Yeah, that's the diagram I started out with, but modified it as I did testing. With only 1 lux everything got really hot really quick. So instead of one luxeon I used two and found that everything still heated up but it was better. So then I threw in a big resister in with the 2 LEDs to dissipate more power and that brought down the temps to a resonable value.

Forward voltage for 1w lux is 3.4V and the 3w is 3.6V. I've done the calcs for the 1w so the problem I run into with wiring up for 3 watt LEDs is the heat. I've got twice the power going through at 700mA so how do I compensate for that? Yeah, I can run 2 LEDs, but I don't know if a big resister in series is going to be enough to keep the temps down. I'm trying to find a different circuit or IC that can handle the current and keep the temps down. I know high end flashlights use pulse modulation but that's too advanced for me.

Alternatively, a company called Cree has come out with a new LED that blows away the 5w luxeon and it only pulls around 300mA. I might have to do some experimenting with those, but they're not cheap at $15 each.

The LM317 is rated for 1 watt without a heatsink and something like 3-4 watts with a heatsink. Use the formula R = 1.23 / I to calculate the resistor you need. R = 1.23 / 0.70 = 1.75 ohm resistor; find whatever is closest to that value. (The 0.70 is the 700 mA 3-watt LED.) Then use P = 1.23 x 1.23 / R to calculate the watts you'll have to dissipate. P = 1.23 x 1.23 / 1.75 = 0.86 watts. As long as you use a resistor with a rating greater than 1.5 watts (to be safe), you will be fine.

The above was used just to calculate the resistor value. Now you have to calculate the power dissipation of the rest of the system. Running two 3.6V LEDs; Vx = max forward voltage with alternator running - LED voltage drop - 1.23 or Vx = 15V - (3.6V + 3.6V) - 1.23 = 6.57 volts. Px = Vx x I or 6.57 x 0.70 = 4.6 watts. Yikes, you're right, I'm not sure how you safely dissipate nearly 5 watts of heat.

I'm sure there is an easy do-it-yourself solution to this problem, but this is certainly one answer to the problem:

This easy to use driver is ideal for powering 3 or 5 watt Luxeon LEDs in automotive applications using a 12VDC power source.

The LuxDriveÖ PowerPuck 700mA LED Drive Module is a true constant current driver designed to drive High-Power LEDs, or strings of 5mm LEDs, from 5-32VDC with efficiency and stability. The different types and numbers of LEDs that can be driven depend on driver current output and circuit configuration. (See the detailed Datasheet for more information).

The 2008 uses filtered switching technologies combined with proprietary drive and sense electronics to provide constant current to LED arrays. The PowerPuck's high efficiency and low cost makes it an excellent choice for automotive and marine applications, portable lighting and solar lighting.

The PowerPuck is an 0.5" x 1.5" dia. encapsulated puck, is resistant to harsh environments and moisture and is supplied with 6", 24AWG colored leads for easy wiring and an aluminium mounting bracket.

Low Voltage DC input power up to 32V
Ideal for Automotive/Marine use
Extremely small form factor* measuring just 1.5"dia x 0.5" tall
Supplied with 24 AWG 6" colored leads for simple connection
Output short circuit protection up to 15 seconds
Output open circuit protection
Designed to power 3 or 5 watt Luxeon LEDs
Will last as long as the LEDs it powers
Fully encapsulated construction
1.5" Diameter x 0.5" Thick

LINK. Price: $17.95

Link for the ones IcySyrup has

http://www.plasmaled.com/high_power_leds.htm