Bulbs: How many watts vs Candle Power rating

bluesman31

Senior Member
I believe there is no direct conversion, but does anyone know what wattage the following bulbs draw at 6 volts?

Candlepower Ratings

32 CP headlight bulb
50 CP headlight bulb
21 CP stop lamp

How many watts each at 6V?
 
I would check out www.donsbulbs.com for the watt ratings for whichever bulb you are using. The database allows you to search for bulbs according to their CP rating.

I don't know if there is a standard wattage for a specific candle power rating though.
 
P = I X E
P(watts) = I (amps/current) X E (Volts)

Example:
E (6 volts) X I (5 amps) = P (30 watts)
 
With a ohm meter measure the resistance of each filament of the bulb.
I = E (volts) over R ( resistance/ohms)
Headlight bulb.
I = E (6 volts) divided by R ( 1.3 ohms) = 4.6 amps.
P= I (4.6 amps) X E ( 6 volts) = 27.6 watts ( 1 filament of a headlight bulb)
 
My GE catalog shows 32/32 = 4.66/4.49 design amps, 50/32 = 7.2/4.66 amps, 21 = 2.69 amps.

Bob
 
I replaced my original GE 1000 bulbs (32/32) with the 50/32 bulbs and not my ammeter reads a discharge when driving with the headlights on. It reads a charge of around 10 AMPs when the lights are turned off. You might want to consider that if you use the higher CP bulbs.
 
With a ohm meter measure the resistance of each filament of the bulb.
I = E (volts) over R ( resistance/ohms)
Headlight bulb.
I = E (6 volts) divided by R ( 1.3 ohms) = 4.6 amps.
P= I (4.6 amps) X E ( 6 volts) = 27.6 watts ( 1 filament of a headlight bulb)


While I'm not 100% sure I am fairly certain the resistance of a filament increases greatly as the filament heats. Cold resistance will not tell you much. This is why incandescent lamps almost always blow when you first turn them on and will last a really long time if you don't cycle them much.
 
I think the short answer is that there is no direct conversion between cp and watts without specifying the construction and operating parameters of the bulb.

The wattage of a blub refers to the power consumed and – for simple resistive circuits (i.e., no inductive or capacitive components) – can be calculated by any of the means posted above (i.e., current in amps squared times resistance in ohms or voltage in volts times resistance in ohms, etc.).

The candlepower of a blub refers to the luminous intensity emitted by the bulb and without getting into filament construction, blub envelope material, etc., the luminous intensity depends on the wavelength of the emitted light, the standard against which it is measured, the use of built-in reflectors (reflective coating integral to the light bulb envelope) and other parameters. In trying to avoid some “Beam me up, Scotty” responses to this reply, the simple example is that the candlepower of a bulb that draws (for example) 2 watts could be the same as the candlepower of a bulb that draws 3 watts, or - to state the converse - two differently constructed bulbs that each draw 3 watts, could have different radiant power, that is, different candlepower.

Now, for the “Beam me up, Scotty” (technically “Scotty, beam us up”) answer: As you will recall from kindergarten, candlepower measures luminous intensity (in candelas, sometimes referred to as ‘candles’). Specifically, one candela is one lumen per steradian (1 cp = 1 lumen/steradian). (I don’t think they covered ‘lumens’ until first grade, and I’m certain they didn’t cover solid geometry and such items as steradians until second grade, but in the event, a lumen is a unit of luminous flux, which is radiant power tempered by the response of the human eye.) As we all know, the average Mark 1 Mod 0 human eye has a peak response to bright light at a wavelength of about 555 nanometers, roughly the wavelength of green light (deep red is about 700 nm and violet is about 400 nm). At this frequency (wavelength being related to frequency by the speed of light) one watt of radiant power equal 683 lumens, whereas one watt of light at other frequencies – colors; i.e., of greater or lesser wavelength - will equal fewer lumens inasmuch as the light is being emitted at a wavelength that the human eye is less sensitive to.

A steradian is a unit of solid angle and as everyone knows, there are 4pi steradians in a sphere. For an isotropic light source (one that radiates equally in all directions, a physical impossibility if you consider the construction of the typical light bulb) a 1 candlepower blub would produce 1 cp x 4pi steradians = 1 lumen/steradian x 4pi steradians = 4pi lumens (about 12.6 +/- lumens) of radiant power. Using the ‘1 watt of radiant power at 555 nanometers = 683 lumens’ from above, our 1 cp bulb producing 4pi lumens = 12.6 lumens/683 lumens-per-watt = 0.0184 watts of radiant power.

In the case of a bulb producing “white” light (that is, light composed of all, more or less, frequencies - or wavelengths – of light) a 1 cp bulb would “equal” only about 0.014 watts, depending on the actual “whiteness” (sometimes referred to as the ‘temperature’ frequently in degrees Kelvin) since – as stated above - the response of the average human eye is less for light produced at wavelengths other than 555 nanometers.

Scotty? Hello? Scotty, are you there…?

And, "yes" it IS a slow day out here on the west coast.
 
Very illuminating! So now you know. Let nobody say that Napa Skip hides his lumens under a bushel basket!
 
With a ohm meter measure the resistance of each filament of the bulb.
I = E (volts) over R ( resistance/ohms)
Headlight bulb.
I = E (6 volts) divided by R ( 1.3 ohms) = 4.6 amps.
P= I (4.6 amps) X E ( 6 volts) = 27.6 watts ( 1 filament of a headlight bulb)

After the filament heats up, the resistance goes up, and you will have much less current draw.

Checking lamps with an ohmmeter is not valid.
 
With apologies to Mr. Ohms and the electronic engineers on this Board, I just went out and looked at my Amp Meter.
I have 32/50 Incandescant bulbs forward and LED's aft.
My brake LED's do not affect the amp meter much and are brighter than the taillites so I think we can discount the taillites.
My ampmeter, engine off, for the two headlights gave me -12 amps at 32 CP
and - 14 amps at 50 cp for TWO headlights.
Ohms law is
Volts times Amps equals Watts
32 cp 6 volts X 6 Amps = 36 Watts
50 CP 6 volts X 7 Amps = 42 Watts

Probably not as scientific as counting Candelas, but I hope this helps you.
Probably delivering 6 volts from my Optima at rest, but I was not about to start checking voltage.

Good Luck !
 
That is good info Bruce, I am trying to figure how much I can get away with (up grade the bulbs) with my stock generator.

I am new to model A's so based on your calculations, if you drive the car only at night with your generator set at normal settings you will eventually kill the battery? How long will it last running on low beams?
 
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I have a chart of all the bulbs and candle power for both 6V and 12V for the Model A. The chart does not have Amps but can add this at a later date. If anyone wants a listing of the bulbs and candle power send me their email address to jrelliott42@gmail .com and will send it to you.
 
"Very illuminating! So now you know..." (post #14 to this thread)

Groooan! (Although in fairness, probably not as big a groan as resulted from post #12...)
 
Suggestion

If you want to keep the generator and get maximum brightness from your headlights you should do 2 things. First install a voltage regulator instead of the cut-out. Then you do not have to mess with the third brush. Second install the new halogen bulbs, the ones that can be run on a generator.

In addition you should also have good reflectors.

Bob
 
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