From f/5.6 to f/22

[Abitconfused]

Check my math. If f/5.6 to f/22 is 4EV, shouldn't there be (2 x 2 x 2 x 2 x 2) or 16 time as much light at f/5.6 than at f/22?

[pnodrog]

x1 = f5.6
x2 = f8
x2 = f11
x2 = f16
x2 = f22

= 4EV

You should not x2 the starting point.

[NorthernFocus]

Yes 4EV is 16x as much light but your math is terrible. You have five sets of 2x in your math which equals 32x

Edit: Posted parallel with Paul's comments.

[xpatUSA]

Check my math. If f/5.6 to f/22 is 4EV, shouldn't there be (2 x 2 x 2 x 2 x 2) or 16 times as much light at f/5.6 than at f/22?

But f/5.6 to f/22 is -4 EV, Ed.

As has already been said, there are only 4 steps, not 5 and thinking in steps can make it a bit less confusing . . .

So:

Step 1: f/5.6 to f/8 = -1 EV.
Step 2: f/8 to f/11 = -1 EV.
Step 3: f/11 to f/16 = -1 EV.
Step 4: f/16 to f/22 = -1 EV.

Total: -4 EV.

[pnodrog]

But f/5.6 to f/22 is -4 EV, Ed.

As has already been said, there are only 4 steps, not 5 and thinking in steps can make it a bit less confusing . . .

So:

Step 1: f/5.6 to f/8 = -1 EV.
Step 2: f/8 to f/11 = -1 EV.
Step 3: f/11 to f/16 = -1 EV.
Step 4: f/16 to f/22 = -1 EV.

Total: -4 EV.

Yep, BUT the statement was "16 times as much light at f5.6 than at f22" .... either way there is a 4 EV difference....we can all be right...

[Abitconfused]

Oh! Yes, should have 4X. So 16 times the light is correct?

[NorthernFocus]

Oh! Yes, should have 4X. So 16 times the light is correct?

Yes.

[xpatUSA]

Yep, BUT the statement was "16 times as much light at f5.6 than at f22" .... either way there is a 4 EV difference....we can all be right...

Oh, a 4 EV difference, now it is clear . .

. . . BUT (in caps) the condition was "If f/5.6 to f/22 is 4EV" - if that's true, then I suppose f/22 to f.5.6 is -4 EV, eh?

[pnodrog]

Ed has defined the question two ways on the same line - we just took different picks....

I blame it all on Ed. Sorry Ed in this modern world we are all looking for someone else to blame. Still if this is the worst thing that happens to you today you are not doing too badly......

[RBSinTo]

Its all just nomenclature.
I never bothered with the EV concept, but just learned it in terms of stops of aperture or shutter speed.
One stop larger aperture, or one stop longer shutter speed represents an increase of twice as much light hitting the film or sensor, while a one stop smaller aperture or faster shutter speed represents a decrease of half as much light hitting the film or sensor. So a three stop decrease in aperture (5.6 to 8 to 11 to 16) means that 1/2 x1/2 x1/2=1/8th as much light will hit the film or sensor, and conversely going the other way (f16 to 5.6) will result in 2x2x2=8 times more light hitting the sensor.
Robert

[xpatUSA]

Its all just nomenclature.<>

So a four stop decrease in aperture (5.6 to 8 to 11 to 16) means that 1/2 x1/2 x1/2 x1/2=1/16th as much light will hit the film or sensor, and conversely going the other way (f16 to 5.6) will result in 2x2x2x2=8 times more light hitting the sensor.
Robert

Going from f/5.6 to f/16 is three steps, not four. (Think of it as one step for each "to" in 5.6 to 8 to 11 to 16).

And "2x2x2x2=8" doesn't seem right somehow.

HTH

[bje07]

I'm not able to explain that in English but you forget that it is only square root of 2 between 2 stops.

[Abitconfused]

I am good with it.

[DanK]

As Ted said, think in steps, not stops. Each step up from any given starting point--an increase of one stop--is a doubling, and each step down is a halving. The starting point doesn't matter.

Why a doubling or halving, when the f/stop numbers aren't multiples of 2? It's because the f/stop number is defined as focal length divided by effective diameter of the aperture. Since the light striking the sensor is a function of the area, not the diameter, the f/stop difference corresponding to a halving or doubling is a function of the square root of two, roughly 1.4.

[RBSinTo]

As Ted said, think in steps, not stops. Each step up from any given starting point--an increase of one stop--is a doubling, and each step down is a halving. The starting point doesn't matter.

Why a doubling or halving, when the f/stop numbers aren't multiples of 2? It's because the f/stop number is defined as focal length divided by effective diameter of the aperture. Since the light striking the sensor is a function of the area, not the diameter, the f/stop difference corresponding to a halving or doubling is a function of the square root of two, roughly 1.4.

Thus starting with an aperture value of 1 and multiplying by 1.4 to get each successive value gives a sequence of numbers that looks like this: 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45, 64... Look familiar?
Robert

[darekk]

function of the square root of two, roughly 1.4.

1.4 or exact square root of 2 (1.4142135623730950488016887242097...) and results rounded down to 2 significant digits ?

 1.4142135623730950488016887242097 →  1.4
 2
 2.8284271247461900976033774484194 →  2.8
 4
 5.6568542494923801952067548968388 →  5.6 instead of  5.7 !!
 8
11.313708498984760390413509793678  → 11
16
22.627416997969520780827019587355  → 22   instead of 23 !!
32
45.25483399593904156165403917471   → 45
64

(the Calculator application in Windows returns so many digits)

The shutter speed looks also like rounded according to some rules:

         not rounded:
30       32
15       16
8
4
2
1
1/2
1/4
1/8
1/15     1/16
1/30     1/32
1/60     1/64
1/125    1/128
1/250    1/256
1/500    1/512
1/1000   1/1024
1/2000   1/2048
1/4000   1/4096
1/8000   1/8192

It looks a little like number (or denominator when shutter speed < 1) was rounded down to 3 significant digits, but third digit can be 0 or 5 only. Or maybe rough values are used to calculate shutter speed ? But in in contrast to aperture several values: 1, (1/)15, (1/)125 ??
In traditional, not digital photography the Schwarzschild law (effect) complicated things.

[DanK]

I suspect the shutter speeds are rounded so as to have all multidigit numbers end in 0 or 5, which does make them simpler to read. The same thing is done in expressing the size of computer memory.

It's mysterious that 5.66 and 22.63 were rounded down, but it doesn't make any practical difference. These differences are too small to matter.

[xpatUSA]

I suspect the shutter speeds are rounded so as to have all multidigit numbers end in 0 or 5, which does make them simpler to read. The same thing is done in expressing the size of computer memory.

It's mysterious that 5.66 and 22.63 were rounded down, but it doesn't make any practical difference. These differences are too small to matter.

By coincidence, I found it necessary today to go into a raw file's meta-data to see the exact aperture and shutter speed used as opposed to what showed in the viewfinder i.e. the "reported" values.

For f/2.8 and 1/20 sec, it said 2.82843 and 0.048194.

[Manfred M]

Of course, no one has gotten into discussing the T-stop, which is used in cinematography. The "T" stands for "transmission".

The f-stop is purely a mathematical ratio of the focal length to input aperture, so it is really an "ideal value". The T-stop (or T-number) goes further and actually measures the drop in light level that comes from some of the light being absorbed by the glass lens elements. This will of course vary from lens to lens. The T-stop / T-number will always be larger than the f-stop; i.e. a f/2.8 lens could be a t/3.5 lens.

[dem]

It's mysterious that 5.66 and 22.63 were rounded down, but it doesn't make any practical difference. These differences are too small to matter.

Which series looks most "natural"?
2.8 5.6 11 22
2.8 5.7 11 22
2.8 5.7 11 23
2.8 5.6 11 23