# Exposure using incident light equation.

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### #1 Israel Romero Ramírez

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Posted 24 November 2014 - 12:33 AM

Have you guys use the next formula?
E=(25f^2)/St

where:

E =  ILLUMINATION IN FOOT CANDLES.

f= T-stop or f stop

t =  exposure time in seconds.
S =  Film speed.

say an example would be like...

How many foot candles are required to expose an ASA, 100 FILM, for 1/50 at f 2.0?

the result would be  50 foot candles.

Would this formula help on anything?  mean, that says how many foot candles we need on asa 100... to expose it... but would you recommend using it?  I mean that way you could use the data sheets of the companyes like mole richardson to get say 50 foot candles, depends on the sequence to make if there are full shots or maybe just MC, CU, etc...

thanks.

Would you relly on a equation like this?  I mean its on the ASC manual.

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### #2 David Mullen ASC

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Posted 24 November 2014 - 12:47 AM

I just use the old rule that you need 100 fc to get an f/2.8 at 100 ASA.  Memorize that and you can figure out what you need.

But that formula is more precise.

Either way, yes, you need some way of converting photometric data for lamps from manufacturers into some idea of what stop you'll be able to shoot at given a certain ASA.

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### #3 Israel Romero Ramírez

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Posted 24 November 2014 - 01:04 AM

I just use the old rule that you need 100 fc to get an f/2.8 at 100 ASA.  Memorize that and you can figure out what you need.

But that formula is more precise.

Either way, yes, you need some way of converting photometric data for lamps from manufacturers into some idea of what stop you'll be able to shoot at given a certain ASA.

Thanks sir  ,

yeah thats what I figured out, and using the light calculator of ARRI    http://calc.arri.de/calculator    using a fresnel TUNGSTEN 650w for example at spot... we would need to move it at least usint  ISO 100 at 2.0  go get that kind of 50fc ... we would place it round  32ft from the subject to frame.  its almost acurrate the formula, using the arri calculator.

for example on the 100 fc you mention.

we would put the same arri 650w at 22 ft from subjet.

Would it be that the right expsure?  I mean thats what the formula says right?, but ofcourse it doesn't says what kind of illumination desing we are setting up right?  I mean, If we want to make a noir illumination for the feature.  It just says what would be the foot candles we need to exposure that ISO on that f/2.0... man I have to many questions in my head... well what I think is that this equation will help knowing that we dont need a 10,000w  for sure, and if we need it, we would know where to put it to get the foot candles that we need right? because the ISO  and the f/stops don't move just because...

Edited by Israel Romero Ramírez, 24 November 2014 - 01:06 AM.

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### #4 Dennis Couzin

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Posted 24 November 2014 - 09:24 AM

But that formula is more precise.

No, Israel's formula isn't more precise than David's rule of thumb.  It is more explicit, since it includes the 't' instead of the commonly assumed 1/50 sec built into David's, but they're otherwise mathematically equivalent.  They're both weakly founded rules of thumb for the simple reason that there is no ASA speed for motion-picture films.

The ASA scale is the linear scale in the ISO film speed system.  But there is no ISO standard for measuring the speed of motion-picture negative films.  ISO 5800:1987 says: "The specifications do not apply to colour negative films for motion-picture and aerial photography.  ISO 7829:1986 handles aerial photography, but there is nothing for cine film speed except ISO 2240:2003 for color reversal film speed.

Probably there is no ISO standard because professional cinematographers are fanatical about exposure and their numerous case-by-case rules of thumb would supercede any standard.

Almost all film speed determinations of negative films have been sensitometric and based on just the toe of the film.  The speed printed on the can is that kind of number: toe speed.  If the film toe is unusual, or if the cinematographer makes an usual use of the toe, or if the rest of the sensitometry is unusual, or if the scene is unusual no general equation based on toe speed will tell you what illumination to put on the scene.

Edited by Dennis Couzin, 24 November 2014 - 09:25 AM.

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### #5 David Mullen ASC

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Posted 24 November 2014 - 11:57 AM

Obviously you may need fewer foot-candles to get an underexposed look or more foot-candles to get an overexposed look; the formula is just for figuring out the amount of light needed for a normal exposure.  If you want to underexpose by one stop, cut the number of foot-candles in half.

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### #6 Israel Romero Ramírez

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Posted 24 November 2014 - 03:37 PM

it says that for simplified for shooting at 24fp with a sutter angle of 172.8º.  which gives an effective exposure of 1/50s.

E=(1250)f^2/s

so, then... thats just only one equation... there is one for exposure using extension tubes or bellows...

=(d+F)^2/F^2

A=Exposure multiplier

d= lenght of extension tube or bellows

F= lens focal length

in a 3 inch lens is moved 150mm further from the film plane by a bellows.  the exposure time before the lens was moved was 1/48s. what is the new exposure time?

Edited by Israel Romero Ramírez, 24 November 2014 - 03:38 PM.

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### #7 Dennis Couzin

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Posted 24 November 2014 - 06:22 PM

What's really going on photometrically is this:  Illumination occurs in two places in the picture taking process: on the scene and on the film in the camera.  There is a simple relation between the two.  When the scene consists of a perfect white card (100% Lambertian reflector) the illumination on the film is 1/(4*N^2) of the illumination on the scene.  N is the f-number of the lens (assumed to be 100% efficient).  If the scene is less reflective than the perfect white card, the illumination on the film is reduced proportionally.

Example: with an 18% grey card (scene) and a lens set at f/2 the illumination on the film is 18%/16 = 1.125% of the illumination on the scene.  You can stick your lightmeter into your film gate to verify this.

For macro-photography, when the lens with focal length F is advanced by amount D from its infinity position (with tubes, bellows, or the focus mount itself) the effective f-number is no longer the marked f-number N but rather N*(F+D)/F = N*(1+D/F).  You would then substitute the effective f-number for N in the 1/(4*N^2) formula above.  A way to understand effective f-number: we think of the f-number as the result of dividing the focal length by the entrance pupil diameter  The effective f-number is result of dividing the imaging distance (F+D) by the entrance pupil diameter.  (Effective f-number is what determines the lens aperture diffraction too.)

Film exposure is the product of the illumination on the film and the exposure time (e.g., lux*seconds).   When the 3 inch lens is advanced 6 inches with a bellows, the effective f-number becomes 3 times as large.  So the illumination on the film becomes 1/9 as large.  Then to maintain the original exposure the exposure time must become 9 times as large.

Edited by Dennis Couzin, 24 November 2014 - 06:23 PM.

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### #8 Israel Romero Ramírez

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Posted 24 November 2014 - 08:58 PM

thanks  Dennis C.  and thanks also to David Mullen.

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