Converting numbers and percentages to stops (Danger: Math!)

[480sparky]

I'm writing a spreadsheet that will allow me to enter in the timed shutter speed of a camera, then compare that to the actual time set. I'm able to easily convert the difference to a percentage, but what I'd like to do is go one step further and and have a formula that converts that to how many stops the shutter speed is deviating from the ideal.

Meaning: If the shutter speed is, say, 1/60, which is 0.01666667. I take a shutter timer and the shutter is actually open 0.02387 sec. It's easy to divide the two and say the shutter is roughly 50% slower than ideal, but how to convert that to an f/stop?

Is it overexposing 1/4 of a stop? 1/3? 1/2? My brain just ain't wired to be able to convert the linear percentages to the non-linear world of f-stops.

I could probably calculate out all the various deviations for each shutter speed (1/4, 1/3, 1/2, 2/3, 3/4 etc.) then nest a boat-load of IF-THEN statements to figure it out, but that seems like the long route. I'd think there's a formula or two that could spit out the result.

I hope this makes sense.

 

[tirediron]

I'm confused... you're talking shutter speed; where does aperture enter in to it? FWIW, f stops are linear in form, it's just the numbers used to express them are a bit whacked.

 

[480sparky]

Shutter speed can be converted to stops, as can ISO.
1/60 is 1 stop from 1/125 and 1/30. ISO 100 is one stop from 50 and 200.

I'm wanting to figure out how many stops 0.02387 sec is from 0.01666667 sec (roughly..... round off to simple fractions like 2/3, 1/2 etc).

(I'm thinking 2, 2.8, 4, 5.6, 8, 11, 16 etc is not linear.)

Maybe EV would be a better term to use.

 

[tirediron]

I get that, but was your reference to 'f stops' a slip of the finger, or are you trying to equate shutter speed exposure to aperture exposure in some way with this project?

 

[480sparky]

I'm trying to express the difference between the actual time a shutter is open to the time set by the camera in stops or EVs.

Is the shutter 'off' by 1/3 of a stop? 1/2? 3/4?

For instance, if you have a camera that you KNOW the shutter is 1/2 stop off for a given speed, then you can easily compensate by adjusting the aperture or ISO accordingly.

 

[table1349]

I'm writing a spreadsheet that will allow me to enter in the timed shutter speed of a camera, then compare that to the actual time set. I'm able to easily convert the difference to a percentage, but what I'd like to do is go one step further and and have a formula that converts that to how many stops the shutter speed is deviating from the ideal.

Meaning: If the shutter speed is, say, 1/60, which is 0.01666667. I take a shutter timer and the shutter is actually open 0.02387 sec. It's easy to divide the two and say the shutter is roughly 50% slower than ideal, but how to convert that to an f/stop?

Is it overexposing 1/4 of a stop? 1/3? 1/2? My brain just ain't wired to be able to convert the linear percentages to the non-linear world of f-stops.

I could probably calculate out all the various deviations for each shutter speed (1/4, 1/3, 1/2, 2/3, 3/4 etc.) then nest a boat-load of IF-THEN statements to figure it out, but that seems like the long route. I'd think there's a formula or two that could spit out the result.

I hope this makes sense.

 

[TCampbell]

"stops" are based on the halving or doubling of light (you knew that) but if you're thinking of actual f-stops (focal ratio) then those are based on the powers of the square root of 2.

If you have a circle and say it has a 10mm diameter. The area of for a circle is: a = Pi * r^2 (Pi times the radius squared). Math "order of operations" requires that you first square the radius before you multiply it by pi. So the area of a circle with a 10mm diameter is: Pi * 5^2. 5^2 = 25. So it's Pi * 25 or 3.1415 * 25 = 78.5 (square millimeters or mm^2).

If you increase the diameter (or radius - doesn't matter you get the same result) by a the square root of 2 then the area of that circle will EXACTLY double. The square root of 2 is roughly 1.41 (it's an irrational number that just keeps going -- just like Pi).

So 10mm x 1.41 * 10 = 14.1. Divide 14.1 by 2 to get the radius of 7.05 and then run the formula for the area and it's 7.05^2 x Pi which is 156 square millimeters (double the previous value.) BTW if you were watching closely you may have noticed a very tiny error -- that's the rounding error because I used 1.41 instead of a longer value for the square root of 2. The more digits you use for Pi and √2 the closer you'll get to a perfect doubling.

If you want to trade off stops of shutter for stops of focal ratio then you'd have to take the percentage difference in time value and convert that to a percentage difference in AREA value and run the formula backward to get the new diameter of your circle. Since the f-stop is the focal length divided by the diameter you can quickly arrive at the effective f-stop.

There is a problem with trying to be so precise... which is that your camera isn't (especially lenses).

Lenses "breathe". Often the lens focal length is reasonably accurate if focused to infinity, but not when focused to a closer subject... and sometimes the error is very large. Since the focal ratio is the ratio of the focal length divided by the area of clear aperture and the aperture isn't changing when you alter the focus -- but the focal length is changing (possibly by a lot) this means you're aperture value may be off by quite a bit. In other words you can't trust that f/8 is really f/8. According to Thom Hogan, the Nikon 70-200mm is a heavy breather and loses as much as 29% of the focal length as you focus closer and approach minimum focus distance. 29% of 200mm = 142mm. A 200mm lens with a 25mm aperture is f/8. But a 142mm lens with a 25mm aperture is only 5.68!!! That's nearly a full stop of light difference (and you would want to manually account for that -- another reason to check those histograms!) All lenses breathe. Very good lenses maybe only breathe by as little as 5% so it's not so much that it makes a stop of light worth of difference.

 

[Dave442]

In your example you are 0.518233 stops slower.

I put the shutter speeds into a table using LOG(1/60,2) then subtracted your measured shutter speed with LOG(0.02387,2) = -5.9068906 - -5.3886576

 

[Dave442]

 

[480sparky]

"stops" are based on the halving or doubling of light (you knew that) but if you're thinking of actual f-stops (focal ratio) then those are based on the powers of the square root of 2.

If you have a circle and say it has a 10mm diameter. The area of for a circle is: a = Pi * r^2 (Pi times the radius squared). Math "order of operations" requires that you first square the radius before you multiply it by pi. So the area of a circle with a 10mm diameter is: Pi * 5^2. 5^2 = 25. So it's Pi * 25 or 3.1415 * 25 = 78.5 (square millimeters or mm^2).

If you increase the diameter (or radius - doesn't matter you get the same result) by a the square root of 2 then the area of that circle will EXACTLY double. The square root of 2 is roughly 1.41 (it's an irrational number that just keeps going -- just like Pi).

So 10mm x 1.41 * 10 = 14.1. Divide 14.1 by 2 to get the radius of 7.05 and then run the formula for the area and it's 7.05^2 x Pi which is 156 square millimeters (double the previous value.) BTW if you were watching closely you may have noticed a very tiny error -- that's the rounding error because I used 1.41 instead of a longer value for the square root of 2. The more digits you use for Pi and √2 the closer you'll get to a perfect doubling.

If you want to trade off stops of shutter for stops of focal ratio then you'd have to take the percentage difference in time value and convert that to a percentage difference in AREA value and run the formula backward to get the new diameter of your circle. Since the f-stop is the focal length divided by the diameter you can quickly arrive at the effective f-stop.

There is a problem with trying to be so precise... which is that your camera isn't (especially lenses).

Lenses "breathe". Often the lens focal length is reasonably accurate if focused to infinity, but not when focused to a closer subject... and sometimes the error is very large. Since the focal ratio is the ratio of the focal length divided by the area of clear aperture and the aperture isn't changing when you alter the focus -- but the focal length is changing (possibly by a lot) this means you're aperture value may be off by quite a bit. In other words you can't trust that f/8 is really f/8. According to Thom Hogan, the Nikon 70-200mm is a heavy breather and loses as much as 29% of the focal length as you focus closer and approach minimum focus distance. 29% of 200mm = 142mm. A 200mm lens with a 25mm aperture is f/8. But a 142mm lens with a 25mm aperture is only 5.68!!! That's nearly a full stop of light difference (and you would want to manually account for that -- another reason to check those histograms!) All lenses breathe. Very good lenses maybe only breathe by as little as 5% so it's not so much that it makes a stop of light worth of difference.

All fine and dandy, but shutters don't focus breathe.

Nor am I dealing with lenses.

Or focal length.


I'm dealing with shutters.

In your example you are 0.518233 stops slower.

I put the shutter speeds into a table using LOG(1/60,2) then subtracted your measured shutter speed with LOG(0.02387,2) = -5.9068906 - -5.3886576

OK, now we're getting somewhere. So this can be 'rounded off' to 1/2 stop. Now that I can convert the difference between the two to a decimal representation of a stop, or EV, now it's how to figure out how to convert it to a simple fraction. I know how to do this with VLOOKUP, but what values do I use to differentiate between, say, 1/3 and 1/2?

 

[TCampbell]

"stops" are based on the halving or doubling of light (you knew that) but if you're thinking of actual f-stops (focal ratio) then those are based on the powers of the square root of 2.

If you have a circle and say it has a 10mm diameter. The area of for a circle is: a = Pi * r^2 (Pi times the radius squared). Math "order of operations" requires that you first square the radius before you multiply it by pi. So the area of a circle with a 10mm diameter is: Pi * 5^2. 5^2 = 25. So it's Pi * 25 or 3.1415 * 25 = 78.5 (square millimeters or mm^2).

If you increase the diameter (or radius - doesn't matter you get the same result) by a the square root of 2 then the area of that circle will EXACTLY double. The square root of 2 is roughly 1.41 (it's an irrational number that just keeps going -- just like Pi).

So 10mm x 1.41 * 10 = 14.1. Divide 14.1 by 2 to get the radius of 7.05 and then run the formula for the area and it's 7.05^2 x Pi which is 156 square millimeters (double the previous value.) BTW if you were watching closely you may have noticed a very tiny error -- that's the rounding error because I used 1.41 instead of a longer value for the square root of 2. The more digits you use for Pi and √2 the closer you'll get to a perfect doubling.

If you want to trade off stops of shutter for stops of focal ratio then you'd have to take the percentage difference in time value and convert that to a percentage difference in AREA value and run the formula backward to get the new diameter of your circle. Since the f-stop is the focal length divided by the diameter you can quickly arrive at the effective f-stop.

There is a problem with trying to be so precise... which is that your camera isn't (especially lenses).

Lenses "breathe". Often the lens focal length is reasonably accurate if focused to infinity, but not when focused to a closer subject... and sometimes the error is very large. Since the focal ratio is the ratio of the focal length divided by the area of clear aperture and the aperture isn't changing when you alter the focus -- but the focal length is changing (possibly by a lot) this means you're aperture value may be off by quite a bit. In other words you can't trust that f/8 is really f/8. According to Thom Hogan, the Nikon 70-200mm is a heavy breather and loses as much as 29% of the focal length as you focus closer and approach minimum focus distance. 29% of 200mm = 142mm. A 200mm lens with a 25mm aperture is f/8. But a 142mm lens with a 25mm aperture is only 5.68!!! That's nearly a full stop of light difference (and you would want to manually account for that -- another reason to check those histograms!) All lenses breathe. Very good lenses maybe only breathe by as little as 5% so it's not so much that it makes a stop of light worth of difference.

All fine and dandy, but shutters don't focus breathe.

Nor am I dealing with lenses.

Or focal length.


I'm dealing with shutters.

Sorry - I misunderstood... I thought you were building a spreadsheet that could convert the shutter speed differences into f-stop differences.

 

[480sparky]

Sorry - I misunderstood... I thought you were building a spreadsheet that could convert the shutter speed differences into f-stop differences.

Basically, I am. But in order to compare two shutter speeds.

 

[Bill The Lurker]

take the ratio of the two shutter speeds and then calculate the base-2 logarithm of that. to calculate the base-2 logarithm of anything take the standard base-10 log (usually just log(x) in whatever) and divide by 0.30103 (not quite but plenty close enough).

so 1/60 divided by 1/120 gives you 2, log(2) is 0.30103 (no, that's not a coincidence) and divide by 0.30103 and you get 1.0 stops.

in your example 0.0166667 divided by 0.02387 is about 0.698, take the log of that and you get -0.156, divide that in turn by 0.30103 and yoiu get -0.51 or about half a stop.

if you go the other way and divide 0.02387 by 0.01666667 you get about 1.43, take the standard log of that to get 0.155, which divided by 0.30103 is about 0.5 or half a stop the other way. which makes sense right?

 

[480sparky]

Yeah, I get the log function.

Now to build a VLOOKUP table that will return whole fractions.

 

[Dave442]

To put in whole fractions I am just thinking it would be simple to use IF and specify 1/3, 1/2 or 2/3 as the result based on whichever of those falls closest to your decimal number. So halfway to 1/3 would go down to the full stop, then halfway between 1/3 and 1/2 display 1/3, then halfway between 1/2 and 2/3 display 1/2, then halfway between 2/3 and 1 display 2/3 and finally above that would display 1. If you need more precision than 1/3 stops then the VLOOKUP is better.

As you could have more than a full stop difference then I would first pull out however many full stops difference and then add that back with the fraction using Concatenate to read out something like: 2 and 2/3 Stops less light.