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VTphotog Apr 1, 2010 9:15 PM

To me, the practical advantage of HDR is that by having at least one of the frames exposed for the shadows, the noise in the darker areas will be reduced. To do this with a single RAW image, as some do, means having to amplify the dark areas more, amplifying noise as well. If you then do noise reduction, you also eliminate some of the detail.
Tone mapping of RAW files is equivalent to using a higher ISO setting for the dark areas, with concomitant noise.

brian

Alan T Apr 2, 2010 6:41 AM

Quote:

Originally Posted by Bynx (Post 1073629)
Well ac, I sure didn't mean all that easy. But there are constants we are dealing with so there is an answer. A jpeg will cover 9 EV of the possible 36 EV which will actually be closer to 30. A Raw file would cover 12 EV?

You have hit the nail on the head here Bynx, my old friend. I may be wrong but I think this question is much simpler than everyone is making it, because we're dealing with digital data recording.

Let us consider for the sake of argument a B&W jpg image. The blackest black it can record has RGB values of (0,0,0), and the whitest white has (255,255,255). What light intensity relates to each of those 256 levels of grey (8-bits) depends on the exposure, and how it's recorded by the particular sensor we happen to be using. The bigger the dynamic range of which the sensor is capable, the more shades of grey could in principle be recorded. But we are going to put them straight into a jpg file with only 256 levels, or a RAW file with however many it can store.

We can record more levels by adjusting the exposure up and down, and doing HDR on them. How many we record altogether depends on how many exposures we include between the longest and the shortest, and how much patience we have.

At the short exposure end, the most light we'll encounter entering our notional lens is, as discussed , a direct shot of the sun's surface from outer space, and there'll be more the closer we get. In practical terms on Earth, I suppose it'll be on top of high mountain. However, our sun isn't a particularly hot star.

At the long exposure end the dimmest we'll get is no photons at all, and to get there we need a subject at absolute zero. The ratio between that and the intensity of any light at all is infinite, and Mark R has already given us the theoretical answer to Bynx's question - infinity percent.

One practical limit would be how long you're willing to wait for your long exposure to end. It'd be best to finish before the sun swallows the Earth, as it will in due course.

We really need a competent astrophysicist plus a current optical astronomer to answer this question realistically. Someone in an observatory on top of a high mountain should be able to tell us what is the intensity ratio between the brightest and dimmest objects they can practically observe. The low limit will be set by the thermal noise in the liquid helium-cooled detectors.

Prof. Brian Cox would do nicely for half the team if anyone wants to drop him a line. He works on the large Hadron Collider at CERN, Geneva, and has recently been presenting a TV series here the Solar System, in which he occasionally waxes lyrical, in his other capacity as a retired rock musician.
( http://en.wikipedia.org/wiki/Brian_Cox_(physicist ).

TCav Apr 2, 2010 7:30 AM

If we stick with B&W (for the sake of simplicity), then an 8 bit image is limited to 256 descrete grays, within the dynamic range of the image. HDR doesn't change that. It just brightens the darker areas of the image, and darkens the brighter areas, so detail in those areas can be seen instead of being lost in the shadows and highlights. The image is still limited to 256 descrete grays; we're just arbitrarily picking and chosing the exposure values we want applied to different areas of the image. We don't record more levels by adjusting the exposure settings; we just choose to position those levels at different places within the dynamic range of the image.

A 12 bit (4096 descrete grays) or 14 bit (16384 descrete grays) image simply splits the dynamic range of the image sensor into finer parts. It doesn't increase it.

ac.smith already gave us a significant list of the brightest and dimmest objects in the sky, in http://en.wikipedia.org/wiki/Apparent_magnitude. But these are all sources of light (either directly or indirectly.) There is no value given for the "apparent magnitude" of interstellar space, so even this doesn't answer the question. The dimmest object in the night sky is still transmitting or reflecting light, so while it is quite dim, it isn't actually "dark".

ac.smith Apr 2, 2010 7:58 AM

Quote:

Originally Posted by Bynx (Post 1073629)
Well ac, I sure didnt mean all that easy. But there are constants we are dealing with so there is an answer. A jpeg will cover 9 EV of the possible 36 EV which will actually be closer to 30. A Raw file would cover 12 EV?

Sorry, depending on memory and mental math. Looked again at DPReviews most recent tests. The best current jpeg engines across the board on 4:3, APS-C and full frame seem to deliver between 8.5 and 8.8 EV range although some still don't reach 8. Again, across the same range of sensors RAW dynamic range is about 12. So we could probably achieve a 36 EV range HDR with 3 raw frames and might just make it with 4 jpeg frames and definately with 5. We would also need ND filters on one end of the exposure range and long exposures on the other.

Granted your question is a hypothetical one but would we want to picture the sun, high in the sky as just a round orb?

A. C.

ac.smith Apr 2, 2010 8:21 AM

Quote:

Originally Posted by TCav (Post 1073845)
ac.smith already gave us a significant list of the brightest and dimmest objects in the sky, in http://en.wikipedia.org/wiki/Apparent_magnitude. But these are all sources of light (either directly or indirectly.) There is no value given for the "apparent magnitude" of interstellar space, so even this doesn't answer the question. The dimmest object in the night sky is still transmitting or reflecting light, so while it is quite dim, it isn't actually "dark".

In a more simplified manner I caveatted the brightness (darkness) as you have. My choice of 9 magnitude as black was was based on the 7-8 being the dimmest object discernible in the darkest sky on earth with the unassisted eye. 9 is therefore half the brightness of the least discernible object but not 0. None the less about as black as we'll encounter in nature.

In the past absolute black was simulated in labs with a closed, curved, tapered tube (specifically a cow's horn) painted glossy black on the inside. Why glossy? The flattening agent in paint (silica - sand) is reflective.

A. C.

ac.smith Apr 2, 2010 8:49 AM

Alan T

Good to see you chime in, thanks. As you can tell my issue with the original post was asking for a linear conversion on quantities we typically measure on a logarithmic scale. Nor has anyone attempted that yet although it's not germane to the real question of quantifying the dynamic range advantages of HDR.

A. C.

TCav Apr 2, 2010 9:34 AM

Quote:

Originally Posted by ac.smith (Post 1073869)
In a more simplified manner I caveatted the brightness (darkness) as you have. My choice of 9 magnitude as black was was based on the 7-8 being the dimmest object discernible in the darkest sky on earth with the unassisted eye. 9 is therefore half the brightness of the least discernible object but not 0. None the less about as black as we'll encounter in nature.

Yes, but that presupposes a Persistence of Vision of 1/25 second, which doesn't apply in still photography. If you use a shutter speed of 1/12 second, then objects of magntitude 9 become visible, and if you use a shutter speed of 1/6 second, magnitude 10 objects become visible, etc. (...presuming an appropriate dynamic range.) And for HDR photography, such shutter speeds are not unheard of.

Alan T Apr 2, 2010 10:06 AM

Thanks, 'Tcav' and 'ac.'

Right, now we're getting somewhere (even if it is rather a discussion about the number of angels that can dance on the head of a pin).

So, to give Bynx a reasonably quantitative answer, I suggest we find out the magnitude of the dimmest astronomical object ever detected with an optical telescope (and therefore probably 'photographed' (so-to-speak) with a fancy digicam for publication somewhere) and give him the number (plus 1.0 so that it's optically invisible) of orders of magnitude between that and the brightest object (presumably the Sun).

Then he can choose how many exposures to take, to cover that spread of dynamic range, and compress them artistically (or auto-HDR them) into a nice picture. I fear it might take a while.

It would be much more fun if we organised a trip to a mountaintop in Hawaii (a) to meet selvin, and (b) to see for ourselves.

ac.smith Apr 2, 2010 10:37 AM

Quote:

Originally Posted by TCav (Post 1073898)
Yes, but that presupposes a Persistence of Vision of 1/25 second, which doesn't apply in still photography. If you use a shutter speed of 1/12 second, then objects of magntitude 9 become visible, and if you use a shutter speed of 1/6 second, magnitude 10 objects become visible, etc. (...presuming an appropriate dynamic range.) And for HDR photography, such shutter speeds are not unheard of.

Absolutely, photographic technology can and has over it's history made visible what is not visible to the unassisted eye.

A. C.

TCav Apr 2, 2010 10:52 AM

But, once again, we wouldn't want to have direct sunlight in a photo, and we certainly wouldn't want to visually compose an image that included the sun, without using at least an ND4 filter. So we should subtract 4 from the magitude of the sun. ... or rather, add 4 to it.


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