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\fancyfoot[CO,CE]{\it Drag Reference Guide by A. A. Tovar, Ph. D., Created Nov. 2009, Amended Nov. 2009.}
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\begin{document}
\title {Air Drag Mathematics Reference Guide}
\author{Anthony A. Tovar, Ph. D.
\\Eastern Oregon University
\\1 University Blvd.
\\La Grande, Oregon, 97850 }
%\date{July 24, 2002}
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\vspace{1 cm}
The purpose of these tutorials is to act as a reference guide for the study of mechanics
which involve air drag, which is a critical component of modeling most macroscopic terrestrial
motion. These tutorial are particularly suited to the study of sports.
This tutorial introduces the following novel ideas:
\begin{itemize}
\item The speed of interest for an object is its speed divided by its net terminal speed, even
for objects that neither rise nor fall. The relevance of air drag is detemined by
``Tovar's Drag Rule" which is introduced.
\item Conservation of QuasiEnergy is introduced.
\item A new conservation law that applies to 2D motion is introduced.
\end{itemize}
\chapter{1D Projectile Motion}
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%\newpage
\section{Terminal Speed}
For onedimensional projectile motion where the size and speed of the
object are such that the quadradicvelocity model dominates,
\EQ
\frac{dv_y}{dt} = a_y = g  \frac{1}{2m} C_d \rho_{air} A v_y v_y
\label{qar1a}
\EN
\EQ
\frac{dy}{dt} = v_y
\label{qar1b}
\EN
where $C_D$ is the drag coefficient, $A$ is the crosssectional area, and $\rho_{air}$ is the air density.
The terminal speed is reached when the acceleration is zero
for a falling object.
Setting Eq. (\ref{qar1a}) to zero yields
\EQ
\boxed{
v_T = \sqrt{\frac{mg}{\frac{1}{2} C_D \rho_{air} A}}
}
\EN
If the mass is constant and the area is increase, then the terminal speed decreases
and the effect of air drag increases. This is what happens when a skydiver
opens her parachute.
The above traditional terminal speed formula can be recast in terms of the object's density.
For a sphere this would become $v_T^2 = (2/3C_D) (\rho_{obj}/\rho_{air}) gr$.
From this new version of the terminal speed formula it can be seen that if the density
is constant, then the larger the object, the larger the terminal speed, and the smaller
the effect of air drag. So, a large rock suffers less drag than a small rock even though
the larger rock has a larger area.
\begin{quote}
{\bf
The larger the object, the SMALLER the effect of air resistance
(for a given density).\\
The larger the object, the LARGER the effect of air resistance
(for a given mass).\\
}
\end{quote}
\subsection{Tovar's Drag Rule}
\begin{quote}
{\bf
If the maximum speed of an object is less than 15\% of its terminal speed, you
can ignore Air Drag.
}
\end{quote}
The percent of energy lost due to drag is approximately $v^2/(2v_T^2)$. If
$v/v_T = 0.15$, then only 1.1 \% of the energy is lost and the effects of air
drag may be neglected.
%\begin{sidewaystable}
%\centering
\begin{center}
\begin{tabular}{c  c c c c c  c c}
\hline
Object & Mass (g) &$C_D$ & Diameter (mm)& $v_T$ (m/s) & $v_T$ (mph) & $v_{typical} (mph)$& Situation\\
\hline
Bullet & 9.72 (150 grn) &.295 &$7.82 $ & 106 &237 & 1984&.3006 gun\\
Shot Put& 7260 (16 lb.) &.5 & $125 $ & 139 &311 & 32 & 20 m throw\\
Baseball & 145 & 0.4 & 75 & 36.6 & 81.8 & 100 & HR Hit\\
Tennis Ball& 57 &0.5 &65 & & & 120 & Serve\\
BasketBall& 624 & 0.3 & 234& & & & Free Throw\\
Football& &0.06 & & & & & Long Pass\\
Golf ball*&46 &0.5 &42 &32 & & & Drive\\
Hail Stone& .48& 0.5&10 &14 & & & \\
Rain Drop& .034& 0.5&4 &9.0 & & & \\
Car & & & & & & & Hwy. Drive\\
\hline
Object & Mass (kg) &$C_D$ & Area ($m^2$)& $v_T$ (m/s) & $v_T$ (mph)\\
\hline
Person (Flat)& 82 & 0.4 & & \\
Person (Pointed)& 82 & 1.0 & & \\
Cyclist & 68 (150 lb)& 0.9& &\\
\end{tabular}
\end{center}
%\end{sidewaystable}
\section{1D Projectile Motion  Dynamical Equations}
\subsection{General 1D Motion}
The governing differential equations are
\EQ
\frac{dv_y}{dt} = g \left( 1 + \frac{v_y v_y}{v_T^2} \right)
\label{ode_gen}
\EN
\EQ
\frac{dy}{dt} = v_y
\label{ode_gen2}
\EN
\subsection{Upward Motion ($v_{y0}>0$)}
The governing differential equations are
\EQ
\frac{dv_y}{dt} = g \left( 1 + \frac{v_y^2}{v_T^2} \right)
\label{ode_up}
\EN
\EQ
\frac{dy}{dt} = v_y
\label{ode_up2}
\EN
and their solutions are
\EQ
\boxed{
v_y = v_T \frac{\left( \frac{v_{y0}}{v_T}\right)  tan \left(\frac{gt}{v_T}\right)}
{1 + \left( \frac{v_{y0}}{v_T} \right) tan \left( \frac{gt}{v_T} \right)}
}
\EN
\EQ
\boxed{
y(t) = y_0 + \frac{v_T^2}{g} ln \left[ \left( \frac{v_{y0}}{v_T}
\right) sin \left( \frac{gt}{v_T} \right) +
cos \left( \frac{gt}{v_T} \right) \right]
}
\EN
\subsection{Downward Motion ($v_{y0}<0$)}
The governing differential equations are
\EQ
\frac{dv_y}{dt} = g \left( 1  \frac{v_y^2}{v_T^2} \right)
\label{ode_down}
\EN
\EQ
\frac{dy}{dt} = v_y
\label{ode_down2}
\EN
and their solutions are
\EQ
\boxed{
v_y = v_T \frac{\left( \frac{v_{y0}}{v_T}\right)  tanh \left(\frac{gt}{v_T}\right)}
{1  \left( \frac{v_{y0}}{v_T} \right) tanh \left( \frac{gt}{v_T} \right)}
=v_T \left[ \frac{
\left( 1+ \frac{v_{y0}}{v_T} \right)
 \left( 1 \frac{v_{y0}}{v_T} \right) e^{2\frac{gt}{v_T}}
}{
\left( 1+ \frac{v_{y0}}{v_T} \right)
+ \left( 1 \frac{v_{y0}}{v_T} \right) e^{2\frac{gt}{v_T}}
} \right]
}
\EN
\begin{equation}
y(t) = y_0  \frac{v_T^2}{g} ln \left[
cosh \left( \frac{gt}{v_T} \right)

\left( \frac{v_{y0}}{v_T} \right) sinh \left( \frac{gt}{v_T} \right)
\right]
\end{equation}
which can also be written
\begin{equation}
\boxed{
y(t) = y_0  \frac{v_T^2}{g} ln \left[
\left( 1 \frac{v_{y0}}{v_T} \right) \frac{e^{\frac{gt}{v_T}}}{2}

\left( 1+ \frac{v_{y0}}{v_T} \right) \frac{e^{\frac{gt}{v_T}}}{2}
\right]
}
\end{equation}
\subsection{Up and Down Motion}
From the equations above, it can be seen that the time it takes for a projectile to
reach the top of its flight is
\EQ
\boxed{
t_{top} = \frac{v_T}{g} tan^{1}\left( \frac{v_{y0}}{v_T}
\right)
}
\EN
then for $0t_{top}$, the height
of the projectile is
\EQ
\boxed{
y(t) = y_{top}  \frac{v_T^2}{g} ln \left[
cosh \left( \frac{g(tt_{top})}{v_T} \right) \right]
=
y_{top}  \frac{v_T^2}{g} ln \left[
\frac{e^{\left( \frac{g(tt_{top})}{v_T} \right)} {e^{\left( \frac{g(tt_{top})}{v_T} \right) }}}
{2}
\right]
}
\EN
and the velocity is
\EQ
\boxed{
v_y = v_T tanh \left(\frac{g(tt_{top})}{v_T}\right)
=
v_T \frac{e^{\left( \frac{2g(tt_{top})}{v_T} \right)} 1}
{e^{\left( \frac{2g(tt_{top})}{v_T} \right)} +1}
}
\EN
\section{1D Projectile Motion  Conservation of QuasiEnergy}
Energy is not conserved. However, one may obtain Conservation Laws relating the position and
velocity in a similar fashion to conservation of energy. The results are the
laws for Conservation of ``QuasiEnergy."
\subsection{Upward Motion ($v_{y0}>0$)}
\EQ
\boxed{\underbrace{mgy}_{Q.P.E} +
\underbrace{\frac{1}{2} m v_T^2 ln \left[ 1 + ( {v_y^2}/{v_T^2}) \right]}_{Q.K.E}
= \mathcal{E}_0}
\EN
\subsection{Downward Motion ($v_{y0}<0$)}
The corresponding quasienergy expression for downward motion can be
obtained by replacing $v_T^2$ with $v_T^2$:
\EQ
\boxed{\underbrace{mgy}_{Q.P.E} +
\underbrace{\frac{1}{2} m v_T^2 ln \left( \frac{1}{1(v_y^2/v_T^2)} \right)}_{Q.K.E}
= \mathcal{E}_0}
\EN
\subsection{Up and Down Motion}
Here we consider a quasienergy expression for a projectile that is launched
upward, and comes down (though the final position can be different than the
initial position). First, the projectile goes up, and reaches a maximum
position, so that
\EQ
mgy_{top}= mgy_{0} + \frac{1}{2} m v_T^2 ln \left[ 1 + ( {v_{y0}^2}/{v_T^2}) \right]
\EN
then the projectile falls some distance. Starting with zero velocity at the
top, the downward motion equation becomes
\EQ
mgy + \frac{1}{2} m v_T^2 ln \left( \frac{1}{1(v_y^2/v_T^2)} \right) =
mgy_{top}
\EN
Substituting $mgy_{top}$ from one of these into the other yields
\EQ
\boxed{mgy + \frac{1}{2} m v_T^2 ln \left( \frac{1}{1(v_y^2/v_T^2)} \right) =
mgy_{0} + \frac{1}{2} m v_T^2 ln \left[ 1 + ( {v_{y0}^2}/{v_T^2}) \right] }
\EN
Note that the quaskinetic energy going up is different than that
going down. It is unusual for kinetic energy to depend on
the direction of the projectile.
\section{Fall Time and Fall Distance}
The ``Fall Time," also called ``characteristic time'' is given by
\EQ
\boxed{
\tau_T \equiv \frac{v_T}{g}
}
\EN
An dropped object that falls 3 characteristic times has reached a speed
over 99.5\% of $v_T$. Put another way, an object has (essentially) reached terminal speed
after a time $3\frac{v_T}{g}$.
After three terminal times, a dropped object falls a distance of approximately
\EQ
\boxed{
d_T = 2.31\frac{v_T^2}{g}
}
\EN
Thus, if an object falls $2.31\frac{v_T^2}{g}$, it has (basically) reached terminal speed.
\chapter{2D Motion with Other Forces}
\section{1D Sliding Motion and Similar Problems}
While the above equations apply to projectiles, they may be easily altered to apply to
{\bf any problem involving only constant forces and drag forces.}
\subsection{Effects of Friction}
Consider the following free body diagram of a sliding (or rolling) object (such as a car)
that has frictional forces, drag forces, gravitational forces, and contact forces:
\begin{center}
\begin{figure}[ht]
\includegraphics[scale=0.4]{graphics/friction.eps}
\caption{A rolling/sliding block experiencing air drag.}
\end{figure}
\end{center}
The car is moving to the right, and if we define that direction to be positive speed, we must also
choose that to be positive acceleration so that Newton's Laws become:
\EQ
N = mg
\EN
\EQ
ma =  F_{fric} F_{drag}
\EN
These may be rewritten and combined as
\EQ
a = \frac{dv}{dt} = \mu g  \frac{1}{2m} C_D \rho_{air} A v^2
\EN
However, if we define
\EQ
\boxed{
g^\prime = \mu g
}
\EN
\EQ
\boxed{
v_T^{\prime 2} = \frac{2 m g^\prime}{C_D \rho_{air} A}
}
\EN
then
\EQ
a = \frac{dv}{dt} =  g^\prime \left( 1 + \frac{v^2}{v_T^{\prime 2}} \right)
\EN
However, this is the same as the ``Downward Motion" Eq. (\ref{ode_down}). Thus, we can use all of
the Dynamical Equations and the QuasiEnergy Equations for Upward Motion to problems that
include buoyancy as long
as we replace $g$ and $v_T$ in those equations with $g^\prime$ and $v_T^\prime$ as shown here!
\noindent
{\bf \large This worked because the frictional force is constant in this problem}.
\subsection{Effects of Buoyancy}
Consider the following free body diagram of a projectile that has significant buoyancy forces:\\
\begin{center}
\begin{figure}[ht]
\includegraphics[scale=0.4]{graphics/buoyancy.eps}
\caption{An object whose density is near that of air experiences a significant buoyancy force.}
\end{figure}
\end{center}
For now, suppose the projectile is falling. Newton's Laws reduce to
\EQ
ma = mg + F_{buoy} +F_{drag}
\EN
This may be rewritten as
\EQ
a = \frac{dv}{dt} = g +\frac{\rho_{air} V}{m} g + \frac{1}{2m} C_D \rho_{air} A v^2
\EN
However, if we define
\EQ
\boxed{
g^\prime = g  \frac{\rho_{air} V}{m}
}
\EN
\EQ
\boxed{
v_T^{\prime 2} = \frac{2 m g^\prime}{C_D \rho_{air} A}
}
\EN
then
\EQ
a = \frac{dv}{dt} =  g^\prime \left( 1  \frac{v^2}{v_T^{\prime 2}} \right)
\EN
However, this is the same as the ``Downward Motion" Eq. (\ref{ode_down}). Thus, we can use all of
the Dynamical Equations and the QuasiEnergy Equations for Downward Motion to the problem of
lateral motion in the presence of friction as long
as we replace $g$ and $v_T$ in those equations with $g^\prime$ and $v_T^\prime$ as shown here!
For simplicity, we considered Downward Motion. However, the results for {\bf Upward Motion} also
apply when the object is going up. In this case, we would use $g^\prime$ and $v_T^\prime$ as in
the Downward case.
\noindent
{\bf \large This worked because the buoyancy force is constant in this problem}.
One should notice however that if the object is lighter than air, the $g^\prime$ will
be negative.
\section{2D Quadratic Model}
\subsection{Equations of Motion}
Thus, the differential equations of motion are
\EQ
\frac{dv_y}{dt} = g \left( 1 + \frac{v_y}{v_T^2} \sqrt{v_x^2 + v_y^2} \right)
\label{dyn_2y}
\EN
\EQ
\frac{dv_x}{dt} = g \left( \frac{v_x}{v_T^2} \sqrt{v_x^2 + v_y^2} \right)
\label{dyn_2x}
\EN
for projectiles going either up or down.
First we note that these are coupled, nonlinear differential equations.
We cannot solve one, and plug the solution into the other, as before.
The procedure to solve these time dependent equations is not straightforward.
While we can divide these equations by their velocity definitions,
\EQ
\frac{dx}{dt} = v_x
\EN
\EQ
\frac{dy}{dt} = v_y
\EN
in an attempt to obtain a
QuasiEnergy expression, the resulting equations do not separate.
It is similarly difficult to obtain the corresponding projectile path
function.
\subsection{A Conservation Law}
One can not solve the dynamical equations for 2D projectile motion analytically. Similarly,
one can not obtain a QuasiEnergy relation. However, one can obtain a relation between
speed and angle:
\EQ
\boxed{
\frac{v_T^2}{v^2 cos^2 \theta} +
\frac{sin \theta}{cos^2 \theta} +
ln \left( \frac{1+ sin \theta}{cos \theta} \right)
=
\frac{v_T^2}{v_0^2 cos^2 \theta_0} +
\frac{sin \theta_0}{cos^2 \theta_0} +
ln \left( \frac{1+ sin \theta_0}{cos \theta_0} \right)
}
\EN
When the angle $\theta$ is zero, the projectile is at the top of it's flight path.
From this it follows that the velocity of a projectile at the
top of its flight is
\EQ
\frac{v_{x,top}}{v_{x0}} =
\frac{1}{\sqrt{ 1 + \left(\frac{v_{x0}}{v_T}\right)^2 \left[
\frac{sin \theta_0}{cos^2 \theta_0} +
ln \left( \frac{1+ sin \theta_0}{cos \theta_0} \right)
\right] }}
\EN
\section{EXAMPLES}
\begin{EX}
You throw a ball straight up into the air. How long does it take
for the ball to return to you? What speed is it going when
it returns to you?
\end{EX}
\begin{SOL}
It takes a time
\EQ
t_{return} = \left( \frac{v_T}{g} \right)
\left[ tan^{1} \left( \frac{v_{y0}}{v_T} \right)
+ cosh^{1} \left( \sqrt{1+ \frac{v_{y0}}{v_T}} \right)
\right]
\EN
to return to you. The projectile has a speed of
\EQ
\boxed{
v_y = \frac{v_{y0}}{\sqrt{{1+
\left( \frac{v_{y0}}{v_T} \right)^2}}}
}
\EN
\end{SOL}
\chapter{Appendices}
\section{APPENDIX A: Maple  Solving Transcendental Equations}
Many of the above equations are transcendental in a variable of interest, or contain
hyperbolic functions. However,
solving one equation in one unknown is relatively straightforward in Maple whether
hyperbolic functions are included or not.
Here is a simple example of how to use the ``solve" function in Maple:
\EQAA
&>&y:=9;\\
&>&ymax:=11;\\
&>&x:=15;\\
&>&R0:=solve(y = ymax  (4*ymax/(R \wedge 2))*(xR/2) \wedge 2,R);\\
&>&R0[2];\\
&>&evalf(\%);\\
\ENAA
Maple can do much more. It can plot an implicit equation as
a function of an arbitrary variable (Note: If you typed in the
above expression into Maple, use the restart command):
\EQAA
&>&restart;\\
&>&f:=solve(v  ln(1+v) = x  ln(1+x),v);\\
&>&plot(f,x=0..2);\\
\ENAA
In this particular case there is more than one solution, but
Maple finds only the wrong one. It finds $v=x$. To get Maple
to find the other solution, change one of the 1's to 1.00000001
like so:
\EQAA
&>&f:=solve(v  ln(1+v) = x  ln(1.0000000001+x),v);\\
\ENAA
This will not change the result in any practical sense, but the
$v=x$ solution is no longer valid, so Maple must try something
else.
\newpage
\section{APPENDIX B: Drag Graphs for Lookup}
Below are some graphs one can use to interpolate the desired information.
However, one needs to keep in mind these formulas. The formula for
the terminal speed is
\EQ
v_T = \sqrt{\frac{mg}{\frac{1}{2} C_D \rho_{air} A}}
\EN
Some No Drag Formulas:
\EQ
R = \frac{v_0^2}{2g} sin^2 (\theta )
\EN
\EQ
y_{max} = \frac{v_0^2}{2g} sin^2 (\theta )
\EN
\EQ
T_{flight} = \frac{2v_0}{g} sin (\theta )
\EN
Interpolation may be used in the following graphs to physics problems that
include air drag:
\begin{enumerate}
\item Optimum Angle ($0 < v0/vT < 1$)
\item Optimum Angle ($v0/vT > 1$)
\item Range  SurfacetoSurface ($0 < v0/vT < 1$)
\item Range  SurfacetoSurface ($v0/vT > 1$)
\item Max. Height Achieved ($0 < v0/vT < 1$)
\item Max. Height Achieved ($v0/vT > 1$)
\item Time of Flight  SurfacetoSurface ($0 < v0/vT < 1$)
\item Time of Flight  SurfacetoSurface ($v0/vT > 1$)
\end{enumerate}
%%%%%%%%%%%%%%%%%
% Optimum Angle %
%%%%%%%%%%%%%%%%%
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/opt_ang.eps}
\caption{Optimum SurfacetoSurface Launch Angle including Drag}
\end{sidewaysfigure}
\end{center}
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/opt_ang_g1.eps}
\caption{Optimum SurfacetoSurface Launch Angle including Drag}
\end{sidewaysfigure}
\end{center}
%%%%%%%%%
% Range %
%%%%%%%%%
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/range.eps}
\caption{SurfacetoSurface Range including Drag}
\end{sidewaysfigure}
\end{center}
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/range_g1.eps}
\caption{SurfacetoSurface Range including Drag}
\end{sidewaysfigure}
\end{center}
%%%%%%%%%%%%%%%
% Max. Height %
%%%%%%%%%%%%%%%
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/ymax.eps}
\caption{Maximum Height of a Projectile including Drag}
\end{sidewaysfigure}
\end{center}
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/ymax_g1.eps}
\caption{Maximum Height of a Projectile including Drag}
\end{sidewaysfigure}
\end{center}
%%%%%%%%%%%%%%%%%%
% Time of Flight %
%%%%%%%%%%%%%%%%%%
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/ftime.eps}
\caption{SurfacetoSurface Flight Time including Drag}
\end{sidewaysfigure}
\end{center}
\begin{center}
\begin{sidewaysfigure}[ht]
\includegraphics[scale=0.8]{graphics/ftime_g1.eps}
\caption{SurfacetoSurface Flight Time including Drag}
\end{sidewaysfigure}
\end{center}
\clearpage
\section{APPENDIX C: Maple  2D Projectile Motion}
Maple can solve the differential equations governing the 2D motion of
a projectile. Here is the first part of a sample worksheet:
\begin{center}
%/\begin{figure}[ht]
\includegraphics[scale=0.7]{graphics/maple_ode0.eps}
%/\caption{Using Maple as an ODE Solver}
%/\end{figure}
\end{center}
Here is the graph produced and the second part of the worksheet:
\begin{center}
%/\begin{figure}[ht]
\includegraphics[scale=0.75]{graphics/maple_ode1.eps}
%/\caption{Using Maple as an ODE Solver}
%/\end{figure}
\end{center}
\clearpage
\section{APPENDIX D: Spreadsheets  Solving ODEs using RK2}
Shown is a spreadsheet that solves the following differential equation
using Second Order RungeKutta:
\EQ
\frac{dy}{dt} = 10 y
\EN
\begin{center}
%/\begin{figure}[ht]
\includegraphics[scale=0.95]{graphics/rk2.eps}
%/\caption{Using Maple as an ODE Solver}
%/\end{figure}
\end{center}
\clearpage
\begin{center}
%/\begin{figure}[ht]
\includegraphics[scale=0.95]{graphics/rk2_formulas.eps}
%/\caption{Using Maple as an ODE Solver}
%/\end{figure}
\end{center}
\clearpage
\section{APPENDIX E: Hyperbolic Sinusoidal Functions}
Maple and Spreadsheet programs recognize hyperbolic sinusoidal functions. Though their
definitions are sometime couched in terms of complex functions, they are real (if real
numbers are put into them).
\EQ
cosh (x) = \frac{e^x + e^{x}}{2} = cos(ix) = cosh(x)
\EN
\EQ
sinh (x) = \frac{e^x  e^{x}}{2} = i sin(ix) = sinh(x)
\EN
\EQ
tanh (x) =\frac{sinh(x)}{cosh(x)}= \frac{e^x  e^{x}}{e^x + e^{x}} = i tan(ix) = tanh(x)
\EN
The hyperbolic sinusoidal functions are similar to their sinusoidal counterparts in the sense
that there exist parallel formulas such as addition formulas and Pythagorean Formula:
\EQ
cosh^2(x)  sinh^2(x) = 1
\EN
Below are the graphs for the hyperbolic trigonometric functions.
\begin{center}
\begin{figure}[ht]
\includegraphics[scale=0.3]{graphics/cosh.eps}
\includegraphics[scale=0.3]{graphics/sech.eps}
\caption{cosh=paraboliclookingthing, sinh=$x^3$lookingthing}
\end{figure}
\end{center}
\end{document}