Film latitude - what does it refer to ?

In short: the quoted film latitude (say Kodak Gold 100 -2/+3) refers

to the exposure or to the film density ?

If it refers to the exposure, then the delta in film density would be

0.55*(2+3) = 2.75 stops (0.55 is the slope of the characteristic

curve, from Kodak Data Sheet).

It does look rather small and would translate into only 6.7x

difference between the shadows and the highlights.

If it refers to the density (5 stops = about 1.5 delta in density,

then the exposure latitude would be 5/0.55 = 9 which is about right

(?).

Any comments ?

Hi Leszek.

 

You are getting way too technical. Film latitude usually refers to how many stops under and over the correct exposure you can go and still get reasonable prints. So you could expose Kodak Gold at 2 stops under (i.e. underexpose it by 2 stops, say by accidentally exposing it at ISO 400) and still get a reasonably good print. You could overexpose it by even more and still get reasonable results.

 

The term has nothing to do with the range from light to dark the film can show when correctly exposed.

 

Be conscious that we are not talking about professional photography here, but snaps for your average Joe.

 

Regards, Ross

Thanks, Ross, but (very respectfully) I have to disagree (at this point of time, that is. When I gain more understanding - it is quite possible that I will agree ;) ).

I have been digging long and hard on many forms, and it seems that there is at least 3 or 4 different opinions as to what "film latitude means".

Your answer presents one of them.

The other was that it is "the difference between the exposure range of the film and the range of brightness of the scene". This one appears to be the same thing as you said, as it allows for exposure errors without losing shadow/highlight details (it still can be saved in print).

Another was that it is the useful range of shadow/highlight that can be captured on film. It appears to me that this may be the correct definition (?).

If you have a look at any print film data sheet, you will see that the density range (in the straight portion of the curve) is about 1.5 to 1.8 which is 5 to 6 stops (0.3 change in density is equivalent to 1 stop change in transparency of the film).

To achieve this range of film density (from the same characteristic curve) one can see that the exposure range is within 9 stops.

 

You may be right in one aspect: since the typical scene has about 4 to 5 stops brightness difference, this leaves about 5 to 4 stops room for error.

 

However, if my scene covers 9 stops, I can still get a perfect print WITH A SPOT-ON exposure. The 9 f-stops of the scene will translate into 5 or 6 stops on the negative without losing details.

BUT, there is no room for error left...

 

If I assume that your definition is correct, then the film latitude would depend on the range of brightness of the scene...

Less range - more latitude. More range - less latitude.

 

My question originated from reading the new Minolta Flash Meter VI specs: it has a latitude scale of the film (user specified, I think) and puts the readings on the scale so one can see if the shadows and highlights are within the film capacity for reproduction.

 

BTW, same is said here: http://www.normankoren.com/zonesystem.html

 

Any thoughts ? It is possible that I got it wrong...

One more thing:

Have a look here.

 

http://creekin.net/films.htm

 

Maybe they are talking about something else, but a latitude of -3/+6 means that you can do whatever you want with the exposure (not true, at least not within 9 stops !

 

Overexposing by 6 stops or underexposing by 3 means that there is no picture.

 

Therefore it may mean something else).

Still, this does not explain what a "latitude" means for the slide film...

Confusing...

Your question is about film latitude not prints. Let's leave prints out of it. How much of the film latitude can be captured on a paper has nothing to do with the film latitude it self. <BR>

Take a gray card and take a normal exposure. Then open up one stop and take another exposure, then open up another stop and take another exposure etc. etc, Then close down one stop and take an exposure, then close down another stop and take another exposure, etc. etc. The number of times that you can do that before the film gets completely clear on one end and completely black on the other is the film latitude.

I believe latitude is related to dynamic range of a film, although

cannot quantify how without a scanner-based Zone System.

<P>

As the creekin.net/films.htm page says, >= 400 films tested (by

Agfa minilab prints and HP-S10 scanning) on <A HREF="http://www.photo.net/bboard/big-image?bboard_upload_id=6722684">this scene</A>, which should be titled "EI 6400".

Batch variation may cause your results to vary.

>> In short: the quoted film latitude (say Kodak Gold 100 -2/+3) refers to the exposure or to the film density ? <<

 

Hi Leszek. I was going to say that the ambiguity was because the way they worded things; they should have perhaps said exposure latitude. But when I checked the Kodak website for Gold 100 film, it turns out they DID say that. Here is a quote, "It [Gold 100] features wide exposure latitude - from two stops underexposed to three stops overexposed.".

 

Hopefully this answers your first question.

 

 

>> I have been digging long and hard on many forms, and it seems that there is at least 3 or 4 different opinions as to what "film latitude means" <<

 

The "Focal Encyclopedia of Photography" is generally considered to be an authoritative reference. It has no entry for "film latitude"; however it DOES have an entry for "exposure latitude"; the Kodak quote above is consistent with the latter. PS; the Focal Encyclopedia correctly points out that exposure latitude varies with subject contrast, among other things.

 

 

>> If it refers to the exposure, then the delta in film density would be 0.55*(2+3) = 2.75 stops (0.55 is the slope of the characteristic curve, from Kodak Data Sheet) <<

 

No, you misunderstand what slope means. IF (I didn't see this on the data sheet; it sounds slightly low to me) the Gold 100 slope is really 0.55, this normally means that the film density changes by 0.55 per 1.00 log exposure units.

 

On the log exposure scale, a change of 0.30 is just like a full f-stop exposure change; thus each 1.00 log exposure change is equivalent to 1.00/0.30 = 3.33 f-stops change. So if the slope is 0.55, this means the film density changes by 0.55 per 3.33 f-stop exposure change; this is also the same thing as film density changing by 0.165 per 1.00 f-stop exposure change.

 

Now you can use your method above, but instead of using 0.55*(2+3), use 0.165*(2+3) = 0.82 film density change per 5 full f-stop exposure change. Note: you can't use the term "stops" to describe a density change. Also, because the film is unlikely to have a perfectly straight response curve, it is not a good idea to calculate a large difference like this; it would be better to read straight from the graph (ie, the characteristic curve).

 

 

>> If you have a look at any print film data sheet, you will see that the density range (in the straight portion of the curve) is about 1.5 to 1.8 which is 5 to 6 stops (0.3 change in density is equivalent to 1 stop change in transparency of the film). <<

 

I think you are having a common misconception here, but I'm not sure; you didn't state things clearly. If you think the density change of 1.5 to 1.8 results from an exposure change of 5 to 6 f-stops, this is NOT necessarily true. If you think that it IS true, you have not yet understood what the slope of the characteristic curve means.

 

 

If I tried to address everything I'd like to, I would have to go one quite a bit longer, so I'm going to quit here. Hopefully I've helped rather than further any confusion.

James, thanks !

Your advice is sound, but the following should be added: "Make the first shot of a grey card in lighting conditions which will produce the film density of about 0.9 to 1.1".

In other words, first shot should be dead center in the middle of the exposure range of the film (or latitude, as you call it).

Otherwise, if the first shot is in very dim lighting, two things will happen:

 

1) Stopping down from the first shot, you may run into the toe region of the film characteristic curve (your first stopped-down shot may be completely black !)

 

2) The lens may run out of the useful range of apertures if the first shot made with either poor or extreme bright lighting.

 

I fully agree with you, that (with the first shot calibrated properly), your test would give us the exposure range of the film. But...most color negative films have the latitude quoted as -2/+3, which is way to small to

be considered as an exposure range (unless we are talking the range of exposed film, which is about right).

 

My suspicion is that "latitude" refers (maybe) to the film ability to be saved by push/pull processing.

Say that the scene brightness range covers full the exposure range of the film, with perfect exposure. No room for error here ! Darkest shadow is almost on the toe of the curve, brightest highligh is far and high on the straight portion of the curve.

 

We can deliberately over/underexpose the film by -2/+2 stops and then push/pull process it, saving the day.

In this case, latitude would mean the range of departure from perfect exposure which can be corrected by post-processing.

Pushing/pulling will change the slope of the curve, but should still keep the shadows/highlights within the straight portion of the curve.

With normal development, however, the film is busted...

 

Bill,

 

What you are saying (and the picture) supports the above.

Pushing the film by 3 stops, however, does not produce exceptional results (as shown in the picture).

So, -2/+2 range seems OK, if it refers to push/pull processing (I have no experience here and never tried to do that with any film below ISO400, so do not quote me).

The "http://creekin.net/films.htm" says that one of the films (Agfa Vista 400) has a latitude of -2.5/+5.

Now, there is something wrong (?) here: I can understand -2/+2 stops (although I haven't seen any recommendations for pushing any ISO100 by 2 stops in the data sheets), but 5 stops ? C'mon...

 

So, we are back to square one.

Oops, maybe I went a bit too far.

 

If we measure the grey card, it will by definition reproduce it properly on the negative (0.9 to 1.1 film density), if the aperture is smaller or equal to max aperture of the lens.

But we will still have to play with aperture/shutter speed if the first shot is made at max. aperture.

The rest what I said remains unchanged.

I'm not talking about push/pull, just that -3 stops underexposure

can produce acceptable results with several films. And yes, Vista 400

exposed +5 at EI 12 looked fine, both prints and scans. Many films

leak dye outside the frame at +5.

"Your question is about film latitude not prints. Let's leave prints out of it. How much of the film latitude can be captured on a paper has nothing to do with the film latitude it self"<BR><BR>Most all C41 films are developed and printed on automated processing machines; that compensate for under exposure or over exposure of the C41 color negatives. The canned "film exposure latitude" statement is written for the users that drop film off at automated labs. <BR><BR>Kodak has been using statements like this since the Flexcolor process arrived; 3 decades ago. The seminar by Kodak used this statement of "better exposure latitude" ; and the much quicker processing time; as a good marketing mission; to replace the old C22 process.<BR><BR>Modern printers many times will make often identical prints; when the exposure is several stops different; when the subject matter is flat.<BR><BR>The film/print good latitude is what makes simple cameras for the masses work. <BR><BR>

very few; if any C41 is actually pushed or pull processed.

Kelly, thanks for your input !

All is good, but can you tell if the auto-lab is fixing the over/underexposed film, or is fixing the print ?

If the exposure error is fixed at the printing stage, then it has not much to do with film latitude (or maybe it has, since the film overexposed by 5 stops supposedly still contains all the info).

 

Anyway, if the scene has say 6 stops range, then the film exposure range would have to be (Agfa Vista 400): 7.5 stops latitude + 6 stops scene range = 13.5 stops.

 

Film density range would be about 0.5+ 0.6*13.5*0.3 = 2.93 where 0.6 is the slope of the characteristic curve and 0.3 is 1 f-stop distance on the log scale.

 

If you take the densitometry tables, the through-film light transmission at this density is about 1/10^2.93, and 6 stops below it would be 1/10^1.13.

So, highlights (on the negative) would be about 0.11% transmission, and shadows would be about 7.5%.

 

Well, it should be possible (?) I am not sure, but maybe...

Bill C,

 

I have spent some more time on your answer. Of course you have been helpful.

I just want to reply to some of your statements:

 

"On the log exposure scale, a change of 0.30 is just like a full f-stop exposure change; thus each 1.00 log exposure change is equivalent to 1.00/0.30 = 3.33 f-stops change. So if the slope is 0.55, this means the film density changes by 0.55 per 3.33 f-stop exposure change; this is also the same thing as film density changing by 0.165 per 1.00 f-stop exposure change"

 

Of course you are right. But the density scale can be converted into the light transmission through the film. It happens that it is a logarythmic scale too, with a base of 10. So, density 0 means 100% transmission (clear film), density 1 means 10% transmission, density 2 means 1% transmission etc. etc.

So, change in film density of 1 means 3.3 stops change in the light intensity that goes through developed negative.

What you said above is true - 3.33 stops change (10x) in exposure equals 0.55 change in density equals 10^0.55=3.54x in transmission through the negative equals 1.82 stops (2^1.82=3.54).

Of course it also should equal 3.33x0.55 = 1.83. All is OK, the density scale can be also expressed in stops (relating to film transmission rate).

 

"Now you can use your method above, but instead of using 0.55*(2+3), use 0.165*(2+3) = 0.82 film density change per 5 full f-stop exposure change. Note: you can't use the term "stops" to describe a density change. Also, because the film is unlikely to have a perfectly straight response curve, it is not a good idea to calculate a large difference like this; it would be better to read straight from the graph (ie, the characteristic curve)."

 

Film density measured in stops of transmitted light - see above.

Reading from the curve - you are right, but generally we try to keep the exposure within the straight portion of the curve, don't we ?

 

 

 

"I think you are having a common misconception here, but I'm not sure; you didn't state things clearly. If you think the density change of 1.5 to 1.8 results from an exposure change of 5 to 6 f-stops, this is NOT necessarily true. If you think that it IS true, you have not yet understood what the slope of the characteristic curve means. "

 

I did not say that that. Density change of 1.5 to 1.8 results in film transmission rate change by 5 to 6 stops. Now,assuming the slope of 0.6 this would result in exposure change of 5 (or 6) divided by 0.6, which brings in 8.33 to 10 stops of exposure.

 

Basically, all my questions revolved around one thing: 10 stops of exposure equals about 5 to 6 stops change in film transmission.

I guess my question could be formulated as: What is the useful range of film density ? And the second question: What is the range of exposure values that can be likely encountered in a single scene ?

How far can we move the range of exposure (keeping in mind the charascteristic curve of the film) without blowing the shadows or the highlights ? In other words without losing the data ?

 

 

"If I tried to address everything I'd like to, I would have to go one quite a bit longer, so I'm going to quit here. Hopefully I've helped rather than further any confusion."

 

You helped a lot. I never thought much about all this, and now I was forced to be a bit more logical in my explanations.

Now things are a bit simpler: 1.5 density change means about 5 stops reproduction on the film, which is a result of 10 stops difference in exposure. Since the useful (?) density of the film can reach 3.2 or so (for the print film) and subtracting the toe area of the curve, the latitude (as measured on the negative) is (3.2-1.5-0.5)=1.2 (density) equals 4 stops, or 4/0.55 = 7.3 stops of exposure. But this is pushing the envelope a lot. For most films it is -2/+3.

Thanks a lot to you and all the others ! Sorry for not being precise in my questions - it is easy to be obscure ;) !

Color negative films can hold a brightness range of 150:1,the bad news is the color paper can only hold 40:1.This is where light ratios,and reflectance ratios come into play.In order to comply with the 40:1 ratio of papers,we have to control & limit the reflectance & light ratios.This is why in most studio work 3:1 or 4:1 light ratios are commonly used.If you have a 9 stop reflectance range off the subject,we have a 36:1 brightness range,which being lower than 40:1 will work with most color paper.

>> Color negative films can hold a brightness range of 150:1,the bad news is the color paper can only hold 40:1 <<

 

Steve, I often see numbers like this quoted, let me try to convince you otherwise.

 

I just picked up a package of 1-hour lab prints off my desk. I then used a densitometer to read the darkest area on the top print. Density of all colors exceeds 2.20. Reading a white area gives densities of about 0.10. Here's how density converts to reflectance: density 2.20 correlates to a reflectance of 0.006 (6/10 percent). Density 0.10 indicates reflectance = 0.79 (79 percent). The ratio of reflectances, 0.79 over 0.006, is more than 130:1.

 

But don't just take my word for it, look up the conversion (optical density = inverse log of transmission, I think, or in this case, reflection) and measure some densities. I can assure you that typical professional color papers won't have any trouble matching these density ranges. If they do, someone is having problems with their developing process.

 

I won't quote density numbers for films, but I can assure you that a professional color negative film like Portra 160 can easily record a subject luminance range exceeding 1000:1. This can also be seen on Kodak characteristic curves.

 

I will grant you that the 1000:1 range is not normally printed; probably more like the 150:1 you said. And I will also grant you that when prints are examined in dim light, someone may not be able to tell the difference between "blacks" of 1.8 density vs 2.2 density and this could appear like a 50:1 range; however the limit is in the viewing condition, not the paper.

Hi again, Leszek. Ok, now I think I see what you were getting at before.

 

Let me say it in my words and see if this is right. You knew that the characteristic curve for a certain film had a slope of 0.55. The characteristic curve is normally in units of log exposure vs optical density. However, you wanted information in terms of f-stop equivalents. So, since you were expressing the initial film exposure in terms of f-stops (actually, the number of full f-stop exposure changes), you simply multiplied slope by the number of full stops exposure change. That is, you said that 0.55*(2+3) = 2.75 stops; in this case, you mean that the 2.75 stops has to do with the transmission of light through the film; specifically that the transmission had been cut in half a total of 2.75 times. Further, if you wanted to know the actual change in film density, you could simply multiply the 2.75 stops times the 0.30 conversion to get an actual density value of 0.825.

 

I agree this COULD be a workable system; in essence, you would be taking a standard characteristic curve and dividing both axes by 0.30 (the base 10 log of 2). But let me try to talk you out of this! All the published curves are in the log E (or log H) vs density format and existing densitometers have readouts in units of density. Also, many people already use the standardized terms of sensitometry. And lastly, using the f-stop term would confuse many people as to whether you mean film exposure vs processed film transmission.

 

You might even have to formally define what a partial f-stop change means. I know that many people explain with an air of authority what ½ f-stop means, but I would like them to point out the source of their authority (in a reputable reference). After quite a few years in the technical side of the industry, I have never seen such. If someone (in a technical manner) ask me the effect of ½ stop change, I always say, "let's agree that this means a log exposure change of 0.15". Otherwise, sometimes they have different ideas what ½ stop is.

 

 

>> ... but generally we try to keep the exposure within the straight portion of the curve, don't we? <<

 

Well, not necessarily so! Most color films have pretty straight line responses; this is important to allow color balancing without causing color crosses. But take a look at some published B&W film curves. Many deviate significantly from straight lines. These deviations (combined with print paper response) have quite a bit to do with nice looking prints. Especially the parts that handle subject detail going into shadow and highlight areas. These don't look like a lot on the graph, but can have a large effect on the print, depending on what the subject is.

 

I'm glad to see you have a better handle on the film response than I originally thought!