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BTW... I seem to recall reading somewhere that if you replace a bulb with an LED, your also need to add a resistor of the appropriate "ohmage"... Is that correct?
Yes, and if you're planning on using a simple circuit where current is limited with a resistor you need to perform the calculation based on the battery voltage. The resistor is wired in series with the LED and is sized to limit the current to the LED's rated forward current. Using this LED as an example, see the Vf (2nd line) in the Electrical Characteristics table. You get both the Forward Voltage (2v typical for Red) and Forward Current (20ma).
Assuming a 12 volt battery, and Vf=2, we need to drop 12-2=10v across the resistor. To limit the current to 20ma, we need R = E / I = 10v / 0.02a = 500 ohms. The power dissipated by the resistor is P=I2R = (0.02)2* 500 = 0.2 watts. Since there's no sense in spending money for exactly this value in this application, you go with a common 510 ohms at 1/4 watt.
Note that it's perfectly save to run this "12v launcher" on 6v--the LED will just be dimmer, and in fact if you happen to run it on 24v you won't let the magic smoke out of the LED since it's maximum forward current is 50ma. However the resistor is going to get hot and could smoke as it's dissipating close to 1 watt. This assumes you're ok with 44ma of continuity current.
Given that the LED is $0.21 and the resistor is $0.17 (ok, multiply by 10 or 20 if you buy at You Do It) I can't imagine playing with incandescent bulbs...