How do you calculate the number of stops between two shutter speeds?

Asked 10/11/2014

1 views

2 answers

0

I want to work out the exposure difference, in stops, between any two shutter speeds. What formula should I use, and is there a simple shortcut for common full-stop shutter speeds?

Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0

Photography Stack Exchange contributor

11y ago

2 Answers

11

It helps to start with exactly what a "stop" means. See What is one "stop"?, but, fundamentally, each stop is a doubling or halving of the exposure.

So, given two shutter durations, you can find the number of stops between them by calculating the binary logorithm (log₂) of each, and subtracting. (If you don't remember your elementary school math very clearly, logorithms may sound complicated, but are really just counting the number of times you multiply by the same number — in this case, two.)

In formula form, that is:

$$\text{# stops} = \log_2{(\text{2nd duration})} - \log_2{(\text{1st duration})}$$

For example, we know that going from ¹⁄₃₀ to ¹⁄₆₀ is down one stop, trivially. But to work it out:

$$\begin{align} \text{# stops} &= \log_2\left({1\over60}\right) - \log_2\left({1\over30}\right) \\ &= {\sim}\,({-5.9})\; - \;{\sim}\,({-4.9}) \\ &=\quad {-1} \end{align}$$

Or for a more extreme example, from ¹⁄₁₀₀₀ to 30 seconds:

$$\begin{align} \text{# stops} &= \log_2\left({30}\right) - \log_2\left({0.001}\right) \\ &= {\sim}\,({4.9})\; - \;{\sim}\,({-9.96}) \\ &=\quad \sim 14.9 \end{align}$$

so, 15 stops. Note that it's incredibly important to not worry about excess precision. The standard scale of shutter speeds follows human-friendly numbers, not log₂, and the closest power of 2 may or may not be used internally. Overall in almost anything in photography, if it's less than a third of a stop, round it off.

If you don't have a calculator supporting \$\log_2\$, you can calculate it in the following manner:

$$ \log_2{x} = \frac{\log x}{\log 2} $$

where "\$\log\$" is logarithm in any base (2, 10, or natural logarithm, or even something else). Very often, \$\log_{10}\$ or \$\log_e\$ (in other words, \$\ln\$, or the natural logarithm) are available when \$\log_2\$ isn't.

If you don't have a calculator handy or don't want to figure out the precise \$\log_2\$, you can simply start with one speed and double or halve in your head, counting the number of times it takes before you reach the other speed. (There is no shame in keeping track on your fingers as you go if you can't both calculate and count at the same time!)

This same formula works for ISO as well, by the way. It doesn't work for aperture as expressed in f/numbers, because those numbers are related to the diameter of the aperture but exposure is based on the area (see Why are f-stops not linear?). You can calculate \$\log_{\sqrt{2}}\$ if you have a fancy calculator handy, or you can double the result after plugging the aperture numbers into the formula (and then it will work).

In practice, though, shutter speed is the only one that where a formula might be handy, because few lenses have more than a 9-stop aperture range (and many much less), most cameras only have a useful ISO range of 6 stops (maybe a bit more these days) — but with bulb mode the range for shutter speed is almost infinite. Still, exponential scales collapse quickly — the difference between ¹⁄₈₀₀₀ second and one hour is only about 25 stops, and the difference between a one hour exposure and a one year exposure only 13!

Originally by user1943. Source · Licensed CC BY-SA 4.0

user1943

11y ago

0

AI Answer

Generated from our catalog & community — verify before relying on it.

A stop is a doubling or halving of exposure time. For shutter speeds, the difference in stops between two durations is:

stops = log2(second shutter time / first shutter time)

Equivalent form:

stops = log2(second shutter time) - log2(first shutter time)

Examples:

  • 1/30 to 1/60 = log2((1/60) / (1/30)) = log2(1/2) = -1, so it’s 1 stop faster (1 stop less exposure).
  • 1/125 to 1/30 = log2((1/30) / (1/125)) ≈ log2(4.17) ≈ 2, so about 2 stops more exposure.

For common full-stop shutter speeds, you can often just compare the ratio:

  • factor of 2 = 1 stop
  • factor of 4 = 2 stops
  • factor of 8 = 3 stops
  • factor of 16 = 4 stops

So if one shutter speed is 8× longer than another, that’s 3 stops. This shortcut works because each stop doubles or halves the exposure time.

UniqueBot

AI

11y ago

Your Answer