Now I get it - DIY twistlocks.... (Good - the 5V & 147R is closed.)
A main issue is how they dim compared to original and other bulbs. But you can do a sample - maybe any LED just to get an idea of any variance.
The question is how the voltage varies with the dimmer.
Ordinary bulbs are much like a resistor - current proportional to voltage.
LEDs are more like a zenor with a series resistor - ie, about 3V plus the iR voltage across is "equivalent" resistor.
Whilst LEDs may still glow with an external 5k resistor, it may still have 3V or 2V across it. Whereas bulbs with 5kR will be low-i = low current and hence a low voltage (not the LEDs 2V or 3V "zenor" plus a similar small voltage).
You might find that current-wise, the LEDs "dim" faster and then suddenly extinguish whereas the bulbs still glow.
But that depends on the LED/bulb balance.....
And I say dimming "current-wise" because bulb and LED intensity may vary with current, but to differing curves etc. (Alas all I remember is that a tungsten bulb is brighter than halogen at lower voltages - the halogen's illumination is a bit like zenor offset - ie, nearly 0 at say 6V, but full brightness at 13.8V etc.)
The gist of the above is - suck it and see. It's too complex to model... (At least with my models - not that I have them - but they often tend to include more than others...)
As to the resistor - still the same - ie, (Vs - Vled) for the LED current.
EG (pedantic):
3.2v - 3.6v, 20mA
Assume max 20mA @ 3.6V
Assume Vs=14.4V (typ alternator; no fuse drops etc)....
Make it 14.6V => 14.6-3.6 = 11V @ 20mA (11 x .02 = .22W = 1/4W)
R = v/i = 11/.02 = 550R = 560 Ohms 1/4W (standard preferred value).
Since .22W is close to .25W (we normally try for say 30% margins in design), a 1/2W resistor would be nicer. (They are probably the same price - maybe cheaper - and 1/2W resistors were traditionally beter quality, and are a bit more robust (physically strong).)
The resistor leads should be as short as possible - or rather, NOT long - to prevent breakage through vibration etc.
[Digression - GoTo next para: I remember when they used to recommend looping resistor leads to act as a spring to absorb vibration and prevent breakage. That was another classic case of some
expert's "logic" being wrong (again!). Silly people like me ignored the advice. We didn't have the increased failure rate that others had. Nor did we have to replace all those resistors once the experts experienced how wrong they were.]
But the resistor may need clearance around the for heat dissipation (less so if using 1/2W due to greater surface area).
(Also short leads usually means more heat loss through PCB track than from longer lead heat dissipation - ie metal tracks beat air)
FYI - 560R @ 20mA = 11.2V. Hence if Vs exceeds (3.6+11.2=) 14.8V, the 20mA is exceeded.
Some manufacturers refer to non-warranty or failure above 20mA. Whether this means it blows at 21mA or merely has a much shorter lifespan....
Welcome to the joys of design!