19

I understand that dynamic range is the difference between the brightest and darkest lighting, but why dynamic? Why not static range?

Jossie Calderon
  • 288
  • 2
  • 6

5 Answers5

44

Because "dynamic range" does not refer to a range that is dynamic, but rather to a range of dynamics. For example: range of luminosities or reflectances in photography, or a range of amplitudes in acoustics.

Note that "dynamic" comes from Greek δύναμις meaning "power".

Kahovius
  • 2,758
  • 12
  • 25
  • 5
    Yes. The term originated in audio engineering and spread to other fields from there. The relation to "dynamics" in music is entirely intended. – hobbs Jul 22 '20 at 02:44
  • 1
    @hobbs: Especially considering that light, like sound, takes the form of waves that have measurable amplitude. – supercat Jul 22 '20 at 12:15
  • 2
    Note that unlike range, dynamic contrast ratio vs static contrast ratio (where "dynamic" and "static" are adjectives) is very much a thing when discussing capabilities of monitors. – Ruslan Jul 22 '20 at 14:26
  • 1
    @supercat I don't think that is a valid comparison. A louder sound is an acoustic wave with a higher amplitude than a lower sound, moving more air. But a brighter light source is not emanating a wave with a higher amplitude, instead it sends out more photons which can or cannot be accurately counted by a sensor (determined by its dynamic range). – smcs Jul 23 '20 at 09:41
  • @smcs actually it does: in the classical regime more photons (on average) means higher amplitude of the EM wave. And it's quite similar to acoustic wave that in a crystal can be described in terms of phonons. – Ruslan Jul 23 '20 at 10:00
  • @Ruslan But in the classical regime, the amplitude of many photons on average is conceptual, there is not an actual wave with crests and troughs that hit a detector, as is the case for acoustic waves. Would you agree? – smcs Jul 23 '20 at 10:12
  • @smcs again, in classical regime Maxwell's theory is complete. There are crests and troughs in electric field that move along the direction of propagation, and the oscillations due to these are the ones that result in excitation of the electrons in the detector. It's typical to treat EM field as background in QM (unlike QED), and in many cases it's sufficient for calculations. – Ruslan Jul 23 '20 at 10:26
  • @Ruslan Of course there are crests and troughs in EM radiation, on a photon level. But they all have the same amplitude, thus the amplitude is irrelevant here for the concept of dynamic range of a system (e.g. a digital camera). – smcs Jul 23 '20 at 11:00
  • @smcs photon level and classical regime are not well compatible: number of photons is not defined when electric&magnetic fields are, so photons don't have amplitude (except for probability amplitude). And when discussing EM radiation as a wave (namely, comparing EMR with sound), photons are irrelevant anyway. They only become significant when we need to account for shot noise or the mechanism of absorption of EMR by the sensor. – Ruslan Jul 23 '20 at 14:35
  • @smcs But camera sensors do not generally measure the amplitude of the wave propagated by a single electron. They measure the amplitude of the entire field strength at a specific location (each photosite on the sensor's surface) over a specific time interval. – Michael C Jul 23 '20 at 23:04
3

It's not the difference, it's the ratio.

A ratio does not have a fixed 'bottom' or 'top', therefore is dynamic rather than static.

Ref: Wikipedia - Dynamic Range

Tetsujin
  • 23,252
  • 3
  • 46
  • 97
  • 6
    To be fair, since DR is usually expressed in logarithms, the logarithm of a ratio is the same as the difference of the logarithms of the parts of the ratio. /nitpick. =) – scottbb Jul 21 '20 at 22:42
  • You sure? The answer below me says different and sounds more persuasive. – Jossie Calderon Jul 21 '20 at 22:56
  • @JossieCalderon which answer are you referring to? (hint, you can click the little 'share' text and copy/paste the actual link the answer, which is not ambiguous) – StayOnTarget Jul 22 '20 at 20:54
  • @JossieCalderon [scottbb]'s statement is a mathematical fact for the type of numbers that are relevant here. – haneefmubarak Jul 23 '20 at 03:26
  • 4
    @JossieCalderon Please remember that the order of answers is a preference set by each user. Every one of us sees a different order, possibly. – jpaugh Jul 23 '20 at 22:36
-3

Is the answer more simple?

Static range suggests a fixed scale regardless of external influence.

Dynamic range suggests the scale is affected by external influence.

So the range at say, 12 noon on a bright sunny day would have a wider difference than 12 noon on a cold dull grey day.

The scale (say 0-99) would still represent the contrast/ratio between darkest and lightest, but the differences between the individual increments of the scale would be dynamic, based on physical lighting conditions.

aml
  • 1
  • 2
  • 1
    The word "static" doesn't really come into play. That is, the term "dynamic range" is a quantifiable property of a signal, image, etc. There is not a corresponding "static range", nor is "dynamic" in opposition to "static" in this context. Analogous to music, we don't talk about the "statics" of a song, passage, etc. We talk about the dynamics, the difference between the loud parts and quiet parts, especially their quick or abrupt change between highs and lows and back. – scottbb Sep 05 '20 at 00:49
  • I wasn't suggesting there was a term 'static range', but I disagree that there isn't one. The full notes on something pitch perfect is a fixed, ie; static value. Middle C is always a specific frequency. What I said was the min/max values of a given scene are dynamic, therefore the increments between those 2 variables will be dynamic. Light is not fixed, so brightness and contrast equates to your loudness/volume within sound. – aml Sep 05 '20 at 09:56
-4

Seemingly-constant illumination in fact represents a mixture of high-frequency waves. If the intensity is low enough, and one exposes a piece of film, these waves will cause a sequence of randomly-distributed changes to the states of the crystals in the emulsion. While some kinds of detection apparatus may seem to regard light as a steady state, what actually happens within the apparatus is that light will knock something in the apparatus out of its rest state, then the apparatus will reset it, then light will knock it again, etc. The processes associated with light are generally so fast that, by the law of large numbers, they appear uniform, but underneath that apparent uniformity things are in fact constantly changing.

supercat
  • 451
  • 2
  • 7
  • This sounds reasonable for the physics / chemistry, but what does it have to do with the range of light intensities that a given film or sensor can actually measure? And more importantly, how does this explain what dynamic range is? e.g. 8-bit linear grayscale luminosity can only represent non-zero pixel intensities from 1 to 255, a pretty pathetic ratio of brightest to dimmest not-fully-black intensity compared to chemical film. – Peter Cordes Jul 23 '20 at 08:02
  • 1
    @PeterCordes: The question was "why not static range". None of the other answers seemed to address the fact that light detection isn't a static process. I didn't see anything in the question about why digital sensors would only count 255 linear values. With light sensing as with audio, a limiting factor of dynamic range is often the noise floor. If a camera supports multiple ISO settings, it may use values from 0-255 to represent either 0 to 100% intensity, or 0 to 25% intensity, but in the latter case end up with much more noise on the picture. If the camera were to try to handle... – supercat Jul 23 '20 at 16:13
  • ...a 4095 or 65535 intensity levels on each pixel instead of 255, the lower bits would end up being mostly noise. Unfortunately, while the useful information content in those bits would be much lower than in the upper bits, the amount of storage needed to record them losslessly would be much greater than for the upper bits. – supercat Jul 23 '20 at 16:15
  • I was just giving a concrete example of dynamic range (which already has numbers built in because it's digital, assuming the noise floor is below 1; good point about noise). But anyway, I don't see how your answer is relevant to dynamic range of a measurement of light intensity averaged over the exposure duration. Normal photography happens in intensity ranges where the number of photons per pixel or per grain of film (times their chance to actually react in a way that makes a diff) are high enough that averaging works very well. – Peter Cordes Jul 23 '20 at 16:31
  • @PeterCordes: Taking a picture with a still camera is fundamentally no different from taking a bunch of pictures with a video camera (possibly at a very fast or very slow frame rate) but then throwing all but one away. As for why camera sensors are different from film, unexposed film seeks to minimize the amount of stuff that photons will interact with without being detected. Camera sensors need to contain a lot of stuff to handle the transport of picture information out of the sensor, thus reducing the fraction of photons that will be usefully detected. – supercat Jul 23 '20 at 16:51
  • @supercat Not necessarily. Very high speed film had gaps between grains that didn't capture every single photon falling on the film, either. – Michael C Jul 23 '20 at 23:10
  • @MichaelC: No film is going to be perfect, and when using high speed films a change to the state of one molecule will likely trigger changes to others nearby during development (which might obscure the fact that those other molecules had also been hit by light) but film is typically much better at recording photons than typical electronic sensors which are designed for high-speed readout. – supercat Jul 24 '20 at 21:11
-6

Multiple frames are captured and extra information (as in more lighting) is added to some frames and reduced in others. The frames are then combined into one to amplify the bright and dark colors. This gives you a high range of saturation which can be changed, so that's why the HDR stands for High Dynamic Range.

Jossie Calderon
  • 288
  • 2
  • 6
  • 9
    I'm confused, you answer your own question, but your answer is only about high dynamic range and not about why it's called dynamic range. Your answer does not really fit to your question. – Matt Jul 21 '20 at 17:54
  • @Matt: Within the about is the why. I literally say why it's called HDR. – Jossie Calderon Jul 22 '20 at 20:04
  • 2
    You explained the High part of HDR, but your question title and body are about the base term "dynamic range". Every sensor and way of representing an image (e.g. 8-bit RGB samples) has a some fixed dynamic range. You don't need to be doing multi-exposure HDR for dynamic range as a concept to be relevant. This answer isn't wrong, it's just not an answer to this question. – Peter Cordes Jul 23 '20 at 07:51
  • 1
    You asked "Why is it called dynamic range?" not "Why is it called high dynamic range?" or "What is high dynamic range?" HDR is a way of increasing your camera's dynamic range by capturing shots at different exposures and combining them. – ProgrammingLlama Jul 23 '20 at 08:34
  • @john HDR is any method that allows capturing details in a scene where the dynamic range of the details in that scene exceed the DR capability of the intended display media. It doesn't necessitate multiple exposures. What Adams did with his Zone System to squeeze high dynamic range scenes onto low dynamic range paper is a form of high dynamic range imaging. The term was around long before digital photography. – Michael C Jul 23 '20 at 23:08