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=============================================================================
A basic set of bright stars -- taken from the xephem program.
Based on the 5th Revised edition of the Yale Bright Star Catalog, 1991, from
ftp://adc.gsfc.nasa.gov/pub/adc/archives/catalogs/5/5050.
Only those entries with a Bayer and/or Flamsteed number are retained
here.
Format: Constellation BayerN-Flamsteed, as available. Bayer is
truncated as requried to enforce a maximum total length of 13
imposed within xephem.
Common names were then overlayed by closest position match from
hand-edited a list supplied by Robert Tidd (inp@violet.berkeley.edu)
and Alan Paeth (awpaeth@watcgl.waterloo.edu)
=============================================================================
Data file format:
name , right assention (radians) , declination (radians) , magnitude (0.0 - 1.0)
=============================================================================
The following information is taken from:
http://www.lclark.edu/~wstone/skytour/celest.html
Please visit the above site, it contains much more complete information.
CELESTIAL MEASUREMENTS
RIGHT ASCENSION AND DECLINATION
Although we know that the objects we see in the sky are of different
sizes and at different distances from us, it is convenient to
visualize all the objects as being attached to an imaginary sphere
surrounding the Earth. From our vantage point, the sky certainly looks
like a dome (the half of the celestial sphere above our local
horizon). The celestial sphere is mapped in Right Ascension (RA) and
Declination (Dec). Declination is the celestial equivalent of
latitude, and is simply the Earth's latitude lines projected onto the
celestial sphere. A star that can be directly overhead as seen from
the Earth's Equator (0 degrees latitude) is said to be on the
Celestial Equator, and has a declination of 0 degrees . The North
Star, Polaris, is very nearly overhead as seen from the North Pole (90
degrees North latitude). The point directly over the North Pole on the
celestial sphere is called the North Celestial Pole, and has a
declination of +90 degrees . Northern declinations are given positive
signs, and southern declinations are given negative signs. So, the
South Celestial Pole has a declination of -90 degrees .
Right Ascension is the equivalent of longitude, but since the Earth
rotates with respect to the celestial sphere we cannot simply use the
Greenwich Meridian as 0 degrees RA. Instead, we set the zero point as
the place on the celestial sphere where the Sun crosses the Celestial
Equator (0 degrees Dec) at the vernal (spring) equinox. The arc of
the celestial sphere from the North Celestial Pole through this point
to the South Celestial Pole is designated as Zero hours RA. Right
Ascension increases eastward, and the sky is divided up into 24
hours. This designation is convenient because it represents the
sidereal day, the time it takes for the Earth to make one rotation
relative to the celestial sphere. If you pointed a telescope (with no
motor drive) at the coordinates (RA=0h, Dec=0 degrees ), and came back
one hour later, the telescope would then be pointing at (RA=1h, Dec=0
degrees ). Because the Earth's revolution around the Sun also
contributes to the apparent motion of the stars, the day we keep time
by (the solar day) is about four minutes longer than the sidereal
day. So, if you pointed a telescope at (RA=0h, Dec=0 degrees ) and
came back 24 hours later, the telescope would now be pointing at
(RA=0h 4m, Dec=0 degrees). A consequence is that the fixed stars
appear to rise about four minutes earlier each day.
=============================================================================
From: steve@mred.bgm.link.com (Steve Baker)
Subject: Re: FG: Fun in the sun ...
Date: Tue, 5 Aug 97 15:37:27 -0500
You probably ought to get the stars right too - there is a database
of the 300 brightest stars in the 'Yale Bright Star Catalog' - which
I enclose below. I'd guess that you could navigate by the stars -
so this might not be a completely useless feature - right?
Anyway, if all else fails - and the flight sim never gets going - we
could at least sell this as a planetarium :-)
The format of the star data is:
Name Right-Ascension Declination Magnitude
(Ascension and Declination are in radians)
We took the magnitude value, scaled it by 0.8 and added 0.2 to make
a 0->1 brightness value. Using the raw data created too many very
dark stars.
Originally, there were constellation names as sub-headings - but I
think I deleted them to make the file easier to parse :-) That makes
the 'name' field pretty pointless.
if you are still talking about the geocentric coordinate system
where the terrain is modelled with Z pointing towards the North
pole, X out of the 0 degree meridian at the equator and Y out at the
Indian ocean at the equator - then you can position the stars using:
star[ X ] = fsin ( ra ) * fcos( decl ) ;
star[ Y ] = fcos ( ra ) * fcos( decl ) ;
star[ Z ] = fsin ( decl ) ;
(which you can precompute at startup)
...and then rotate them about the Z axis using GMT*two_pi/24.0
#
Put them all in a display list - use GL_POINTS as the primitive...
glNewList ( ...whatever... )
glBegin ( GL_POINTS ) ;
for ( int i = 0 ; i < num_stars ; i++ ) {
glColor3f ( star_brightness[i], star_brightness[i], star_brightness[i] ) ;
glVertex3f ( star_x[i], star_y[i], star_z[i] ) ;
}
glEnd () ;
glEndList () ;
You need to draw them out by the far clip plane so they don't occult
anything. Then you need to translate them using the same x/y/z as
the eyepoint so that you can never fly any closer to them.

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%
% `Stars.tex' -- describes our procedure for drawing stars
%
% Written by Curtis Olson. Started December, 1997.
%
% $Id$
%------------------------------------------------------------------------
\documentclass[12pt]{article}
\usepackage{anysize}
\papersize{11in}{8.5in}
\marginsize{1in}{1in}{1in}{1in}
\usepackage{amsmath}
\usepackage{epsfig}
\usepackage{setspace}
\onehalfspacing
\usepackage{url}
\begin{document}
\title{
Flight Gear Stars Representation and Rendering.
}
\author{
Curtis L. Olson\\
(\texttt{curt@me.umn.edu})
}
\maketitle
\section{Introduction}
Flight Gear attempts to render the top several hundred stars in the
correct position in the sky for the current time and position as will
as rendering these stars with the correct ``magnitude''.
This document will give a quick overview of our approach.
\section{Resources}
\subsubsection{XEphem}
The set of bright stars was extracted from the xephem program. For a
full list of features and the latest news, please see the xephem home
page at \url{http://iraf.noao.edu/~ecdowney/xephem.html}. The XEphem
star list was itself taken from the Yale Bright Star Catalog.
\begin{quote}
Based on the 5th Revised edition of the Yale Bright Star Catalog,
1991, from ftp://adc.gsfc.nasa.gov/pub/adc/archives/catalogs/5/5050.
Only those entries with a Bayer and/or Flamsteed number are retained
here.
Format: Constellation BayerN-Flamsteed, as available. Bayer is
truncated as required to enforce a maximum total length of 13
imposed within xephem.
Common names were then overlayed by closest position match from
hand-edited a list supplied by Robert Tidd (inp@violet.berkeley.edu)
and Alan Paeth (awpaeth@watcgl.waterloo.edu)
\end{quote}
The author of XEphem, Elwood Downey (ecdowney@noao.edu), was very
instrumental in helping me understand sidereal time, accurate star
placement, and even contributed some really hairy sections of code.
Thanks Elwood!
\section{Terminology and Definitions}
The following information is repeated verbatim from
\url{http://www.lclark.edu/~wstone/skytour/celest.html}: If you are
interested in these sorts of things I urge you to visit this site. It
contains much more complete information.
\subsection{Celestial Measurements}
Although we know that the objects we see in the sky are of different
sizes and at different distances from us, it is convenient to
visualize all the objects as being attached to an imaginary sphere
surrounding the Earth. From our vantage point, the sky certainly
looks like a dome (the half of the celestial sphere above our local
horizon). The celestial sphere is mapped in Right Ascension (RA) and
Declination (Dec).
\subsubsection{Declination}
Declination is the celestial equivalent of latitude, and is simply the
Earth's latitude lines projected onto the celestial sphere. A star
that can be directly overhead as seen from the Earth's Equator (0
degrees latitude) is said to be on the Celestial Equator, and has a
declination of 0 degrees . The North Star, Polaris, is very nearly
overhead as seen from the North Pole (90 degrees North latitude). The
point directly over the North Pole on the celestial sphere is called
the North Celestial Pole, and has a declination of +90 degrees .
Northern declinations are given positive signs, and southern
declinations are given negative signs. So, the South Celestial Pole
has a declination of -90 degrees .
\subsubsection{Right Ascension \& Sidereal Time}
Right Ascension is the equivalent of longitude, but since the Earth
rotates with respect to the celestial sphere we cannot simply use the
Greenwich Meridian as 0 degrees RA. Instead, we set the zero point as
the place on the celestial sphere where the Sun crosses the Celestial
Equator (0 degrees Dec) at the vernal (spring) equinox. The arc of
the celestial sphere from the North Celestial Pole through this point
to the South Celestial Pole is designated as Zero hours RA. Right
Ascension increases eastward, and the sky is divided up into 24
hours. This designation is convenient because it represents the
sidereal day, the time it takes for the Earth to make one rotation
relative to the celestial sphere. If you pointed a telescope (with no
motor drive) at the coordinates (RA=0h, Dec=0 degrees ), and came back
one hour later, the telescope would then be pointing at (RA=1h, Dec=0
degrees ). Because the Earth's revolution around the Sun also
contributes to the apparent motion of the stars, the day we keep time
by (the solar day) is about four minutes longer than the sidereal
day. So, if you pointed a telescope at (RA=0h, Dec=0 degrees ) and
came back 24 hours later, the telescope would now be pointing at
(RA=0h 4m, Dec=0 degrees). A consequence is that the fixed stars
appear to rise about four minutes earlier each day.
\subsection{Implementation}
Here is a brief overview of how stars were implemented in Flight Gear.
The right ascension and declination of each star is used to build a
structure of point objects to represent the stars. The magnitude is
mapped into a color on the gray/white continuum. The points are
positioned just inside the far clip plane. When rendering the stars,
this structure (display list) is rotated about the $Z$ axis by the
current sidereal time and translated to the current view point.
\subsubsection{Data file format}
The star information is stored in a simple data file called
``Stars.dat'' with the following comma delimited format: name, right
ascension(radians), declination(radians), magnitude(smaller is
brighter). Here is an extract of the data file:
\begin{verbatim}
Sirius,1.767793,-0.266754,-1.460000
Canopus,1.675305,-0.895427,-0.720000
Arcturus,3.733528,0.334798,-0.040000
Rigil Kentaurus,3.837972,-1.032619,-0.010000
Vega,4.873563,0.676902,0.030000
Capella,1.381821,0.802818,0.080000
Rigel,1.372432,-0.136107,0.120000
Procyon,2.004082,0.091193,0.380000
Achernar,0.426362,-0.990707,0.460000
\end{verbatim}
\subsubsection{Building the display list}
The display list is built up from a collection of point objects as the
star data file is loaded. For each star, the magnitude is mapped into
a brightness value from 0.0 to 1.0 with 1.0 being the brightest. Our
coordinate system is described in the coordinate system document: Z
points towards the North pole, X out of the 0 degree meridian at the
equator, and Y out at the Indian ocean at the equator. Given this
coordinate system, the position of each star at 0:00H sidereal time is
calculated as follows:
\begin{align}
x &= \mathrm{distance} * \cos(\mathrm{rightascension}) *
\cos(\mathrm{declination}) \\
y &= \mathrm{distance} * \sin(\mathrm{rightascension}) *
\cos(\mathrm{declination}) \\
z &= \mathrm{distance} * \sin(\mathrm{declination})
\end{align}
\subsubsection{Transformations and Rendering}
The tricky part about rendering the stars is calculating sidereal time
correctly. Here's where Elwood Downey saved my butt. 0:00H sidereal
time aligns with 12:00 noon GMT time on March 21 of every year. After
that they diverge by about 4 minutes per day. The solar day is
approximately 4 minutes longer than the side real day. Once you know
the proper sidereal time, you simply translate the center of the star
structure to the current view point, then rotate this structure about
the Z axis by the current sidereal time.
The stars are drawn out by the far clip plane so they don't occult
anything. They are translated using the same x/y/z as the eye point
so that you can never fly any closer to them.
\end{document}
%------------------------------------------------------------------------
% $Log$
% Revision 1.1 1999/02/14 19:10:47 curt
% Initial revisions.
%

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%
% `Sky.tex' -- describes the sky rendering procedure
%
% Written by Curtis Olson. Started December, 1997.
%
% $Id$
%------------------------------------------------------------------------
\documentclass[12pt]{article}
\usepackage{anysize}
\papersize{11in}{8.5in}
\marginsize{1in}{1in}{1in}{1in}
\usepackage{amsmath}
\usepackage{epsfig}
\usepackage{setspace}
\onehalfspacing
\usepackage{url}
\begin{document}
\title{
Flight Gear Sky Representation and Rendering.
}
\author{
Curtis L. Olson\\
(\texttt{curt@me.umn.edu})
}
\maketitle
\section{Introduction}
No flight simulator should be without a nice sky that smoothly
transitions into haze at the horizon. Such a sky should also be able
to render sunrise and sunset effects. This document describes how we
have implemented such a sky.
\section{Overview}
The sky is represent as a 12 sided dome (or upside down bowl if you
prefer.) Figure \ref{fig:dome} shows how a 6 sided dome might be
constructed.
\begin{figure}[hbt]
\centerline{
\psfig{file=dome.eps}
}
\caption{Simplified (6 Sided) Sky Dome}
\label{fig:dome}
\end{figure}
The center section can be constructed with a triangle fan. The inner
and outer ``skirts'' can be constructed with triangle strips.
The colors of each vertex can be independently controlled to achieve
sky to haze transitions, sunrise/sunset effects with a pinkish/oranges
glow, and one side of the sky can easily be made brighter than the
other. By enabling smooth shading in OpenGL, the colors will be
blended together for a very nice effect.
\section{Implementation}
This sections describes how the sky has been implemented in OpenGL.
\subsection{Vertex Generation}
The sky dome structure is intended to be centered over the current
view point at sea level. This way we could paste cloud textures on
the dome if we liked. So, we simply have to generate vertices for a
fixed dome, and then use OpenGL calls to transform and rotate it to
the desired place. Please refer to the actual code
(.../Src/Scenery/sky.c) for specifics, but
to generate the vertices we simply create a 12 element array for the
inner set of vertices, another 12 element array for the middle set of
vertices and a last 12 element array for the outer set of vertices.
\subsection{Vertex Coloring}
For each vertex position array, there is a corresponding vertex color
array. This way we don't have to compute each vertex color every
iteration. Also, by being able to individually control the color at
each vertex, we can do all sorts of nice sky to haze blending with
dusk and dawn effects. Again, please refer to the source
(.../Src/Scenery/sky.c) for specific details on how the coloring is
implemented. However, here's the quick overview.
\subsubsection{Day and Night Coloring}
For the general middle of the day, or middle of the night sky, we
already know the desired sky color, and the haze color. This is
computed elsewhere based on the current sun position. During the
night these colors are both nearly black. During the dawn they are
smoothly transitioned to day time colors. And, during the dusk they
are smoothly transitioned back to night time colors.
The center of the dome is assigned the current sky color. The color
of the first inner ring of vertices is weighted 70\% towards the sky
color and 30\% towards the fog color.
Then color of the middle ring of vertices is weighted 10\% towards the
sky color and 90\% towards the fog color.
The the outer ring of vertices are assigned the current fog color.
\subsubsection{Dusk and Dawn Effects}
Dusk and dawn effects can be accomplished by controlling the color of
the vertices. Rather than trying to figure out which vertices are
near the current sun position, I just rotate the dome so the 0'th
vertex of each ring (and the center fan) align with the sun. This
makes it easier to calculate vertex colors. But, there is a fair
amount of work involved in calculating the proper dome rotation.
\begin{figure}[hbt]
\centerline{
\psfig{file=earth.eps}
}
\caption{Overview of Earth}
\label{fig:earth}
\end{figure}
Figure \ref{fig:earth} shows an overview of the setup. $P$, the
current view position, and $\mathbf{n}$, the local ``up'' vector,
define the plane which is tangent to the Earth's surface at point $P$.
Just for a quick review of your linear algebra, given $\mathbf{v_0}$,
the position vector of $P$ and $\mathbf{v}$, the position vector of
some other arbitrary point on the plane, and $\mathbf{n}$, the normal
to the plane, then the vector $\mathbf{n}$ and the vector $(\mathbf{v}
- \mathbf{v_0})$ are orthogonal (perpendicular.) If the two vectors
are orthogonal then their dot product will be zero, so the following
must be true:
\begin{equation}
\mathbf{n} \cdot ( \mathbf{v} - \mathbf{v_0} ) = 0
\end{equation}
This is the vector equation of the plane and can be rewritten as:
\begin{align}
a(x - x_0) + b(y - y_0) + c(z - z_0) &= 0 \\
ax + by + cz - (\mathbf{n} \cdot \mathbf{v_0}) &= 0
\end{align}
We want to find a vector $\mathbf{v}$ representing the
direction along the current tangent plane towards the position on the
Earth where the Sun is directly overhead. The vector $\mathbf{u}$ is
defined as $\vec{\mathbf{PS}}$.
\begin{figure}[hbt]
\centerline{
\psfig{file=local.eps}
}
\caption{Vectors and Points in Local Coordinate System}
\label{fig:local}
\end{figure}
Figure \ref{fig:local} shows a more detailed ``local'' view of the
points and vectors involved. The point, $P$, is the current view
point. The vector, $\mathbf{n}$, is the local up vector. $S$
represents the current position on the Earth's surface where the Sun
is directly overhead. We want to find the vector, $\mathbf{v}$ which
is a projection of $\mathbf{u}$ onto the plane defined by $P$ and
$\mathbf{n}$.
To do this we first calculate $\mathbf{u_1}$ which is the shortest
distance from point $S$ to the tangent plane.
\begin{equation}
\mathbf{u_1} = \frac { \mathbf{n} \cdot \mathbf{u} }
{ {\| \mathbf{n} \|}^2 } \mathbf{n}
\end{equation}
Armed with $\mathbf{u_1}$ we can now calculate
$\mathbf{v}$ which is the local surface direction on the tangent
plane towards the sun, $S$.
\begin{equation}
\mathbf{v} = \mathbf{v_0} + \mathbf{u} - \mathbf{u_1}
\end{equation}
Ok, so now we have $\mathbf{v}$, but the fun doesn't stop here. Now
we need to calculate a rotation angle $\theta$ about $\mathbf{n}$ to
align our dome with $\mathbf{v}$. The origin of the dome always
aligns with a vector pointing directly South. So, we need to repeat
the above procedure to map a vector pointing straight down $( 0, 0,
-\mathbf{z} )$ onto our tangent plane to produce the local, surface,
south vector $\mathbf{w}$. We then take the $\arccos()$ of the dot product
of $\mathbf{v}$ with $\mathbf{w}$.
\begin{equation}
\theta = \arccos( \mathbf{v} \cdot \mathbf{w} )
\end{equation}
Whew, that gives us the angle we want. Well almost, not quite. The
problem is that the dot product returns a number in the range of
$(-1.0 \ldots 1.0)$. Thus, the $\arccos()$ function returns a $\theta$
in the range of $(0.0 \ldots 180.0)$. But this is not enough
information to determine if $\mathbf{v}$ is in the east hemisphere or
west hemisphere and if this angle should be positive or negative.
So, to get that last piece of information we need, we can rotate the
vector $\mathbf{w}$ by 90 degrees about $\mathbf{n}$. This gives us
the local surface east vector on the tangent plane. Taking the dot
product of $\mathbf{v}$ and the local east vector tells us which
hemisphere $\mathbf{v}$ is in. And, from this, we can uniquely
determine the proper angle for the sky dome rotation.
\end{document}
%------------------------------------------------------------------------
% $Log$
% Revision 1.1 1999/02/14 19:12:21 curt
% Initial revisions.
%

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#FIG 3.2
Portrait
Center
Inches
Letter
100.00
Single
0
1200 2
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 7
4800 2700 3375 3450 4050 4500 6600 4500 7725 3450 6750 2700
4800 2700
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 8
3375 3450 2700 4500 2400 5400 3300 7200 7500 7200 8700 5400
8400 4500 7725 3450
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 4
2700 4500 3450 5850 7200 5850 8400 4500
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 3
3300 7200 3450 5850 4050 4500
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 3
6600 4500 7200 5850 7500 7200
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
3375 3450 3450 5850
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
4050 4500 7200 5850
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
6600 4500 8400 4500
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
2700 4500 3300 7200
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
3450 5850 7500 7200
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
7200 5850 8700 5400
2 1 2 1 0 7 0 0 -1 4.000 0 0 -1 0 0 4
2700 4500 4125 3600 7275 3600 8400 4500
2 1 2 1 0 7 0 0 -1 4.000 0 0 -1 0 0 3
6750 2700 7275 3600 7350 4350
2 1 0 1 0 7 0 0 -1 4.000 0 0 -1 0 0 3
4050 4500 5625 3075 6750 2700
2 1 0 1 0 7 0 0 -1 4.000 0 0 -1 0 0 3
4800 2700 5625 3075 6600 4500
2 1 2 1 0 7 0 0 -1 4.000 0 0 -1 0 0 4
2400 5400 4050 4350 7350 4350 8700 5400
2 1 2 1 0 7 0 0 -1 4.000 0 0 -1 0 0 3
4050 4350 4125 3600 4800 2700
2 1 0 1 0 7 0 0 -1 4.000 0 0 -1 0 0 3
3375 3450 5625 3075 7725 3450

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#FIG 3.2
Landscape
Center
Inches
Letter
100.00
Single
-2
1200 2
5 1 0 2 0 7 0 0 -1 0.000 0 1 0 0 4800.000 -4650.000 2400 4800 4800 5100 7200 4800
5 1 0 2 0 7 0 0 -1 0.000 0 0 0 0 14250.000 4800.000 4800 7200 4500 4800 4800 2400
5 1 2 1 0 7 0 0 -1 3.000 0 1 0 0 -4650.000 4800.000 4800 7200 5100 4800 4800 2400
5 1 2 1 0 7 0 0 -1 3.000 0 0 0 0 4800.000 14250.000 2400 4800 4800 4500 7200 4800
1 3 0 2 0 7 0 0 -1 0.000 1 0.0000 4800 4800 2400 2400 4800 4800 7200 4800
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 4800 4800 25 25 4800 4800 4820 4815
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 6300 4200 25 25 6300 4200 6320 4215
2 1 2 1 0 7 0 0 -1 3.000 0 0 -1 0 0 2
4500 5100 5100 4500
2 1 0 1 0 7 0 0 -1 4.000 0 0 -1 0 0 2
6600 5400 7800 4200
2 1 2 1 0 7 0 0 -1 3.000 0 0 -1 0 0 2
2400 4800 7200 4800
2 1 0 2 0 7 0 0 -1 6.000 0 0 -1 0 0 5
6600 4200 6600 6600 7800 5400 7800 3000 6600 4200
2 1 0 1 0 7 0 0 -1 4.000 0 0 -1 0 0 2
7200 3600 7200 6000
2 1 0 1 0 -1 0 0 20 4.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
7200 4800 6750 4200
2 1 0 1 0 -1 0 0 20 4.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
7200 4800 6300 4200
2 1 2 1 0 -1 0 0 20 3.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
6750 4200 6300 4200
2 1 0 1 0 -1 0 0 20 4.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
7200 4800 8100 4800
4 0 0 0 0 0 14 0.0000 4 105 105 8175 4875 n\001
4 0 0 0 0 0 14 0.0000 4 150 120 7275 5025 P\001
4 0 0 0 0 0 14 0.0000 4 105 105 6825 4200 v\001
4 0 0 0 0 0 14 0.0000 4 195 645 5625 4275 S (Sun)\001
4 0 0 0 0 0 14 0.0000 4 105 105 6375 4500 u\001
4 0 0 0 0 0 14 0.0000 4 195 525 4800 5025 Origin\001

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#FIG 3.2
Landscape
Center
Inches
Letter
100.00
Single
-2
1200 2
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 4500 8700 21 21 4500 8700 4515 8715
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 6900 8100 21 21 6900 8100 6915 8115
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 6900 6600 21 21 6900 6600 6915 6615
1 3 0 1 0 -1 0 0 20 0.000 1 0.0000 5700 6000 21 21 5700 6000 5715 6015
2 1 0 2 0 7 0 0 -1 0.000 0 0 -1 0 0 5
1800 7200 4200 4800 9600 4800 7200 7200 1800 7200
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
4500 7200 6900 4800
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 0 0 2
3000 6000 8400 6000
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
5700 6000 5700 4200
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
5700 6000 6900 8100
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
5700 6000 6900 6600
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 1 0 2
0 0 1.00 60.00 120.00
6900 6600 6900 8100
2 1 0 1 0 7 0 0 -1 0.000 0 0 -1 1 0 2
1 1 1.00 60.00 120.00
4500 8700 5686 5994
4 0 0 0 0 0 14 0.0000 4 105 105 6825 6450 v\001
4 0 0 0 0 0 14 0.0000 4 195 645 6750 8325 S (Sun)\001
4 0 0 0 0 0 14 0.0000 4 105 105 6975 7500 u\001
4 0 0 0 0 0 10 0.0000 4 105 75 7080 7568 1\001
4 0 0 0 0 0 14 0.0000 4 195 525 4200 8925 Origin\001
4 0 0 0 0 0 14 0.0000 4 150 120 5475 5850 P\001
4 0 0 0 0 0 10 0.0000 4 105 75 5550 5925 0\001
4 0 0 0 0 0 14 0.0000 4 105 105 5625 6375 v\001
4 0 0 0 0 0 10 0.0000 4 105 75 5700 6450 0\001
4 0 0 0 0 0 14 0.0000 4 105 105 6450 7725 u\001
4 0 0 0 0 0 14 0.0000 4 105 105 5625 4125 n\001