# Calculating required light intensity

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### #1 David Edward Keen

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Posted 20 October 2015 - 02:28 PM

Just curious, is there a mathematical way to decide how much light to bring, given that the talent will be standing 10 feet from the key light, outside at night? I realize there are many variables, which may simply mean I will need to bring a bunch anyway...some method of ballparking it you guys have learned along the way?
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### #2 JD Hartman

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Posted 20 October 2015 - 02:33 PM

Yes, ARRI has a calculator of their website.  No doubt someone has written a smartphone ap as well.

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### #3 David Mullen ASC

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Posted 20 October 2015 - 04:44 PM

There is photometric data from the manufacturer but it all assumes hard lighting, you'd have to guess that you'd lose roughly half of that by soft-lighting techniques, and that's a pretty crude generalization.

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### #4 John E Clark

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Posted 20 October 2015 - 04:53 PM

Just curious, is there a mathematical way to decide how much light to bring, given that the talent will be standing 10 feet from the key light, outside at night? I realize there are many variables, which may simply mean I will need to bring a bunch anyway...some method of ballparking it you guys have learned along the way?

The 'standard' motion picture exposure is Light at subject 100 footcandles, = f/2.8 @ 1/48 sec. @ ISO 100 will yield a 'correct' exposure for 18% grey...

The next piece of information is what lamp your are using (or set), and at this point it is recommended to get a incident meter and measure the actual amount of light falling on the subject from the lights.

However, lighting manufacturers like ARRI will list the lamp, and the amount of light at various distances. So you could work out some estimates before hand.

Most 'cheap' manufacturers of the 'chinese' variety do not list out the amount of light at various distances, and only list 'max wattage'... which doesn't really indicate how much is falling on the subject.

The math equation for a 'point source', is the amount of light falls off as the inverse square of the distance that is 1/r^2. So if one has 100 Footcandles at 10 feet, one will expect to measure 25 Footcandles at 20, that is 1/4 the light, or 2 stops lower...

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### #5 Satsuki Murashige

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Posted 21 October 2015 - 01:47 AM

To further add to everyone's points, you need to start with some idea of what ISO, f-stop, and shutter speed you expect to be working at. Assuming 800 ISO, T2.8, 1/48, then using the equation John provided you would need three stops less light than 100fc, or 12fc (every stop is +/- 50%). But let's say you want to shoot 48fps with a 180 degree shutter, or 1/96 shutter speed. Then you would require an extra stop of light, or 25fc. If you want to shoot 1/96 and also at 400 ISO, then add another stop of light, or 50fc. Then you would figure out what lighting units you want to use and check out the photometric charts to decide if you have enough light. When you gather enough experience with the common lighting units you'll eventually be able to estimate what you need without having to rely on the charts. Or you'll just depend on your gaffer to know what is needed.
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### #6 David Edward Keen

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Posted 21 October 2015 - 01:31 PM

Thanks everyone...this is great. I'm seeing a method now. All that remains is the madness!
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### #7 David Edward Keen

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Posted 21 October 2015 - 01:50 PM

Satsuki, I thought 180 degree shutter angle is equal to 1/48 shutter speed? Confusing. If you can recommend a book to help learn that which is learnable through reading, perhaps it'll help me when I'm experimenting with the lights and camera
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### #8 David Mullen ASC

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Posted 21 October 2015 - 01:58 PM

180 degree shutter angle only gives you 1/48th shutter time when the camera is running at 24 fps.

The spinning mechanical shutter is a half-circle in the case of a 180 degree angle (180 is half of 360) so if the camera is taking 24 pictures a second, and the shutter is closed for half that time (half of 1/24th of a second), then the exposure time is 1/48th of a second per frame.

Satsuki was saying that at 48 fps, your time then would be 1/96th of a second per frame.

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### #9 John E Clark

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Posted 21 October 2015 - 02:35 PM

180 degree shutter angle only gives you 1/48th shutter time when the camera is running at 24 fps.

The spinning mechanical shutter is a half-circle in the case of a 180 degree angle (180 is half of 360) so if the camera is taking 24 pictures a second, and the shutter is closed for half that time (half of 1/24th of a second), then the exposure time is 1/48th of a second per frame.

Satsuki was saying that at 48 fps, your time then would be 1/96th of a second per frame.

This also implies that one needs 2x the light for a 48 fps @ 180 deg. shutter, than at 24 fps. And even higher light levels for higher frame rates.

Since my 'vision system' is such that I don't really notice the difference between 24 fps or 30 fps or 'faster'... at some point for a high frame rate, one may not maintain the 180 deg. 'rule'...

In regard to 30 fps... just for the sake of discussion... I find that 'TV-ness' is almost always due to the style of lighting, color saturation levels, the amount of detail in sets, etc. (and then there's the radio like dialog... but I digress...), rather than some subtle impact of photons on my eyes due to the frame rate.

Edited by John E Clark, 21 October 2015 - 02:36 PM.

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### #10 David Edward Keen

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Posted 21 October 2015 - 07:28 PM

thanks folks! i'm gonna see if i can't learn myself this.

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