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Why are my pictures over exposed when taking pictures of running water? Answered

My camera keeps telling me the 'subject is too bright' when trying to take pictures of running water. To get thet 'smokey look' the shutter needs to be left open for say 3 seconds or more but my camera says the subject is too bright consequently the results are over-exposed. Even on a very dull day with ISO at a minimum and a dark filter on my lense. What am I doing wrong?


The answers below are helpful but I think there could be easier ways to solve this problem:
1 - Turn off the flash. If you can't, block it with your finger or black electrical tape.
2 - Take the pictures from different angles.
3 - Use a polarized lens.
4 - Block a light source with a black sheet stapled to a frame.

Bracket your picture ie normal exposure,underexposure and overexposure and fine tune from there.

Put it on shutter priority and the camera should work it out for you.

If it's messing around go on manual, shutter at three seconds, minimum ISO then change the aperture down until it looks good, I'm guessing it's a digital camera, if not you don't have the luxury of giving a guess then looking at the resulting picture.

One thing remember aperture goes backwards, F4.0 really means F4.0/1 so a high number means small aperture and less light.

Sorry I just caught the bit about ISO in yours there, a really small aperture like, F32 might be needed, well maybe not that far but try it and see.

Part of the problem is that the water is transparent, a lot of light gets through it and reflects off it, depending on angle, meaning it's bright as hoohah

If playing with the aperture ruins any depth of field effects you want try doubling up on filters.

Also the cameras light meter may be metering wrongly, are the photos actually turning out overexposed?

One thing remember aperture goes backwards, F4.0 really means F4.0/1 so a high number means small aperture and less light.

Correct, in part...but not the math, or the F stop relationship. F4.0/1 = 4, so F8.0/1 would equal 8, a higher number.

The "F" stop number means the focal length (F) divided by the "stop" number. That's why F stops are written with a "slash", like "F/8". So the diameter of the aperture for a particular lens gets smaller as the F stop # gets bigger. For a 50mm lens, for instance:

F/16 = 50mm / 16 = 3.125mmF/8  = 50mm / 8  = 6.8mmF/4  = 50mm / 4  = 12.5mm

As killerjackalope says, the larger the F stop number, the smaller the aperture, and the less light transmitted to the film / image sensor.

It's possible "Guest's" camera just can't handle long time exposures, possibly a failure of it's programming--and the assumption of the manufacturer that most people will not use a tripod, etc., a necessity for long exposures.

Sorry... Poor wording and I didn't think to do the maths that would have quickly shown that bad wording up - higher number was referring to stop number rather than the end result of the equation... The point you bring up about programming could be it, though it may just be a badly calculated relationship between shutter speed and amount of light, making it overexpose, I had a camera that managed to permanently overexpose or take ridiculously long shutter speeds in medium light but exposed accurately in bright light, granted it only cost 20 quid...

Yep, I know the "end result" wasn't intended to mean anything in that context...

F stops are a simple ratio, and I think it benefits people to understand what the "F" stands for...and what the ratio represents.

Note that the F/stop ratio can represent lots more:

For instance, F stops illustrate the inverse square law -- a doubling of the F stop (4 to 8, 8 to 16) represents not twice the exposure, but four times the exposure (or quantity of light.)

So the in-between stops: 4 - 5.6 - 8 or 8 - 11 - 16, are the half-way points. F/5.6 is twice the exposure as F/4, and one half of F/8.

The cool thing is that this is dictated by the inverse square law: Since the relationship is a squared number, the intermediate steps are the square root of 2, or approx 1.414:

4 X 1.414 = ~5.6
5.6 X 1.4 = ~8

Use 1/2 the square rt of 2 for descending values:

8 X .707 = ~5.6

So what good is all this math stuff?

Well, that sequence of numbers (2.0, 2.8, 4, 5.6, 8, 11, 16, etc.) can be used with studio lighting, for instance.

If the main light is 8 feet from the subject, and you want a fill light ratio of one half, then place the fill light 11 feet away (assuming both lights are identical, of course.) Lighting ratios are "prefigured" and printed on any lens!

Same with black and white enlarging: My profs used to say that Fstops represent a step of "significant change" when adjusting printing times. If 20 seconds is too light (not really light, and not "close enough"), then the next "significant" step is 28 seconds.

Of course, with enlarging the printing times won't conform directly to the Fstop numbers, they are just a guide.

It works with sizing prints, too. Sizing an 8 in. print up to 11 in. would require twice the exposure.

(all this semi-useful information is what a bachelor's degree in photo illustration will get ya...)

I would have thought that a decent camera with a dark filter (commonly 3 stops) would be able to accommodate a 3 second exposure on a dull day so something is a bit strange. Possible reasons are that you have a cheap and nasty camera on which the light sensing is not off the CCD so the filter would not affect the exposure. I doubt if this is the case as even the most inexpensive compact camera does this nowadays. Another possibility is that the background is very dark in colour. This can fool the camera into thinking the scene is a lot darker than it is, and overexpose the shot (the opposite of under-exposure in snowscapes). If your camera has a spot-metering setting, try getting a reading off a mid-toned feature in the frame and locking this value to take the shot. If your camera has a manual mode, play around with this to get a feel for what the correct exposure should be. BTW, what make / model of camera do you have?