Lets say an object is lit to 1000 foot-candles from a light source ten feet away. How many fc would there be 15ft away?

500fc????

Lets discuss

# The inverse square law of lighting

Started By Daniel Madsen, Feb 17 2006 05:22 PM

6 replies to this topic

### #1

Posted 17 February 2006 - 05:22 PM

### #2

Posted 17 February 2006 - 05:52 PM

I'd rather say, "by nose", more likely 400, but let's calculate...

You first need to calculate the equivalent intensity of the source, assuming it's punctual :

Sorry I must transfer fc to lux (1fc = 10 lux) and therefore, I is in Candela.Ten feet is about 3 meters...

E(illumination) = I (source intensity)/d²

Then I = E.d² = 10000.3² = 90.000 cd

Then, at 15 ft (4.58 m) you'd get

90.000/4.58² = 4286 lux = So let's say 430 cd

You first need to calculate the equivalent intensity of the source, assuming it's punctual :

Sorry I must transfer fc to lux (1fc = 10 lux) and therefore, I is in Candela.Ten feet is about 3 meters...

E(illumination) = I (source intensity)/d²

Then I = E.d² = 10000.3² = 90.000 cd

Then, at 15 ft (4.58 m) you'd get

90.000/4.58² = 4286 lux = So let's say 430 cd

### #3

Posted 17 February 2006 - 08:06 PM

Thanks

### #4

Posted 19 February 2006 - 06:20 PM

WIth the information you give, it's possible to avoid some of the conversions that Laurent explained.

The inverse square law says that the illumination of a subject from a light source is inversely proportional to the square of its distance from the light source. You can do this in fc, or candelas, or any other unit of illumination, so long as you are consistent. And you can use any units of distance, so long as you are consistent.

So if you double the distance, (x2) then the illumination is a quarter (x1/4).

If the distance is increased by (15/10), then the illumination is decreased by (10/15)^2.

That's two thirds squared: which is four ninths, or 0.4444

So your illumination of 1000 fc becomes 444fc.

That's assuming the light source behaves as a point source (i.e. punctual as Laurent calls it): a narrowly focussed spot won't follow this rule, neither will a large bank of soft light.

The inverse square law says that the illumination of a subject from a light source is inversely proportional to the square of its distance from the light source. You can do this in fc, or candelas, or any other unit of illumination, so long as you are consistent. And you can use any units of distance, so long as you are consistent.

So if you double the distance, (x2) then the illumination is a quarter (x1/4).

If the distance is increased by (15/10), then the illumination is decreased by (10/15)^2.

That's two thirds squared: which is four ninths, or 0.4444

So your illumination of 1000 fc becomes 444fc.

That's assuming the light source behaves as a point source (i.e. punctual as Laurent calls it): a narrowly focussed spot won't follow this rule, neither will a large bank of soft light.

### #5

Posted 03 March 2006 - 07:16 AM

Thanks Dominic. I "felt" there was a way to avoid the intensity calculation, but did not find it at once. You did.

### #6

Posted 03 March 2006 - 08:59 AM

Inverse square law is exact when the rays of light are diverging from a point. If the rays are exactly parallel to each other than there is no drop off in intensity with distance.Thanks Dominic. I "felt" there was a way to avoid the intensity calculation, but did not find it at once. You did.

A softlight's beam has a central area where the light rays are effectively parallel, therefore less drop off of intensity with distance. Practically, when the physical size of the light is considerably smaller than the area being lit, then expect inverse square behaviour (the "u" is for our UK forum participant's comfort).

The actual physics of a softlight are more complex. Imagine standing close to a soft light and forming a framing box with your fingers while looking at the light. If your framing box is small enough that you see only a portion of the softlight and then move farther away, your fingers enclose a larger and larger area of the light's face. So what's happening is inverse square is still operating but as you move away your fingers are enclosing more of the light's face area and therefore more of the total light coming from the fixture. The converse happens if you walk closer to the light, you enclose less area as you move closer, but also less of the light flux. When you walk away, the enclosed area is increasing by the square of distance, while the light is dropping off with the square of distance, therefore no change in the light flux through your fingers, the opposite happens if you walk closer.

If you take a 20K fresnel, scrim it way down, and place talent very close to it (don't scrim it way down and I think we all would expect vfx - hair on fire), expect a flat beam. Conversely, the same light lighting up a building at night will be operating pretty much in the inverse square region.

If you placed talent inside a box built out of a diffusion material and then evenly lit all six sides from the outside, there would be no change in light intensity anywhere in the box. I may be shooting a "heaven" scene for a short film this summer - I've been thinking about how to construct a practical set that would create six sided lighting. I could throw a pile of CGI money at it - if I had the budget - but it wouldn't look right. The car chases in "Bullitt" were done at 24fps, no undercrank, no VFX - that's why they look so darn good - and why that film is so revered (IMHO). I just read that in one of the chases, a stunt driver lost it and trashed an ARRI 2C second or third camera, which can be seen lying on the road in pieces in one shot. Anyone got a clip of that scene? (but I digress, as usual)

A large constellation of small lights is another way to "beat" inverse square. There was a still photographer working in the LA area years ago who used myriads of very small lights to light portraits. I regret I can't remember his name, I had an elderly friend in Gadsden who had a portrait of herself in her youth shot by him. It looks similar to, but not identical to, a modern portrait shot with umbellas, tents, etc. It's quite striking, having that slightly different look that fascinates the eye.

Symbolically inverse square can be expressed as: new intensity = original intensity x (old distance/new distance)^2

For instance: move a physically small light (like a little Lowell open face) from 25 feet to 35 feet that has 100 footcandles intensity at 25 feet.

100fc x (25/35)^2 = 51.02fc

Move same light from 25 feet to 10 feet.

100fc x (25/10)^2 = 625fc

Edmond, OK

### #7

Posted 03 March 2006 - 10:26 AM

In simple terms, if the light source is focused in any way (shaped reflector, fresnel, etc.), you will have less fall-off with distance than the inverse square law would predict. It's the total AREA illuminated by the light source that really determines the light level. The light beam from an arc spotlight with a parabolic reflector diverges very little, and "carries" very far, but a bare lamp would follow the inverse square law.