Thanks for keeping me straight guys. Glad to see everyone came out so quickly to correct this!!
So, the revised formulation is:
($1300 for f/1.2 - $315 for f/1.4)/0.5 = $1970 per stop. Still expensive...but more reasonable.
Erich
From the link:
"f/stop Graph... The diameter and area values of each f/stop are based on a lens with a focal length of 50mm. Note that an f/stop value is the ratio of the focal length of the lens to the diameter of the aperture. Thus, f/1 has an aperture diameter of 50mm, f/2 a diameter of 25mm, f/8 a diameter of 6.25mm, etc. The percentage value shown next to the aperture area is the aperture area as a percentage of the total f/1 aperture area. The amount of light entering the camera is directly governed by the area of the aperture opening. Thus, f/1.4 is 50% or half the area of f/1 and therefore cuts the light in half -- a "full stop" less light. f/2 is 25% or 1/4 the area of f/1 and is 2 stops less light. "
One third stop differences would be:
f/1.0, 1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.8, 2.0, 2.2, 2.4, 2.8, 3.2, 3.4, 3.6, 4.0, 4.5, 4.8, 5.0, 5.6...
Actually there should only be two 1/3-stop steps between f/1 and f/1.4, and those would be f/1.1 and f/1.3. Ditto for f/1.4 and f/2 (1.6, 1.8); f/2.8 and f/4 (3.2, 3.5); and f/4 and f/5.6 (4.5, 5.0).
The math isn't hard; each one-stop difference is calculated by multiplying by 2 to the 1/2 power (the square root of 2 or about 1.414). Fractional-stop differences are calculated by multiplying by 2 to the (1/2 * differences in stops); for example, third-stop differences are mutliplied by 2 ^ (1/2 * 1/3) or 2 ^ 1/6 or about 1.122.
I could try to derive the formula to determine how many stops faster (or slower) x is than y for arbitrary values of x and y, but it's getting late and my algebra is rusty.
"f/stop Graph...
The diameter and area values of each f/stop are based on a lens with a focal length of 50mm. Note that an f/stop value is the ratio of the focal length of the lens to the diameter of the aperture. Thus, f/1 has an aperture diameter of 50mm, f/2 a diameter of 25mm, f/8 a diameter of 6.25mm, etc. The percentage value shown next to the aperture area is the aperture area as a percentage of the total f/1 aperture area. The amount of light entering the camera is directly governed by the area of the aperture opening. Thus, f/1.4 is 50% or half the area of f/1 and therefore cuts the light in half -- a "full stop" less light. f/2 is 25% or 1/4 the area of f/1 and is 2 stops less light. "
It's interesting that given the above definition going from f/1.4 to f/1.0 will give you twice as much light (one full stop) but going from f/1.4 to f/1.2 only gives you 36% more light even though it is technically "half a stop" (and that you get ~75% of the light at f/1.4 that you had at f/1.2).
Ahh yes, the graphs are very instructive. I did go to the page the first time, but I didn't see the graphs because they seem to work only in Internet Explorer (I usually use Firefox). Anyway, just thought some of the "math heads" out there might appreciate a more algebraic explanation.
It's interesting that given the above definition going from f/1.4 to f/1.0 will give you twice as much light (one full stop) but going from f/1.4 to f/1.2 only gives you 36% more light even though it is technically "half a stop" (and that you get ~75% of the light at f/1.4 that you had at f/1.2).
It's correct, although somewhat inaccurate due to rounding errors.
If we assume that there is a true half stop between the indicated aperture settings, opening up half a stop from f/1.4 to f/1.2 actually lets about 41% more light in, or put another way, we now have 141% of the light we originally had. Opening up another half stop to f/1 gives us 141% of the light we had at f/1.2. If we had opened up from f/1.4 to f/1 we would expect to have twice as much light, or 200% of the original amount of light (let's call that x). Does it work out? Let's see:
x * 141% * 141%
= x * 1.41 * 1.41
= x * 1.99~= 2x
36% is close to 41%, but this makes sense when you consider that a half stop faster than f/1.4 is actually ~ f/1.189 (also keep in mind that one stop faster than f/1 is actually ~ f/1.414)--so f/1.2 isn't quite half a stop faster than f/1.4. Of course irrational numbers like the square root of 2 are hard to deal with, so we typically round to nearest tenths for small f/numbers and don't bother to round at all for larger f/numbers.
It's correct, although somewhat inaccurate due to rounding errors.
If we assume that there is a true half stop between the indicated aperture settings, opening up half a stop from f/1.4 to f/1.2 actually lets about 41% more light in, or put another way, we now have 141% of the light we originally had. Opening up another half stop to f/1 gives us 141% of the light we had at f/1.2. If we had opened up from f/1.4 to f/1 we would expect to have twice as much light, or 200% of the original amount of light (let's call that x). Does it work out? Let's see:
x * 141% * 141%
= x * 1.41 * 1.41
= x * 1.99~= 2x
36% is close to 41%, but this makes sense when you consider that a half stop faster than f/1.4 is actually ~ f/1.189 (also keep in mind that one stop faster than f/1 is actually ~ f/1.414)--so f/1.2 isn't quite half a stop faster than f/1.4. Of course irrational numbers like the square root of 2 are hard to deal with, so we typically round to nearest tenths for small f/numbers and don't bother to round at all for larger f/numbers.
Cheers,
Jeremy
OK. So when a camera/lens reports an aperture of f/1.4 does that really mean f/1.414 and f/1.2 is actually f/1.189? Ziggy also recently posted that the f-stops in camera systems are actually *not* exactly focal length/diameter because the stops are calibrated to ISO standards and therefore dependent on the transmission of the optics and # of elements making up the lens....
By the way, I think the lens sounds and looks like it's going to be a collector's item but will certainly only appeal to a specialized group. I'll stick with my 50mm f/1.4.
OK. So when a camera/lens reports an aperture of f/1.4 does that really mean f/1.414 and f/1.2 is actually f/1.189? Ziggy also recently posted that the f-stops in camera systems are actually *not* exactly focal length/diameter because the stops are calibrated to ISO standards and therefore dependent on the transmission of the optics and # of elements making up the lens....
All the math is theory--I'd expect actual lenses not to have aperture settings that are so precise. Does f/1.4 mean f/1.400? I doubt it--it could be f/1.41 or f/1.39 or anything else for that matter. Only the lens manufacturer knows for sure (and then there are manufacturing tolerances as well!)
Then of course there's the optical transmission thing. No optical element, no matter how advanced the coating, will transmit 100% of the light hitting it--there will always be some loss. However, modern coatings are pretty darned good, transmitting (at least according to some manufacturer claims) in the neighborhood of 99.7% of the light. It's not uncommon to see lenses with 12 groups of elements, which if I'm not mistaken means 24 air/glass interfaces (presumably coated). If we have 99.7% transmission at each of these interfaces, then the transmission through the entire lens will be 0.997 ^ 24 or about 93%. 7% loss is a small fraction of a stop--I'd say that's pretty good.
Are aperture settings calibrated to compensate for (in the above case) 7% light loss? In other words, would that lens' f/2.8 setting actually be somewhat larger than f/2.8 to allow an additional 7% light through? My guess is no, but perhaps someone else has a more authoritative answer.
The only confusing thing about all this is that
there are the exact (mathematical) fstop numbers
and the "rounded" fstop numbers.
The mathematical values are needed for example
when creating your hyperfocal distance chart.
Those values can easily be derived by looking
the area of the aperature and dividing it by 2
to reach the next full fstop.
The easiest way to calculate the full fstops is
to take the powers of the suqare root of two:
The "rounded" fstops are not always derived
from the mathematical values by rounding them
mathematically. Sometimes they are only "the
next pretty number, close to the exact value".
f/22 for example is mathematically f/22.6274.
It is also incorrect that a f/1.4 lens is two times
brighter than a f/2.0 lens. It only has a opening
twice as large. How much light passes through the
lens though is determined by the number and
characteristics of the glass elements in the lens,
and not (only) on its aperature.
Back in the old days ppl actually tried to give
lenses an additional value for their "light passing
ability". But this wasn't as convenient as using the
fstop scale instead.
“To consult the rules of composition before making a picture is a little like consulting the law of gravitation before going for a walk.”
― Edward Weston
In my world, an variation of 1/8th of a stop or even 1/6th of a stop is insignificant to the final image.
As for accuracy of aperture reading by manufacturers - they vary all over the map. See below.
Look at some of your telephoto lenses and measure the diameter in millimeters ( or centimeters if you prefer ). Take this diameter in millimeters, and divide it into the focal length of the lens and see how that compares to the labled fstop. You may find more variation than you think. This is more true for long glass than short focal length zooms.
Canon 300mm f2.8 IS L measures approximately 105mm across the front objective lens.
300mm / 105mm = 2.857 That's an honest f2.8
Sigma 120-300 f2.8 measures approximately 97mm across the front objective.
300 / 97 = 3.09 Not f2.8, but more like f3.1, unless the lenses focal length is shorter than 300 which I suspect it is. This is significantly smaller than the Canon 300 f2.8, and they cannot both be 300mm focal length and f2.8. Do the math.
The Tamron 200-500 f5.6-6.3 measures approximately 80mm across the front lens element ( It uses 86mm filters)
500 / 80 = 6.25 pretty close to the spec'd f6.3 I thought it might be less. But it does AF on a 20D so it must be fairly close to f5.6
The Canon 500 f4 measures 122mm across the front element.
500 / 122 = 4.098 Pretty close to f4.
Tamron 180 f3.5 Di macro measures 60mm across the front element.
180 / 60 = 3.0 Better than the stated f3.5, but that may be to allow for the changes that occur when shooting 1:1
Not sure what all this means, but it is worthwhile to measure the front optics diameter to verify the stated aperture.
The measured diameters of the front optics may vary by a millimeter or so, because the optics are recessed, and I cannot get a ruler exactly to the surface, but must measure with the ruler about 1 cm above the surface of the optics due to the curvature of the lens.
The front element of the 50mm f1.2 must be approximately 42mm in diameter then.
Comments
Thanks for keeping me straight guys. Glad to see everyone came out so quickly to correct this!!
So, the revised formulation is:
($1300 for f/1.2 - $315 for f/1.4)/0.5 = $1970 per stop. Still expensive...but more reasonable.
Erich
From the link:
"f/stop Graph...
The diameter and area values of each f/stop are based on a lens with a focal length of 50mm. Note that an f/stop value is the ratio of the focal length of the lens to the diameter of the aperture. Thus, f/1 has an aperture diameter of 50mm, f/2 a diameter of 25mm, f/8 a diameter of 6.25mm, etc. The percentage value shown next to the aperture area is the aperture area as a percentage of the total f/1 aperture area. The amount of light entering the camera is directly governed by the area of the aperture opening. Thus, f/1.4 is 50% or half the area of f/1 and therefore cuts the light in half -- a "full stop" less light. f/2 is 25% or 1/4 the area of f/1 and is 2 stops less light. "
The math isn't hard; each one-stop difference is calculated by multiplying by 2 to the 1/2 power (the square root of 2 or about 1.414). Fractional-stop differences are calculated by multiplying by 2 to the (1/2 * differences in stops); for example, third-stop differences are mutliplied by 2 ^ (1/2 * 1/3) or 2 ^ 1/6 or about 1.122.
I could try to derive the formula to determine how many stops faster (or slower) x is than y for arbitrary values of x and y, but it's getting late and my algebra is rusty.
Cheers,
Jeremy
Jeremy Rosenberger
Zeiss Ikon, Nokton 40mm f/1.4, Canon 50mm f/1.2, Nokton 50mm f/1.5, Canon Serenar 85mm f/2
Canon Digital Rebel XT, Tokina 12-24mm f/4, Tamron 28-75mm f/2.8, Sigma 30mm f/1.4, Canon 50mm f/1.4
http://ubergeek.smugmug.com/
In my haste, I typed the 1/3 stop increments and accidently also included the 1/2 stop increments.
The graphic is quite good helping to explain the area relationship with the f-stop numbers.
"You miss 100% of the shots you don't take" - Wayne Gretzky
It's interesting that given the above definition going from f/1.4 to f/1.0 will give you twice as much light (one full stop) but going from f/1.4 to f/1.2 only gives you 36% more light even though it is technically "half a stop" (and that you get ~75% of the light at f/1.4 that you had at f/1.2).
Cheers,
Jeremy
Jeremy Rosenberger
Zeiss Ikon, Nokton 40mm f/1.4, Canon 50mm f/1.2, Nokton 50mm f/1.5, Canon Serenar 85mm f/2
Canon Digital Rebel XT, Tokina 12-24mm f/4, Tamron 28-75mm f/2.8, Sigma 30mm f/1.4, Canon 50mm f/1.4
http://ubergeek.smugmug.com/
If we assume that there is a true half stop between the indicated aperture settings, opening up half a stop from f/1.4 to f/1.2 actually lets about 41% more light in, or put another way, we now have 141% of the light we originally had. Opening up another half stop to f/1 gives us 141% of the light we had at f/1.2. If we had opened up from f/1.4 to f/1 we would expect to have twice as much light, or 200% of the original amount of light (let's call that x). Does it work out? Let's see:
= x * 1.41 * 1.41
= x * 1.99~= 2x
Cheers,
Jeremy
Jeremy Rosenberger
Zeiss Ikon, Nokton 40mm f/1.4, Canon 50mm f/1.2, Nokton 50mm f/1.5, Canon Serenar 85mm f/2
Canon Digital Rebel XT, Tokina 12-24mm f/4, Tamron 28-75mm f/2.8, Sigma 30mm f/1.4, Canon 50mm f/1.4
http://ubergeek.smugmug.com/
OK. So when a camera/lens reports an aperture of f/1.4 does that really mean f/1.414 and f/1.2 is actually f/1.189? Ziggy also recently posted that the f-stops in camera systems are actually *not* exactly focal length/diameter because the stops are calibrated to ISO standards and therefore dependent on the transmission of the optics and # of elements making up the lens....
Erich
Erich
"You miss 100% of the shots you don't take" - Wayne Gretzky
Then of course there's the optical transmission thing. No optical element, no matter how advanced the coating, will transmit 100% of the light hitting it--there will always be some loss. However, modern coatings are pretty darned good, transmitting (at least according to some manufacturer claims) in the neighborhood of 99.7% of the light. It's not uncommon to see lenses with 12 groups of elements, which if I'm not mistaken means 24 air/glass interfaces (presumably coated). If we have 99.7% transmission at each of these interfaces, then the transmission through the entire lens will be 0.997 ^ 24 or about 93%. 7% loss is a small fraction of a stop--I'd say that's pretty good.
Are aperture settings calibrated to compensate for (in the above case) 7% light loss? In other words, would that lens' f/2.8 setting actually be somewhat larger than f/2.8 to allow an additional 7% light through? My guess is no, but perhaps someone else has a more authoritative answer.
Cheers,
Jeremy
Jeremy Rosenberger
Zeiss Ikon, Nokton 40mm f/1.4, Canon 50mm f/1.2, Nokton 50mm f/1.5, Canon Serenar 85mm f/2
Canon Digital Rebel XT, Tokina 12-24mm f/4, Tamron 28-75mm f/2.8, Sigma 30mm f/1.4, Canon 50mm f/1.4
http://ubergeek.smugmug.com/
there are the exact (mathematical) fstop numbers
and the "rounded" fstop numbers.
The mathematical values are needed for example
when creating your hyperfocal distance chart.
Those values can easily be derived by looking
the area of the aperature and dividing it by 2
to reach the next full fstop.
The easiest way to calculate the full fstops is
to take the powers of the suqare root of two:
sqrt(2)/sqrt(2) = 1.0
sqrt(2) = 1.41
sqrt(2)*sqrt(2) = 2.0
sqrt(2)*sqrt(2)*sqrt(2) = 2.83
...
The "rounded" fstops are not always derived
from the mathematical values by rounding them
mathematically. Sometimes they are only "the
next pretty number, close to the exact value".
f/22 for example is mathematically f/22.6274.
It is also incorrect that a f/1.4 lens is two times
brighter than a f/2.0 lens. It only has a opening
twice as large. How much light passes through the
lens though is determined by the number and
characteristics of the glass elements in the lens,
and not (only) on its aperature.
Back in the old days ppl actually tried to give
lenses an additional value for their "light passing
ability". But this wasn't as convenient as using the
fstop scale instead.
― Edward Weston
A nice graphic representation of the law of diminishing returns! I hate to think where a new 1.0L would fall now. :uhoh
http://www.chrislaudermilkphoto.com/
As for accuracy of aperture reading by manufacturers - they vary all over the map. See below.
Look at some of your telephoto lenses and measure the diameter in millimeters ( or centimeters if you prefer ). Take this diameter in millimeters, and divide it into the focal length of the lens and see how that compares to the labled fstop. You may find more variation than you think. This is more true for long glass than short focal length zooms.
Canon 300mm f2.8 IS L measures approximately 105mm across the front objective lens.
300mm / 105mm = 2.857 That's an honest f2.8
Sigma 120-300 f2.8 measures approximately 97mm across the front objective.
300 / 97 = 3.09 Not f2.8, but more like f3.1, unless the lenses focal length is shorter than 300 which I suspect it is. This is significantly smaller than the Canon 300 f2.8, and they cannot both be 300mm focal length and f2.8. Do the math.
The Tamron 200-500 f5.6-6.3 measures approximately 80mm across the front lens element ( It uses 86mm filters)
500 / 80 = 6.25 pretty close to the spec'd f6.3 I thought it might be less. But it does AF on a 20D so it must be fairly close to f5.6
The Canon 500 f4 measures 122mm across the front element.
500 / 122 = 4.098 Pretty close to f4.
Tamron 180 f3.5 Di macro measures 60mm across the front element.
180 / 60 = 3.0 Better than the stated f3.5, but that may be to allow for the changes that occur when shooting 1:1
Not sure what all this means, but it is worthwhile to measure the front optics diameter to verify the stated aperture.
The measured diameters of the front optics may vary by a millimeter or so, because the optics are recessed, and I cannot get a ruler exactly to the surface, but must measure with the ruler about 1 cm above the surface of the optics due to the curvature of the lens.
The front element of the 50mm f1.2 must be approximately 42mm in diameter then.
Moderator of the Technique Forum and Finishing School on Dgrin