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I'd like to know how do battery-powered system flashes compare to studio units in terms of light output.

I know that AlienBees B400 has 400 "effective"  Ws, and that Canon Speedlite 580EX II has guide number 42 (meters) at ISO 100 and 50 mm zoom setting.

But how do I compare these two? Is there any way to convert guide numbers to watt-seconds?

che
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3 Answers3

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No, there is no way to convert guide numbers to watt-seconds. Watt-seconds is a measurement of how much energy is used by the flash, not how much light is put out. A significant portion of this energy is wasted as heat, infrared, ultraviolet, etc.

A 4 watt-second flash that is 100% efficient will put out the same amount of light as a 400 watt-second flash that is 1% efficient.

"Effective watt-seconds" are ill-defined as well, so are basically just as useless as watt-seconds.

By contrast, guide numbers are fairly well defined. They can be directly compared at a given beam spread (assuming manufacturers aren't stretching a bit. Ken Rockwell seems to think most flashes are over-rated by about a stop.)

However, the most accurate way to compare two flashes is through an actual scientifically defined unit like the lumen-second. From Wikipedia,

The lumen (symbol: lm) is the SI derived unit of luminous flux, a measure of the power of light perceived by the human eye.

As photographers, we're all keenly aware that time plays an important role in exposure. The longer the shutter is open, or the longer the light is on, the higher the exposure. Thus, lumen-seconds more directly translate to exposure than lumens do.

Here's a page from AlienBees' website which includes the specs on your B400 (7000 lumen-seconds) as well as a paragraph or two about how bogus "effective watt-seconds" are.

The Effective Wattseconds rating, however, is rather arbitrary and cannot be easily proven true or untrue, as it is merely used as a basis for inflated comparison of different flash systems.

I've looked around for a lumen-second rating of the 580 ex II, but can't seem to find one.

EDIT: David Hobby, master of the speed light, keeps saying 60 watt-seconds. (Note: that last link is an April Fools joke.)

Evan Krall
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    FWIW, I think Ken Rockwell isn't too off base here. The Japan Camera Inspection Institute (which, as I understand it, manufacturers take seriously) allows +/- 1 GV from a metered reading. And of course there's not much point in quoting on the minus side. – mattdm Dec 12 '10 at 05:21
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    Wouldn't a Watt-second be a joule? Watts is joules per second. – Nick Bedford Dec 12 '10 at 11:48
  • one other thing to be aware of, especially with lower-end strobes (I know, for example, that the Vivitar 285HV does this; I haven't tested a 580EX -- and perhaps this even applies with studio units??) is that it's possible for different firings of the flash to offer different amounts of light, especially if you're shooting in rapid succession, and increasingly so the higher power you have set. I think some of the higher-end units have ways to avoid this (firing at the same intensity or not at all), and I'm fairly certain that some units at least do a better job of being uniform than others. – lindes Dec 12 '10 at 17:22
  • @Evan +1, great answer. But why would a lumen, which apparently measures human visual perception, be appropriate for an electronic sensor? For a sensor with linear response and equal sensitivity across the visible spectrum, it seems we should be interested in actual light flux without any adjustment for our vision. – whuber Dec 12 '10 at 18:05
  • Turns out I was right. A Watt-second is just a joule which is a more quantitative measurement. – Nick Bedford Dec 12 '10 at 23:22
  • @Nick That's correct. Strobes probably use the term watt-seconds because people are more familiar with watts. – Evan Krall Dec 12 '10 at 23:46
  • @whuber Because the range of wavelengths called "visible" is defined by human visual perception. Our electronic sensors, films, printing techniques, and displays are all designed with human perception in mind. Red, Green and Blue are the primary colors of light because those are the center wavelengths of each of the different cone cells humans have. Without compensating for what humans can see, you would measure all of the infrared, ultraviolet, X-ray, radio frequency, and other types of radiation coming from the flash. See http://en.wikipedia.org/wiki/Luminous_flux – Evan Krall Dec 13 '10 at 00:12
  • @Evan You seem to have read too much into my question. It did not concern what the visible spectrum is; it only asked about the logic of using a human-based criterion for a non-human sensor. But an adequate answer is buried there: you claim that the sensors attempt to emulate human vision (which is true). However, to the extent to which sensors do not respond like human eyes (and none of them do exactly), it seems more logical--and more accurate--to measure light output in terms of the sensor response, not in terms of our visual response. The sensor creates the photo, not the eye. – whuber Dec 13 '10 at 00:39
  • @whuber - It's important to remember that some of these came about before digital. Also, if film and digital are to capture what we see, then they need to respond to light with a close approximation of human response. – Joanne C Dec 13 '10 at 01:33
  • @John Actually, films (especially B&W) tend to deviate from human visual responses more strongly than digital sensors do. The distinction between the film's sensitivity and our sensitivity used to be important. One's choice of film was part of the art of making a photo precisely because film does not capture what we see, but under the control of a competent photographer it can capture what we imagine. – whuber Dec 13 '10 at 15:08
  • @Evan: Thank you for detailed explanation. Is it possible to find lumen specs for any small flash at all? I don't really need to compare these specific models, what interests me more is how battery vs. studio units fare in general. – che Dec 13 '10 at 21:42
  • I have very little trust in Ken Rockwell's recommendation. Ever since I read his opinions about RAW vs JPEG, I can't really put any faith in anything he thinks. – Fake Name Dec 15 '10 at 04:49
  • @Nick The same holds for kWh (3.6*10^6 Joule) ... but people got accustomed to it. @Fake Name Aside: I just read the article of Ken about RAW and concur. I yet have to see a series (!) of pictures, where only RAW could save the day. See http://photo.stackexchange.com/questions/2627/good-examples-of-raws-advantages-over-jpeg/6034#6034 – Leonidas Jan 14 '11 at 06:35
  • @Nick I used to laugh when I heard photographers talking about Watt-seconds but it does kind of make sense as you can compare a one second exposure from a 60 Watt light bulb to a 60 Ws flash (of course a scientist would have been able to figure this out from Joules) – Matt Grum Jan 14 '11 at 10:31
  • @Matt How does this comparison help you? I usually don't take (a lot of) exposures of one second, so this figure does not help me at all. Model-Light with 60 Watts? Just qudaruple it to 1/25 and 240 Watts? That comparison does not even take the difference in efficiency between the sources into account - it is at least factor 2 off. – Leonidas Jan 14 '11 at 12:41
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    Again FWIW, I don't mean to imply that Ken Rockwell is generally a good source of information. In fact, I think it's worth making a special note because I think he happens to be right on manufacturer-stated guide numbers. That is, although not everything he says is correct, not everything he says is wrong either. May be the stuck-clock-twice-a-day effect, but still. :) – mattdm Jan 14 '11 at 13:47
  • @Leonidas It helps you precisely because you don't take a lot of one second exposures, even indoors under 60W bulbs, therefor a 60Ws flash is far brighter! I'm not suggesting you use the figure for calculating exposure, it just provides a means to gauge the ballpark output compared to regular bulbs. – Matt Grum Jan 14 '11 at 14:19
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FYI, Paul Buff does list an approximate guide number on his AlienBees spec page. You would have to set your 580 to a similar angle of coverage to compare the two.

che
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The only worthwhile way to compare lights, IMO, is to place them at the same distance from a subject, turn them all the way up then measure their power with an incident meter. This not only lets your record their effective, useful (for a photography) power output, but lets you easily compare them to figure out flash ratios. You can even use this method with various modifiers, which is how strobes in the real world are usually used.

Jędrek Kostecki
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  • But what if you don't have both lights available to do this (e.g., when making a purchase)? Essentially, you're rejecting the use of standard units. – Reid Jan 14 '11 at 14:17
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    Rentals. But really, as with oh so many things in photography, the proof is in the pudding. – Jędrek Kostecki Jan 14 '11 at 14:40
  • If you do this make sure you have a comparable beam spread, otherwise you wont be measuring the useful power output. The best approach would be to use the light modifiers as you suggest at the end! – Matt Grum Jan 14 '11 at 15:31