I need to run as many LEDs as possible using a generator putting out
110 volts AC.  The LEDs are rated at 2 volts 30 mA.  I will use a full
wave bridge to get the most light as possible.  How many LEDs should I
put in series and what resistance should I add?  They will be encased
in a 2 inch plastic tube.  Do I need to wory about heat buildup?  I'd
like to run 3 circuits of lights in the tube.  These will be mounted
where I cannot get to them so I need as much reliability as possible. 
What kind of protection filtering should I put in the circuit since it
will be running on a generator?  Is ther anything else I should know?

j4bbbbb...

LEDs require DC voltage, so you'll have to convert your AC
to DC with some form of power supply. This earlier, related
question and answer may also be helpful:
http://answers.google.com/answers/main?cmd=threadview&id=208170

That is what the full wave bridge is for.  If I just run it on ac it
will fire on half the cycle.  I have a friend who has done this.  I am
more concerned with how many LEDs to put in the circuit.  My friend
says I should run about 75 to be safe.  This would cover the peak
voltages of 150 volts.  I'm also concerned about heat buildup since
they will be contained in a plastic tube.

you may have a bigger problem though
running 110 volts DC can be more difficult to handle - insulation,
switches, may be different, check specs, and be aware that personal
contact will zap you a LOT worse than AC, ikely to be fatal.

And if one led dies, they all go out, or try to arc over and burn
severely.
AC / DC have very different characteristics

Howabout a combo serial/parallel circut on a 12 volt DC line (many
generators supply this too)

1 pos 12V   1 neg12V 
and as many strips of 6x2v leds as needed

A lot easier to build & fix

What you want to do is to have a rc filter after the bridge.  That
will smooth the voltage waveform and allow you to maximize the light
output from the LEDs.

Something to keep in mind is that LEDs being diodes have a very
non-linear I-V curve.  Once you start conducting, not too much more
voltage with give a very large increase in current.  Your goal is to
make the LED string conduct all the time right in the middle of the
safe operating zone.

Secondly, the threshold voltage of each LED will be slightly different
from it's spec'd number.  However, with that many in series, it will
average out to be very close to what it should be.  More important
however is the fact that there is a temperature coefficient on that,
so the threshold voltage will be high when the diodes are cold, but
lower when they are hot - exactly backwards from what you want.

None of this should be too big a problem if you are careful.  The
reason for the RC is twofold.  First, the R provides the stability you
want by creating negative feedback, the higher the current flowing,
the lower the voltage across the diodes.  They will therefore find a
stable operating point for a given bridge voltage.  Second, the filter
smoothes out the voltage fluctuations in the cycle, so their is less
of a variation in the operating point, and therefore less stress on
the diodes.

First, you need to size the resistor so that the rms average voltage
from the output of the bridge equals the voltage across all the diodes
that you want it to run at, + the product of the current you want
times the resistor value.  Or,

Vrms = N*Vd + Id * R, where N is the number of diodes in series, Id is
the total current you want flowing and R is the resistor value in
ohms.

Next, you need to size the capacitor for the RC filter.  If you are
using 60Hz as the input source, the time constant should be somewhere
in the neighborhood of 30Hz (it is a low pass filter, so that means
that by 30Hz, only 50% of the original variation is going to pass, and
it goes down by 3dB every octave.)  To do that just take the inverse
of 30 to get the time constant in seconds, or .033sec.  That is equal
to RC, so divide the time constant by the resistor you used and that
gives the value of the capacitor to use.

 All that is left is to make sure that peak current the diodes carry
is within their safe operating zone, and that they never shut off -
mainly because that will look bad.  Just recalculate the governing
equation using the maximum voltage they will see (That is the average
plus the ripple of ~25% of the difference between the average and the
peak of the source.) makes sure everything is okay.

Hope that helps - I was bored at work and figured this was halfway
interesting.

Can I buy a RC filter to put in the circuit?  What size should it be
if I am trying to run as many LEDs as possible so that I would not be
using a resistor in the circuit..  60 would give me a drop of 120v,
but do I need to worry about the peak voltages or just the RMS?  Or
does the RC filter handle this problem?  I probably want to err toward
the safe side.

The LEDs need 50 mA.  Does this mean I should have a 3k resistor in
the circuit addition to the 75 LEDs?  I was a physics major and it's
been a while since my classes in electricity.

You say the LEDs need 50mA.  What is the manf and model #, 50mA sounds
very high as the needed current.  It sounds more like an upper limit
to me.  (2V * .05 = .1W, since the efficiency of these things in
absolute terms is low, .1W sounds like it is near the case dissipation
limit.)

How did you get to 75 LEDs?  It seems you are missing the whole point
in what I said, since the point of the RC is to reduce the focus on
the peak voltage and put it on the rms avg.

Say you knew the voltage of the LEDs was exactly 2V, lit up at 20mA,
burnt up at 50mA and was just perfect at 30mA.  The solved equation
for 50 LEDs gives a R of 667 Ohms, with a parallel capacitor of
anything over 60 microfarads.  The capacitor should be rated for 200V
or more.

If you used 75-2V LEDs, giving 150V, and had .05A going through 3k,
you would need at average voltage of 150 + .05 * 3k = 300V.  The idea
is to have the total voltage of the load equal the total voltage of
the source, at the operating point you want.  What you did was to have
the two parts of the load EACH equal some voltage.  I guess the good
part is that would hurt you, but on the other hand, it wouldn't do
anything useful either.

I don't know what I was thinking about the resistor.  I went to the
spec sheet for the LED.  You are right. It's #ET6-5TS630-30.....30Ma -
1.84v to 2.4v, typical 1.94v.    Here is the circuit as I understand
it:  I can use 60 LEDs in series with a capacitor in parallel of at
least 60 microfarads rated 200v or more and a full wave bridge
rectifier and no resistor.  Does this sound right?  Thank you for
following up on this.

lol, this is more than I bargained for timewise, but there you go.  

I would need to know the manufacturer in order for the part number to
mean anything to me.

The problem with deleting the resistor is twofold.  First, in order
for the RC to do anything, you need both the R and the C.  If you
delete the R, you are just using the line resistance, and the bridge
resistance to work with.  I seriously doubt that that add up to more
than a few ohms.  So, there will be no filtering of the AC ripple. 
Second, without the filter you have an input voltage that will swing
from 0 to 120*1.414, or ~170V.  At 170 volts you are going to have
nearly 3 volts per LED.  Unless they are able to handle that, you are
going to have fireworks.

Speaking of fireworks, I didn't think about it before, make sure the
resistor is sized for the dissipation level with margin.  Also, during
the charge up of the capacitor, the resistor with dissipate a huge
slug of heat (1/2*C*V^2), so you need to be aware of that.  A 3 watt
wire wound would be safe, I believe.  Is this a home project?  If not,
have you considered just having an engineer design up a real circuit
for you.  The correct way to do this is with a tiny pwm circuit to
actively control the current though the diodes.  The parts would
probably cost a total of $10 in quantity.  Just be careful any time
you are playing around with rectified AC.  Good luck.

Thank you, leoj.  You have been incredibly helpful.  This is a home
project.  I'm building a large heart that will be in a 60 ft wooden
man for the Burning Man festival as a special suprise.  I will use the
circuit for 50 LEDs that you described.  Thank you very much again. 
My knowledge of EE has gotten fuzzy and was not that great to begin
with.  I'm glad that I was at least good for a laugh.