Fuji Acros 100 For Long Exposures

IanBarber · May 29, 2016

I am planning a ini vacation over at the coast and want to do some lone exposure sea scape photography during the day using the Mamiya 645.

In order to get the effect of the water i want, i shall be using my 16 or 10 stop ND filter with a total exposure time of around 4 minutes.

Questions:

Has anyone had experience using 10 or 16 stop ND filters with Acros 100
Will I need to add 1/2 stop to the 4 minutes for reciprocity failure ?
At 4 minutes would adding 1/2 stop result in a 6 minute exposure ?

howardpan · May 29, 2016

Yes, you will need to add 1/2 stop per the datasheet. A 4 minute exposure becomes a 6 minute exposure.

MattKing · May 29, 2016

IanBarber said:
At 4 minutes would adding 1/2 stop result in a 6 minute exposure ?

Actually 4 minutes + 1/2 stop = 5.6 minutes, but 6 minutes will be pretty close.

Think of the progression 4, 5.6, 8, 11, 16, 22 .....

IanBarber · May 29, 2016

Thanks for the clarification

aiborshiuan · May 31, 2016

MattKing said:
Actually 4 minutes + 1/2 stop = 5.6 minutes, but 6 minutes will be pretty close.

Think of the progression 4, 5.6, 8, 11, 16, 22 .....

Matt, I'm not sure I understand. The progression you mention represents the size of the hole through which the light enters in relationship to the focal length. The numbers aren't spaced evenly because they represent circle size. Each stop represents doubling the size of the previous hole, thus doubling the amount of light passing through, assuming the shutter speed remains constant.

Time, on the other hand, is a straightforward progression. Double the time, double the exposure. An increase from 4 minutes to 8 minutes would double the amount of exposure, which is the equivalent of one stop in terms of the film's exposure to light. In this case, a half stop correction for reciprocity failure would be 6 minutes. If I'm wrong, please correct me.

Thanks.

MattKing · May 31, 2016

The common f/stop progression is of whole stops when it comes to lens apertures, but conveniently is also a progression of half stops when it comes to exposure times.

I use that progression all the time in the darkroom when I want test strips with half stop progressions.

It works great with camera exposure times, as long as you don't try to apply it to fractions of seconds. For those, use a calculator and multiply each time by the square root of 2 to get the next one. Which is exactly what the common f/stop progression yields (in an easy to remember form).

MattKing · May 31, 2016

aiborshiuan said:
An increase from 4 minutes to 8 minutes would double the amount of exposure, which is the equivalent of one stop in terms of the film's exposure to light. In this case, a half stop correction for reciprocity failure would be 6 minutes. If I'm wrong, please correct me

The first part of your statement is right, but the second is incorrect.
The half stop point between 4 seconds and 8 seconds is 5.6 seconds because we are using stops to determine exposure, not seconds. The progression uses "square root of 2" steps, and you multiply rather than add to get from one place in the progression to the next.

This is due to the fact that exposure calculations are logarithmic rather than linear.

aiborshiuan · May 31, 2016

MattKing said:
The first part of your statement is right, but the second is incorrect.
The half stop point between 4 seconds and 8 seconds is 5.6 seconds because we are using stops to determine exposure, not seconds. The progression uses square root of 2 steps, and you multiply rather than add to get from one place in the progression to the next.

This is due to the fact that exposure calculations are logarithmic rather than linear.

Thanks very much for the clarification.

aiborshiuan · Jun 5, 2016

MattKing said:
Actually 4 minutes + 1/2 stop = 5.6 minutes, but 6 minutes will be pretty close.

Think of the progression 4, 5.6, 8, 11, 16, 22 .....

Matt, i appreciate your posting the clarification. However, I am deeply embarrased to say that I still do not understand. I've been doing it wrong for 44 years, Without your help I would never have known. Could you possibly point me to a website or book that talks about time in terms of the numbers you've mentioned? I've searched the interwebnet thing but haven't found anything about the time progression that you so thoughtfully described.
If it's too much time and effort, I understand.
Thanks a lot for your help.

MattKing · Jun 5, 2016

aiborshiuan said:
Matt, i appreciate your posting the clarification. However, I am deeply embarrased to say that I still do not understand. I've been doing it wrong for 44 years, Without your help I would never have known. Could you possibly point me to a website or book that talks about time in terms of the numbers you've mentioned? I've searched the interwebnet thing but haven't found anything about the time progression that you so thoughtfully described.
If it's too much time and effort, I understand.
Thanks a lot for your help.

This is hard to answer with a clear reference, because so many of the references work mostly in the range of fractions of a second - the part of shutter speeds that are found on the usual shutter speed dial.

But this reference does at least say the same as I do. That when working with shutter speeds (and EI/ISO sensitivity), you use cube roots of "2" to calculate 1/3 stop changes in shutter speed (and EI/ISO), and square roots of "2" to calculate 1/2 stop changes in shutter speed (and EI/ISO).

Be sure to read down the page to the Shutter speed and ISO section. http://www.scantips.com/lights/fstop.html

NortheastPhotographic · Jun 6, 2016

"There's an app for that."

There really is though, and it's called Reciprocity Calculator. It has Acros among it's list of films and I've not been let down yet by it. I have it for iPhone but I think it's also available for Android. Makes life a little easier.

aiborshiuan · Jun 6, 2016

You guys are incredibly helpful. Matt, your explanation, although perfectly clear, for some reason did not provide an "Aha!" moment for me. The explanation with the time chart finally made me see the light. Thanks once again for your enlightenment.
Sperdynamite, thanks for the app info. Actually, I had already downloaded it, but stubbornly refused to use it until I fully understood the math behind it. Love that Acros. Hope it remains available forever.

Old-N-Feeble · Jun 6, 2016

aiborshiuan said:
You guys are incredibly helpful. Matt, your explanation, although perfectly clear, for some reason did not provide an "Aha!" moment for me. The explanation with the time chart finally made me see the light. Thanks once again for your enlightenment.
Sperdynamite, thanks for the app info. Actually, I had already downloaded it, but stubbornly refused to use it until I fully understood the math behind it. Love that Acros. Hope it remains available forever.

Let's start with 2 seconds and do the math to increase the time 1/2 stop and 1 full stop. The square root of 2 is 1.414 and here we go...

2 x 1.414 = 2.8
2.8 x 1.414 = 4

and a bit further...

4 x 1.414 = 5.6
5.6 x 1.414 = 8

The relationship between f/stop adjustment and time adjustment are mathematically identical... 2, 2.8, 4, 5.6, 8 and so on. I wouldn't concern myself with smaller numbers because no matter whether one errs to the high or low side, if you read your meter properly and do the math correctly, you'll always expose within 1/2 f/stop, all other variables discounted. So if you're worried about underexposure after your calculations, just open 1/2 stop (if shooting negs) and be happy.

Old-N-Feeble · Jun 6, 2016

And BTW, the same math applies to lighting too. If you want to decrease exposure by 1 f/stop then move the light away from the subject 1.414 times the original distance. To decrease exposure 2 f/stops then move the light away an additional 1.414 times that second distance. Or, to decrease the light falling on the subject 2 f/stops, move the light 2 times the original distance.

*** Look up "inverse square law".

Original distance = 4 feet.
Desired exposure decrease = 2 f/stops.

(4 feet) x 2 = (8 feet)

Same progression as both f/stop and time...

(4 feet) x 1.414 = (5.6 feet) = 1 f/stop decrease in light falling on the subject
(5.6 feet) x 1.414 = (8 feet) = 2 f/stops (total) decrease in light falling on the subject

*** Look up "inverse square law".

In other words, 2x distance = 1/4 the light.

Inverse Square Law:

Inverse of 2 = 1/2
Square of 1/2 = 1/4

So, 2x the distance of the light and have 1/4 the light (decrease of 2 f/stops).

*** Look up "inverse square law".

timekeeper · Jan 15, 2019

Hi guys,
Sorry for the necro bump. I was researching acros100 and came upon this thread. There is a misconception being promoted here so I thought it would be worth adding a correction to avoid the thread confusing future readers.

I've attached an image which is a graph of a circle's radius, x, vs its area, y.
When the radius is 1, the area is ~3 (pi). If the area is doubled to 6, the radius is ~1.4. Double the area again to 12 and the radius is now ~2. An area of ~24 gives a radius of 2.8, and so on. Pretty clear?

Same principle applies for inverse square law. Imagine a light emitting from the centre of a sphere. As the sphere's radius is doubled, it's surface area gets 4x bigger. That light or the surface will appear 4x dimmer as it is spread onto a larger area.

This doesn't apply to time though. If the aperture is constant, the light entering will be too. An analogy is water going through a pipe. If you double the amount of time the pipe is open from 4min to 8min, you will get twice as much water. If you only want 50% more water, you open it for 6min. Same as shutter duration, provided the other variables remain the same. Simple as that!

Nodda Duma · Jan 15, 2019

timekeeper you’re very correct!

Stop increases ultimately track irradiance at the image plane or radiant flux through the system in terms of aperture diameter for convenience. Remember, an optical system is a linear system. Increase the stop by 1/2, and by definition you increase the irradiance at the film plane by ~1.5x. The throughput (radiant flux, think water through a firehose) is actually dependent on aperture area, not aperture diameter. The stop numbers refer to relative diameter (stop # f/# = f/D), but throughput is proportional to area and thus proportional to stop^2. f/4 -> f/2.8 = 1.414x the diameter but 2x the aperture area and thus 2x the light coming through.

In the linear region of the film characteristic curve, you are thus compensating for throughput by reducing exposure or reducing radiant flux by ~1.5x.... not 1.414x.

You can verify this relationship with any type of imaging media or detector, so it’s not a function of the imaging media itself. Which makes sense because, as a linear system, the behavior of the system is independent of the output.

-Jason

Rich Ullsmith · Jan 22, 2019

Long time since I used Acros for pinhole, too expensive . . .it was more predictable further out on the curve than others.
Problem is, there are always anomalies and errors in metering (usually too small to matter for "normal" exposures) but when those errors however small are plugged into the perfect math detailed above, and multiplied several times, the small discrepancy becomes a large one, not necessarily rendering the whole exercise useless but obviously bracketing is called for, when you get minutes onto the reciprocity curve. The math is linear and the curve is not.

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