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Ok so I was reading one of the other threads and I was really confused.
I know that crop factor effectively multiplies the focal length, but in regards from full frame to crop bodyhow does it effect f/stop? I was trying to figure it out looking at effective focal lengths and apertures and I could not figure it out. Can you please explain??
Quote:
Originally Posted by powers_brent
Yes, you need to multiply the f-stop by the FOVCF as well. The reasonfor this is to achieve the same framing with a 35mm format camera (given the same focal length) you would need to get considerably closer to your subject. The closer you are to your subject the thinner the DOF.
DOF is dictated by: focal length, aperture, subject distance
I'll give an example. I have no idea how far or close you would need to be to achieve the same subject framing so I'll make that part up, the rest should apply though:
If you took a picture of a subject that was 10 feet away with a 100mm lens at f/2.8 on a 1.6X FOVCF camera you may need to move in to 7 feet to take that same picture (same framing) with a 35mm format camera. Since you are now closer to your subject, the DOF is now thinner. In order to achieve the same DOF as you would have with the 1.6X FOVCF camera you would need to stop down to f/4.48 (2.8 * 1.6). Make sense?
Mark's right. I would add that moving closer or further away will change a vital element of composition: perspective. Many times it's not desirable to change this element of composition, and sometimes it's even impossible. The only way to keep perspective the same between two sensor sizes is to change the focal length by the crop factor.Quote:
Originally Posted by Mark Elberson
If any two lenses have the same iris diameter, the DoF, diffraction, and light gathering power will be the same, no matter what the focal length, f-number, or sensor size. Given the same field of view, and non-macro focus distance.
ahhhhhhhhhh ok! Now I get it. Perfect. Thanks Guys.
What do you mean, "how does it effect f/stop?" In terms of exposure, it doesn't affect it, at all. The same amount of light per square centimeter gets through the lens, regardless of the camera or sensor.
If you're concerned about depth of field, don't use the "crop factor." Instead, go to DOFMaster and check out his calculator. It takes into account the "circle of confusion," which depends somewhat upon the magnification when printed. (Depth of Field is not a linear function of the aperture NOR the focal length--you can't just multiply the DOF numbers by the crop factor.)
Agreed. But I would argue that "total amount of light" matters far more than "light per area".Quote:
The same amount of light per square centimeter gets through the lens, regardless of the camera or sensor.
F-number is the established paradigm of photographers; how most of us think and work day in and day out. This mode of thought was necessary for film, since the response curve was intrinsically tied to the exposure (no matter what the sensor size), and it continues to work just fine in any discussion of a raw digital sensor of a single size.Quote:
Originally Posted by George Slusher
However, in a discussion of raw digital sensors of multiple sizes, the old paradigm is suboptimal and even misleading. Now that we have a linear capture medium, it doesn't matter what the intensity of light per area is: all that matters is the *total* amount of light.
In film, if you go from 35mm to MF, you can't reduce exposure to keep DOF the same, because the image would be underexposed; that's why many MF/LF shooters used tripods and slow shutter speeds. But that's not true of digital: the size of the sensor compensates perfectly for reduced exposure, so any larger sensor can get the same image as a smaller sensor.
If you think in terms of f-number only, then it may seem that larger sensors have less noise, thinner DOF, less diffraction, higher MTF, more weight, etc. But that's not true in all circumstances: only when f-number is kept the same.
For example, people often lament that there is no FF equivalent to the 17-55 f/2.8 IS. But there is, and it's even better! The only reason they didn't see it is because they were held back by their f-number-centric frame of mind. The 24-105 has lower light intensity per area (exposure) but it has so much larger area that it more than makes up for it and actually collects more total light and has less total noise. You actually have to stop down the 24-105 by a third stop to get images similar to the 17-55.
As another example, some think that the f/2 lenses on four thirds are a great advantage over the f/2.8 lenses on full frame: they think they can get faster shutter speeds. In fact, it only takes an f/4 lens on FF35 to get the same shutter speed, noise, DoF, and diffraction as an f/2 lens on Four Thirds.
It's not always even possible to keep f-number the same when going to a larger sensor. Most medium format lenses are only f/2.8 or slower, and actually have deeper DOF (and lower total light gathering power in low light) than the f/1.2 lenses on FF35.
A tiny 5x4mm digicam at f/2.8 has the same light intensity as a 56x41mm f/2.8 medium format digital back. But the larger camera gathers for more light in total, and when printed at the same size, it has far less noise, less diffraction, and thinner DOF. However, if we apply the crop factor to f-number: f/31 on the MFDB will result in the same total light, same noise, same diffration, and same DOF.
When comparing multiple format sizes, keeping f-number is same is not necessary, desirable, or even possible in all circumstances.
Crop factor explains equivalence between two different sized sensors. Multiply the focal length by the crop factor to get the same angle of view. But angle of view is not the only thing that may be compared between two camera systems.
Noise, DOF, diffraction, MTF, MP, weight, etc. can also be compared, and these comparisons should be done by applying the crop factor to the f-number. The effect is to keep the iris diameter the same. Then, no matter what the sensor size, noise, DOF, and diffraction remains the same.
For non-macro focus distances, I find that the crop factor does provide equivalent depth of field to a very close approximation, so I find it highly useful.Quote:
Originally Posted by George Slusher
I've done the calculations from 1/3" sensors up to 6x17 cm sensors and found that crop factor precisely correlates with DOF. Have you found any examples where it does not? As Mark stated, 50mm f/2.8 has the same DOF on APS-C as 80mm f/4.5 on FF35.Quote:
Originally Posted by George Slusher
To correct a bit what I wrote before, rather than editing the previous post:
What do you mean, "how does it effect f/stop?" In terms of exposure,
it doesn't affect it, at all. The same amount of light per square
centimeter gets through the lens, regardless of the camera or sensor.
If you're concerned about depth of field, don't use the "crop factor." Instead, go to DOFMaster
and check out his calculator. It takes into account the "circle of
confusion," which depends somewhat upon the magnification when printed,
if I recall correctly.
The reason I said that is that which
camera's f/stop gets multiplied by the FOVCF depends upon what you are
comparing. If you want the same framing with the same lens,
then Mark is right (or, at least, pretty close--the details are a
little bit different). Here's data from DOFmaster for the 5D (full
frame) and 30D (1.6x).
30D, 100mm, 80 ft, f/2.8: 70.8 - 92 ft, total 21.2 ft, 9.2 ft in front, 12 ft behind
5D, 100mm, 50 ft (to get the same framing), f/2.8: 44.3 - 57.4 ft, total 13.1 ft
To get the same DOF total, stop the 5D down to f/4.5:
5D, 100mm, 50 ft, f/4.5: 41.5 - 62.8 ft, total 23.1 ft, 8.5 ft in front, 12.8 ft behind
(Do note the difference in the front and behind numbers--not great, but different.)
That's only one comparison, however. What about comparing the DOF at the same distance
and focal length? In that case, it's the 30D's aperture that would have
to be stopped down--the reverse of the above. Start with the 50 ft,
f/2.8 data above.
30D, 100m, 50 ft, f/2.8: 46.2 ft - 54.4 ft, 8.19 ft total
Stop down the 30D to f/4.5 and you get 44.3 - 57.4 ft, just like the 5D at f/2.8.
Why
is that important? Distance, not focal length, determines perspective.
If you want the same perspective, you'll have to have the cameras at
the same point.
Now, get the same perspective--i.e., distance--and the same framing, by changing the focal length. We could use either camera as the standard, so choose the 5D.
5D, 100mm, 50 ft, f/2.8: 44.3 - 57.4 ft, total 13.1 ft, 5.7 ft in front, 7.4 ft behind
30D, 63mm, 50 ft, f/2.8: 41.5 - 62.9 ft, total 21.5 ft, 8.5 ft in front, 12.9 ft behind
So, to get the same DOF along with the same perspective and framing, the 5D would have to be stopped down to f/4.5, as before.
Quote:
Originally Posted by George Slusher
And this is because DOF/bokeh is amplified by the focal length and thus to reduce that you stop down right?
*DOF shortened, Bokeh amplified
I can't quote Daniel's message--my browser won't show the formatting tools, etc.
I corrected my error in my second message. The crop factor is close, though one has to figure out which way to apply it, as I showed. It all depends upon what you want to be the same: framing, perspective, or both. You only have three variables: distance, focal length, and aperture.
Applying it to point-and-shoot cameras in another example, using the Canon S3 IS (6x FOVCF).
S3, 16.7mm (equivalent to 100mm full-frame), 50 ft, f/2.8: 28.1 - 228.6 ft, total 200.5 ft, 21.9 ft in front, 178.6 ft behind
That would have the same perspective and framing as my last 5D example:
5D, 100mm, 50 ft, f/2.8: 44.3 - 57.4 ft, total 13.1 ft, 5.7 ft in front, 7.4 ft behind
Now, stop down the 5D to 6 x 2 = f/16.8. DOFMaster only has f/16, at 29 - 183 ft, and f/18, 27.5 - 273 ft, so f/16.8 would probably be about right.
I can't argue with your analysis, Daniel--I certainly don't know enough to argue, except for one thing: for me, shutter speed, itself, is important--not for "camera shake," but for moving subjects. In that case, the light per unit area per unit time does matter. If all you photograph are static scenes, you're probably right, though I expect that there's some effect of the size of the "pixels." (The 1.6x bodies generally use smaller pixels, especially the high-density bodies like the 50D.) However, photograph a galloping horse with a 5x7 view camera. For the same perspective (distance) and framing, you'll need the same shutter speed as a full-frame, 1.6x body or 6x point-and-shoot to stop the action. What counts there is the angular speed of the moving parts, which translates into motion blur. Theoretically, it wouldn't depend upon focal length/framing, but only upon distance. However, in practical terms, a shorter focal length lens gives a smaller image of the horse, so the motion blur isn't as noticeable. As I recall, that's also the reason for the 1/FL guideline, essentially the angular rate as a fraction of the field of view, though correct me if I'm wrong. (It won't be the first time.)
I can't help but be completely confused by this thread, and by the general frequency with which people on this forum multiply the maximum aperture of a lens by a 1.6x crop factor when it's used on a small sensor body. For instance saying the 17-55mm f/2.8 IS is effectively a 27-88mm f/4.5.
My confusion is this. Say I go outside on a sunny day, with an incident light meter, and get an exposure reading of ISO100 at 1/3200 second at f/2.8 And let's say I'm shooting with the three different bodies: an EOS 1V film camera, an EOS 5D mark II, and an EOS 50D. I'll say I'm using the 16-35mm f/2.8 L on the full frame bodies, and the 17-55mm f/2.8 IS on the 50D.
My best guess is that if I shoot a photo with each camera, lens wide open, at 1/3200 second shutter speed and ISO 100, all three would produce a correct (and identical) exposure on all three cameras, despite the variation in field of view, and depth-of-field. Thus, I would say for the purpose of exposure the lens is an f/2.8 lens regardless of film or sensor size.
So, what am I missing? Is the aperture adjustment only for depth-of-field calculations?
Quote:
Originally Posted by powers_brent
You could put it that way.
One other thing: The noise comparisons between different bodies with the same size sensors (e.g., APS-C, like the 40D vs 50D) but different pixel densities do show a marked difference in noise for the same shutter speed, aperture and ISO. A lot depends upon the sensor technology, as well, i expect, plus the internal processing by the camera.
Quote:
Originally Posted by nrdavis
Short answer: yes.
For most of us, in practical terms, the FOVCF has little to do with "correct" exposure. Daniel got into a very deep technical analysis that is concerned with total image quality--diffraction, noise, etc., not just simple shutter speed + aperture + ISO exposure determination.) Do remember that the camera takes care of a lot of this in its processing, e.g., in what "ISO 100" means in terms of the sensor output can vary from camera to camera.
I'll get back--have to leave right now.
Quote:
Originally Posted by Daniel Browning
Does this mean that a 16-35/2.8L on a crop body will also have to be stopped down to F4.5 to get the same exposure as with the EF-S 17-55/2.8 at F2.8 on the same body?
Wow! I love this forum. It is like going to school, but for an avocation. I have to go, too. I need to popsome popcorn and pour some tea! This is fascinating.
Quote:
Originally Posted by George Slusher
Agreed. Everything in my post assumed the same shutter speed in all cameras; I wouldn't have it any other way. Shutter speed, like perspective and angle of view, is a critical and fundamental aspect of composition and must be fixed for any comparison to be equivalent.
Quote:
Originally Posted by George Slusher
Here's why I think it doesn't matter: the larger sensors can be underexposed (increase ISO) and yield the same image as the smaller sensor.
I'm glad you used the S3 in your example. Take a look:
http://photos.imageevent.com/sipphot...re example.jpg
Looks pretty close to me. The 5D is getting much lower light intensity per area, but this is fully compensated by the much higher amount of total area. (Keep in mind that the ISO 80 setting on the S3 would be equivalent to ISO 160 on a 5D, because it only meters for 2.5 stops of highlight headroom. The important thing is that they both have the same shutter speed and DOF.)
Quote:
Originally Posted by George Slusher
If you use the same iris diameter, then it doesn't matter *what* size sensor or f-number you use: the images will always have the same light, same noise, same DOF, and same diffraction. But there would be no sense in paying more for large sensors if that's how it would be used. So to get any benefit from a larger sensor, one must use a slower shutter speed (that will keep your depth of field, but improve noise, light, etc.), or use thinner DOF.
In ample light, you can shoot the 5D with the same DOF as a digicam, but get more light, less noise. In low light, you have the choice of using thinner DOF than the digicam to get more light.
Quote:
Originally Posted by George Slusher
If it was a digital 5x7 view camera, it would be no problem to stop it down to match the DOF of the digicam, use the same shutter speed, and still get the exact same amount of light and noise.
Quote:
Originally Posted by George Slusher
I kindly disagree: the motion blur would be the same for all the cameras as long as they have the same distance, same angle of view, and same print size. The smaller focal length on the S3 is only less blurred on the sensor, but once you magnify the sensor 50 times for a large print, the blur becomes visible. The large sensor is magnified much less, so the motion blur comes out equal in the end. (If you print them at different sizes, though, all bets are off, of course.)
Quote:
Originally Posted by George Slusher
Yes, agreed.
Quote:
Originally Posted by nrdavis
Exposure is the light intensity (scene luminance, ND filters, shutter speed, f-number, etc.). ISO is not part of the exposure, but it guides you to select the exposure.
By definition, they all have the same exact *exposure*, because they have the same shutter and f-number.
Now, that doesn't mean they all have the same noise. The 5D2 will have less noise than the 50D.If you stop down the 5D2 by 1+1/3 stops (f/4.5), it will have 1.3 stops less exposure than the other cameras. And the brightness on the LCD screen will be less. If you increase the brightness by changing ISO (remember that ISO is not exposure, but it does affect the brightness and noise) to ISO 260, then the 5D2 will again have the same brightness and same noise as the 50D.
Quote:
Originally Posted by nrdavis
FWIW, I don't think it's very realistic to compare different field of view. If I have a 50D and 50mm lens, then upgrade to a 5D2, I'm not going to stop shooting "telephoto" (80mm-equivalent) focal lengths forever and restrict myself to normal only: I will buy an 80mm lens so I can continue to shoot short tele. What I'm saying is that photographers don't generally let their camera format decide what angle of view they're going to shoot.
Quote:
Originally Posted by nrdavis
True, but that has nothing to do with the purpose of crop factor. The purpose is to understand what's equivalent.
For the purpose of angle of view, a 50mm is a 50mm regardlness of film or sensor size.
For the purpose of angle of view, a 50mm-on-APS-C is equivalent to 80mm-on-FF35.
For the purpose of depth of field, f/2.8 on APS-C is equivalent to f/4.5 on FF35.
For the purpose of noise, diffraction, total light gathering power, etc., f/2.8 on APS-C is equivalent to f/4.5 on FF35.
Quote:
Originally Posted by nrdavis
It's not just depth of field, but also noise, total amount of light, diffraction, approximate lens weight, approximate MTF (in the case of EF-S vs EF), and probably correlated with other factors as well. Size matters to a lot of things.
Quote:
Originally Posted by George Slusher
I don't think so. See the recent thread I started:
Myth busted: smaller pixels have more noise, less dynamic range, etc.
Quote:
Originally Posted by George Slusher
JPEG does for sure, but the raw data itself (before conversion) does not vary much with pixel size.
Quote:
Originally Posted by ShutterbugJohan
No. The 16-35 f/2.8 does indeed project far more total light than the 17-55 f/2.8, but if you put it on a crop body, all that light will fall on dead spead instead of a sensor, so it is completely wasted.
Quote:
Originally Posted by HiFiGuy1
I'm enjoying myself too. :)
The paradigm shift I'm trying to promote is that the *total amount of light* is what really matters, everything else (focal length, f-number, ISO, sensor size, etc.) are just the factors that affect light.
In fact, there is already a very useful shortcut for determining the total amount of light. It used to be called aperture, but the meaning of that word was usurped and lost to most photographers, so now I use a new word: "iris diameter". For a given angle of view and perspective, the iris diameter correlates perfectly to the total amount of light, no matter what the sensor size, f-number, focal-length, or ISO.
Iris diameter also correlates with diffraction, depth of field, and lens weight (loosely).
A 6mm iris diameter on S3 digicam has the same angle of view, focus distance, light gathering power, depth of field, and diffraction as a 6mm iris diameter on fourth thirds, as well as 6mm iris diameter on APS-C, FF35, Medium Format, and even Large Format.
It all comes down to iris diameter.
Quote:
Originally Posted by Daniel Browning
I think I'm missing a line or few between some dots. how does the iris diameter determine the angle of view? I thought that was a combination of the focal length and the sensor size... Same lostness on the focus distance....
Did you mean that given the same angle of view and focus distance, a 6mm iris has the same light gathering power, depth of field, and diffraction, regardless of the sensor format, or do I need a rewind on something?
Quote:
Originally Posted by Colin
It doesn't; what I said was misleading.
Quote:
Originally Posted by Colin
Precisely.
Thanks for the correction.
Quote:
Originally Posted by HiFiGuy1
Haha in this thread I feel like I'm in a physics class when I've only just passed Algebra 1.
Yeah, "We're way beyond birthdays now". (Clear and Present Danger)
Daniel,
Does this mean that the iris diameterchanges when zooming a 70-200 @ f/2.8 from 70 to 200?
If I were to shoot an evenly illuminatedwhite screen (16 ft/L)at 70mm and at 200mm, both shots at f/2.8, it seems like there would be less total light available at 200mm since I would be shooting a smaller area of screen, thus less available Lamberts. For the total light to be equal, given the same ISO and shutter speed, it seemsit would make sense ifthe focal length and iris diameter correlate inversely to maintain an equivalent f-stop. If the iris diameter is constant, then wouldn't something have to give? Shutter speed?
I'm not Daniel--nowhere near his technical knowledge, but this is an easy one: Yes. The definition of "f/stop" is the lens focal length divided by the effective aperture diameter. Thus, the effective aperture diameter of the 70-200mm lens (or any zoom lens) changes with the focal length for a particular f/stop. Most zoom lenses have a range of maximum apertures (lowest/fastest f/stop). My 100-400mm lens has maximum aperture ranging from f/4.5 to f/5.6, for example. (It's probably hard to make a lens with that wide a zoom range--4:1--with a constant maximum aperture.) Those with a constant maximum aperture tend to be expensive and large, especially if the aperture is fast, like f/2.8.
Thanks George. I felt like it must, but my concept of f-stop and irisup until now has been simply "lower f-stop means larger iris". I guess it does hold true, but obviously focal length must be considered when comparing f-numbers. I never really thought about it in the context of zooms, but now this thread is making me!!
In terms of taking the picture and setting exposure within a single camera, the f/X number is really what we want to know. If the F/X number is the same, the same amount of light hits the sensor. the actual iris diameter will be larger at a longer focal length, but the angle of view is less, so it collects less light from an angular sense, so makes it up with a larger iris diameter.
Quote:
Originally Posted by Daniel Browning
Daniel, are you saying that it is *not* true for macro?
Yes, that's what I'm saying; at a minimum, bellows factor has to be considered as well.
would you mind explaining bellows factor?
I *thought* bellows factor refered to the fact that effective f number is f number times (1 + magnification). (Ie, if you're shooting at f/10 and magnification 1:1 then you must expose as if you were shooting f/20). With ttl metering, we don't have to worry about that much, except to know we'll need more light for macros.
Effective f number also gives rise to more diffraction. I don't think you use this "effective f number" in dof calculations. dof is proportional to f number (the real one, not effective f) and inversely proportional to magnification.
But I don't see how this changes the fovcf rule. That is, it seems to me that a 100mm lens on a 1.6fovcf camera shooting at f/10 and 1:1 magnification still acts like 160mm f/16 at 1:1 would on ff (in terms of exposure and dof).
I'm probably wrong, though [^o)]