5
\$\begingroup\$

Let say that we have the following values, as an example:

Shutter speed: 1/40
Aperture: 2
ISO: 1000

How can I get the EV number for this exposure, extract (or add) an EV value (0.33, 0.66 etc.) and calculate back the new shutter speed (or ISO) value?

In fact I want to calculate how the Exposure Compensation will affect my values.

Improve this question
\$\endgroup\$
5
  • \$\begingroup\$ possible duplicate of What is the EV scale? \$\endgroup\$
    – mattdm
    Commented Aug 15, 2012 at 20:00
  • \$\begingroup\$ Also: What is one "stop"?. \$\endgroup\$
    – mattdm
    Commented Aug 15, 2012 at 20:02
  • \$\begingroup\$ Those aren't literally the same question, but if you understand what the scale represents, the math just falls out naturally. Plus, you'll see that it's conveniently designed so that you can make natural adjustments by just counting rather than calculating. (That's the point of using such a scale.) \$\endgroup\$
    – mattdm
    Commented Aug 15, 2012 at 20:04
  • \$\begingroup\$ Let me me try my interpretation of the question. If my camera is in manual mode and I change the exposure compensation, what are the new actual settings that the camera uses? Does the shutter speed change? Does the aperture change? BTW, does the ISO change? Same question, but with camera in full auto mode (yeah there are different auto modes), or shutter priority mode, or aperture priority mode. (Does the ISO ever change with exposure compensation?) \$\endgroup\$
    – Jim
    Commented Apr 25, 2013 at 18:59
  • 1
    \$\begingroup\$ @Jim, that will depend on camera model and the way auto ISO is implemented, as well as the specific mode chosen on that camera. If your camera is in full manual mode, changing EV compensation will not change any of the parameters, but may change the displayed meter reading depending on the camera. \$\endgroup\$
    – mattdm
    Commented Apr 25, 2013 at 19:35

5 Answers 5

Reset to default
6
\$\begingroup\$

The Exposure value can be calculated with the given Aperture and shutter speed using the following formula:
enter image description here (src: wikipedia)
where,
N: Aperture value
t: Shutter Speed

The above EV is for ISO 100 called as EV100. EV For subsequent ISO values is calulated with help of EV100 using the following formula:
enter image description here
where,
EV100: EV from the first formula
S : required ISO value

The Exposure compensation works by adding or subtracting the required Exposure compensation value with the current EV to get the new EV.
This new EV can be reverse engineered with the above formulas to get the Aperture and Shutter speeds.

Applying the values you have given:

EV100 = 7.32  //At ISO 100
EV1000 = 7.32 + 3.32 = 10.64  //At ISO 1000

On applying Exposure Compensation of 0.33:

EV(new) = 10.62 +0.33 = 10.95

Now reverse engineer this value to get N and t accordingly.

Improve this answer
\$\endgroup\$
8
  • \$\begingroup\$ Honest question: Does anyone actually understand any of that? Looks like an equation only Good Will Hunting would understand ;-) \$\endgroup\$
    – Mike
    Commented Aug 15, 2012 at 8:19
  • 2
    \$\begingroup\$ These are simple logarithm and squares. I will apply the given gives and add it. \$\endgroup\$
    – vivek_jonam
    Commented Aug 15, 2012 at 8:24
  • 2
    \$\begingroup\$ Nope, I'm not Good Will Hunting. :-) However I think I understand. :-) \$\endgroup\$
    – John Thomas
    Commented Aug 15, 2012 at 8:53
  • \$\begingroup\$ @Mike Yup - log2 is just the opposite of taking two to the power of some other number, everything else is just multiplication and division (n^2 is just n x n) \$\endgroup\$
    – Matt Grum
    Commented Aug 15, 2012 at 9:17
  • \$\begingroup\$ @Mike Also you can make use of the link for log calculations easily. 1728.org/logrithm.htm \$\endgroup\$
    – vivek_jonam
    Commented Aug 15, 2012 at 13:57
4
\$\begingroup\$

[I can neither comment nor vote (missing rep), but both the other answers with calculations are wrong]

Exposure Value (EV) is desribed as:

EV

where N is the relative aperture, t the exposure time (shutter speed) and S the ISO speed.

An aperture N=f/1.0, a shutter speed of t=1s and an ISO speed of S=100 corresponds to EV=0. An EV of +1 means half the light, and EV of -1 means double the light of EV=0 will be exposed to the recording medium.

Source: Wikipedia

[Note: This seems counterintuitive, but a brighter scene requires a higher EV to be correctly exposed, while a darker scene requires a smaller EV to be correctly exposed. ]

Given your numbers your total EV will be:

enter image description here

Now exposure compensation works the opposite as EV. A positive exposure compensation (brighten up picture) will decrease EV, a negative exposure compensation (darken picture) will increase EV. So you need to subtract your compensation from the EV.

Source: Wikipedia

Applying your exposure compensation of +0.33:

EV = 4 - (+0.33) = 3.66

You can reverse engineer this value in order to get either a new N, t or S. For example applying an exposure compensation of +0.33 (brighten up picture) will have the same effect as increasing your exposure time to t=1/32.

Improve this answer
\$\endgroup\$
3
  • 1
    \$\begingroup\$ @mattdm I am not sure what your point is. A high EV means bright sunlight (thats why smaller apertures and shorter exposure times correspond to higher EV values), while a low or even negative EV describes a very dark scene (wide aperture and long exposure time). EV=0 is a very dark scene. By exposure compensation of +0.33 let more light in (t=1/32 is longer than t=1/40) \$\endgroup\$
    – michaelk
    Commented Apr 18, 2013 at 22:31
  • 1
    \$\begingroup\$ Too often we say "a scene is EV=0" when we mean "a scene that would be properly exposed at EV=0." The EV is an expression of how much light we are allowing to expose our recording medium, not the brightness value of the scene. A higher EV # is a lower exposure setting that is appropriate for a brighter scene. \$\endgroup\$
    – Michael C
    Commented Apr 19, 2013 at 4:43
  • \$\begingroup\$ @MichaelClark and mattdm: thx for your feedback. I tried to update and clarify that. Feel free to edit my post as well to improve it. My main intention was to give a correct formula and calculation for the question asked, which is still wrong in the other two answers and I cannot comment on them (due to missing rep). \$\endgroup\$
    – michaelk
    Commented Apr 19, 2013 at 6:59
1
\$\begingroup\$

0 EV = f/1 at ISO 100 for shutter speed of 1 sec. See here.

  • f/2 is 2 stops less than f/1 = 2EV
  • ISO 1000 is appx. 3 1/3 stops more than ISO 100 = -3 1/3 EV
  • Shutter speed of 1/40 is approximately 5 1/3 stops less than 1 sec = 5 1/3 EV

So total EV for Shutter speed of 1/40, Aperture f/2 and ISO 1000

= 2 EV - 3 1/3 EV + 5 1/3 EV
= 4 EV

To add 1/3 EV for shutter speed, you just add 1/3 of a stop after 1/40 sec i.e. 1/50 sec.

To add 1/3 EV for ISO 1000, it becomes ISO 1250

Improve this answer
\$\endgroup\$
1
1
\$\begingroup\$

The entire "EV" system is designed to remove the need to do real calculation while shooting. A change of one stop in EV compensation means that one of the exposure factors — aperture, shutter, or ISO — will be doubled (to let in more light, for positive EV adjustments) or halved (do darken the exposure with a negative EV adjustment).

For shutter and ISO, that's a straightforward doubling or halving, for aperture, since the opening is a two-dimensional area, the change is by the square root of two, which you also don't need to calculate because there's a basic sequence that we all just memorize– f/1.4, f/2.0, f/2.8, f/4, f/5.6, f/8, f/11, f16.

So, if you have shutter speed 1/40, aperture f/2, and ISO 1000 and you dial in +1 EV compensation, one of these happens: shutter speed drops to 1/20, aperture opens to f/1.4, or ISO increases to 2000.

If you dial in -1 EV, one of these happens: shutter speed increases to 1/80, aperture closes to f/2.8, or ISO drops to 500.

If you use fractional EV compensation, like the 1/3rd or 2/3rd stops given in your example, it's slightly more complicated because the change is less than a full doubling or halving, but rather than worrying about the exact math, I think it's most helpful to just think of it as "okay, less than a full stop of change". That way, you don't have to fill your mind with arithmetic and can concentrate on your photograph and the actual impact of changing the exposure.

More on the EV scale at: What is the EV scale? and What is one "stop"?

Improve this answer
\$\endgroup\$
3
  • \$\begingroup\$ Re "one of these happens": Is it purely random which of the three parameters will change or by what is it determined? Also, how does one find out ex-post what happened? I just shot the same scene in manual mode (M) once with no exposure compensation, once with (-3 EV), and none of the three values appears to have changed. \$\endgroup\$
    – nutty about natty
    Commented Jan 10, 2022 at 19:50
  • \$\begingroup\$ photo.stackexchange.com/a/38472/31300 seems to answer this a bit... \$\endgroup\$
    – nutty about natty
    Commented Jan 10, 2022 at 19:53
  • \$\begingroup\$ Right, so it appears the ISO was blinking, as the (dark) scene would have required a higher ISO value than I had permitted... now I see the ISO value change (M) in response to the EV setting... question more-or-less answered then :) \$\endgroup\$
    – nutty about natty
    Commented Jan 10, 2022 at 19:58
1
\$\begingroup\$

Although the implementation of many features such as this is likely different across manufacturers, market classes and even generations of equivalent models:

As observed on a Nikon D3100, assuming that ISO is fixed:

  • In semi-auto modes (A and S) there is only one variable, which will be adjusted as requested.

  • In P mode, the program diagram shown in the manual will be followed, just with an adjusted EV.

  • In M mode, compensation can only be set from the menu, and it shifts the meter scale normally.

When auto ISO is enabled, it will behave as if the difference in final EV was due to an actual difference in subject brightness. Therefore, it can be assumed that exposure compensation is applied directly to the meter reading, not influencing the camera any differently from a variation in the physical LV being metered.

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.