OK, let's see. It is actually really simple, once you understand the principals. Firstly, LEDs have an almost constant voltage drop across them. Usually it is around 1.6V for a red LED and 1.8V for a green LED, something else for an amber LED and a lot more for a white LED (like 2.2V). The different voltages is due to the materials used to make the colour. Whne you buy an LED, it will specify the "forward voltage", which I have called the voltage drop. It is called "forward", because when you connect an LED in reverse it will not work. There is also a limiting reverse voltage, which can be as low as 5V. When you apply more than the rated voltage, the LED will sustain damage. If you go below the forward voltage, then the LED turns off.

LED brightness is determined by mainly 3 factors, and is rated in mcd (millicandessa). A really bright LED will be a few thousand mcd, the ones you find on almost every consumer product is a few hundred mcd. The brightness is rated at a certain current, and is measured directly in front of the LED. The 3 factors are 1) the material it is made of, 2) the current through the LED, and 3) the viewing angle (often brightness is increased by simply narrowing the viewing angle). Usually a clear lens means is assosiated with a narrow viewing angle, a cloudy lens gives a diffused light. But virtually all ultra bright LEDs are narrog angle, which is around 15 degrees off center.

So now we know there is a "fixed" forward voltage drop, and more current = more brightness. But there is also a limit in the forward current, often around 40mA or so (brightness is often specified at 25mA for comparison). So supposed the LEDs we want to use have a forward voltage drop of 1.6V and 40mA maximum current. If we have 10 in series, then we need 10 x 1.6V = 16V just to turn them on. That will not work, we have only a 12V battery. Let's say the battery will never go below 10V, so we can have 10V / 1.6V = 6.25 LEDs in series. Obviously that means 6 LEDs. So 6 x 1.6V = 9.6V required just to turn them on. The maximum voltage from the battery would be about 14.2V, so we need a resistor to drop the voltage to 14.2V - 9.6V = 4.6V. With a resistor you will have the maximum current at the maximum voltage drop, so to calculate the resistor value it is simply 4.6V / 40mA = 115 Ohm (40mA = 0.04A). The nearest value in the so called E12 range (12 values per decade) is 120 Ohm, so let's pick 120 Ohm. As a check, let's calculate the current when the battery supplies only 12V i.e. with the engine off: (12V - 9.6V) / 120 Ohm = 20mA which would be plenty bright enough. It is only when you get to 15mA to 10mA that they get dim. Now you need to do one more calculation for the resistor, which is the power rating. Most poer will be dissipated when you have maximum voltage and maximum current, so that is (14.2V - 9.6V) * (14.2V - 9.6V) / 120 Ohm = 176 mW. Hobby shops will sell stock 250mW, 400mW, 1W, and higher power resistors. So a 250mW resistor will be fine.

So now we have a string of 6 LEDs and a resistor, all to be connected in series. The LED has an Anode (+) and Cathode (-). The LED body has a round body with a flat spot next to the Cathode. If you view the string of LEDs and resistor as if it was one LED with one Anode and one Cathode, then the Anode goes to the battery positive and the cathode goes to the battery negative (hope this makes sense).

So now we have a string of 6 LEDs and one resistor, but want to more. Just make another string of 6 LEDs and one resistor and connect the 2 strings in parallel. Someone might ask if you can't just add LEDs parallel to each of the LEDs in the one string, but that will not work because the voltage drop of each LED is slightly different and you end up with only the ones with the lowest voltage drop turning on and the others not. (Sorry for the complex explanations, just want to be complete instead of having to answer lots of individuel questions.)

As for the indicators, the standard electromechanical flasher unit needs a high current load, else it will flash at about 3 times the normal rate. The way that an auto electrician will fix it, is usually to fit a resistor parallel to the LED clusters so that you draw more current. This resistor will also heat up, and it is likely you will need like a 2 Ohm 50W resistor then somehow deal with the generated heat. I >think< that a so called electronic flasher unit does not need this load, otherwise you can look for a flasher unit kit at your local electronics hobby shop and build it yourself (but would recommend an electronic flasher unit if it will work, because it is rugged).

Now regarding the comments about heat generated by LEDs, it is minimal. A string of 6 LEDs drawing (14.2V - 9.6V) / 120 Ohm = 38mA max will dissipate a maximum of 14.2V x 38mA = 540mW. So 60 LEDs will dissipate about 5.4W of heat over a surface of about 2" by 2". The lens on your car is happy with bulbs dissipating 21W over a much smaller surface, you will not have any problems with heat.

I would recommend finding out about a product called Veroboard and soldering the components onto that. It has copper strips and lots of holes on it, and you use a small drill bit to cut the strips. Just the drill bit in your hand, no drill required. This will also allow a nice, even layout of LEDs. I also recommend that you pot the circuit in epoxy, to protect it and to prevent any electrical short circuits. It is not nice to see the damage from a short circuit lead acid battery!

Cost wise, LEDs are much more expensive than bulbs to get the same light output. But if properly designed they will never fail and they also turn on much faster than a bulb so the driver behind you will react earlier when your brake lights come on... They also use less power, so your battery lasts longer if you have to leave your park lights on in an emergency.

All my calculations above are based on phantom values for forward voltage drop and operating current specifications, you must use the correct values for the LEDs you will be using.

For further info on LEDs, have a look here: http://en.wikipedia.org/wiki/LED