¿What is the luminous efficacy of a lamp?

Discussion in 'Lighting Equipment' started by paco_rosso, Jun 23, 2014.

  1. I have some problems with the luminous data of several lamps makers companies.
    This table resume the situation: http://pacorossofoto.files.wordpress.com/2013/08/par641.jpg
    (Commas are used in the way it is used in Europe, as decimal separator, not as thousands, so 11,00 is 11.00 not 11 000).
    The first column "grados" is the plane anle of emmision in degrees. (From catalogue)
    Second column is the solid angle, calculated from w = 2 * PI * (1-cos (a/2)) where PI is 3.141599, w is the solid anle in stereorradians and a is the plane angle.

    Third column "candela" is the light intensity in candles (data from catalogue). (This lamps are PAR 64).
    Fourth column "lumen" is light flux in lumen. It is calculated multiplying the light intensity (3th column) for the solid angle (2nd column) because the light intensity is J = F / w Where F is the flux, J is the intensity and w is the solid angle. This data (flux) is calculated from catalogue data.
    Fifth column "W" is the electric power in watts (from catalogue).
    Sixth column "R" is the luminous efficacy calculated dividing the lumen (fourth column) over the power (fifth column). R = F / w. It is, R is lumen per watts.
    Seventh column is the lamp. OSR stands for Osram, Phil stands for Phillips. The lamps are PAR64 conceived for stage lighting and photographic studio (in wide sense: cinema, TV, etc).
    My problem... See the R number. About 11lm/w.
    In the bibliography it is said the typical R for tungsten is about 25lm/w...

    ¿What do you think about it?
    (If we talk about flash lamps (strobes, not lanterns) the R is as low as 14)

    I think I am wrong but do not know where I fail.
    (Another day we'll talk about why the real BCPS is much lower than the catalogue says).
  2. If you are saying that watt to watt-second , electronic flash is less efficient at turning electricity into light, try this simple
    test. Illuminate a target with a 500 watt light and repeat the same with a 500 watt-second (AKA Joule) electronic flash and
    set your camera to ISO 100 and a shutter speed of 1/125.
  3. Very, very interesting ?
  4. http://en.wikipedia.org/wiki/Incandescent_light_bulb
    According to a table at the bottom of that page there seems nothing above 16lm/W in the 220V realm while 110V light bulbs go higher up to 20.7lm/W.
    I have no clue what you are reading "In the bibliography it is said the typical R for tungsten is about 25lm/w..." but I guess if the author was a 110V guy writing about short lived photo bulbs -"Nitraphot" and you are looking at longer lasting 220V stuff it all makes sense?
    Sorry I have no clue why Voltage might matter that much, but considering a trend towards 12V halogenic spot lights in households I guess it all makes sense.
  5. My problem... See the R number. About 11lm/w. In the bibliography it is said the typical R for tungsten is about 25lm/w...​
    Hi, I think you're comparing two different things; you should probably look deeper into the bibliography item.
    As I think you realize, the 'R' value essentially tells how efficient the lamp is in turning a certain amount of electrical power into light power. In principle, you'd want to measure ALL of the light. Since the tungsten filament essentially sends out light in (essentially) all directions, you'd have to measure all of this. My guess is that this is what your bibliography item (R ~ 25) is based on, at some given color temperature (probably about 3200 K).
    However, the lamps being specified have measured light output in a much smaller area. So I think the numbers in your table are also stating how effective the reflector is, with respect to capturing all of the light and redirecting it into that smaller area (only the smaller area is being measured). (You might noticy a tendency for the smaller beam angles to have a lower efficacy value.)
    Does this make sense?
  6. J = F / w​
    I don't think you can just divide the luminous flux by the solid angle and come back with a luminous power figure. That seems to take no account of the fact that the area of the circle of illumination is proportional to the square of its radius, which in turn is proportional to the tangent of half the beam angle. It's the area of illumination at a fixed distance that you need to work out for each lamp and then normalise by multiplying the Candela figure by the area of illumination.
    Therefore the formula I'd use would be: (Tan illumination angle/2)^2 * Candela figure; to arrive at a figure proportional to the total light output power. From that you can work out the relative efficiency of each lamp. A bit more work then needs to be done to arrive at an absolute Lumen value, but you only need to do that for one lamp and then use the relative efficiency figure for the rest.
    As bill says, the filament is emitting light pretty much into a 360 degree solid angle - (I'm not sure why you're converting to Steradians, that's just complicating things unnecessarily. Simple degrees will work just as well and can be more easily normalised from the first column of data.) Therefore the efficiency of the reflector/diffuser will come into play.
    By working out a theoretical area of illumination, I actually got a bigger variation in efficiency than you've shown; with the 57 degree lamp being over 2.7 times more efficient than the 12 degree lamp in the same series. However, that assumes that the illumination will be absolutely even across the solid angle, which of course it can't be. So the given Candela figure is presumably taken in the centre of the beam, where it will be at a maximum. A bit of inverse square law then needs to be applied to find the (theoretical) level of illumination at the edge of the beam. After working out the integrated light level, you'll probably find that the R figures are much more even. Lesson 101: Never take manufacturer's figures at face value.
    And when it comes down to it, we're talking about extremely inefficient old hot filament technology, giving out far more heat than light. So who really cares? All you need to know is that it'll burn your hand badly if you touch it while it's lit! And for some time after it's switched off as well.

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