See note 4 of page 2 of this data sheet: http://www.avagotech.com/docs/AV02-0373EN
It says that "θ1/2 is the off-axis angle where the luminous intensity is half the on-axis intensity. The viewing angle is given as 2θ1/2, which would be total angle, θ1/2 being half that, or 7.5° θ1/2 for a 15° stated viewing angle (2θ1/2).
Reverse voltage is the max. reverse voltage that they guarantee current flow not to exceed a stated safe level - beyond that reverse voltage, you could be damaging the LED. They say 10uA max. reverse current (at 5 volts reverse voltage). Normally you wouldn't put a reverse votlage across it, but they're saying you can put up to 5 volts in the reverse bias direction and not risk damaging it. Beyond that, the risk goes up and they offer no guarantees.
Continuous forward current is the max. continuous current that they are designed for. They fail to tell you over how many hours and to what percentage of intensity it would drop to given that continuous forward current rating. With the cut throat competition among the manufacturers, if they are a manufacturer of unkown reputation (i.e., typical ebay sellers), I wouldn'g go much above 1/2 their stated current if you don't want failure or dimming over time. Even with a reputable manufacturer, if the intensity level is important, you stay below the max. rating. If really critical, you find out how the ratings are determined statistically and design accordingly.
For stacking LED's and sizing resistors, below is a past post of mine. It is poor design to put a fixed voltage across an LED or stack of LED's based on a nominal forward voltage spec. because the tolerance variations around that and small power supply variations will create *huge* variations in the current (and thus brightness and life). *Always* design in current regulation or resistors to reduce the current sensitivity to supply voltage variations.
When you read of LED's failing, either poor or no current regulation was designed in, or the manufacturer pushed the current to exceed a reasonable rating so that they could advertise some high brightness spec. so people who didn't know better would purchase their product vs. the competition who may be more honest in their ratings and/or more conservative in the design. No skin off their teeth when the LED's fail after a few months - an outrageously positive ebay rating was given long before that.
You always need resistors with LED's to limit current when a constant voltage source is powering them (like when replacing regular light bulbs). Some LED's come with built in resistors, but their packaging or spec. sheet will say that if that's the case. Assume it has no resistor built in if it doesn't say.
Decide on how many milliamps you want going thru the LED at full voltage. (mcd = millicandellas is the measure of brightness at a given current).
R = (V(b) - V(LED)) ÷ C
R is in ohms
V(b) is your car's power system voltage (assume 13.8 with it running)
V(LED) is the voltage drop across the LED (assume 1.2 unless the spec. sheet tells you)
C is the current you want in amps. 20 mA = 0.02 amps
So for example: For V(b) = 13.8 volts, V(LED) = 1.2 volts, and desired current is 20 mA (0.020 amps), the resistor would be 630 ohms - find the nearest value to that - 620 is a standard value I think.
Lower resistance will give you more current, more brightness, shorter LED life.
Some of these bulb locations may have lower source voltages, so for a given location, measure the voltage with the brightness turned all the way up with the original bulb installed, and use that number for your V(b).
For multiple LED's in series, use their total voltage for V(LED) in the formula. For example, if one LED voltage drop is 1.2 volts, and you are putting 2 LED's in series, use 2.4 volts for V(LED) in the formula. For 3 LED's, put 3.6 volts in the formula. Do not stack so many LED's that their total voltage gets more than 4 or 5 volts from the source voltage. Because the source voltage varies some, you need to have a resitor in there so that the current (and the LED brightness) is not going to be super-sensitive to those variations. If you use a voltage regulator, you still need a resistor, but you can get a little closer to the source voltage (in that case, the output voltage of the regulator) because the regulator will give less variation in its output voltage than the vehicle's system.
To determine the wattage needed in the resistor, use the formula:
P = C^2 x R
P = power in watts
C = current in amps
R = actual resistor value in ohms
So for our above example of 20 mA (milliamps) and 620 ohms, the resistor would dissipate 0.248 watts. That's right at 1/4W, but you would not use a 1/4W - you want to go up one size which would be 1/2W. Always try to use at least 50% safety margin on the wattage (for one thing, the wattage rating gets de-rated at higher ambient temperatures). You need some safety factor on that so your resistor lasts forever. If your wattage calculation came out to, say, 0.15 watts, you could safely get by with 1/4W. Any more than that, move up to 1/2W. If wattage comes out more than 0.33 or so, use more than 1/2W (or parallel two 1/2 watters of twice the resistance value each for the same net resistance value).