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I hear people talking about vibration reduction giving you extra 'stops' on a lens, thus giving you extra shutter speed to play with.

How do people work this out into an amount of stops?

E.g. by using the VR on this lens it gives you about an extra 1/4 stop in shutter speed.

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I don't know anyone who calculate those kind of thing, people just pull numbers out of their .. hemm .. I mean people produce approximate numbers based on their rich experience.

If you did want to calculate this the method would be:

  1. Test what is the slowest shutter speed that give you sharp hand-held images with IS turned off
  2. Test what is the slowest shutter speed that give you sharp hand-held images with IS turned on.
  3. Calculate the difference in stops.

Obviously the "slowest shutter speed that give you sharp hand-held images" depends on a lot of external factors (from camera holding technique to how tired you are at the moment) so that is why the numbers are mostly made up, sorry, I meant to say approximation based on experience.

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  • \$\begingroup\$ Experience does count for a lot though as the benefit of optical stabilisation is entirely situation dependent as well as photographer dependent. Dropping from 1/60 to 1/15 might be ok for an architecture shot where the subject doesn't move, but it's not going to help at all if your subject is a fidgety person like me. \$\endgroup\$ Commented Oct 3, 2014 at 8:01
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First we should explain what a 'stop' (or more correctly, an 'f-stop', the ratio of the lens's focal length to the diameter of the entrance pupil.) is: it means a doubling or halving of the light gathered during an exposure by opening or closing your aperture. But more generally, the same exposure effect (purely in terms of the amount of light, ignoring other effects) can be achieved, via the infamous exposure triangle, by changing iso, aperture or time. Double one of these, and you increase exposure by a stop.

Now, if you test a feature like VR, you take test shots with VR off and find the point where they start to get blurry. Now you turn VR on and repeat. You'll may find that you achieve the same quality with an exposure time 4 times as long. This means you gained two f-stops. You can close your aperture by two stops, and still get a sharp shot, thanks to the longer time.

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  • \$\begingroup\$ "...and still get a sharp shot." As long as the subject is not moving. \$\endgroup\$
    – Michael C
    Commented Oct 3, 2014 at 0:18
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The "vibration reduction", "image stabilization", "optical stabilization" or whatever the manufacturer calls it is a function that uses small optical elements to counter the small erratic movements that the photographer introduces when handholding the camera.

When handholding a camera without this stabilizing function a rule of thumb is to not use a shutter speed slower than 1/[focal length in mm] seconds. Using a slower shutter speed is likely to induce blur from the movement of the camera during the exposure. A stabilized lens counteract this to a certain degree.

Assume that a lens with stabilization turn of gives sharp results hand held at a shutter speed of a 1/60th of a second. Turning it on could very well push that limit to 1/15th of a second. This allows you to gather 4 times as much light that can be used for other things (such as closing down the aperture). This is what people mean when they tell you that the vibration reduction is giving you the extra playroom.

Of course the number of fstops that you gain by turning on the stabilization is not calculated by rather a result of the photographers experience as it depends on the shooting techique and the resolution of the camera among other things.

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  • \$\begingroup\$ I know this is what it means but how can they calculate that extra 1/45th of a second like you mentioned? \$\endgroup\$
    – connersz
    Commented Oct 2, 2014 at 10:06
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    \$\begingroup\$ @connersz First of by turning on the IS you dont gain an extra 1/45th of a second. It let's you increase the shutter speed by a factor of 4 (which is not the same thing mathematically). Regardless, it is usually not calculated but rather a matter of experience. Usually the IS of a modern camera let's you lower the shutter speed by a few stops (compared to IS turned off) before the shots get blurry again. This is not a hard limit since it depends on the resolution of the camera, the steadiness of the photographer etc. \$\endgroup\$
    – Hugo
    Commented Oct 2, 2014 at 10:14
  • \$\begingroup\$ @connersz The 1/45 second is simply the difference between 1/60 second and 1/15 second, but it is more accurate to say two stops, because the difference between 1/500 second and 1/125 second is also two stops (four times the amount of light strikes the sensor), even though the difference in time is only 1/166 second. The stop scale is logarithmic, not linear. \$\endgroup\$
    – Michael C
    Commented Oct 3, 2014 at 0:22
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    \$\begingroup\$ @MichaelClark It is true that the difference between 1/125 second and 1/500 second is 1/167 second, but I have to clarify (for other readers) that when you refer to connersz "1/45 second" you're not saying that it actually is the difference between 1/60 second and 1/15 second (it is 1/20), but rather explaining what commersz though it represented. \$\endgroup\$
    – Hugo
    Commented Oct 3, 2014 at 0:41
  • \$\begingroup\$ @Hugo Yeah,I totally didn't catch that. Still, the concept remains: two stops difference is not a linear unit of time, but rather a unit exponentially proportional to the beginning value. \$\endgroup\$
    – Michael C
    Commented Oct 4, 2014 at 0:54

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