Pablito
coco frío
Given the fascination with fast lenses and the recent announcement of a very fast 50mm M-mount lens by you-know-who, I wondered if some of our members who really understand optical engineering could enlighten us about the ACTUAL speed advantage of the faster-than-f1.4 lenses. For example, does f1.1 offer a TRUE FULL STOP advantage over f1.4? Or only half a stop, or some other fraction of a stop? What about f1.2 compared to f1.4?
mfogiel
Veteran
dexdog
Veteran
you might try the table at this link.
http://www.cinemagraphica.com/tech/stop_scale.html
1.1 looks to be about 2/3 stop faster than 1.4, and about 1/3 stop between 1.2 and 1.4. Also, I recall that ferider once posted a table for f stops for 50mm lenses- I have it stuck on my bulletin board at work. Don't know where it comes from, but ferider's table gives this sequence of 1/6 f stops: 1.00, 1.06, 1.12, 1.19, 1.26, 1.33, 1.41, 1.50...
http://www.cinemagraphica.com/tech/stop_scale.html
1.1 looks to be about 2/3 stop faster than 1.4, and about 1/3 stop between 1.2 and 1.4. Also, I recall that ferider once posted a table for f stops for 50mm lenses- I have it stuck on my bulletin board at work. Don't know where it comes from, but ferider's table gives this sequence of 1/6 f stops: 1.00, 1.06, 1.12, 1.19, 1.26, 1.33, 1.41, 1.50...
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Each F-Stop is a factor of a square root of 2, or 1.4142 (A geek uses it as the combo of their luggage). Focal Length/ F-Stop is the diameter of the "light collecting area" of the lens. Except in modern lenses where light is squisheed up here and expanded out there and you never know where those designers stuck the physical aperture stop...
So: F0.7, F1.0, F1.4, F2, F2.8, F4, F5.6, F8, F11, F16, F22, F32, F45 are all Full-stop differences.
F1.2 is "about" 1/2 stop faster than F1.4, F1.1 is "about" 1/4 stop slower than F1. F0.95 is "about" 1/6th stop faster than F1.0. In real terms of gain in light hitting the film: not so easy to figure because of transmission loss through the optics.
So: F0.7, F1.0, F1.4, F2, F2.8, F4, F5.6, F8, F11, F16, F22, F32, F45 are all Full-stop differences.
F1.2 is "about" 1/2 stop faster than F1.4, F1.1 is "about" 1/4 stop slower than F1. F0.95 is "about" 1/6th stop faster than F1.0. In real terms of gain in light hitting the film: not so easy to figure because of transmission loss through the optics.
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teo
Well-known
I'd like if the still picture cameras had lens brightness showed as T (actual transmission, like movie cameras), instead of F.
But it would ruin all the marketing about fast lenses (especially with the "constant aperture" zooms)...
But it would ruin all the marketing about fast lenses (especially with the "constant aperture" zooms)...
I have lens tests that show actual measured F-Stop and T-Stop. It was surprising how many manufactureres had "F1.3" lenses marked as F1.2.
In terms of T-Stop, the Nikkor 55/1.2 Multicoated was the fastest of all the Nikkors(made for photography).
In terms of T-Stop, the Nikkor 55/1.2 Multicoated was the fastest of all the Nikkors(made for photography).
Sparrow
Veteran
Brian; is the t-stop the just in the centre or an average of the whole frame.
ferider
Veteran
This is a sequence of 1/6th of f-stops. You can pick 1/3rd or half stops as you like:
The actual transmission of a lens depends on the manufacturer (marketing vs. engineering), nr. elements, coating, etc. Dante has a nice write-up on this for classic superfast lenses:
http://www.dantestella.com/technical/fast.html
Cheers,
Roland.
The actual transmission of a lens depends on the manufacturer (marketing vs. engineering), nr. elements, coating, etc. Dante has a nice write-up on this for classic superfast lenses:
http://www.dantestella.com/technical/fast.html
Cheers,
Roland.
Pablito
coco frío
THANKS to all.
uhligfd
Well-known
Simple: just square the f number and compare then; why? because you are comparing the area of the circles formed by the aperture. And circles have the area of pi times r^2.
So:
f/1 => 1^2 = 1
f/1.1 => 1.1^2 = 1.21
f/1.2 => 1.2^2 = 1.44
f/1.3 => 1.3^2 = 1.69
f/1.4 => 1.4^2 = 1.96
f/1.5 => 1.5^2 = 2.25 etc
Thus comparing f/1.1 with f/1.4 gives an f/1.1 lens about (1.4)^2/(1.1)^2 = 1.96/1.21 ~ 1.61 times the exposure that a f/1.4 lens would., or roughly 61 % more. If it were precisely 1.666.., it would mean 2/3 (66 %) of a stop more light.
As simple as pie!
So:
f/1 => 1^2 = 1
f/1.1 => 1.1^2 = 1.21
f/1.2 => 1.2^2 = 1.44
f/1.3 => 1.3^2 = 1.69
f/1.4 => 1.4^2 = 1.96
f/1.5 => 1.5^2 = 2.25 etc
Thus comparing f/1.1 with f/1.4 gives an f/1.1 lens about (1.4)^2/(1.1)^2 = 1.96/1.21 ~ 1.61 times the exposure that a f/1.4 lens would., or roughly 61 % more. If it were precisely 1.666.., it would mean 2/3 (66 %) of a stop more light.
As simple as pie!
Al Kaplan
Veteran
Sparrow, yes, the center of the image is getting more light than the corners. The light going to the corners, expecially with an extreme wide angle lens can easily be travelling twice as far to get to the film. That's one big advantage of retrofocus designs, with the entire lens being further away from the focal plane. Another factor is that with most lenses there's vignetting due to the fact that when you look directly through the lens (off the camera) the aperture goes from a full circle at small f/stops to an elipse as you aproach full aperture.
There are ways to design a lens to somewhat mitigate these things. The classic 8.5cm f/2 Nikkor has a front element about twice the diameter of the rear element. If you stop it down to say f/11 and look straight through the rear of the lens, then tilt it towards the side, the diaphragm opening not only becomes eliptical but appears to grow larger, thus giving the corners of the picture a bit more light.
The classic "long" 90mm f/2.8 Elmarit also looks like the aperture from the rear is larger than when viewed from the front, but only slightly larger. Also it stays round, which would result in less light fall-off in the corners.
T/stops tell you what the effective aperture is for the center only. Wide open the corners might get a half stop less light with the Elmarit but probably in the neighborhood of a full stop with the Nikkor. Internal reflections, especially in older single coated optics, don't result in all that much loss of light reaching the focal plane. Some does get lost being reflected back out the front of the lens or into the black baffling of the inside of the lens barrel, but a lot of it just gets bounced around and ends up on the focal plane as non image forming flare, and it still adds light to the exposure.
Lastly, we still have absorbtion by the glass itself. That f/2 Nikkor has some very thick elements compared to the Elmarit, while the 90mm f/4 Elmar only has three fairly thin elements.
There are ways to design a lens to somewhat mitigate these things. The classic 8.5cm f/2 Nikkor has a front element about twice the diameter of the rear element. If you stop it down to say f/11 and look straight through the rear of the lens, then tilt it towards the side, the diaphragm opening not only becomes eliptical but appears to grow larger, thus giving the corners of the picture a bit more light.
The classic "long" 90mm f/2.8 Elmarit also looks like the aperture from the rear is larger than when viewed from the front, but only slightly larger. Also it stays round, which would result in less light fall-off in the corners.
T/stops tell you what the effective aperture is for the center only. Wide open the corners might get a half stop less light with the Elmarit but probably in the neighborhood of a full stop with the Nikkor. Internal reflections, especially in older single coated optics, don't result in all that much loss of light reaching the focal plane. Some does get lost being reflected back out the front of the lens or into the black baffling of the inside of the lens barrel, but a lot of it just gets bounced around and ends up on the focal plane as non image forming flare, and it still adds light to the exposure.
Lastly, we still have absorbtion by the glass itself. That f/2 Nikkor has some very thick elements compared to the Elmarit, while the 90mm f/4 Elmar only has three fairly thin elements.
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Ed S.
Member
I did an Excel exercise on this subject. Lots of typing and checking, to catch any mistakes. I did this mostly out of curiosity: is f/0.95 1/8- or 1/6-stop below f/1? An 11% overall light loss makes an f/1.1 effectively an f/1.2 (but neither were 100% to begin with, . . . . . .). It looks like the sharpness of the various 50mm f/1.7 - f/1.8 primes usually makes them a better choice than the costly f/1.4s (except when I visit museums and historic buildings and can’t use my tripod).
The 1/8-stop increments I’ve seen on digital cameras has slightly better granularity (+9%open, -8.3%close) than 1/6-stops (+12.25%open,-9.1%close), but I wonder how much of an advantage this difference is.
On one worksheet, I created a list of apertures (areas), from f/0.5 to f/32, in whole stop (1-stop) increments. I then interpolated between each stop, in ¼-stop increments and in 1/6-stop increments.
I set the table up so that I could enter a focal length on the f/1.0 row, and the area and radius would be calculated, displayed and used as a basis for other area calculations. This simplifies things a bit (a 50mm f.1,0 has a nominal diameter of 50mm), and lets me change the focal length in one place, and see the results.
I repeated this exercise a few columns over, I repeated the exercise using the formula M=2^(0.5*stop). I used the resulting f-stop numbers to calculate the area for this part of the table (based on the shared f/1.0 base value), so there is a question of accuracy.
Both tables agree, to 5 decimal places, only at whole stop rows (f/1, f/1.4, f/2. F/2.8, . . .). Otherwise, they disagree, varying up to about 7%.
The formula generates f-stop numbers, and they seem to agree with published values (Wiki, etc.). There are minor questions, such as whether f/1.3 is a 3/4 -stop number (1.29684) or a 5/6-stop number (1.33482), but I assume that this is simply a rounding and convenience agreement (2/3-stop: 1.25989).
The areas generated do increase (or decrease) in constant percentage amounts for both 1/4- and 1/6-stop increments, which is expected, because of the function. The whole-stop relationships work (increase from f/1.4 to f/2.0, decrease area by 50%, etc), but the rest are off.
Is f/1.5 91.667% of f/1.4, or only 83.333%? Is f/1.7 75% of f/1.4, on only 70.7%? This is a small matter, as it’s only a few percent (but it does repeat for every whole-stop interval). I can’t find any errors in my table.
The interpolated area values show the expected relationships: f/1.8 is 33.33% larger than f/2; f/1.2 is 25% smaller than f/1.0, etc, The calculated f-stop numbers are off (except at whole-stops).
I don’t have good explanation(s) for my results. I think that I caught all of my errors. I suspect legacy reasons, and that possibly there is a correction factor that the formula covers nicely. In pre-PC days, simply doing a lookup in a book might have been very a good choice.
Additional information and comments are welcome.
The 1/8-stop increments I’ve seen on digital cameras has slightly better granularity (+9%open, -8.3%close) than 1/6-stops (+12.25%open,-9.1%close), but I wonder how much of an advantage this difference is.
On one worksheet, I created a list of apertures (areas), from f/0.5 to f/32, in whole stop (1-stop) increments. I then interpolated between each stop, in ¼-stop increments and in 1/6-stop increments.
I set the table up so that I could enter a focal length on the f/1.0 row, and the area and radius would be calculated, displayed and used as a basis for other area calculations. This simplifies things a bit (a 50mm f.1,0 has a nominal diameter of 50mm), and lets me change the focal length in one place, and see the results.
I repeated this exercise a few columns over, I repeated the exercise using the formula M=2^(0.5*stop). I used the resulting f-stop numbers to calculate the area for this part of the table (based on the shared f/1.0 base value), so there is a question of accuracy.
Both tables agree, to 5 decimal places, only at whole stop rows (f/1, f/1.4, f/2. F/2.8, . . .). Otherwise, they disagree, varying up to about 7%.
The formula generates f-stop numbers, and they seem to agree with published values (Wiki, etc.). There are minor questions, such as whether f/1.3 is a 3/4 -stop number (1.29684) or a 5/6-stop number (1.33482), but I assume that this is simply a rounding and convenience agreement (2/3-stop: 1.25989).
The areas generated do increase (or decrease) in constant percentage amounts for both 1/4- and 1/6-stop increments, which is expected, because of the function. The whole-stop relationships work (increase from f/1.4 to f/2.0, decrease area by 50%, etc), but the rest are off.
Is f/1.5 91.667% of f/1.4, or only 83.333%? Is f/1.7 75% of f/1.4, on only 70.7%? This is a small matter, as it’s only a few percent (but it does repeat for every whole-stop interval). I can’t find any errors in my table.
The interpolated area values show the expected relationships: f/1.8 is 33.33% larger than f/2; f/1.2 is 25% smaller than f/1.0, etc, The calculated f-stop numbers are off (except at whole-stops).
I don’t have good explanation(s) for my results. I think that I caught all of my errors. I suspect legacy reasons, and that possibly there is a correction factor that the formula covers nicely. In pre-PC days, simply doing a lookup in a book might have been very a good choice.
Additional information and comments are welcome.
antiquark
Derek Ross
To compare the light gathering power of two f-stops, fA and fB, just use the equation:
(fA/fB)^2
The result will be the advantage of fB. (So if fB is the slower aperture, the result should be lower than 100%.)
So to compare (1.4/1.5)^2 = 88%
Or to compare (1.4/1.7)^2 = 68%
So your numbers look similar to mine. However I just used the "face value" of the aperture, it sounds like you constructed numerically accurate aperture values to many decimal places. That could be the source of the difference.
Also to convert a light gathering percentage into stops, use the formula:
stops = log(percent/100)/log(2).
So in the comparison above between the f1.4 and f1.5 lenses, the percentage difference was 88. Converting that to stops:
log(88/100)/log(2) = -0.18 stops or about 1/5th of a stop.
Finally, to completely combine the formulas to give the stop difference between two apertures, fA and fB, use:
difference in stops = log[(fA/fB)^2]/log(2)
hope that helps.
amateriat
We're all light!
I think that would be pi.
This is a small reason why I gave up big SLRs with zooms for RFs with little fixed-focal-length optics (and only three or so of them...).
- Barrett
Ed S.
Member
AntiQ:
The situation that my Excel spreadsheet shows me is this:
If a 50mm f/1.0 lens has a nominal diameter of 50mm, then it has a clear aperture area of 1963.495mm2.
So, interpolating:
50/1.1 (1/3-stop) 1636.246 mm2 = 0.8333 area of f/1.0. or 1.6667 area of f/1.4
50/1.1 (3/8-stop) 1595.340 mm2 = approximately same area as f/1.1 below
50/1.2 (1/4-stop) 1472.622 mm2 = 0.7500 area of f/1.0. or 1.500 area of f/1.4
50/1.3 (1/3-stop) 1308.997 mm2 = 0.6667 area of f/1.0. or 1.333 area of f/1.4
50/1.4 (1/2-stop) 981.748 mm2 = 0.5000 area of f/1.0
The math seems to work, but there may be errors or incorrect assumptions on my part..
Using the equation M = 2.0^(0.5*f) , then dividing the f/1.0 area by the result:
f/1.0 = 2.0^(0.5*0.0000) = 1.0000 1963.495 mm2
f/1.? = 2.0^(0.5*0.2500) = 1.09051 1651.096 mm2
f/1.1 = 2.0^(0.5*0.3333) = 1.12245 1588.463 mm2
f/1.2 = 2.0^(0.5*0.5000) = 1.18921 1388.401 mm2
f/1.3 = 2.0^(0.5*0.6666) = 1.25989 1236.982 mm2
f/1.? = 2.0^(0.5*0.7500) = 1.29684 1167.501 mm2
f/1.4 = 2.0^(0.5*1.0000) = 1.41421 981.748 mm2
A couple of problems, that I cannot resolve, appear.
Your calculation of about 1/5 stop looks good to me. My interpolated calculations show f/1.511, 859.029mm2, as being 1/4-stop past f/1.4.
The equation shows f/1.5 = 2.0^(0.5*1.1666) = 1.49830 as 1/6-stop past f/1.4, and I’ve seen this suggested in print.
I do not doubt your formulas, and the film I developed last week says your formulas are correct, but the math in front of me does not add up. I still do not see any mistakes on my spreadsheet.
My reference books on optics tell me that adjustable aperatures are good things to have, and to watch out for diffraction issues when using small apertures. One of them weighs more than my Kiev-6C+50mm Flek, yet says nothing more. Which animated graphic should I use here?
Is it possible that, due to legacy considerations, f/1.1, 1.2, etc., are accurate in the sense that they mark the correct stop (because they work and do not cause confusion), but do not link to calculated values?
The situation that my Excel spreadsheet shows me is this:
If a 50mm f/1.0 lens has a nominal diameter of 50mm, then it has a clear aperture area of 1963.495mm2.
So, interpolating:
50/1.1 (1/3-stop) 1636.246 mm2 = 0.8333 area of f/1.0. or 1.6667 area of f/1.4
50/1.1 (3/8-stop) 1595.340 mm2 = approximately same area as f/1.1 below
50/1.2 (1/4-stop) 1472.622 mm2 = 0.7500 area of f/1.0. or 1.500 area of f/1.4
50/1.3 (1/3-stop) 1308.997 mm2 = 0.6667 area of f/1.0. or 1.333 area of f/1.4
50/1.4 (1/2-stop) 981.748 mm2 = 0.5000 area of f/1.0
The math seems to work, but there may be errors or incorrect assumptions on my part..
Using the equation M = 2.0^(0.5*f) , then dividing the f/1.0 area by the result:
f/1.0 = 2.0^(0.5*0.0000) = 1.0000 1963.495 mm2
f/1.? = 2.0^(0.5*0.2500) = 1.09051 1651.096 mm2
f/1.1 = 2.0^(0.5*0.3333) = 1.12245 1588.463 mm2
f/1.2 = 2.0^(0.5*0.5000) = 1.18921 1388.401 mm2
f/1.3 = 2.0^(0.5*0.6666) = 1.25989 1236.982 mm2
f/1.? = 2.0^(0.5*0.7500) = 1.29684 1167.501 mm2
f/1.4 = 2.0^(0.5*1.0000) = 1.41421 981.748 mm2
A couple of problems, that I cannot resolve, appear.
- The area ratios from the equation do not have the same relationships to one another as the calculated ones do. The interpolated values show f/1.2 having 50% more area than f/1.4, and f/1.0 having 50% more area than f/1.2. This sounds right.
- Using the formula values, I can’t tell where f/1.1 and f/1.3 should be, i.e., which areas f/1.1 and f/1.3 represent.
Your calculation of about 1/5 stop looks good to me. My interpolated calculations show f/1.511, 859.029mm2, as being 1/4-stop past f/1.4.
The equation shows f/1.5 = 2.0^(0.5*1.1666) = 1.49830 as 1/6-stop past f/1.4, and I’ve seen this suggested in print.
I do not doubt your formulas, and the film I developed last week says your formulas are correct, but the math in front of me does not add up. I still do not see any mistakes on my spreadsheet.
My reference books on optics tell me that adjustable aperatures are good things to have, and to watch out for diffraction issues when using small apertures. One of them weighs more than my Kiev-6C+50mm Flek, yet says nothing more. Which animated graphic should I use here?
Is it possible that, due to legacy considerations, f/1.1, 1.2, etc., are accurate in the sense that they mark the correct stop (because they work and do not cause confusion), but do not link to calculated values?
Chris101
summicronia
The precision of the last few posts surpasses that of the lens manufacturing process. I read those magazines that test lenses, and one of the things they report is the actual vs marked maximum f-stop. I've noticed they can be up to a half stop different, usually with the error on the slower side.
antiquark
Derek Ross
I think the numbers don't match because the first F-numbers are rounded off.
For example, at the top you use F1.3, and below you show that it is equal to 1.25989. (They should be the same.)
However I agree with Chris101, that level of precision is outweighed by manufacturing realities.
Ed S.
Member
I suppose that I should clarify a bit.
I generated all of my data using Excel, and Excel defaults to 7 or 8 decimal places. Just create a formula, and voila, unachievable amounts of precision are provided. The third decimal place (xxx.yyy square millimeters) is clearly meaningless for my purposes, but it does suggest that roundoff is not the issue. BUt the numbers do look good.
I too have spent time in manufacturing. Getting a human to achieve 0.01" accuracy by hand (no leadscrew, or other type of assistance permitted) is tough.
But the original question was about f-stops and actual differences between f/1.1 and f/1.4 (and f/1.2), and I thought I’d generate a spreadsheet on the subject. Questions answered; no fuss, no muss, except it didn’t quite work out.
I realize that numbers on the aperture ring are abbreviated, if for no other reason than there usually isn’t room for 3 digits, much less "f/" in front of al numbers. And there’s something to be said for the simplicity of mostly linear numbers: 1.1,1.2, 1.3, . . . 2.0, 2.8, 4. . . . and clicks in between.
But a look-up table would be nice. To some degree, it is simply a buyer’s guide.
Nikon makes, or made, a 135mm/2.3. Where is f/2.3; is it 1/3 or 1/2 stop past f/2? If you have the opportunity to buy, is a 1/3-stop decrease acceptable? Is ½-stop too much? There are a few 200mm and 300mm lenses in the f/3.5 to f/4.5 range; just how much difference is there between f/3.5, f/4 and f/4.5? I have some Konica equipment. Konica made a 135/3.5, /3.2 and /2.5. The f/3.2 is the sharpest of the lot, and only 2/3 stop slower than the 2.5. I chose the 3.2.
I generated all of my data using Excel, and Excel defaults to 7 or 8 decimal places. Just create a formula, and voila, unachievable amounts of precision are provided. The third decimal place (xxx.yyy square millimeters) is clearly meaningless for my purposes, but it does suggest that roundoff is not the issue. BUt the numbers do look good.
I too have spent time in manufacturing. Getting a human to achieve 0.01" accuracy by hand (no leadscrew, or other type of assistance permitted) is tough.
But the original question was about f-stops and actual differences between f/1.1 and f/1.4 (and f/1.2), and I thought I’d generate a spreadsheet on the subject. Questions answered; no fuss, no muss, except it didn’t quite work out.
I realize that numbers on the aperture ring are abbreviated, if for no other reason than there usually isn’t room for 3 digits, much less "f/" in front of al numbers. And there’s something to be said for the simplicity of mostly linear numbers: 1.1,1.2, 1.3, . . . 2.0, 2.8, 4. . . . and clicks in between.
But a look-up table would be nice. To some degree, it is simply a buyer’s guide.
Nikon makes, or made, a 135mm/2.3. Where is f/2.3; is it 1/3 or 1/2 stop past f/2? If you have the opportunity to buy, is a 1/3-stop decrease acceptable? Is ½-stop too much? There are a few 200mm and 300mm lenses in the f/3.5 to f/4.5 range; just how much difference is there between f/3.5, f/4 and f/4.5? I have some Konica equipment. Konica made a 135/3.5, /3.2 and /2.5. The f/3.2 is the sharpest of the lot, and only 2/3 stop slower than the 2.5. I chose the 3.2.