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\section{Introduction}

Astronomy is a physical science. Just like biology, chemistry, geology, and
physics, astronomers collect data, analyze that data, attempt to understand
the object/subject they are looking at, and submit their results for 
publication. Along the way astronomers use all of the mathematical techniques
and physics necessary to understand the objects they examine. Thus, just
like any other science, a large number of mathematical tools and concepts
are needed to perform astronomical research. In today's introductory lab,
you will review and learn some of the most basic concepts necessary 
to enable you to successfully complete the various laboratory exercises
you will encounter later this semester. When needed, the weekly laboratory
exercise you are performing will refer back to the examples in this 
introduction---so keep the worked examples you will do today with you at all 
times during the semester to use as a reference when you run into these
exercises later this semester (in fact, on some occasions your TA might
have you redo one of the sections of this lab for review purposes).

\section{The Metric System}

Like all other scientists, astronomers use the metric system. The
metric system is based on powers of 10, and has a set of measurement units
analogous to the English system we use in everyday life here in the US. In the
metric system the main unit of length (or distance) is the meter, the unit of 
mass is the kilogram, and the unit of liquid volume is the liter. A meter is 
approximately 40 inches, or about 4" longer than the yard. Thus, 100 meters is 
about 111 yards. A liter is slightly larger than a quart (1.0 liter = 1.101 
qt). On the Earth's surface, a kilogram = 2.2 pounds. In the Astronomy 110
labs you will mostly encounter units of length/distance (variations on the 
meter). 

As you have almost certainly learned, the metric system uses prefixes to change
scale. For example, one thousand meters is one ``kilometer''. One thousandth
of a meter is a ``millimeter''. The prefixes that you will encounter in
this class are listed in Table 1.

\begin{center}
        Table 1: Metric System Prefixes\\
\begin{tabular}{c|c|c}
\hline
Prefix Name & Prefix Symbol & Prefix Value\\
\hline
Giga	&	G	& 1,000,000,000 (one billion)\\
Mega	&	M	& 1,000,000 (one million) \\
kilo	&	k	& 1,000 (one thousand) \\\
centi	&	c	& 0.01 (one hundredth) \\
milli	&	m	& 0.001 (one thousandth) \\
micro	&	$\mu$	& 0.0000001 (one millionth) \\
nano    &	n	& 0.0000000001 (one billionth)\\
\hline
\end{tabular}
\end{center}

In the metric system 3,600 meters is equal to 3.6 kilometers; while 0.8 meter is
equal to 80 centimeters, which in turn equals 800 millimeters, etc. In the lab
exercises this semester we will encounter a large range in sizes and
distances. For example, you will measure the sizes of some objects/things in 
class in millimeters, talk about the wavelength of spectral lines in 
nanometers, and measure the sizes of features on the Sun that are larger than 
100,000 kilometers.
\subsection{Beyond the Metric System}
When we talk about the sizes or distances to those objects beyond
the surface of the Earth, we
begin to encounter very large numbers. For example, the average
distance from the Earth to the Moon is 384,000,000 meters or 384,000 kilometers
(km). The distances found in astronomy are usually so large that we
have to switch to a unit of measurement that is much larger than the
meter, or even the kilometer. In and around the solar
system, astronomers use ``Astronomical Units''. An Astronomical Unit is the 
mean distance
between the Earth and the Sun. One Astronomical Unit (AU) = 149,600,000 km.
For example, Jupiter is about 5 AU from the Sun, while Pluto's average
distance from the Sun is 39 AU. With this change in units, it is easy
to talk about the distance to other planets. It is more convenient to say that
Saturn is 9.54 AU away than it is to say that Saturn is 1,427,184,000 km from
Earth. 

When we talk about how far away the stars are in our own Milky Way galaxy, we 
have to switch to an even larger unit of distance to keep the numbers 
manageable. One such unit is the ``light year''. A light year (ly) is the 
distance light travels in one year. The speed of light is enormous: 300,000 
kilometers per second (km/s) or 186,000 miles per second. Since one year
contains 31,536,000 seconds, one ly = 9,460,000,000,000 km! The nearest star, 
Alpha Centauri, is 4.2 ly away. The Milky Way galaxy is more than 150,000 light 
years across.  The nearest galaxy with a size similar to that of the Milky Way,
the Andromeda Galaxy (see the sky chart for November online at
http://astronomy.nmsu.edu/tharriso/skycharts.html
for a picture and description of the Andromeda galaxy),
is 2.2 million light years away!

In the Parallax lab we will introduce the somewhat odd unit of ``parsecs''. 
For now, we will simply state that one parsec (``pc'') = 3.26 ly. Thus, Alpha 
Centauri is 1.28 pc away. During the semester you will frequently hear the term
parsec, kiloparsec (1 thousand pc), Megaparsec (1 million pc), and even
the term Gigaparsec (1 billion pc). Astronomers have borrowed the prefixes
from the metric system to construct their own shorthand way of describing
extremely large distances. The Andromeda Galaxy is at a distance of 700,000 pc 
= 0.7 Megaparsecs (``Mpc'').

\subsection{Changing Units and Scale Conversion}

Changing units (like those in the previous paragraph) and/or scale conversion 
is something you must master during this semester. The concept is fairly
straightforward, so let's just work some examples.

\begin{flushleft}

1. Convert 34 meters into centimeters\\
\vspace{0.2in}
Answer: Since one meter = 100 centimeters, 34 meters = 3,400 centimeters.\\
\vspace{0.3in}
2. Convert 34 kilometers into meters:\\
\vspace{0.3in}
3. If one meter equals 40 inches, how many meters are there in 400 inches?\\
\vspace{0.3in}
4. How many centimeters are there in 400 inches?\\
\vspace{0.3in}
5. How many parsecs are there in 1.4 Mpc?\\
\vspace{0.3in}
6. How many AU are there in 299,200,000 km?
\end{flushleft}

One technique that you will use this semester involves measuring a photograph
or image with a ruler, and converting the measured number into a real unit of 
size (or distance). One example of this technique is reading a road map. In 
the next figure is a map of the state of New Mexico. Down at the bottom left 
hand corner is a scale in Miles and Kilometers.

\begin{figure}[!ht]
%\centering
\leavevmode
\epsfysize=0.7\textheight
%\epsfbox{COMET/Asteroid460.eps}
\epsfbox{LAB0/new-mexico-map.ps}
%\epsfbox{new-mexico-map.ps}
\caption{A map of New Mexico}
\end{figure}

\begin{flushleft}
{\bf Map Exercises} (using a ruler determine):\\
\vspace{0.3in}
1) How many kilometers is it from Las Cruces to Albuquerque?\\
\vspace{0.3in}
2) What is the distance in miles from the border with Arizona to the border
with Texas if you were to drive along I40?\\
\vspace{0.3in}
3) If you were to drive 100 km/hr (kph), how long would it take you to
go from Las Cruces to Albuquerque?\\
\vspace{0.3in}
4) If one mile = 1.6 km, how many miles per hour (mph) is 100 kph?\\
\vspace{0.3in}
\end{flushleft}

\section{Squares, Square Roots, and Exponents}

In several of the labs this semester you will encounter squares, cubes, and
square roots. Let us briefly review what is meant by such terms as squares,
cubes, square roots and exponents. The square of a number is simply that
number times itself: 3 $\times$ 3 = 3$^{\rm 2}$ = 9. The exponent is the
little number ``2'' above the three. 5$^{\rm 2}$ = 5 $\times$ 5 = 25. The
exponent tells you how many times to multiply that number by itself: 
8$^{\rm 4}$ =  8 $\times$ 8 $\times$ 8 $\times$ 8 = 4096. The square of a
number simply means the exponent is 2 (three squared = 3$^{\rm 2}$), and the 
cube of a number means the exponent is three (four cubed = 4$^{\rm 3}$).
Here are some examples:

\begin{flushleft}
1) 7$^{\rm 2}$ = 7 $\times$ 7 = 49\\
\vspace{0.2in}
2) 7$^{\rm 5}$ = 7 $\times$ 7 $\times$ 7 $\times$ 7 $\times$ 7 = 16,807\\
\vspace{0.2in}
3) The cube of 9 = 9$^{\rm 3}$ = 9 $\times$ 9 $\times$ 9 = 729\\
\vspace{0.2in}
4) The exponent of 12$^{\rm 16}$ is 16\\
\vspace{0.2in}
5) 2.56$^{\rm 3}$ = 2.56 $\times$ 2.56 $\times$ 2.56 = 16.777\\
\vspace{0.35in}
{\bf Your turn:}\\
\vspace{0.3in}
7) 6$^{\rm 3}$ = \\
\vspace{0.3in}
8) 4$^{\rm 4}$ = \\
\vspace{0.3in}
9) 3.1$^{\rm 2}$ = \\
\end{flushleft}

The concept of a square root is easy to understand, but
is much harder to calculate (we usually have to use a calculator).
The square root of a $number$ is that number whose square is the 
$number$: the square root of 4 = 2 because
2 $\times$ 2 = 4. The square root of 9 is 3 (9 = 3 $\times$ 3). The 
mathematical operation of a square root is usually represented by the
symbol ``$\surd$~'', as in $\surd$9 = 3. But mathematicians also represent
square roots using a $fractional$ exponent of one half: 9$^{\rm 1/2}$ = 3.
Likewise, the cube root of a number is represented as 27$^{\rm 1/3}$ = 3
(3 $\times$ 3 $\times$ 3 = 27). The fourth root is written as 
16$^{\rm 1/4}$ (= 2), and so on. We will encounter square roots in the
algebra section shortly. Here are some examples/problems:

\begin{flushleft}
1) $\surd$100 = 10\\
\vspace{0.3in}
2) 10.5$^{\rm 3}$ = 10.5 $\times$ 10.5 $\times$ 10.5 = 1157.625\\
\vspace{0.3in}
3) Verify that the square root of 17 ($\surd$17 = 17$^{\rm 1/2}$) = 4.123\\

\end{flushleft}
\section{Scientific Notation}

The range in numbers encountered in Astronomy is enormous: from the size
of subatomic particles, to the size of the entire universe. You are certainly
comfortable with numbers like ten, one hundred, three thousand, ten million,
a billion, or even a trillion. But what about a number like one million
trillion? Or, four thousand one hundred and fifty six million billion? Such
numbers are too cumbersome to handle with words. Scientists use something
called ``Scientific Notation'' as a short hand method to represent very large
and very small numbers.
The system of scientific notation is based on the number 10. For example,
the number 100 = 10 $\times$ 10 = 10$^{\rm 2}$. In scientific notation
the number 100 is written as 1.0 $\times$ 10$^{\rm 2}$. Here are some
additional examples:\\
\begin{flushleft}
~ ~\\
Ten = 10 = 1 $\times$ 10 = 1.0 $\times$ 10$^{\rm 1}$
~ ~\\
One hundred = 100 = 10 $\times$ 10 = 10$^{\rm 2}$ = 1.0 $\times$ 10$^{\rm 2}$
~ ~\\
One thousand = 1,000 = 10 $\times$ 10 $\times$ 10 = 10$^{\rm 3}$ = 1.0 $\times$ 10$^{\rm 3}$\\
One million = 1,000,000 = 10 $\times$ 10 $\times$ 10 $\times$ 10 $\times$ 10 $\times$ 10 = 10$^{\rm 6}$ = 1.0 $\times$10$^{\rm 6}$\\
\end{flushleft}

Ok, so writing powers of ten is easy, but how do we write 6,563 in scientific 
notation? 6,563 = 6563.0 = 6.563 $\times$ 10$^{\rm 3}$. To figure out the 
exponent on the power of ten, we simply count-up the numbers to the left of 
the decimal point, but do not include the left-most number. Here are some
other examples:
\begin{flushleft}
~ ~\\
1,216 = 1216.0 = 1.216 $\times$ 10$^{\rm 3}$\\
~ ~ \\
8,735,000 = 8735000.0 = 8.735000 $\times$ 10$^{\rm 6}$\\
~ ~ \\
1,345,999,123,456 = 1345999123456.0 = 1.345999123456 $\times$ 10$^{\rm 12}$\\
\end{flushleft}

Your turn! Work the following examples:

\begin{flushleft}
\vspace{0.2in}
121 = 121.0 = \\
\vspace{0.2in}
735,000 = \\
\vspace{0.2in}
999,563,982 = \\
\end{flushleft}

Now comes the sometimes confusing issue: writing very small numbers. First, 
lets look at powers of 10, but this time in fractional form. The number 0.1 = 
1/10.  In scientific notation we would write this as 1 $\times$ 10$^{\rm -1}$. 
The negative number in the exponent is the way we write the fraction 1/10.
How about 0.001? We can rewrite 0.001 as 1/10 $\times$ 1/10 $\times$ 1/10 = 
0.001 = 1 $\times$ 10$^{\rm -3}$. Do you see where the exponent comes from?
Starting at the decimal point, we simply count over to the right of the first 
digit that isn't zero to determine the exponent. Here are some examples:

\begin{flushleft}
\vspace{0.2in}
0.121 = 1.21 $\times$ 10$^{\rm -1}$ \\
\vspace{0.2in}
0.000735 = 7.35 $\times$ 10$^{\rm -4}$ \\
\vspace{0.2in}
0.0000099902 = 9.9902 $\times$ 10$^{\rm -6}$ \\
\vspace{0.4in}
{\bf Your turn:}\\
\vspace{0.3in}
0.0121 =    \\
\vspace{0.3in}
0.0000735 =    \\
\vspace{0.3in}
0.0000000999 =    \\
\vspace{0.3in}
$-$0.121 = \\
\end{flushleft}

There is one issue we haven't dealt with, and that is when to write numbers
in scientific notation. It is kind of silly to write the number 23.7 as 
2.37 $\times$ 10$^{\rm 1}$, or 0.5 as 5.0 $\times$ 10$^{\rm -1}$. You use
scientific notation when it is a more compact way to write a number to insure
that its value is quickly and easily communicated to someone else. For example,
if you tell someone
the answer for some measurement is 0.0033 meter, the person receiving that
information has to count over the zeros to figure out what that means.
It is better to say that the measurement was 3.3 $\times$ 10$^{\rm -3}$ meter.
But telling someone the answer is 215 kg, is much easier than saying
2.15 $\times$ 10$^{\rm 2}$ kg.  It is common practice that numbers bigger
than 10,000 or smaller than 0.01 are best written in scientific notation.

How do we multiply and divide two numbers in Scientific Notation? It is
a three step process: 1) multiply (divide) the numbers out front, 2)
add (subtract) the exponents, and 3) reconstruct the number in Scientific
Notation. It is easier to just show some examples:

\begin{flushleft}
(2 $\times$ 10$^{\rm 4}$) $\times$ (3 $\times$ 10$^{\rm 5}$) = (2 $\times$ 3) 
$\times$ 10$^{\rm (4 + 5)}$ = 6 $\times$ 10$^{\rm 9}$\\
\vspace{0.3in}
(2.00 $\times$ 10$^{\rm 4}$) $\times$ (3.15 $\times$ 10$^{\rm 7}$) = (2.00 $\times$ 3.15) 
$\times$ 10$^{\rm (4 + 7)}$ = 6.30 $\times$ 10$^{\rm 11}$\\
\vspace{0.3in}
(2 $\times$ 10$^{\rm 4}$) $\times$ (6 $\times$ 10$^{\rm 5}$) = (2 $\times$ 6) 
$\times$ 10$^{\rm (4 + 5)}$ = 12 $\times$ 10$^{\rm 9}$ = 1.2 $\times$ 10$^{\rm 10}$\\
\vspace{0.3in}
(6 $\times$ 10$^{\rm 4}$) $\div$ (3 $\times$ 10$^{\rm 8}$) = (6 $\div$ 3) 
$\times$ 10$^{\rm (4 - 8)}$ = 2 $\times$ 10$^{\rm -4}$\\
\vspace{0.3in}
(3.0 $\times$ 10$^{\rm 4}$) $\div$ (6.0 $\times$ 10$^{\rm 8}$) = (3.0 $\div$ 6.0) 
$\times$ 10$^{\rm (4 - 8)}$ = 0.5 $\times$ 10$^{\rm -4}$ = 5.0 $\times$ 10$^{\rm -5}$\\
\vspace{0.3in}
Your turn:\\
\vspace{0.3in}
(6 $\times$ 10$^{\rm 3}$) $\times$ (3 $\times$ 10$^{\rm 2}$) =\\
\vspace{0.3in}
(8.0 $\times$ 10$^{\rm 18}$) $\div$ (4.0 $\times$ 10$^{\rm 14}$) =  \\

\end{flushleft}
Note how we rewrite the exponent to handle cases where the number out front
is greater than 10, or less than 1. 
\section{Algebra}

Because this is a freshman laboratory, we do not use high-level mathematics.
But we do sometimes encounter a little basic algebra and we need to briefly
review the main concepts. Algebra deals with equations and ``unknowns''.
Unknowns, or ``variables'', are usually represented as a letter in an equation:
$y$ = 3$x$ + 7. In this equation both ``$x$'' and ``$y$'' are variables. You 
do not know what the value of $y$ is until you assign a value to $x$. For 
example, if $x$ = 2, then $y$ = 13 ($y$ = 3$\times$2 + 7 = 13).
Here are some additional examples:

\begin{flushleft}
y = 5x + 3, if x=1, what is y? Answer: y = 5$\times$1 + 3 = 5 + 3 = 8\\
\vspace{0.3in}
q = 3t + 9, if t=5, what is q? Answer: q = 3$\times$5 + 9 = 15 + 9 = 24\\
\vspace{0.3in}
y = 5x$^{\rm 2}$ + 3, if x=2, what is y? Answer: y = 5$\times$(2$^{\rm 2}$) + 3
= 5$\times$4 + 3 = 20 + 3 = 23\\
\vspace{0.3in}
What is y if x = 6 in this equation: y = 3x + 13 =\\
\end{flushleft}

These problems were probably easy for you, but what happens when you have this 
equation: y = 7x + 14, and you are asked to figure out what x is if y = 21?
Let's do this step by step, first we re-write the equation:
\begin{flushleft}
y = 7x + 14\\
\vspace{0.2in}
We now substitute the value of y (y = 21) into the equation:\\
\vspace{0.2in}
21 = 7x + 14\\
\vspace{0.3in}
Now, if we could get rid of that 14 we could solve this equation! Subtract
14 from both sides of the equation:\\
\vspace{0.2in}
21 $-$ 14 = 7x + 14 $-$ 14 \hspace{0.25in}(this gets rid of that pesky 14!)\\
\vspace{0.2in}
7 = 7x \hspace{0.25in}(divide both sides by 7)\\
\vspace{0.2in}
x = 1\\
\end{flushleft}
Ok, your turn: If you have the equation y = 4x + 16, and y = 8, what is x?
\vspace{0.4in}

We frequently encounter more complicated equations, such as y= 3x$^{\rm 2}$ +
2x $-$ 345, or p$^{\rm 2}$ = a$^{\rm 3}$. There are ways to solve such 
equations, but that is beyond the scope of our introduction. However, you
do need to be able to solve equations like this: y$^{\rm 2}$ = 3x + 3 (if 
you are told what ``x'' is!). Let's do this for x = 11:
\begin{flushleft}
Copy down the equation again:\\
\vspace{0.2in}
y$^{\rm 2}$ = 3x + 3 \\
\vspace{0.2in}
Substitute x = 11:\\
\vspace{0.2in}
y$^{\rm 2}$ = 3$\times$11 + 3 = 33 + 3 = 36\\
\vspace{0.2in}
Take the square root of both sides:\\
\vspace{0.2in}
(y$^{\rm 2}$)$^{\rm 1/2}$ = (36)$^{\rm 1/2}$\\
\vspace{0.2in}
y = 6
\end{flushleft} 

Did that make sense? To get rid of the square of a variable you have to
take the square root: (y$^{\rm 2}$)$^{\rm 1/2}$ = y. So to solve for y$^{\rm 2}$,
we took the square root of both sides of the equation.

\section{Graphing and/or Plotting}

The last subject we want to discuss is graphing data, and the equation
of a line. You probably learned in high school about making graphs. Astronomers
frequently use graphs to plot data. You have probably seen all sorts
of graphs, such as the plot of the performance of the stock market shown
in the next figure (\ref{fig:dji}). A plot like this shows the history of the stock market
versus time. The ``x'' (horizontal) axis represents time, and the ``y'' 
(vertical) axis represents the value of the stock market. Each place on
the curve that shows the performance of the stock market is represented
by two numbers, the date (x axis), and the value of the index (y axis).
For example, on May 10 of 2004, the Dow Jones index stood at 10,000.
\begin{figure}[!ht]
\centering
\leavevmode
%\epsfysize=0.5\textheight
%\epsfbox{COMET/Asteroid460.eps}
\epsfbox{LAB0/big.chart.ps}
\caption{The change in the Dow Jones stock index over one year (from April 
2003 to July 2004).}
\label{fig:dji}
\end{figure}

Plots like this require two data points to represent each point on the curve
or in the plot. For comparing the stock market you need to plot the
value of the stocks versus the date. We call data of this type an ``ordered 
pair''. Each data point requires a value for $x$ (the date) and $y$
(the value of the Dow Jones index). In the next table is
the data for how the temperature changes with altitude near the Earth's
surface. As you climb in altitude the temperature goes down (this is why
high mountains can have snow on them year round, even though they are located
in warm areas). The data in this table is plotted in the next figure.
\clearpage
\begin{center}
Temperature vs. Altitude\\
\begin{tabular}{c|c}
\hline
Altitude & Temperature\\
(feet) & $^{\rm o}$F\\
\hline
0       &  59.0\\
2,000   &  51.9\\
4,000   &  44.7\\
6,000   &  37.6\\
8,000   &  30.5\\
10,000  &  23.3\\
12,000  &  16.2\\
14,000  &  9.1\\
16,000  &  1.9\\
\hline
\end{tabular}
\end{center}

\begin{figure}[!ht]
%\centering
\leavevmode
\epsfysize=0.7\textheight
%\epsfbox{COMET/Asteroid460.eps}
\epsfbox{LAB0/tempvsalt.ps}
\caption{The change in temperature as you climb in altitude with the
data from the preceding table. At sea level (0 ft altitude) the surface
temperature is 59$^{\rm o}$F. As you go higher
in altitude, the temperature goes down.}
\end{figure}

Looking at the plot of temperature versus altitude, we see that a straight 
line can be drawn through
the data points. We can figure out the equation of this straight line and
then predict the temperature at any altitude. In high school you learned
that the equation of a line was y = $m$x + b, where ``$m$'' is the ``slope''
of the line, and ``b'' is the ``y intercept''. The y intercept is simply
where the line crosses the y-axis. In the plot, the y intercept is at 59.0,
so b = 59. So, we can rewrite the equation for this line as y = $m$x + 59.0.
How can we figure out $m$? Simple, pick any other data point and solve the
equation--let's choose the data at 10,000 feet. The temperature (y) is
23.3 at 10,000 feet (= x): 23.3 = 10000m + 59. Subtracting 59 from both
sides shows 23.3 $-$ 59 = 10000x + 59 $-$ 59, or $-$35.7 = 10000m. To find
$m$ we simply divide both sides by 10,000: m = $-$35.7/10000 = $-$0.00357.
In scientific notation, the equation for the temperature vs. altitude is
y = $-$3.57$\times$10$^{\rm -3}$x + 59.0. Why is the slope negative? What
is happening here? As you go up in altitude, the temperature goes down.
Increasing the altitude (x) decreases the temperature (y). Thus, the slope
has to be negative.


\begin{flushleft}
Using the equation for temperature versus altitude just derived, what is the 
temperature at 20,000 feet?
\end{flushleft}

Ok, your turn. On the blank sheet of graph paper in Figure \ref{fig:graph} plot
the equation y = 2x + 2 for x = 1, 2, 3, and x = $-$1, $-$2, and $-$3. What 
is the y intercept of this line? What is its slope?

\begin{figure}[!ht]
%\centering
\leavevmode
\epsfysize=0.7\textheight
%\epsfbox{COMET/Asteroid460.eps}
\epsfbox{LAB0/graph.ps}
\caption{Graph paper for plotting the equation y = 2x + 2.}
\label{fig:graph}
\end{figure}

While straight lines and perfect data show up in science from time to time,
it is actually quite rare for $real$ data to fit perfectly on top of a line. One
reason for this is that all measurements have error. So, even though
there might be a perfect relationship between x and y, the noise of the
measurements introduces small deviations from the line. In other cases,
the data are $approximated$ by a line. This is sometimes called a best-fit
relationship for the data. An example of a plot with real data is shown in 
Figure \ref{fig:mvporb}. In this case, the data suggest that there is a general trend between
the absolute magnitude (M$_{\rm V}$) and the Orbital Period
in certain types of binary stars. But some other factor plays a role
in determining the final relationship, so some stars do not fit very well,
and hence their absolute magnitudes cannot be estimated very well from
their orbital periods (the vertical bars associated with each data point
are error bars, and represent the measurement error).
\begin{figure}[!ht]
%\centering
\leavevmode
\epsfysize=0.7\textheight
%\epsfbox{COMET/Asteroid460.eps}
\epsfbox{LAB0/mvporbforlab.eps}
\caption{The relationship between absolute visual magnitude (M$_{\rm V}$)
and Orbital Period for cataclysmic variable binary stars.}
\label{fig:mvporb}
\end{figure}
