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missunderstanding, lightmeter


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#1 Michael Althaus

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Posted 22 June 2006 - 08:24 AM

I have something that I don't understand about using a lightmeter.: Let's make an example. I have a huge (18%) grey wall. I use an incident lightmeter and measure the light falling onto the wall and set the aperture accordingly. It doesn't matter how far away from the wall I place the camera (as long as the wall is all over the frame), the setting for the aperture will always be the same, right?

The intensity of reflected light decreases with the distance so how should I get a normal exposed negative if I move the camera away from the wall and don't compensate for the loss in light intensity. I mean, measuring the light falling on the subject doesn't take into account how far away the camera is.

I hope you can understand what I'm talking about. Sorry for my bad English. It's quite difficult to explain.

thanx
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#2 Adam Frisch FSF

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Posted 22 June 2006 - 01:46 PM

That doesn't matter - that object has got the same amount of light on it regardless if you're 1" from it or a thousand miles away. You are photographing the object, it's reflection at just that point, not the beams it sends away.

However, if you do your measurement closer to the source, then your value will increase by the square. Obviously, that's why one measures close to the object one is photographing.

Make sense?
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#3 Stuart Brereton

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Posted 22 June 2006 - 02:23 PM

Actually, Richard Kline, ASC talks about this issue in an interview in AC (Jan 06) . He describes a shot from King Kong, looking straight down from the top of the World Trade Center. He says that although the meter was reading 50fc down on the ground, the quarter mile distance to the camera rendered this reading useless, and he had to make an educated guess (and push the stock).
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#4 Laurent Andrieux

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Posted 22 June 2006 - 03:12 PM

So we shouldn't be able to see the stars in the sky then nor even the moon, should we ?
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#5 Michael Nash

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Posted 22 June 2006 - 03:28 PM

So we shouldn't be able to see the stars in the sky then nor even the moon, should we ?


No, you just don't have to squint from the brightness ;)
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#6 Laurent Andrieux

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Posted 22 June 2006 - 03:33 PM

Who tells you I do ? :angry: :D
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#7 Stuart Brereton

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Posted 22 June 2006 - 03:40 PM

So we shouldn't be able to see the stars in the sky then nor even the moon, should we ?


Take it up with Mr Kline - I'm just telling you what he said.....
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#8 Laurent Andrieux

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Posted 22 June 2006 - 03:43 PM

:lol:
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#9 Michael Althaus

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Posted 22 June 2006 - 03:59 PM

That doesn't matter - that object has got the same amount of light on it regardless if you're 1" from it or a thousand miles away. You are photographing the object, it's reflection at just that point, not the beams it sends away. ...

Make sense?


Doesn't make sense to me. Sure, the object has got the same amount of light on it regardless if you are 1" from it or a thousand miles away but what you see on the film is the reflected light (the "beams" it reflects) which decreases by the square. So it should make a difference if the reflected lights travels a long distance or just a short one. Am I wrong?
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#10 Hal Smith

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Posted 22 June 2006 - 04:14 PM

If the wall stays completely within the frame (and is evenly lit) there is no divergence of light, therefore no dropoff of intensity. Inverse square is based on the idea of a point source with a solid angle of divergence. If the gray card has stayed within the frame then you have the practical equivalent case of parallel rays of light - no inverse square.

Another way of looking at is that as you back away from the wall, and for a given lens focal length (fixed angle of acceptance), your viewed area gains as the square of distance exactly counteracting the inverse square dropoff of intensity of each individual infinitesimal point source of light.

To research the above subject (assuming you've got some serious math chops) look at divergence in a Physics Electricity and Magnetism text, they usually have a section on basic Physical Optics.

If one shoots through 1200' of atmosphere in a city like New York there's going to be a lot of light loss - sort of like a smog filter factor.
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#11 Michael Nash

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Posted 22 June 2006 - 04:19 PM

So it should make a difference if the reflected lights travels a long distance or just a short one. Am I wrong?


Don't overthink it. A gray wall is the same brightness 10 feet away or 100 feet away. You use the same exposure. And actually, very distant objects in daytime (like distant mountains) can actually read brighter because there's atmospheric haze between the lens and subject.

The best thing for you to do is test and see for yourself. That's how you'll learn to use the meter anyway.

If one shoots through 1200' of atmosphere in a city like New York there's going to be a lot of light loss - sort of like a smog filter factor.


But then the smog is reflecting light also, so you pick your exposure. But we are talking about the same thing.
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#12 Laurent Andrieux

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Posted 22 June 2006 - 04:19 PM

When your camera stands futher from the object, the surface of the object looks smaller. So the whole light that comes out from it in the camera's direction is fewer (it splits in many directions that your lens doesn't catch anymore), but the appearing surface has the same lumination. Fewer rays reach your camera, but each ray has the same energy.

Imagine a surface of an infinite dimension. The fact that you go further has no effect. It's lumination is still the same and the surface catched by the lens is always covering the entire frame according to Hal's explanation.

Now considering a limited object, when you go far from the object, other objects enter the frame. So the question becomes : will the setting that is good for how much light is reached by the object (let's say from your metering in fc) will give me a proper stop setting when the camera sees this object as only a small part of my image, wich is also made of many other objects.

If all objects in the frame have the same illumination (your fc reading) the answer should be yes.

One point can be the atmospheric pollution, but apart from that, the fact you went further has no impact.

Since your viewing field becomes wider, then comes the question of an "average" stop value that lets see the highlights and low lights corectly. The required stop for this wide shot won't necessarly be just as good for a close shot in the same field, depending on the constrast of the different objects in it...
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#13 Rik Andino

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Posted 22 June 2006 - 06:02 PM

Doesn't make sense to me. Sure, the object has got the same amount of light on it regardless if you are 1" from it or a thousand miles away but what you see on the film is the reflected light (the "beams" it reflects) which decreases by the square. So it should make a difference if the reflected lights travels a long distance or just a short one. Am I wrong?


That's why you should also take a spot reading from the camera.
It'll give you an accurate reading of the light reflecting from the subject.

But let's be practical...
Most shot are less than a 100' away from the subject
So this issue isn't really that big of a deal.

If the majority of shots were taking from 1000' or more--away from the subject
Then maybe you could argue that issue.

Ultimately most of us aren't theoritical physicist
So what we discuss here comes from our practical real-world experience...
If we tell you it works to just measure the light falling on the subject...
Trust us--it works.


Good Luck
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#14 Alessandro Machi

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Posted 22 June 2006 - 09:10 PM

If the wall stays completely within the frame (and is evenly lit) there is no divergence of light, therefore no dropoff of intensity...........

...............If the gray card has stayed within the frame then you have the practical equivalent case of parallel rays of light - no inverse square.


"If the gray card has stayed within the frame" excellent visual explanation.

Lets say you're a block away and you're doing a wide shot. Take multiple incident readings within the frame, not just the point you are interested in, just so you have an overall idea about the entire frame's contrast range. If the spot in the frame that you are interested in is two stops hotter than anything else, you have to make a decision based on that reading, do you reduce the contrast by lowering the light source at the key spot in the frame, keep it as is but perhaps underexpose the entire shot one f-stop which creates a clear exposure of the important spot in the shot while creating solid contrast in the rest of the frame, and so on. You are dealing with two equally important issues, contrast range and then the strategy of how much to over or underexpose the most important part of the shot, multiple incident readings will give you critical information towards your decision.


What if your shot was a telephoto shot from a block a way? In that instance, a spot reading might make more sense, especially one taken through the lens because that will take into account any loss of light that the zoom lens might induce, but apparently that is primarily a super-8 option. It would be interesting to see if carrying a Super-8 camera around because of its "through the lens" spot metering might be a quick fix to figuring out telephot exposure, even for 16mm and 35mm projects. Of course one would have to create the necessary offsets that relate to shutter angle differences between the formats and the super-8 viewfinder splitting some of the exposure. But once the proper exposure offset were found, a super-8 camera could be a handy tool for quickly determining telephoto exposures in scenarios in which the object is far away.

-----------------------------------


A follow up topic question you might want to post could be, "how much contrast do you like between the background objects and your primary subject", followed up with, "Once you establish the contrast you like, do you always overexpose the primary subject by one stop and if not, when do you stray from that decision?"
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#15 Alan Duckworth

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Posted 22 June 2006 - 11:39 PM

You probably shouldn't use a Super 8 cam as a spot meter for other cameras, because the meter in most [if not all] of them has been compensated for that individual model to deal with the loss of light due to the beam-splitter used for reflex viewing. [and the opposite is also true for most Super 8's - you shouldn't use the reading from a handheld meter to set a Super 8 cam unless you know by prior testing exactly what that loss is].

To deal with the original question: the subject is reflecting light, and therefore its surface luminance remains constant - hence the exposure for that surface does not change, regardless of camera distance. However, as the distance increases usually other subject matter enters the field of view, and this starts to complicate matters, as other subject matter will likely be over or under exposed in relation to the original subject, and this introduces the contrast issues and creative decisions that must then be made [following a zillion new light readings!].

The other complication is that anything in the frame that is a light source [eg streetlights, stars, headlights etc] will obey the inverse square law of propagation of radiation. In english, that means that the light intensity falls off twice as fast as you think it is going to! If you double the distance, you don't halve the the intensity, you quarter it. The inverse square law is why you always need more lights than you thought you could get away with, and is also why most lights used photographically for illumination have parabolic reflectors and/or fresnel lenses - these are valiant attemts to defy the laws of physics.

BTW, nobody ever said that exposure was easy. It can take a while to really get a handle on understanding it.
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#16 Alessandro Machi

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Posted 23 June 2006 - 12:49 AM

You probably shouldn't use a Super 8 cam as a spot meter for other cameras, because the meter in most [if not all] of them has been compensated for that individual model to deal with the loss of light due to the beam-splitter used for reflex viewing. [and the opposite is also true for most Super 8's - you shouldn't use the reading from a handheld meter to set a Super 8 cam unless you know by prior testing exactly what that loss is].


Which is why I stated that once the correct "offset" is determined, it could be viable. Offset means that the camera reading is consistenly off the same amount, so everytime a reading is taken the reading is adjusted by that same amount. If one knew the Super-8 reading would always be off by one f-stop, it doesn't take much to do the additional math. Polaroid cameras have been used for the past 25 years for exposure evaluation by calculating various offsets such as shutter speed, f-stop and ASA, then snapping a picture, and then evaluating the result. Note of caution, probably a film cartridge would need to be kept in the super-8 camera so the camera's ASA setting would be locked in. Then by locking in a frame rate speed, two of the super-8 camera's variables would be locked in. The 85 filter and the beam splitting viewfinder are the remaining variables, and those variables are easily calculable.


To deal with the original question the subject is reflecting light, and therefore its surface luminance remains constant - hence the exposure for that surface does not change, regardless of camera distance. However, as the distance increases usually other subject matter enters the field of view, and this starts to complicate matters, as other subject matter will likely be over or under exposed in relation to the original subject, and this introduces the contrast issues and creative decisions that must then be made [following a zillion new light readings!].


It doesn't bother me that you pretty much said what I said, but when you preface it with "to deal with the original question", that kind of implies that I didn't, and that strikes me as inaccurate portrayal of what I wrote since you basically corroborated what I said.

Now when I quoted someone else in my previous post, I actually complemented them, but when someone refers to my posts, I get the reverse treatment, and it's INACCURATE to boot. This happened TWICE tonight, I find this exasperating, especially after I put in over an hour for both responses.
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#17 Adam Frisch FSF

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Posted 23 June 2006 - 04:35 AM

You guys are really overcomplicating something that is quite straightforward on a practical level. You measure, you make a decision, you shoot it.

Sure that white house over yonder is 3 stops brighter than the house next to it that's brown, but it's supposed to be be that - it's white. If you don't want to overexpose just the white facade, then don't. But it won't be white anyomre, it'll be grey.

All this confusion stems from using spotmeters in conjunction with incident meters. If you measure the incident light, then no mistakes can be made. If I'm on a hilltop on a sunny day, shooting down on a city with lots of different houses in different colours, what is the point of spotmeasuring anything? All I do is stick my little incident lightmeter up into the light, make a decision, and shoot. The light that hits me is obviosly exactly the same as the light hitting that city below.

I'm on a crusade against spotmeters (either built in or in a meter) - they're the reason all this confusion comes along. I can think of one, and only one instant when a spotmeter is important: and that's the same scenario as above, but at night. There you'd want to be able to get some readings from pools of lights under streetlamps and stuff WITHOUT having to run down there and get it. But if you're close enough to make an incident reading, then I wouldn't need a spotmeter. Taking readings off backdrops in studios, you say? Just flip the dome aside and get a reflected reading on your incident meter - done deal.

Save your money and buy yourself a nice holiday instead of a spotmeter.
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#18 Alessandro Machi

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Posted 23 June 2006 - 04:44 AM

You guys are really overcomplicating something that is quite straightforward on a practical level. You measure, you make a decision, you shoot it.

Sure that white house over yonder is 3 stops brighter than the house next to it that's brown, but it's supposed to be be that - it's white. If you don't want to overexpose just the white facade, then don't. But it won't be white anyomre, it'll be grey.

All this confusion stems from using spotmeters in conjunction with incident meters. If you measure the incident light, then no mistakes can be made. If I'm on a hilltop on a sunny day, shooting down on a city with lots of different houses in different colours, what is the point of spotmeasuring anything? All I do is stick my little incident lightmeter up into the light, make a decision, and shoot. The light that hits me is obviously exactly the same as the light hitting that city below.

I'm on a crusade against spotmeters (either built in or in a meter) - they're the reason all this confusion comes along. I can think of one, and only one instant when a spotmeter is important: and that's the same scenario as above, but at night. There you'd want to be able to get some readings from pools of lights under streetlamps and stuff WITHOUT having to run down there and get it. But if you're close enough to make an incident reading, then I wouldn't need a spotmeter. Taking readings off backdrops in studios, you say? Just flip the dome aside and get a reflected reading on your incident meter - done deal.

Save your money and buy yourself a nice holiday instead of a spotmeter.


I guess what I disagree with is that not all film stocks behave the same. So knowing the contrast differences between different objects in a scene do serve a purpose if you know what to do with the information. Spot meters also can help differentiate how much contrast exists in the scene without having to leave the camera, and that can be useful and safe as well. Holding up an incident meter by the camera versus the location where one is shooting will not always reveal the same brightness because black objects can subtract light even if they are not creating shadows within the frame, and white objects can reflect additional light that might not be metered when standing by the camera.
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#19 Sam Wells

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Posted 23 June 2006 - 07:59 AM

If one knew the Super-8 reading would always be off by one f-stop, it doesn't take much to do the additional math.


If one knows the f 2.0 "telephoto" is really a T 2.3 at that stop, then you can just use your spot meter etc.

If you _don't_ know the T stop, using the S8 camera won't help you, except by coincidence.nYou need to know how much loss is happening at that aperature regardless.

If the lens is marked accurately in T stops, no problem here.

In truth it's not going to be a profound difference usually (big & especially old zooms are another story)

And the difference will diminish somewhat as you stop down.

-Sam
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#20 Alan Duckworth

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Posted 23 June 2006 - 07:00 PM

Alessandro - I was simply referencing back, no disrespect intended. I was trying to come at it from another angle, and in fact I have quite enjoyed reading all the various submissions here - their diversity illustrates the reason why the question was asked in the first place.
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