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Lumens question....


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#1 Tony Brown

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Posted 29 August 2007 - 11:26 AM

If I project white from an LCD Projector (3000 Lumens) roughly what stop will I read at say, 10 feet on my incident meter.....

Ball park will do fine.....

Thanks
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#2 Kieran Scannell

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Posted 29 August 2007 - 03:26 PM

Tony What on earth are you shooting? whats your stock rated at?

Kieran.
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#3 e gustavo petersen

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Posted 29 August 2007 - 04:16 PM

Your question is a bit incomplete, but I'll give you some information to try to piece it out.

One lumen uniformly distributed over one square foot of surface provides an illumination of 1 foot-candle. So, 3000 lumen at one foot away from equals approximately 3000 foot-candles. I say approximately because I don't know if you're 3000 figure takes into account the light loss from your projector's lens. Some projectors claim to be, say, 10,000 lumen but you then have to factor in light loss from the various lenses you can rent for that projector. Since you're questioning a 3000 lumen projector, I"m gonna guess it's a fixed lens unit.

FC = lamp lumens / area sq.

I had a project two years ago where we needed at least 40 foot-candles at a 15' x 20' screen. The 40 foot-candles took into account the light loss.

15x20 = 300 sq. ft.
300x40 = 12,000 lumens needed.

If you can find your foot-candles, you can then find your stop.
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#4 Tony Brown

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Posted 30 August 2007 - 05:28 AM

Your question is a bit incomplete, but I'll give you some information to try to piece it out.

One lumen uniformly distributed over one square foot of surface provides an illumination of 1 foot-candle. So, 3000 lumen at one foot away from equals approximately 3000 foot-candles. I say approximately because I don't know if you're 3000 figure takes into account the light loss from your projector's lens. Some projectors claim to be, say, 10,000 lumen but you then have to factor in light loss from the various lenses you can rent for that projector. Since you're questioning a 3000 lumen projector, I"m gonna guess it's a fixed lens unit.

FC = lamp lumens / area sq.

I had a project two years ago where we needed at least 40 foot-candles at a 15' x 20' screen. The 40 foot-candles took into account the light loss.

15x20 = 300 sq. ft.
300x40 = 12,000 lumens needed.

If you can find your foot-candles, you can then find your stop.


Too complicated to me anyone know what stop I'd get? Any asa, I'll do the easy maths from there....

Thanks
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#5 Phil Rhodes

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Posted 30 August 2007 - 10:00 AM

We need to know the area you're trying to illuminate. Giving a distance isn't enough because we don't know what lens you have on the projector. The inverse square law is not constant over variable collimation.

But go via exposure values.

Assume F=4.0 at 1/50s (assuming this is a PAL TV project). Don't bother with the mathematics, just look it up: EV 10 at 100ASA. Now go look up what EV 10 is in footcandles - about 240.

As we saw above, fc = lamp lumens / area. So:

235 = 3000 / area

I don't know what your area is. Rearrange:

area = 3000 / 235

Ergo, in order to shoot 100ASA stock at 25fps and F=4.0, you may illuminate an area up to 12.77 square feet.

Phil

PS: the mathematics, should you care, are: 0.46*2^(log2(4^2/(1/50)))
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#6 e gustavo petersen

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Posted 30 August 2007 - 02:08 PM

Too complicated?! Seriously!?

There's one last thing not taken into account by the math and that's what the projector is projecting. The math provides assumptions based on projection of a solid white field. If you're projecting dark images you're going to get a different reading on set.

Simple questions rarely come with simple answers. I'll say it again the thing no one wants to hear... test, test, test.
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#7 Paul Bruening

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Posted 30 August 2007 - 02:32 PM

Hello Tony,

I'm going to assume you're gonna' shoot that screen or include it in a shot. Therefore, correctly shoot a gray card with video and project it. Then, take a reading of the gray card image from the spots where your camera will be shooting the screen and get your readings based on films you might use. Write it all down. The angle matters because all screens have strong and weak reflecting angles. As well, distances of the projector to the screen and screen to the camera can affect readings. If you're including the projected image in a shot of other lit things, then balancing the subject light so that the screen will appear correctly exposed will take some juggling. I'd use a video cam and monitor to set that up and judge. Any light hitting that screen other than the projected light will wash out the projected image to varying degrees.
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#8 Tony Brown

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Posted 30 August 2007 - 11:30 PM

Shooting nights, nights that have a habit of turning into days, tired, so yes way too complicated

No screen, just walls, faces, no fixed size

Thanks Phil, next time I get 6 hours sleep I'll digest your answer, cheers.
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