Or is the best method for figuring out an LEDs voltage drop what Duane explained above?
You can't reliably 'figure out' or 'predict' an LEDs voltage drop and you really don't usually have to since this value is a characteristic, not a rating.
If you are thinking along the lines of supplying 'voltage' to an LED you are doomed to failure. This is because you do NOT apply some specific voltage voltage to an LED to get it to work. You apply some arbitrary voltage that is greater than the expected forward voltage of a typical LED of the type that you propose to use. You then use a series resistor to limit the current to a value that is less than the forward current rating of your LED.
It is in the calculation of the series resistance that you use the forward voltage characteristic. Here is an example that I did for a previous post.
To deal with an LED you start with the forward current that you need, typically about half it's maximum rated value (lets use 20 mA).
Next you estimate what the forward voltage drop will be with that current flowing through the LED. If you have a datasheet for the LED you may be able to get a fairly accurate value, otherwise you take a guess based upon your experience or the experience of others (I usually use 1.7v for a red LED).
Next you pick out a supply voltage which must be higher than the voltage you just determined (I'll use 5v).
Now you can use Ohm's law to determine the required resistance. The voltage across the resistor will be the difference between the supply voltage you decided to use and the voltage that you guessed would be across the LED (5v - 1.7v = 3.3v). The current through the resistor will be the same as the current through the LED (20mA). Ohm's law for the resistor says that R = V/I (R = 3.3/0.020 = 165 ohms).
You then pick the closest value resistor that you happen to have and stick that in your circuit. Most likely the current won't be exactly what you desired and the LED voltage won't be what you guessed would be there but you won't see any smoke either and you will see light from the LED.
Don