Flash Theory

I have heard that if one moves a flash along a straight line using

standard f-stop distances the lights intensity will change by the

number of stops. ie if a flash is eight feet from the subject and

one moves it to eleven feet from the subject the light will decrease

by one stop. I don't get it. First, Using the inverse square law

moving the flash three feet from eight to eleven feet will decrease

the flash by one-ninth (3 sqaured), not one half (one stop). Second,

f-stops are baased on a cirular lenses opening not a straight line.

So I am confused. Can anyone explain how the inverse square law

relates to moving a flash f-stop distances along a straight line and

why a staight line would work since f-stops are based on a cirular

lenses?

 

Thanks,

dazed and confused

You don't take the 3 feet and square that, you take the fraction of moving the distance from 8 to 11 feet and square that instead.

 

Example, 8 feet to 11 feet is not doubling (1/4 stop) or tripling (1/9 stop) the distance.

 

It is:

 

11/8 = 1.375, 1.375^2 = 1.89 which is almost 2 so you have a ratio of 1/1.89 which is almost half the light.

 

Inverse square law states something like double the distance then take the inverse of the ratio and square it. So moving light 2x further away gives you 1/4th the light hitting any given area.

 

Triple the distance and you get 1/9th

 

But you aren't doubling or tripling the distance. you're only moving the light 1.375 further away when you move from 8 to 11 feet.

 

BTW, yes, a quick way to set lighting ratios is to use the f-stop scale and set your lights accordingly, cool huh?

 

Hope this helps.

 

BTW, try all the f-stop numbers and relate them to each other this way and you can see they all fall pretty close to 1/2 the light when you move the distances further like above.

The line stating 1.375 further away should read something like 1.375 times (1.375x) away.

 

Sorry for the confusion.

You can indeed set ratios with equal powered lights,based on their distances from the subject,(use a flash meter to see for yourself).For example the main light at 8' and the fill 16',will produce a 2 stop difference or a 5:1 ratio.Remember that we are dealing with square roots as we move nearer/farther away with the lights.Light from any point source(tungsten or flash),falls off abiding by the inverse square law.The light doesnt fall off at the same actual rate over distance from the subject.For example 2.8,4,5.6,8,11 are all within 9 feet of each other.But 11,16,22,32 are over a 20 foot area!A len's aperture halves & doubles the amount of light based on diameter and focal length,moving a light does the same thing,this is simple physics.

You can also determine bellows factor in close-up situations by converting distances to f stops. You focus on your subject at the magnification that you want then measure the distance from the lens to the film plane and compare that larger distance to the focal length of the lens. But how do you know where to measure the focal length at the lens ?

 

Most cameras have a mark on the camera body for the location of the film plane. If you focus your lens at infinity you can measure from the film plane to the lens. At the focal length distance of your lens, you can mark the lens barrel with tape and that's the point on the lens which you use to re-measure the lens to film distance when you are shooting close-ups.

 

You can also use this "distance equals f-stop" technique to determine exposure in the darkroom when you change magnification and enlargment sizes.

And picking up on that last point, you'll notice that standard enlargement sizes are about a stop apart: 4x5, 5x7, 8x10, 11x14, 16x20, 20x24, 30x40, where the short dimensions are just about equal to 4, 5.6, 8, 11, 16, 22, 32.

Brave attempt at an explanation but bad mathematics, Mr. Stephen Ratzlaff.

 

You should consult a basic college physics text like Resnick. It is a factor of Square Root of 2 which produces the numbers. If you look hard at the Inverse <B>Square</B> Law, you will find that it is so.

Actually, Mr. Ratzlaff is correct in pretty much everything he said. The Inverse Square Law states that the intensity of a light is inversely proportional to the square of its distance.

<br><br>

You are correct in saying that the f-stops are based on square root of two.<br>

sqrt(2) = f/1.4<br>

sqrt(4) = f/2<br>

sqrt(8) = f/2.8<br>

sqrt(16) = f/4<br>

sqrt(32) = f/5.6<br>

sqrt(64) = f/8<br>

sqrt(128) = f/11.3 or f/11<br>

sqrt(256) = f/16<br>

sqrt(512) = f/22.6 or f/22<br>

sqrt(1024) = f/32<br>

<br>

By squaring the distance ratio 11/8, Mr. Ratzlaff took care of the proper math and came up with the correct intensity ratio of 1/1.89. Had he taken a longer explanation route he might have said that the square of 8 feet is 64, and the square of 11 feet is 121. 121 sq. ft. is nearly twice as much as 64 sq. ft, and so the intensity at 11' is approximately half that of the intensity at 8'.

<br><br>

The key is that 11 feet is (like Mr. Ratzlaff mentioned) 1.375 times farther away than 8 feet. 1.375 times farther is very close to 1.414 times farther, which is what would give exactly half the light.

<br><br>

Incidentally, 1.414 times farther away 8 feet is 11 feet, 3.76 inches.

My god, LOL, you all are giving me a headache!! I did not realize I had to be a math wiz to figure lighting! Geesh, LOL... I read to learn, true enough, but I feel as though I need a college course to figure out the math on this one.. Is there a simple "lay-explaination" for the "layman shooter"? You guys are really way over my head but I think if the above could be explained in more simple terms, alot of us who follow this thread could learn alot! Thanks in advance to those who bring it down a notch....

 

Happy Shootin'...

 

Marylou

It's not really that terrible ;)<br><br>

 

You already know that as you get farther away from a light, it shines less and less light on you. The amount that you receive corresponds to how far away from the light you are.<br><br>

 

The "Inverse Square Law" states that as you get farther away from a light, its brightness will decline in proportion to the <b>square</b> of the distance you moved. So, if you move <b>twice</b> as far away from a light, its apparent brightness will decline by a factor of <b>four</b> (two, squared). If you move three times as far away, it will suddenly only have 1/9 the brightness.<br><br>

 

All this business with the ugly numbers has to do with how to get just <b>half</b> the light. We know how to figure how much less light we're getting if we know how far we moved - just take the relative distance and square it (ie, moved to 8 feet from 4, 8/4 = 2, squared is 4 times less light). But what do we do if we know how much less light we want, but don't know how far to move?<br><br>

 

In that case, just take how much less light you want (in this case, half, or 2x less), find its square root, and move that much farther away. The square root of 2 is 1.4, so we move 1.4x farther away to get half the light.

 

As for an easy way to calculate lighting, the best way would probably to go off the f-stops on your lens. They're not just those same numbers I listed there because of coincidence ;) If you have a light 4 feet away from your subject, and you move it to 11 feet, that's a 3-stop difference, so you'll get an eighth the light as before.

Mr. Sarbora, you present that argument in college physics and I guarantee you an F outright. It is bad mathematics anyway you try to justify it.

Looks like I started a Row. But it is interesting:}

 

Matthew

"I did not realize I had to be a math wiz to figure lighting!" <p>You don't... t

Well, if you'd give me an F on it then I'd honestly like it explained to me correctly. I've taken college-level physics courses and they come pretty much second-nature to me. If I could draw a diagram perhaps I could make my explanation clearer, but you seem to think my fundamental math is wrong. In that case, please show me the correct application. I'm not daring you or anything, if I'm truly wrong then I'd like to be righted.

The matter of distances above is correct. You can use the f-numbers as distances (inches/feet/cm/meters/yards/etc.) to judge lighting ratios as I described.