[jspi]

I don't consider myself a beginner, more of an intermediate but I feel this is a question I should have grasped (memorized?) long ago. If I'm wrong on any of my information, feel free to educate me.


1 stop of adjustment is either halving or doubling the amount of light hitting the sensor. This can be done by adjusting the shutter speed or aperture. Increasing the sensitivity of the sensor (ISO) doesn't increase the amount of light but has the same effect (although it does increase noise).


Let's say I'm taking a photo at f8, 1/640s, and ISO 6400 but it's "1 stop" too dark.

I could increase ISO to 12,800 (1 stop) but it'll have more noise than I like.

Or I could slow the shutter speed to 1/320s (1 stop) but the action will be blurry.

So I must adjust the aperture. Going with halving/doubling to adjust one stop, the aperture would be f4 but I believe that would be closer to 2 stops.

I understand the math behind 1 stop of aperture but I can't calculate that on the fly. My assumption is I need to memorize which f stops are one full stop from the one below/above (I've been refusing to do that). But I though I'd ask if there's a quick way to figure it out without having to memorize a chart. For now, I just assume f8 to f4 is 2 stops so f6.3 is probably close to one stop? f4 to f2 is likely 2 stops so f4 to f3.2 is probably 1 stop. Assume... Probably... Just set to M, look at the expose meter, and hope it's close. Yeah, there's a bit of guessing and I'm most likely wrong but I have a big SD card so I'm faking my way to descent photos... Although a bit of memorization would be a small price to pay for throwing away less photos. Not to mention the extra time I'd have to be more creative while framing shots.

[cmohr]

In terms of Aperture, 1 stop from f8 is f5.6.

Best way is to look at a chart and remember them.

1.4 / 2 / 2.8 / 4 / 5.6 / 8 / 11 / 16 / 22 are the ones that matter .

[wadge22]

I don't have much insight on what a stop actually represents. I'd guess it's just a human designed unit of measurement, like how a foot is just a foot because someone once said it was and we've all agreed on it.

I can possibly help memorize the apertures for each stop: for me it's fairly easy to remember some of the common f stop numbers, specifically 2.8 or 5.6, which are commonly the widest aperture for lenses. Well, double 2.8 and you get 5.6, and they also happen to be two stops apart. What number can you double to get 2.8? 1.4 which happens to be two stops brighter than 2.8. So doubling the number equals two stops. The ones that fall between these are easier to remember: f2, double it (two stops) to f4, double that (two more stops) to f8, double that (you guessed it, two more stops) gets you f16.

So... f1.4 -> f2 -> f2.8 -> f4 -> f5.6 -> f8 -> f11 (why not 11.2?... I guess I don't know) -> f16. Those are the main ones to remember, and the doubling 2.8 thing makes it easy, at least for me. I'll never forget f2.8, that's the speed of all the fancy pro zooms. The rest of the hard part just goes from there.

[RobA_Oz]

Remember that the f-stop number represents the ratio of the lens focal length to the diameter of the entrance pupil (diaphragm aperture), but the total amount of light entering depends on the area of the pupil. Assuming the pupil is circular (near enough, usually) that amount of light is therefore dependent on the square of the diameter, so you have to square the f-number to obtain a measure of the relative illumination.

Hence, f2, f2.8, f4, f5.6 when squared become 4, 8, 16, 32 etc. Obviously, the f-numbers are approximations in many cases.

Is that helpful?

[UncleVanya]

Each stop of the square root of two apart.

1.414213562373095 * 5.6 = 7.919595949289332 (8).

The numbers are rounded and approximated.

[jspi]

Best way is to look at a chart and remember them.

1.4 / 2 / 2.8 / 4 / 5.6 / 8 / 11 / 16 / 22 are the ones that matter .

Yes, memorizing a chart is what I'd like to avoid but I figure it's what I need to do. I've kind of developed a feel for it but I'm also often wrong (close, but wrong).

---------- Post added 05-12-22 at 03:58 AM ----------

To summarize other comments... Doubling the light means doubling the surface area (aperture) letting light pass. Assuming squares (yes, apertures are round and requires pi but that's not as simple)... 2x2 = an area of 4 but doubling that to 4x4 = an area of 16 which is 4 times the area/light (2 stops), not double the light. Doubling the area of 2x2 (4) would be 8. So I need the square root of 8... I need a calculator... 2.82 x 2.82 = 7.95. So f2 is 1 stop faster (twice the light) than f2.8.

I need to memorize a chart... Or work on my mental math skills.

[cmohr]

I noticed you mention 640th as a shutterspeed, this suggests you have your camera set for 1/3ev stops. to make it a bit simpler , change that to 1/2 stops, then each click of the wheel is a half stop, so then it's easy to go one stop, just two clicks of the wheel.

Or, use exposure compensation, and set that to 1 stop over or under, that way, the exposure will stay that amout adjusted, even when you change subjects.

[acoufap]

If you want to really dig deeper to understand how the relative "photographic system" works, have a look at wikipedia - The APEX System and here: Exposure Value.

Interesting infos but for sure partly hard to understand without some basics in physics and mathematics. I'd suggest have a try - all the best!

Within the first article also Canon, Pentax and Leica are mentioned.

The easy way is to simply to know the main aperture steps mentioned above and count your up and down clicks when change the parameters of the system.

An addition:

You Tube

You Tube

explained.

[UncleVanya]

I need to memorize a chart... Or work on my mental math skills.

Just remember f1.4 and f2. Everything is comes from there. One stop more than f2? Double f1.4 (2.8). One stop more than that? Double f2 (4). Another stop? Previous stop times two. (F4’s previous stop is f2.8, next stop is 5.6… )

[cmohr]

Remember that the f-stop number represents the ratio of the lens focal length to the diameter of the entrance pupil (diaphragm aperture), but the total amount of light entering depends on the area of the pupil. Assuming the pupil is circular (near enough, usually) that amount of light is therefore dependent on the square of the diameter, so you have to square the f-number to obtain a measure of the relative illumination.

Hence, f2, f2.8, f4, f5.6 when squared become 4, 8, 16, 32 etc. Obviously, the f-numbers are approximations in many cases.

Is that helpful?

This is correct. The Aperture stop comes from the relationship between focal length and physical size of the diametre of the pupil

IE, if you have a 100mm lens, and set to F8, the dia of the pupil is 12.5mm, and that makes an area of 122.7 sq mm. To move 1 stop you need to double that area, about 254 sq mm, which is a pupil of about 17.8mm, which when you divide 100mm with, gives you 5.6.

It's all relative, 100mm, F4 pupil 25mmDia, 49.8 sq mm, twice the area of the 17.8mm dia pupil. so on and so on.

[allanmh]

I learned back in the mists of time that the aperture is the denominator in a handy, but only approximate, formula that shows how much light actually gets through a lens from front surface to back surface. The formula reads like this: Light(L) = 1/AxA. f/1.0 gives you 100% of light passage, f/2.0 gives you 25%, f/1.4 gives you 50%. The easy (for me) way to memorise this is two lines of apertures: 1, 2, 4, 8, 16, 32, 64. Ansel Adams was a member of an association of photographers who called themselves F64. The second line of apertures: 1.4, 2.8, 5.6, 11, 22, 45. The formula is only approximate, as we all know know of f/0.95 lenses, and I read somewhere that for the film 'Tom Jones' back in the '70s, special lenses with f/0.7 were built.
Fun factoid: the formula means that at f/8 only about 1.6% of the light that hits the front of the lens actually gets through to the sensor.
I don't know it this helps. It helps me understand why the smallest f-numbers give the most light.

[dms]

Going beyond your question, there is another possibility. This applies to Pentax DSLRs from K-5 to K-1, which are basically "isoless" [for each doubling of iso, the DR is reduced by 1 stop]. Take the picture at the:
-- desired shutter speed (to avoid too much motion blur)
-- a desired f-stop (to avoid too little DOF)
-- and a desirable low iso
thus taking the shot with you might think less light than optimum (for the shadows), but not so. By not increasing the iso, you:
(1) avoid possibly blowing the highlights, and
(2) as regards noise in the shadows, there will be no more noise than if you increased the iso.

But to answer your question: two stops is doubling (or halving the aperture), and thus one stop is about 1.5 times (or 1.5 divide into) the aperture (or simply 3/2 or 2/3). If you remember any f-stop you know the others. So for example f/4. Two stops is f/8 or f/2, and 1 stop is f/4 x 3/2 = f/6 (close enough to f/5.6) or 4 x 2/3 = 8/3 ~ f/3 (close enough to f/2.8). Actually the f-stop progression is not mathematically correct, as if so they would all be related by 1.414.. [square root of 2], or 1/1.414, and they aren't. (So there is no rule to remember those intermediate f-stops. f/1, f/2, f/4, f/8, etc. are fine, but not the in between ones.)

[Alex645]

But I though I'd ask if there's a quick way to figure it out without having to memorize a chart.

No, you just have to memorize. As others have said, it helps to know for "full stops" to start with f/1 and f/1.4 and then double every other f/stop to know the next f/stop.

Shutter speeds get funky and also require memorization because if you start with 1 second, how do we end up at 1/15" instead of 1/16"? And why does it go from 1/60" to 1/125"? In reality the numbers should be 1/16", 1/32", 1/64", 1/128", etc. And have you ever timed a 15" or 30" exposure? Yup, it's really 16" and 32".

As a teacher working with students with some cameras that shoot in increments of 1/2 stops and 1/3 stops, you just have to memorize those numbers. Back in the 20th century, we would call f/6.7 a "split f/5.6-8", because it was some unknown value between the written f/stops. For 1/3 stops, instead of f/6.3 we'd call it a 'hot 5.6" and f/7.1 a 'cool 8'.

And even ISO's are not immune. ISO 50, 100, 200, etc, great! But what's up with ISO 125 and 160 between 100 and 200? Why not 133 and 166?

The brain is an amazing thing, and we got to the moon and back with pencil and paper and slide rulers with some rudimentary help from computers. And I've known cinematographers that never used a light meter and could look at the light and nail the EV as well as a spot meter on an 18% gray card.

[Papa_Joe]

This thread really makes me feel old. When I started photography no one would have mused about memorizing the f stops. Maybe the half stops for those bean counters. You had a lens with the f-stops on the aperture ring and a wheel with the exposure time on it and you had the values in your head quicker than you could think. Pentax owners were a little disadvantaged of course, they had to look through the viewfinder to see the shutter times on their M-series camera *).

*) Except the proud owners of a MX.

[GUB]

I don't consider myself a beginner, more of an intermediate but I feel this is a question I should have grasped (memorized?) long ago. If I'm wrong on any of my information, feel free to educate me.


1 stop of adjustment is either halving or doubling the amount of light hitting the sensor. This can be done by adjusting the shutter speed or aperture. Increasing the sensitivity of the sensor (ISO) doesn't increase the amount of light but has the same effect (although it does increase noise).


Let's say I'm taking a photo at f8, 1/640s, and ISO 6400 but it's "1 stop" too dark.

I could increase ISO to 12,800 (1 stop) but it'll have more noise than I like.

Or I could slow the shutter speed to 1/320s (1 stop) but the action will be blurry.

So I must adjust the aperture. Going with halving/doubling to adjust one stop, the aperture would be f4 but I believe that would be closer to 2 stops.

I understand the math behind 1 stop of aperture but I can't calculate that on the fly. My assumption is I need to memorize which f stops are one full stop from the one below/above (I've been refusing to do that). But I though I'd ask if there's a quick way to figure it out without having to memorize a chart. For now, I just assume f8 to f4 is 2 stops so f6.3 is probably close to one stop? f4 to f2 is likely 2 stops so f4 to f3.2 is probably 1 stop. Assume... Probably... Just set to M, look at the expose meter, and hope it's close. Yeah, there's a bit of guessing and I'm most likely wrong but I have a big SD card so I'm faking my way to descent photos... Although a bit of memorization would be a small price to pay for throwing away less photos. Not to mention the extra time I'd have to be more creative while framing shots.

Sympathies with the struggle but it gets better with practise.
cmohr in #7 has the best practical advice.
Set your adjustment wheels to half stop increments and it gets easy to count them off.
And it is worth noting that despite the units being somewhat confusing they are what they are for good mathematical reasons - and also so they won't be confused with each other.