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Filter factors.


jim_gardner4

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<p>Hi all. I am getting confused with filter factors and how to equate them to aperture changes or shutter speed changes but mainly how the factor numbers were assigned in the first place.<br>

Lets say a filter is assigned a factor of 8, why is it the makers dont just say "this will require a 3 stop increase"?</p>

<p>Also, staying with a filter with a factor of 8 (lets say a dark red) how is it that a 3 stop ND filter will say 0.9 on it, a 2 stop ND will say 0.6 and a 1 stop 0.3?<br>

My head hurts.</p>

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<p>OK but I have read in Steve Simmons book, Using the view camera, that he would meter through the filter to get an exposure<em> then</em> apply the filter factor.<br>

His meter (through the filter) would already account for the loss of light that the filter absorbs so wouldnt applying a factor of 3 after metering mean his shot would be 1 1/2 stops over exposed?</p>

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<p>Jim-</p>

<p>That does not make sense to me either. When using a filter, I just stick it on the DSLR/SLR and off I go, fat, dumb & happy letting the camera's meter do the head scratching. I may chose to override by a 1/3 or 1/2 stop, but I would do that anyway even without filtration if I were after a certain 'feel' to the shot.</p>

<p>Filter factors are sort of holdovers from the day when in-camera metering was rare or absent. One used a handheld meter, then applied the filter factors to set up the camera. I still do that when shooting my old Graflex, as it is meterless of course.</p>

<p>The only area where I might agree with the author is with graduated ND filters, where to some degree the half lite/half dark filtration is sort of "odd". That is somewhat messing with smarts of the camera and outcomes are sometimes still surprises. But, today with digital, it's trivial to fire and peek, then alter and re-do. All it takes is time.</p>

<p>Write the author. He may respond and shed some light. Maybe it's just "Jims" that don't get it.<br>

:o) <br>

Jim</p>

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<p>I don't quite get the idea of metering through the filter and then adding the whole filter factor either, but there are some older CdS meters that are not entirely panchromatic, which may require compensation. For example, my old Nikon F adds two stops for a red filter, though the factor is actually 3.</p>
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<p>On another page he goes on to say he was using an orange filter. From the text I think he is saying that the filter absorbs a certain amount of light but shadow areas which are illuminated bt blue light will receive even less light than the rest of the shot.<br>

Its a good book but that isnt explained very well.</p>

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<p>>> Lets say a filter is assigned a factor of 8, why is it the makers dont just say "this will require a 3 stop increase"? <<</p>

<p>Hi Jim. This is one of those situations where you can look at it from a distance, and you can see clearly how it works, no two ways and no doubt about it. But when you try to look closer, you find that things you once knew turn out to be not quite so right.</p>

<p>In this case, you may find that the exact, correct filter factor of 8 is not actually assigned to the filter, but rather to a certain film and lighting situation when that filter is used.</p>

<p>Here's an example - Wratten #25 red filter: Using Kodak film data sheets, for Tri-X film in daylight, factor = 8 (same as the 3 stops you mentioned). But in tungsten light (ie, the sort of lightbulb that can burn your fingers), factor = 5. If you switch over to T-max film, the tungsten factor = only 4. The real case is a bit deeper, with respect to the scene content, but this should help get your feet wet!</p>

<p>With respect to metering, it seems like the other guys ought to be right - either meter normally plus apply the filter factor, OR meter through the filter and be done. I wondered how a guy like Steve Simmons could be so wrong, so I did some internet searching. Apparently, he is NOT saying to use the regular filter factor - he is saying to meter THRU THE FILTER, then for SOME FILTERS, additionally apply a special "Hutchings Filter Factor." But someone must reveal to you what the Hutchings factors are... a guy in this thread listed a half-dozen of them: <a href="http://www.largeformatphotography.info/forum/showthread.php?t=30847">http://www.largeformatphotography.info/forum/showthread.php?t=30847</a></p>

<p>So there must be some situations, where the spectral responses of the film and meter, or perhaps certain scene situations don't match up well, and I am presuming that Steve Simmons is specifically compensating for these. But I'm just guessing about this. I just thought this information might help to get you on another track.</p>

<p>BTW, those neutral density filters are rated with their "optical density", same kind of number you would get by reading them with a densitometer. There are good reasons to do this, but I agree with you, for use on a camera, the maker ought to specify an exposure correction.</p>

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<p>Thank you all for the information. I can follow it easily enough but after doing EI tests, making sure meters are accurate etc then measuring a scene very carefully with a spot meter, adjusting camera movements , metering through a filter (if that is the way we are taking readings) and determining exposure to the nearest 1/2 stop, we then just add "+2 stops" depending on time of day and colour of the light.</p>

<p>Bracketing with a LF camera is time consuming and costly enough but with a 5 foot long 11x14 camera,,well lets not even go there.</p>

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<p >Exposure is the product of the light energy hitting the film/sensor and the time the light energy is allow to play on the film/sensor. The equation is e=it. This is called the law of photographic blackening. Where e = exposure and I = intensity of the light energy and t = time of exposure. </p>

<p >When we compensate for a loss of either i or t we use math to calculate the revision. </p>

<p >If we mount a neutral density filter that attenuates (reduces) light transmission 50%. This filter will be labeled filter factor 2 (two). The 2 (two) tells us we must multiply the exposure time by 2 to compensate for the light loss which is one half. </p>

<p >How do we apply? Assume exposure without filter was f/11 at 1/50 second. </p>

<p >1. Multiply time of exposure by 2. Example 1/50 sec. x 2 = 1/25 sec. We use f/11 at 1/25 sec.</p>

<p >2. Open up the aperture to allow twice as much light to play on film/chip. Example: Change aperture to f/8 now the exposure is f/8 at 1/50 sec. </p>

<p >Table of filter factors:</p>

<p >2 = 50% reduction -- open up 1 f/stop or multiply exposure time by 2.</p>

<p >4 = 25% reduction -- open up 2 f/stops of multiply exposure time by 4.</p>

<p >8 = 12.5% reduction -- open up 3 f/stops of multiply exposure time by 8.</p>

<p >16 = 6.25% reduction -- open up 4 f/stops of multiply exposure time by 16.</p>

<p >Note: For the number of f/stops required to compensate we count on our fingers in powers of 2 thus: 2 - 4 - 8 - 16 - 32 - 64. Whereas 2 = 1 finger = 1 f/stop. 4 lands on second finger thus 2 f/stops. 8 lands on third finger thus 3 f/stops. 16 lands on fourth finger thus 4 f/stops. </p>

<p > </p>

<p >This is true because the f/stops are based upon each being a 2x change from the next i.e. a doubling. </p>

<p > </p>

<p >When using a hand-held light meter we can use this method; divide the ISO by the filter factor and reset the meter to this value. This forces the meter to do all the math. </p>

<p > </p>

<p >If we mount a filter on a camera with a built-in thru-the-lens metering system, it is likely that the camera with its chip logic will correctly compensate for the filter.</p>

<p > </p>

<p >If we allow a hand-held meter to measure the scene through the filter, likely the reading will be correct and no further compensation is required. </p>

<p > </p>

<p >A pitfall may be encounter. The sensitivity of a meter sensor may differ from the film/chip sensitivity and this can induce an exposure error. Likely such an error will be miniscule. </p>

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