New answers tagged exposure
3
This comes down to understanding how logarithms work. Specifically in photography, the log base-2 of a doubling ratio equates to a difference of 1 stop.
We use logarithms to reduce repeated multiplication and division by growth factors to simpler linear addition or subtraction of stop values. If you are familiar with the decibel scale, this is the same ...
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As you know, cameras and other optical applications frequently use the f-stop system. The focal ratio (f/#) set in common usage is:
1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 -22 -32 -45 -64
The f-number set is grounded on the geometry of circles. The f/# is obtained by dividing the focal length by the working diameter of the aperture. The resulting ratio ...
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Why are not taken the real Area increments of 1.5, 1.33, 1.66 using the radius √(3/2), √(4/3), √(5/3) as the real fractional stops of +1/2EV; +1/3EV; +2/3EV?
Because of addition property.
Surely, you would like adding +1/2EV twice to produce +1EV which means 2*light.
Now, if +1/2EV was 1.5*light, then +1EV would be 2.25*light.
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I can't load the photo, but the statement 'vertical' marks has me thinking you might be seeing 'surge holes'.
They occur and line up with sprockets, and occur during over-agitation in development. Typically film is lightly rotated to ensure fresh chemistry is constantly in contact with the surface of the film. The film itself will deplete the chemicals ...
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I would add that it could be a sticky aperture problem. Whilst in DOF mode the aperture blades did close, during the fast shutter moment, it may not have enough time to react. I had a similar problem with one of my yongnuo lenses. I could temporarily remedy the problem by repeatedly doing a DOF preview until the aperture blades had loosened. When I took ...
2
If you use a directional light, and we make some assumptions, we can roughly calculate the shutter speed for the ISO and f stop in your comment.
As a sanity check, here is the calculation for my test setup. It is a stop off, so some of the assumptions aren't correct. This is only a ballpark calculation anyway.
More comments on my test setup. A typical ...
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Sunlight is exposure value = 15. It's about 100 000 lux.
So, with e.g. about 100 lux you get exposure value = 5, because every doubling of brightness increases exposure value by 1 and every halving of brightness reduces it by 1. 1000 is approximately 10 doublings (well, ok, to be exact, 10 doublings is 1024).
However, LED lights have their output specified ...
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