Hi del,
I'm out of my comfort zone here; but as a confirmed tinkerer would be up for the below.... think if you grab a beer & go through it slowly it may turn up what you're after.
Until, of course, Ash simply tells you the 'number'...
Sorry can't be of more help,
Simon
http://www.ngineering.com/led_calculators.htm
The load resistor for a LED is influenced by the forward volt drop (FVD)of the LED and this varies for the different colours.For the first RED LEDs this was 1.8 volts but its about 3.5 volts for a WHITE one. So what you do is subtract the FVD from the supply voltage and then use that voltage figure in OHMS Law to calculate the required series resistance. So for a WHITE LED and 12 v supply and say a 50 milliamp desired current (0.050 A) you get 12 - 3.5 v = 8.5v So, applying Ohms law R=V/I ... the resistor value comes out as 8.5/0.05 =170 Ohms.
However,it's not as simple as this,because with with modern LED 'arrays' they have multiple LED's embedded into the same device and connected in a series and parallel arrangement to optimize minimum power drop in the series resistor and also some even have integrated 'constant current' circuitry in addition to the LEDs to give a constant brightness irrespective of any supply voltage fluctuations (if you think about it a 12 volt bike battery can reach 14v)
Best thing may be to just get an wirewound 'pot' (variable resistor) of say 100 Ohms and connect this in series with the LED and adjust it until you get the desired brightness then measure its value with a DVM and replace it with a fixed resistor. If you sent me a LED bulb I could easily do this for you as I have all of the gear to do it quickly.
Sorry about the long winded answer .. my kids/family have no interest whatsoever in anything like this so I have to pass the knowledge onto someone
