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Thread: Flash output question

  1. #1

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    Flash output question

    I have a question, probably dumb, about the way that flash output is characterized. It is more of a curiousity than anything. The output for any kind of photo lighting is given in W.s (watt-seconds). Since power (watts) is energy per unit time the implication is that the output is given in energy. Does this mean that for each firing of a 1000 W.s flash (set on max) a 1000W.s 1kJoule of energy will be given off? If so, what soes it imply for continuous sources? I guess what I am wondering about is how do you characterize a device in units of energy.

  2. #2

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    Re: Flash output question

    Exposure is an energy thing (lux-seconds).

    Here's something I'd already written that seems to be about right as the rest of an answer:

    Tungsten lamps are about half as efficient as strobes in terms of converting electrical power to light. Typical strobes achieve around 40 lumens per watt (eg GE tubes and White Lightning units). That's less than HMI or fluorescent continuous lamps, but more than tungsten lamps. The efficiency of tungsten lamps depends on the temperature they run at so the 3200 kelvin lamps used in the studio are more efficient than general service household lamps*. Nonetheless they achieve around 20 lumens per watt (lm/W).
    As I'm sure you are aware, few studio strobes have a flash duration longer than 1/125 second, so in 1/125 of a second effectively all the flash duration will be captured. So a 1000 Ws flash will put out about 40,000 lumen-seconds (lm-s) while the shutter is open.

    To achieve the same energy output from a 1000 W tungsten light would require a two-second exposure, ie 1000 watts times 20 lumens per watt times two seconds = 40,000 lm-s. That's a difference of eight stops in this case (choosing a different shutter speed will change the relationship). Even if the tungsten lamp had the same lm/W conversion as the flash tube, the difference would still be seven stops at 1/125. Of course there will be differences, possibly dramatic, in how those lumens are used and that will depend on the lighting instrument and the modifiers.

    There's also a small complication because light measurements (eg lumens) are done using a particular spectral weighting (usually the standard photopic curve for the eye) and that may not represent the response of your film or your meter, but that doesn't affect the general sense of the above. Two sources could have the same output when measured in lumen-seconds, but have different effects on a meter cell or on film - or put another way, the colour temperature of the source also has an effect on how useful the lumens are photographically.

    *This also applies to flash tubes, and that is why they are usually designed to operate at a higher colour temperature than daylight, then get filtered to reduce the colour temperature - it is more efficient than producing the light at the correct CT to begin with.

    Is that any help?

    Best,
    Helen
    Last edited by Helen Bach; 8-Nov-2006 at 10:39.

  3. #3

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    Re: Flash output question

    A 1000 W lamp left on for 1 second uses 1000 WS or 1 KJoule. But note that these are the energy or power that the lamp consumes, most of the energy is lost as heat or output in the infrared or ultraviolet. Tungsten lamps are 12-17 lumens per watt, I see one source stating xenon flashlamps at 15-50 lumen-seconds per watt-second. So the xenon is 2-3 times more efficient. So to match your 1000 w-s flash, you migh have to leave a tungsten 1000W lamp on for 2-3 seconds.
    To return, 1 watt-second equals 1 Joule. A 1 watt lamp consumes 1 Joule in 1 second.

  4. #4

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    Re: Flash output question

    Norm, do you have anything practical in mind, as for example the aperture to set given flash-to-subject distance and flash power? If so, power going into the flash is nice to know but irrelevant.

  5. #5

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    Re: Flash output question

    Quote Originally Posted by Helen Bach View Post
    Exposure is an energy thing (lux-seconds).

    Here's something I'd already written that seems to be about right as the rest of an answer:

    Is that any help?

    Best,
    Helen
    Hi Helen,

    Yes, of course what you write makes sense. I still find the units an interesting choice for a characterization. I mean, irrespective of efficiencies, geometries, spectra, etc...,
    the unit is that of integrated power. To be extra picky, what is left out here is the "shape" of the output. For a constant source that one can imagine fuctuations in the supplied voltage to the tube, or even more subtle effects. Over a long exposure this wont matter because the fluctuations are certainly small compared to the total integral, i.e. the sum of light falling on the film won't be noticibly affected. For short exposures this isn't so obvious, certainly when working with a flash. Since a flash is generated by the discharge of one or many capacitors there is an inherent shape to the voltage vs time behaviour. It is possible that during the length of the flash the light output varies considerably. So for a 1/125 second flash if you have your shutter speed set to 1/250 s (appropriately sync'd) you will get different results than what you might estimate as being 1/2 of the total flash output. The unit of W-s just seems meaningless to me. Yes I know we assume that flash output scales in an understandable (linear) way and it is sufficient to just determine what you need relative to what works for other situations. As you say, efficiencies and other technicalities dominate so a relative measure is just fine.

    It just seems to me that lighting should be characterized in Watts, not Watt-seconds. Now this wouldn't help in the example I have given but it would make me feel better .

    Norm

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    Re: Flash output question

    Quote Originally Posted by Dan Fromm View Post
    Norm, do you have anything practical in mind, as for example the aperture to set given flash-to-subject distance and flash power? If so, power going into the flash is nice to know but irrelevant.
    Hi Dan,

    Yes it is totally irrelevant. I have nothing particular in mind. My approach to flash is brute force anyhow - take a digital capture with the flash and when I get a good result I bring in the film cameras. I don't concern myself with technical details in practice.

    I recently bought a few flash heads and I just estimated what I would need by talking to people and adding 50% more than was recommended. This method worked perfectly. I just got hung up on the tecnical specs and like to understand things. Regular house lighting is characterized in Watts after all.

    Norm

  7. #7

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    Re: Flash output question

    Hi Norm,

    I guess that the best answer is that with electronic flash there is an assumption that the shutter is open longer than the duration of significant light emission, so the energy measurement is valid. 1/125 is, of course, very long for an electronic flash. That’s not the case for good old flash bulbs, which could be used with shutter speeds shorter than the duration of the light emission.

    As you say, the output varies considerably during the flash – a rapid rise followed immediately by a slower decay. I’m not sure how that could be usefully quantified by a power measurement. As an aside, there are ways of flattening out the curve or of breaking it into short pulses, but I don’t know whether they have been used for run-of-the-mill photographic flash units apart from the portable units that can operate with suitable FP shutters at faster-than-normal sync speed.

    Best,
    Helen
    Last edited by Helen Bach; 8-Nov-2006 at 11:56.

  8. #8
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    Re: Flash output question

    Ratings are also somewhat variable from manufaturer to manufacturer so that the same rating from manufaturer A and manufacturer B doesn't ncessarily equate to equal output in the real world.
    Last edited by Ted Harris; 8-Nov-2006 at 12:49.

  9. #9

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    Re: Flash output question

    Then you change the reflector and output changes again.

  10. #10

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    Re: Flash output question

    Quote Originally Posted by Norm Buchanan View Post
    ....
    I recently bought a few flash heads and I just estimated what I would need by talking to people and adding 50% more than was recommended. This method worked perfectly. ...Norm
    It means than what the people told you was basically wrong... How did you come to the 50% bigger figure?

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