by damned truths

submit your photo

Hall of Fame
View past winners from this year

Please participate in Meta
and help us grow.

Take the 2-minute tour ×

I have the following formula:

DR=(1.4L_Sat G_i)/N_D

where DR is the dynamic range;
L_sat is the luminance at the “margin level”，70.1% of the maximum recordable （saturation）luminance（yes，L_sat is a peculiar symbol for that);
G_i is the incremental gain at the reference black luminance;
N_D is the measured digital noise at the reference black luminance

But L_sat is 100/140(1/1.4) of the actual saturation luminance (which we call here L_clip ).Thus we can recast equation 2 thus:

DR=(L_clip G_i)/N_D

I cannot understand this L_clip and L_Sat and the relationship between them.

share|improve this question
Hi, I've attempted to reformat your question as it's very hard to understand at the moment. Please check I haven't changed the meaning. –  ChrisF Jul 31 '12 at 11:20
This question has strayed off-topic for this site. It's no longer about photography as such; it has become a question belonging in the physics, math or electronics SE sites, so I have voted to close. –  user2719 Jul 31 '12 at 12:21
Where did you found these formulas? What exactly you don't undestand? Is this related to photography? –  Francesco Jul 31 '12 at 18:08
sorry for the obvious typo: did you found -> have you found :-) –  Francesco Jul 31 '12 at 18:32
Hi，Chrisf. the definition i mentioned from this article "The ISO Definition of the Dynamic Range of a Digital Still Camera",i cannot understand the numerator "the maximum luminance that receives a unique coded representation (the “saturation” luminance)", L_clip in your notation. can you help me? thank you very much? –  petalse Aug 1 '12 at 9:19

1 Answer 1

up vote 2 down vote accepted

I have found the following interesting reference "The ISO Definition of the Dynamic Range of a Digital Still Camera" which, at pages 6, clarifies your formula.

It is based on the assumption that a useful signal is one for which the Signal-to-Noise ratio is >= 1.

This luminance level is your denominator N_D. The numerator is "the maximum luminance that receives a unique coded representation (the “saturation” luminance)", L_clip in your notation. But digital luminance is not equal to luminance (The famous "gamma correction"). This slope factor is the G_I (the incremental gain).

L_sat and L_clip are directly related: if one is known, the other follows by simply multiplying (or dividing) by a factor. It is a matter of definition.

The 1.4 factor (roughly the inverse of 70%, as you note) provides a sort of a buffer (the paper calls it the well-known so-called “half stop margin” against overexposure).

The ratio of:

  • (numerator) Gamma corrected maximum luminance
  • (denominator) lowest luminance level where the SNR is >= 1

gives the output, the dynamic range of the sensor. It is (as it must be) based on a series of convention: but if you apply them consistently to various sensor you can numerically describe them according to this metric.

share|improve this answer
Thank you Francesco. These formulas from the article you mentioned above.Now i am learning the Dynamic Range of DSC. about its definition i feel confused. thanks for your explain,but i still can't understand completely. I again understand it. –  petalse Aug 1 '12 at 2:15
Hi，Francesco. the input (L_clip) through the gamma correction(camera sensor in-built) get to the output( L_sat) ,can i understand it? if so, how to calculate the L_sat ? –  petalse Aug 1 '12 at 3:16
@petalse I tried to expand on that a bit –  Francesco Aug 1 '12 at 9:48
perhaps i can question like this::" Now if i give you a mobile phone, how do you measure the dynamic range of his camera depend on the ISO definition ?" –  petalse Aug 2 '12 at 2:51
the sensor of a mobile phone camera is a sensor all the same. So: find out (or measure, if you have the adequate capabilities) the values for L_sat, G_i, N_d, and compute the ratio. Just to be sure, have you read the numerous other questions and answers on the topic of Dynamic Range on this same site? –  Francesco Aug 2 '12 at 6:44

Your Answer


