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Thread: What is effective focal length after cropping? SAT-like question.

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    What is effective focal length after cropping? SAT-like question.

    Feel like going back in time and taking an SAT test? Ok, this is a techie question that is tough for me to frame.... so here goes...

    If I crop the image from my 5D Mk II taken with a 200mm lens, what is the focal length equivalent? This question is like the crop sensor vs full sensor with the same lens length issue.

    My 5D Mk II put out a JPG with these dimensions: 5616x3744 pixels. If I crop it down (same proportions and not resizing) to where the final image is 960x640 pixels, that is as if I had a ???mm lens.

    Some more background..... a friend asked me if they should jump to a 500mm lens, a very expensive proposition, and it got me thinking.... with my 5D Mk II and my 200mm lens, I get fantastic results cropping down a bird, small in the frame, down to 100% assuming I get the exposure just right. That is just like having a bigger lens. What would this effective lens be?
    OK, for additional points, if I did the same using a 500mm lens and a 105mm lens, what would the effective focal lengths be? I am assuming this a single formula I could use... no?

    Thanx so much.
    Bruce
    My photo and humor blog: http://travelthroughpictures.com/

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    Quote Originally Posted by Bruce in Philly View Post
    My 5D Mk II put out a JPG with these dimensions: 5616x3744 pixels. If I crop it down (same proportions and not resizing) to where the final image is 960x640 pixels, that is as if I had a ???mm lens.

    The 35mm-equivalent focal length is 1170mm (crop factor is 5.85X). Here's how you calculate it:
    crop factor = 5616 / 960
    35mm-equivalent focal length = 200 * crop factor


    It is almost exactly the same as using a smaller sensor. In fact, a 960x640 crop from the 5D2 (6.1 x 4.1mm) is about the same size as the sensors in most compact digicams (e.g. DMC-FZ150). The FZ150 lens only goes to 105mm though, so you'd have to crop the digicam down from 12 MP to 4 MP in order to get the same field of view as the 5D2. Of course, 4 MP is a whole lot more than 0.6 MP.


    Quote Originally Posted by Bruce in Philly View Post
    That is just like having a bigger lens.

    Well, it's like having a longer, slower lens. It's like an optically perfect teleconverter.


    A 200mm f/2.8 lens on a teeny tiny 6.1mm sensor (which is what your 5D2 is after you crop it down to 960 pixels) is equivalent to a 1170mm f/16.4 lens on the 5D2.


    Quote Originally Posted by Bruce in Philly View Post
    OK, for additional points, if I did the same using a 500mm lens and a 105mm lens, what would the effective focal lengths be?

    First, let me rant a little bit about the word "effective". There is an important difference between "equivalent" and "effective". Let me illustrate with some examples:


    * A tiny boat requires 10 horsepower to go 10 MPH.
    * A large cruise ship requires 100,000 horsepower to go 10 MPH.


    Now, which of the following is correct?


    * The engine on the tiny boat is effectively 100,000 horsepower.
    * The engine on the tiny boat is equivalent to 100,000 horsepower on a cruise ship as it pertains to MPH.


    Here is another example:


    * To fill my field of view, I sit 5 feet away from a 10-foot-wide display.
    * To fill my field of view, I sit 50 feet away from a 100-foot-wide display.


    Which of the following is correct?


    * If I sit 5 feet from the 10-foot-wide display, it effectively becomes 100 feet wide.
    * If I sit 5 feet from the 10-foot-wide display, it is equivalent to the field of view of a 100-foot-wide display seen from 50 feet.


    Now with photography:


    * "An 80mm lens on APS-C is effectively 130mm."
    * "An 80mm lens on APS-C is equivalent to 130mm on full frame for angle of view."


    The difference is important. Comparing two things (such as angle of view between formats) doesn't "effectively" change anything, so effective is not the right word to use. Discussing the "equivalent" is very useful, though.


    Furthermore, in most lenses, the focal length printed on the lens is only true when focused at infinity. When you focus closer, the "effective" focal length changes. It may say 100mm on the box, but at nearest focus it is really only 85mm. Similarly, adding a teleconverter truly changes the "effective" focal length. Those are both cases where the focal length truly does change, so it would make sense to say "effective focal length" if you desired to make a distiction between the number printed on the lens and the actual property itself.


    Friends don't let friends use "effective" ineffectively.


    The reason why I consider it important is that I think using the word effective instead of equivalent causes a lot of confusion and misconceptions. Use of the term "effective" is very widespread, including Canon's own camera manuals, but I still think it's incorrect and misleading. Every new photographer who sees that term is going to assume the honest-to-goodness focal length truly changes. If we used the term "equivalent focal length" instead, then they wouldn't get that misconception.


    We should reserve "effective" for cases where it is appropriate, such as teleconverters and lens breathing. Take these three examples:
    The effective focal length of a 200mm macro focused to 1:1 is 170mm.
    The effective focal length of a 200mm lens with 2X TC is 400mm.
    The effective focal length of a 200mm lens on APS-C is 320mm.
    To me, the first two are sensible and normal uses of the word effective. Internally-focusing lenses really do change the honest-to-goodness focal length when focusing. Similarly, adding a TC changes the focal length. Using the word "effective" here is useful to highlight that the actual focal length is not what you might expect -- it's different from the number printed on the lens. They're different from the third usage.




    Quote Originally Posted by Bruce in Philly View Post
    OK, for additional points, if I did the same using a 500mm lens and a 105mm lens, what would the effective focal lengths be?

    Now, to answer your question, a 960px crop with a 500mm f/4 lens would be equivalent to 2925mm f/23.4. At 105mm f/2.8 it would be equivalent to 614mm f/16.4


    Hope that helps.

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    Great stuff... Thanx so much Daniel

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    Dan, your comment on "Friends don't let friends use "effective" ineffectively." reminds me of my nearly constant rant ( in real life I consult to health care entities) that effective must precede efficient - there is no such thing as an efficient ineffective process, care guide, etc. The only way to make efficient something that is ineffective is to stop doing it.

    Back to the post - why is the f-stop jumped to 16.4? the f-stop doesn't jump when we take our FF lens and stick on a crop sensor? What am I missing?
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    Daniel, you really should combine all your posts here on the technical aspects of digital photography. It won't sell a million copies, but it will be definitely very interesting and in my opinion understanding the equipment helps in taking better pictures.

    So thanks again for the explanation

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    You'd sell extra copies if you truncated your name, and sold it as "Dan Brown".

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    The f-stop thing depends on what you're doing.

    If you are in the same place, taking a picture with FF and crop, the crop will appear to have a longer focal length. Since the sensor is denser, you'll blur sooner at the pixel level. Since you're looking at a smaller part of the image, you'll see more overall blur, as the blur appears magnified just as much as the in-focus parts.

    If the crop image is taken from further away, in order to maintain the same framing as the FF image, the results change. The increased distance results in less blur compared to the FF image, as though you were using a smaller aperture.

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    Ahhhhh I get it now.... I can see the point as it relates to the pin hole effect (diffraction?) and DOF items would be in the same process. Would light gathering be a separate process. i.e. it is still a f-x.x from a light gather standpoint same number of photons striking the same surface area - just a tiny surface area? I know that each pixel would be getting fewer as the density is higher.

    Be gentle Dr. C and others already caused my brain to explode w/ the electronics discussion pixel quality, bit depth, etc. it is Sunday morning and I can feel the pressure already beginning to increase
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    Quote Originally Posted by Busted Knuckles View Post
    Back to the post - why is the f-stop jumped to 16.4?
    Same reason the focal length "jumped" to 1170mm. Which is to say, because it gives the "equivalent" for just about everything that matters.

    Quote Originally Posted by Busted Knuckles View Post
    the f-stop doesn't jump when we take our FF lens and stick on a crop sensor? What am I missing?
    Well, the focal length doesn't jump when take our FF lens and stick it on a crop sensor either. But the *equivalent* focal length does.

    In the same way, the f-number doesn't change with sensor size, but the *equivalent* f-number does.

    That's why I say that 200mm, f/2.8, 1/60, ISO 100 on a digicam is equivalent to 1170mm, f/16.4, 1/60, ISO 5800 on FF35. But those are obviously different exposures, so how can they be equivalent? Let me answer that question with another question: why does exposure matter? If you think of all the ways that exposure matters, you'll find that in almost every case it *does* come out equivalent. Brightness? The ISO setting takes care of that. Noise? No problem. Given equal sensor performance, ISO 5800 on FF35 is the exact same as ISO 100 on a digicam.

    Quote Originally Posted by Busted Knuckles View Post
    Ahhhhh I get it now.... I can see the point as it relates to the pin hole effect (diffraction?) and DOF items would be in the same process.
    Precisely.

    Quote Originally Posted by Busted Knuckles View Post
    Would light gathering be a separate process. i.e. it is still a f-x.x from a light gather standpoint same number of photons striking the same surface area - just a tiny surface area? I know that each pixel would be getting fewer as the density is higher.
    Basically, the total number of photons falling on sensor area is the same between f/16.4 on FF35 and f/2.8 on the digicam. The only difference is that one has a strong intensity of light on a small area and the other is a low intensity of light over a large area.

    It might help you to think of both sensors (large and small) as having the same pixel count (e.g. 1 MP), so that the large sensor has large pixels and the small sensor has small pixels. (Such circumstances are not actually required, but it helps to make things easier to think about.)

    Hope that helps.

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