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I have a medium size church room for doing standard wedding photographs and I'm trying to learn how to do the math for lighting. Over 1k of continuous lighting isn't near as bright as my two little Neewer speedlights at full power via a trigger for 1/200s. Basically, strobes put much more light out than continuous lighting for still photos because it's a sub-second burst.

All this has me wondering how to calculate lumens (or something else) that I can use to gauge the amount of light needed to fill certain areas. I read this article from tutsplus network about strobe power calculations but I'm not totally sure I understand how they figured out the 10,000,000 lumens number or where the math came from.

Take a strobe, with its xenon gas discharge tube, which attains somewhere in the region of 300 lumens per watt. Let's use a relatively low-power 60Ws speedlight, and assume the manufacturer isn't fudging the numbers and the electronics are high-efficiency. If we multiply the lm/W by the Ws, we cancel the Watts and end up with lumen-seconds. So the lumen-second output is around 18,000lm-s. Yes, but, remember: all of those lumen-seconds from the strobe are being discharged in around 1/2500 second. So we take the lumen seconds, divide by seconds to leave lumens, and what do we have?

18000/ 1/2500 = 4800*2500 = 45,000,000 lumens! Realistically the output from flashes is more like 10,000,000 lumens, due to optical and electronic inefficiencies, but still. They're all hitting your subject almost instantaneously, allowing you to very briefly overpower the sun, very briefly light up whole rooms or hillsides or waves.

I've tried to understand guide numbers (GN) but they seem to be a different lengths on my lights so I'm not sure how to compare them.

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    \$\begingroup\$ What, exactly, are you asking? \$\endgroup\$
    – Michael C
    May 11, 2016 at 22:46
  • \$\begingroup\$ Math for what? 1k of what? Why are you looking at studio strobe calculation instructions, when you're using speedlights? And you do understand that published guide numbers for many speedlights is deceptive, because they're only given at maximum zoom (to make the flash look more powerful). Have you considered looking into getting/learning to use a handheld light meter, so you can just measure the light instead of trying to calculate it? \$\endgroup\$
    – inkista
    May 12, 2016 at 1:15
  • \$\begingroup\$ "I've tried to understand guide numbers (GN) but they seem to be a different lengths on my lights so I'm not sure how to compare them." What do you mean "different lengths"? Do you mean your two Neewer speedlights have different Guide Numbers? \$\endgroup\$
    – Mike Sowsun
    May 12, 2016 at 1:45
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    \$\begingroup\$ @MikeSowsun He means the guide number for each is measured at two different focal length flash zoom settings. \$\endgroup\$
    – Michael C
    May 12, 2016 at 3:58

1 Answer 1

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I love photo math but that approach will drive you crazy and nothing will come of it. The Guide Number method is tried and true.

Once you know the guide number for your flash or combination of flashes, you divide the subject distance into that value.

Suppose the guide number is 200 and the subject is 18 feet from the camera. The math is: 200 ÷ 18 = 11. You set the aperture to f/11 and go to work. One pitfall is: published guide numbers are normally tailored for small rooms of average ceiling height. In a high ceiling environment, likely you will need to open up a stop or a little more.

If a guide number is not published: Place a test subject in the environment at a distance of 10 feet. Shoot a series of shots at different f/number settings. Inspect the results for best exposure. Say the f/11 shot was best. Multiply f/# used by footage distance thus 11 X 10 = 110. That’s the guide number for this arrangement at the ISO setting used. This will work for a single or multiple flash setup.

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    \$\begingroup\$ Guide number work great until you put modifiers between the flash and the scene. Then you have to figure out how much light the modifier is eating. \$\endgroup\$
    – Michael C
    May 12, 2016 at 4:33
  • \$\begingroup\$ @MichaelClark, in that case, it sounds like the last paragraph would do the trick, would it not? \$\endgroup\$
    – FreeMan
    May 12, 2016 at 14:17
  • \$\begingroup\$ It depends on how the modifier shapes the light. The inverse square rule only applies in its basic form if the light beam is allowed to spread with distance. Some modifiers project colimated light only rather than a cone of light. \$\endgroup\$
    – Michael C
    May 12, 2016 at 23:29
  • \$\begingroup\$ strobecalc.eu5.org This simple calculator makes for the hard part of it. \$\endgroup\$
    – user68970
    Oct 7, 2017 at 9:48

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