s, for an insect ({\displaystyle \mathbf {} d} ~ 0.01 m) {\displaystyle \mathbf {} v_{t}}~ 9 m/s, and so on. Terminal velocity for very small objects (pollen, etc.) at low Reynolds numbers is determined by Stokes law.

Terminal velocity is higher for larger creatures, and thus potentially more deadly. A creature such as a mouse falling at its terminal velocity is much more likely to survive impact with the ground than a human falling at its terminal velocity. A small animal such as a cricket impacting at its terminal velocity will probably be unharmed. This, combined with the relative ratio of limb cross-sectional area vs. body mass (commonly referred to as the Square-cube law), explains why very small animals can fall from a large height and not be harmed.[17]

## Very low Reynolds numbers: Stokes’ drag

Trajectories of three objects thrown at the same angle (70°). The black object does not experience any form of drag and moves along a parabola. The blue object experiences Stokes’ drag, and the green object Newton drag.

The equation for viscous resistance or linear drag is appropriate for objects or particles moving through a fluid at relatively slow speeds where there is no turbulence (i.e. low Reynolds number, {\displaystyle R_{e}<1}).[18] Note that purely laminar flow only exists up to Re = 0.1 under this definition. In this case, the force of drag is approximately proportional to velocity. The equation for viscous resistance is:[19]

{\displaystyle \mathbf {F} _{d}=-b\mathbf {v} \,}

where:

{\displaystyle \mathbf {} b} is a constant that depends on the properties of the fluid and the dimensions of the object, and
{\displaystyle \mathbf {v} } is the velocity of the object

When an object falls from rest, its velocity will be

{\displaystyle v(t)={\frac {(\rho -\rho _{0})Vg}{b}}\left(1-e^{-bt/m}\right)}

which asymptotically approaches the terminal velocity {\displaystyle \mathbf {} v_{t}={\frac {(\rho -\rho _{0})Vg}{b}}}. For a given {\displaystyle \mathbf {} b}, heavier objects fall more quickly.

For the special case of small spherical objects moving slowly through a viscous fluid (and thus at small Reynolds number), George Gabriel Stokesderived an expression for the drag constant:

{\displaystyle b=6\pi \eta r\,}

where:

{\displaystyle \mathbf {} r} is the Stokes radius of the particle, and {\displaystyle \mathbf {} \eta } is the fluid viscosity.

The resulting expression for the drag is known as Stokes’ drag:[20]

{\displaystyle \mathbf {F} _{d}=-6\pi \eta r\,\mathbf {v} .}

For example, consider a small sphere with radius {\displaystyle \mathbf {} r} = 0.5 micrometre (diameter = 1.0 µm) moving through water at a velocity {\displaystyle \mathbf {} v} of 10 µm/s. Using 10−3 Pa·s as the dynamic viscosity of water in SI units, we find a drag force of 0.09 pN. This is about the drag force that a bacterium experiences as it swims through water.

## Drag in aerodynamics

### Lift-induced drag

Induced drag vs.lift[21][22]

Lift-induced drag (also called induced drag) is drag which occurs as the result of the creation of lift on a three-dimensional lifting body, such as the wing or fuselage of an airplane. Induced drag consists of two primary components, including drag due to the creation of vortices (vortex drag) and the presence of additional viscous drag (lift-induced viscous drag). The vortices in the flow-field, present in the wake of a lifting body, derive from the turbulent mixing of air of varying pressure on the upper and lower surfaces of the body, which is a necessary condition for the creation of lift.

With other parameters remaining the same, as the lift generated by a body increases, so does the lift-induced drag. This means that as the wing’s angle of attack increases the lift coefficient increases (up to a limit called the stall point) so too does the lift-induced drag. At the onset of stall, lift is abruptly decreased, as is lift-induced drag, but viscous pressure drag, a component of parasite drag, increases due to the formation of turbulent unattached flow on the surface of the body.

### Parasitic drag

Parasitic drag is drag caused by moving a solid object through a fluid. Parasitic drag is made up of multiple components including viscous pressure drag (form drag), and drag due to surface roughness (skin friction drag). Additionally, the presence of multiple bodies in relative proximity may incur so called interference drag, which is sometimes described as a component of parasitic drag.

In aviation, induced drag tends to be greater at lower speeds because a high angle of attack is required to maintain lift, creating more drag. However, as speed increases the angle of attack can be reduced and the induced drag decreases. Parasitic drag, however, increases because the fluid is flowing more quickly around protruding objects increasing friction or drag. At even higher speeds (transonic), wave dragenters the picture. Each of these forms of drag changes in proportion to the others based on speed. The combined overall drag curve therefore shows a minimum at some airspeed – an aircraft flying at this speed will be at or close to its optimal efficiency. Pilots will use this speed to maximize endurance (minimum fuel consumption), or maximize gliding range in the event of an engine failure.

### Power curve in aviation

The power curve: form and induced drag vs. airspeed

The interaction of parasitic and induced drag vs. airspeed can be plotted as a characteristic curve, illustrated here. In aviation, this is often referred to as the power curve, and is important to pilots because it shows that, below a certain airspeed, maintaining airspeed counterintuitively requires more thrust as speed decreases, rather than less. The consequences of being “behind the curve” in flight are important and are taught as part of pilot training. At the subsonic airspeeds where the “U” shape of this curve is significant, wave drag has not yet become a factor, and so it is not shown in the curve.

### Wave drag in transonic and supersonic flow

Qualitative variation in Cd factor with Mach number for aircraft

Wave drag (also called compressibility drag) is drag that is created when a body moves in a compressible fluid and at speeds that are close to the speed of sound in that fluid. In aerodynamics, wave drag consists of multiple components depending on the speed regime of the flight.

In transonic flight (Mach numbers greater than about 0.8 and less than about 1.4), wave drag is the result of the formation of shockwaves in the fluid, formed when local areas of supersonic (Mach number greater than 1.0) flow are created. In practice, supersonic flow occurs on bodies traveling well below the speed of sound, as the local speed of air increases as it accelerates over the body to speeds above Mach 1.0. However, full supersonic flow over the vehicle will not develop until well past Mach 1.0. Aircraft flying at transonic speed often incur wave drag through the normal course of operation. In transonic flight, wave drag is commonly referred to as transonic compressibility drag. Transonic compressibility drag increases significantly as the speed of flight increases towards Mach 1.0, dominating other forms of drag at those speeds.

In supersonic flight (Mach numbers greater than 1.0), wave drag is the result of shockwaves present in the fluid and attached to the body, typically oblique shockwaves formed at the leading and trailing edges of the body. In highly supersonic flows, or in bodies with turning angles sufficiently large, unattached shockwaves, or bow waves will instead form. Additionally, local areas of transonic flow behind the initial shockwave may occur at lower supersonic speeds, and can lead to the development of additional, smaller shockwaves present on the surfaces of other lifting bodies, similar to those found in transonic flows. In supersonic flow regimes, wave drag is commonly separated into two components, supersonic lift-dependent wave drag and supersonic volume-dependent wave drag.

The closed form solution for the minimum wave drag of a body of revolution with a fixed length was found by Sears and Haack, and is known as the Sears-Haack Distribution. Similarly, for a fixed volume, the shape for minimum wave drag is the Von Karman Ogive.

Busemann’s Biplane is not, in principle, subject to wave drag at all when operated at its design speed, but is incapable of generating lift.

In 1752 d’Alembert proved that potential flow, the 18th century state-of-the-art inviscid flow theory amenable to mathematical solutions, resulted in the prediction of zero drag. This was in contradiction with experimental evidence, and became known as d’Alembert’s paradox. In the 19th century the Navier–Stokes equations for the description of viscous flow were developed by Saint-Venant, Navier and Stokes. Stokes derived the drag around a sphere at very low Reynolds numbers, the result of which is called Stokes’ law.[23]

In the limit of high Reynolds numbers, the Navier–Stokes equations approach the inviscid Euler equations, of which the potential-flow solutions considered by d’Alembert are solutions. However, all experiments at high Reynolds numbers showed there is drag. Attempts to construct inviscid steady flow solutions to the Euler equations, other than the potential flow solutions, did not result in realistic results.[23]

The notion of boundary layers—introduced by Prandtl in 1904, founded on both theory and experiments—explained the causes of drag at high Reynolds numbers. The boundary layer is the thin layer of fluid close to the object’s boundary, where viscous effects remain important even when the viscosity is very small (or equivalently the Reynolds number is very large).[23]

https://en.wikipedia.org/wiki/Parabola

## References

[Wikipedia_1]

1.  “New Quantum Theory Separates Gravitational and Inertial Mass”. MIT Technology Review. 14 Jun 2010. Retrieved 3 Dec2013.
2.   Jacob Aron (10 Jan 2013). “Most fundamental clock ever could redefine kilogram”. NewScientist. Retrieved 17 Dec 2013.

[Wikipedia_2]

1.   “Definition of DRAG”. www.merriam-webster.com.
2.  French (1970), p. 211, Eq. 7-20
3.  “What is Drag?”.
4.   G. Falkovich (2011). Fluid Mechanics (A short course for physicists). Cambridge University Press. ISBN 978-1-107-00575-4.
5.   Eiffel, Gustave (1913). The Resistance of The Air and Aviation. London: Constable &Co Ltd.
6.   Marchaj, C. A. (2003). Sail performance : techniques to maximise sail power (Rev. ed.). London: Adlard Coles Nautical. pp. 147 figure 127 lift vs drag polar curves. ISBN 978-0-7136-6407-2.
7.   Drayton, Fabio Fossati ; translated by Martyn (2009). Aero-hydrodynamics and the performance of sailing yachts : the science behind sailing yachts and their design. Camden, Maine: International Marine /McGraw-Hill. pp. 98 Fig 5.17 Chapter five Sailing Boat Aerodynamics. ISBN 978-0-07-162910-2.
8.   “Calculating Viscous Flow: Velocity Profiles in Rivers and Pipes”(PDF). Retrieved 16 October 2011.
9.   “Viscous Drag Forces”. Retrieved 16 October 2011.
10.  Hernandez-Gomez, J J; Marquina, V; Gomez, R W (25 July 2013). “On the performance of Usain Bolt in the 100 m sprint”. Eur. J. Phys. IOP. 34 (5): 1227. Bibcode:2013EJPh…34.1227H. doi:10.1088/0143-0807/34/5/1227. Retrieved 23 April 2016.
11.   Note that for Earth’s atmosphere, the air density can be found using the barometric formula. It is 1.293 kg/m3 at 0 °C and 1 atmosphere.
12.   Size effects on drag, from NASA Glenn Research Center.
13.   Wing geometry definitions, from NASA Glenn Research Center.
14.   Roshko, Anatol (1961). “Experiments on the flow past a circular cylinder at very high Reynolds number”. Journal of Fluid Mechanics10 (3): 345–356. Bibcode:1961JFM….10..345R. doi:10.1017/S0022112061000950.
15.   Batchelor (1967), p. 341.
16.   Brian Beckman (1991). “Part 6: Speed and Horsepower”. Retrieved 18 May 2016.
17.   Haldane, J.B.S., “On Being the Right Size”
18.  Drag Force Archived April 14, 2008, at the Wayback Machine.
19.   Air friction, from Department of Physics and Astronomy, Georgia State University
20.   Collinson, Chris; Roper, Tom (1995). Particle Mechanics. Butterworth-Heinemann. p. 30. ISBN 9780080928593.
21.  Clancy, L.J. (1975) Aerodynamics Fig 5.24. Pitman Publishing Limited, London. ISBN 0-273-01120-0
22.  Hurt, H. H. (1965) Aerodynamics for Naval Aviators, Figure 1.30, NAVWEPS 00-80T-80
23.  Batchelor (2000), pp. 337–343.

## Bibliography

• French, A. P. (1970). Newtonian Mechanics (The M.I.T. Introductory Physics Series) (1st ed.). W. W. Norton & Company Inc., New York. ISBN 978-0-393-09958-4.
• G. Falkovich (2011). Fluid Mechanics (A short course for physicists). Cambridge University Press. ISBN 978-1-107-00575-4.
• Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed.). Brooks/Cole. ISBN 978-0-534-40842-8.
• Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 978-0-7167-0809-4.
• Huntley, H. E. (1967). Dimensional Analysis. Dover. LOC 67-17978.
• Batchelor, George (2000). An introduction to fluid dynamics. Cambridge Mathematical Library (2nd ed.). Cambridge University Press. ISBN 978-0-521-66396-0. MR 1744638.
• L. J. Clancy (1975), Aerodynamics, Pitman Publishing Limited, London. ISBN 978-0-273-01120-0

[Wikipedia_3]

## References

1.  The Principles of Physics by Rohit Metha, Chapter 11 Page 378 Para 3