How does shutter timing work on an FE?

[Ariston]

To find the difference in stops ∆ from an initial time t1 to a second time t2, calculate

∆ = 1.4427*ln(t2/t1)

Example 1: Moving from t1 = 4 seconds to t2 = 16 seconds, we have

∆ = 1.4427*ln(16 seconds/4 seconds) = 2.00, a 2-stop increase

Example 2: For t1 = 16 seconds to t2 = 2 seconds

∆ = 1.4427*ln(2 seconds/16 seconds) = - 3.00, a DECREASE of 3 stops (indicated by the negative sign).

For your question of posts #1 and #23, t1 = 8 seconds and t2 = 5.5 seconds, so

∆ = 1.4427*ln(5.5 seconds/8 seconds) = - 0.54, a decrease of 0.54 stops.

This doesn't make sense to me (and I'm not just talking about my mathematical inadequacies, which are considerable!). If changes in time affect exposure in a linear fashion, i.e., every doubling of the time increases exposure by one stop, and every halving decreases exposure by one stop, then how does decreasing time by 1/3 not decrease exposure by 1/3 stop?

I was so confused I even pulled out my D700 and set it on 1/3 stop for exposure adjustment, and decreasing an 8-second exposure by 1/3 stop gave me 6 seconds, which is close enough to the 5.5 seconds I was talking about. One-half exposure change from 8 seconds made it 4 seconds on my D700, so it still seems linear to me.

I'm not being argumentative or stubborn. I genuinely don't understand.

I understand the reasoning behind the f/stop differences Matt was explaining above, because that is dealing with a diaphragm, so I don't expect it to be linear.

 

[MattKing]

I understand the reasoning behind the f/stop differences Matt was explaining above, because that is dealing with a diaphragm, so I don't expect it to be linear.

But you are missing the fact that the difference between 8 seconds exposure and 11 seconds exposure is 1/2 a stop, not 3/8 of a stop.
Exposure changes logarithmically, not linearly.

 

[Ian C]

It has to do with the nature of the f-stop. Here is how logarithms got involved.

When Hurter and Driffield investigated the unknown relationship between exposure and the developed density of photographic emulsions they needed to construct a simple LINEAR function that would make the calculations easy. Finding such a function is useful in constructing a mathematical model that corresponds to the observations. If such a function can be found (not always possible), it provides a simple, useful tool with which to analyze and describe the phenomenon under study. Much of elementary physics (which forms the basis of engineering applications) exploits this idea of finding a linearizing function of the independent variable.

H&D found that they could construct a linear model when they plotted developed density as a function of the LOGARITHM of exposure. The relationship isn’t perfect and the linearity begins to fail with very low exposures, forming the “toe” at the left end of the curve, and again with overly high exposures, which forms the “shoulder” of the curve at the right.

https://en.wikipedia.org/wiki/Hurter_and_Driffield

https://en.wikipedia.org/wiki/Sensitometry

https://encyclopedia2.thefreedictionary.com/Hurter-Driffield+curves

Note: In the formula of my post #25 in this thread, I simplified the equation by substituting the approximate value 1/ln(2) = 1.4427 into the equation

Δ = ln(t2/t1)/ln(2)

The numbers generated by the camera in post #26 are due to round-off and the limitations of the system. The difference from 8 seconds to 6 seconds is a decrease of 0.42 stops. A 1/3-stop decrease is, of course, 0.33 stops. These differ by roughly 1/10 stop, probably not enough to cause a problem.

 

[Chan Tran]

This doesn't make sense to me (and I'm not just talking about my mathematical inadequacies, which are considerable!). If changes in time affect exposure in a linear fashion, i.e., every doubling of the time increases exposure by one stop, and every halving decreases exposure by one stop, then how does decreasing time by 1/3 not decrease exposure by 1/3 stop?

I was so confused I even pulled out my D700 and set it on 1/3 stop for exposure adjustment, and decreasing an 8-second exposure by 1/3 stop gave me 6 seconds, which is close enough to the 5.5 seconds I was talking about. One-half exposure change from 8 seconds made it 4 seconds on my D700, so it still seems linear to me.

I'm not being argumentative or stubborn. I genuinely don't understand.

I understand the reasoning behind the f/stop differences Matt was explaining above, because that is dealing with a diaphragm, so I don't expect it to be linear.

But if you use a stopwatch and time the shutter of your D700 you will find out that when set at 8 it's 8 but when set at 6 it's isn't 6.

 

[Ariston]

My question is why is it linear at full stops, but not half stops?

(People keep mentioning f-stops. I am talking about shutter speeds only.)

 

[shutterfinger]

My question is why is it linear at full stops, but not half stops?

https://en.wikipedia.org/wiki/F-number
For reference download the free third stop chart at http://www.photographyuncapped.com/articles/photography/iso-shutter-speeds-f-stops/

Basically its the math involved to accurately record a given light level to a given density on film. The characteristics of light and how it works determines the math involved. The f stops and shutter speeds are rounded or truncated to an easy to remember and use numbers rather than the actual mathematical fractional or odd number that occurs from the math. Would you like to use 1/9s, 1/16s, 1/31s, and similar for shutter speeds?

 

[MattKing]

My question is why is it linear at full stops, but not half stops?

(People keep mentioning f-stops. I am talking about shutter speeds only.)

I'm not referring to f/stops when I'm talking about exposure times.
I'm referring to stops (actually half stops) when I'm talking about exposure times.
It just happens that the sequence we are all familiar with - 2.8, 4, 5.6, 8, 11, 16, 22, 32 etc. - works with both exposure times (in a half stop progression) and apertures (in a full stop progression).
If you want to do it 1/3 stops, just read the progression of ISO values on a meter that actually lists them by 1/3 stops, and use them.

 

[Chan Tran]

My question is why is it linear at full stops, but not half stops?

(People keep mentioning f-stops. I am talking about shutter speeds only.)

If you reason your way then how much different between 4 sec and 8 sec?

 

[Chan Tran]

1/3 stop less exposure from 8 seconds is 6.35 seconds.
2/3 stop less exposure from 8 seconds is 5.04 seconds
1 stop less exposure from 8 seconds is 4 seconds.
5.5 seconds is 0.54 less exposure than 8 seconds.

 

[Ian C]

In answer to “My question is why is it linear at full stops, but not half stops?” of post 30:

f-stops and shutter speeds are intrinsically connected. You have apparently assumed that because the standard shutter-speed sequence is formed by halving the speeds when successively reducing exposure by 1 stop at a time and by doubling for successive 1-stop increases that this is evidence of a linear relationship.

It certainly looks that way numerically at whole-stop differences, but this is misleading. Perhaps the following will help.

The equation that describes the difference in f-stops (near the end of post 28) can be reworked into an equivalent form as

t2 = (2^Δ)t1

In this form we have the new time t2 as a function of the difference in stops Δ and the initial time t1. It’s non-linear, since the coefficient 2^Δ is an exponential function of Δ. The main point is: Time is a NON-LINEAR function of the difference in stops Δ.

When Δ takes on positive or negative integer values, you get the familiar doublings and halvings of time. When Δ is not an integer you get intermediate values that might seem “odd”, even though they’re correct.

For the numerical examples of post #34,

t2 = [2^(-1/3)]*8 seconds = 6.35 seconds (negative exponent denotes a decrease)

t2 = [2^(-2/3)]*8 seconds = 5.04 seconds

t2 = [2^(-1)]*8 seconds = 4 seconds

t2 = [2^(-0.54)]*8 seconds = 5.5 seconds

At whole-stop differences using two examples of post #25

t2 = [2^(2)]*4 seconds = 16 seconds

t2 = [2^(-3)]*16 seconds = 2 seconds