Coverting from Lumens, Candlepower, Watts, Etc...

Is it possible to convert a lights output candlepower or lumens into actual total output wattage (solar energy)?

I'm trying to determine just how much actual light energy in watts is being released by these 1, 2, 3, 5, and 10 million candlepower spotlights (or similar devices) on the market. I'm aiming for a pump device capable of driving a powerful non-coherent emission through a system of focused lens's. I'd like to do a project for college which aims towards battery derived energy, a wood-burning demonstration, and then demonstrate how pure daylight emission from the sun has a much higher potential when utilized properly (in theory, and relative to what is obtained easily for "free" in a sense).

My goal is high-output light focused to an extremely small point which then slides across the wood via a channel once it's in focus to demonstrate it's potential to smoke an object.

Thanks guys.

No, those are all incommensurate units. Besides, you shouldn't take the manufacturers' claims about lumens, candlepower, etc. at face value. Often these are absurdly exaggerated. Just use the power rating of the lamp. Most spotlights are 50 or 100 watts.

asdalton said:

No, those are all incommensurate units. Besides, you shouldn't take the manufacturers' claims about lumens, candlepower, etc. at face value. Often these are absurdly exaggerated. Just use the power rating of the lamp. Most spotlights are 50 or 100 watts.

Click to expand...

But the power rating of the lamp is just what it is rated at, like a 500mw diode. Once the light passes the reflector and lens, how would I calculate accurately for light output? What kind of efficiency losses do these mega-candlepower spotlights incurr?

I need approximately 15-25 useable watts of directed light OUTPUT energy (by "directed", I mean light able to be focused by a circular lens system upon output), or more.

But, what you are saying is that I should just look at the rated bulb input and assume, say, 15% loss on that number upon output?

EDIT: What would be a good way (since I don't need to first spread the light within the reflector assembly, as it's not going to be used as a spotlight) to omit the mirror reflector and lens, and replace with a much smaller reflector assembly to pump the initial lens's with smaller diameter light output? This, way, I won't need to adapt a 10" lens to a spotlight, which then focuses through a 5", then a 2"@70x etc. (more efficiency lost of coarse doing it this way than by simply using less and smaller lens's).

Thanks a lot.

Sorry to bump...

Did I choose the correct place for this thread?

This is more a project for a physicist than a question for a flashaholic. You have lots of homework to do.

Volts times Amps = Watts

A tungsten (or other) lamp will convert electricity to light, but the conversion ratio or equation is not linear, or even first order.

Lumens = MSCP x 4 Pi. Or, 1 MSCP=12.57 Lumens ( MSCP is Mean Spherical Candlepower)

1 watt for 1 second = 1 Joule.

A joule is the electrical unit for work And, the International system unit of energy, equal to the work done when the point of application of a force of 1 newton is displaced 1 meter in the direction of the force.

Irradiance is the direct, diffuse, and reflected solar radiation that strikes a surface, usually expressed in kilowatts per square meter. The light falling on a solar panel is sometimes also measured in "Suns" so we have a relationship there.

Irradiance multiplied by time equals insolation And, Insolation is the solar power density incident on a surface of stated area and orientation, usually expressed as Watts per square meter or Btu per square foot per hour. Now we are talking some stuff!

This should get you started looking up some information on the physics. Bring your calculator, and good luck.

Thank you!

With that correlation info I can do the rest. My field of study is astronautical eng. (not photonics, but hey, bring on the physics!), I think I can pull it off from here.

Appreciated.