1.6x-whatever demystified?

[vvizard]

Could we please get this "whatever" demystified now? If it's already been done, please post the topicid for me to read.

This "Whatever" is called different things each time I've seen it, and the one I first saw write it "whatever" is the best I've found so far =) (don't remember who though). So how does this work? It's a smaller sensor. Therefore, it's a crop of a fullframe. No magic focal-length multiplier involved. Right? But, if you take the exact same picture from a 1Ds and a 10D, and print it in a given size (A4), the image from the 10D is then "stretched" to fit the media it's printed on. And when you stretch a crop, what do you get? Yeah, you get magnification. And what does Canon call this in a 10D? They call it "focal-length-multiplier". What do they call it in P&S? Digital-zoom!

Is the 1.6x-whatnot just a digital-zoom with a fancy name? Could it be they figured their sensor was so good that it could handle 1.6x magnification without quality-loss, and therefore they shrunk it 1.6x and called it focal-length-multiplier?

Am I totally off now? Or is this a decent way to put it? If it is, I'll be gratfull, cause this thing have chewed on my mind from the day I understood I was going to get a 10D or 300D. If it isn't, somebody please explain.

[john_houghton]

A 100mm lens on a 10D gives the same field of view as a 160mm (1.6 x 100mm) lens on a standard 35mm camera. Hence the 1.6 focal length factor. That's all there is to it, really.

John

[J.A.F. Doorhof]

Hi,

I have ignored the other threads but here is my view on this.

It's a 1.6x magnification factor for me, not cropped or whatever.

Why ?

You have 6.3 MP effective resolution. This 6.3 MP's are located on a chip, when you use the FULL chip you will have a 160mm view with a 100mm lens, so it's 1.6x magnification factor.

Why not digital zoom.
Simple,
With digital zoom you use less MP's in the final result than full sensor.

The difference between full sensor and the smaller sensor.
Now that's a bit more difficult to grasp, but it's also very simple explained.
if BOTH have 6.3 MP's than you WILL get a more zoomed in picture with the smaller sensor than with the full sensor, meaning you HAVE a REAL 1.6x magnification factor.

Take a 100mm lens and mount it on a 6.3MP FULL sensor and take a shot, and print this on A4 (somewhat native for 6.3MP).
Now mount that same lens on a 10D and take the same shot, and print this also on A4.

Both pictures use EXACTLY the same pixelcount, the 10D however will have a smaller ammount of the scene in the picture but with the same MP's as the fullsensor version with the same scene.

This is for me a 1.6 magnification and not digital zoom, crop factor or whatever.

The story is totally different when we are speaking about a 14MP sensor and a 6.3MP sensor, in that case you can talk indeed about cropping, but that's not a fair conclussion, because than you can also talk about the full sensor being a cropped version compared to medium or even large format camera's.

Hope this helps a bit :D.

Greetings,
Frank

[Parias]

And i have a question within the question... Does the x1.6 have any effect on Depth of field for any given Lens at the same aperture?

i.e with a 50mm Lens on a 300D, the FOV is the same as a 80mm Lens... At the same aperture (say f=5.6) will the DOF still be the one for 50mm or 80mm? (We know a 50mm lens has more DOF than a 80mm Lens)

[David Wild]

The practical effect of depth of field does have something to do with magnification, so if a picture from a full-size sensor camera and one from a 10D are both printed to the same size there will be slightly less **effective** depth of field in the picture which has been magnified more.

For most purposes, though, it doesn't matter much.

[EXA1a]

Parias wrote:
And i have a question within the question... Does the x1.6 have any effect on Depth of field for any given Lens at the same aperture?

i.e with a 50mm Lens on a 300D, the FOV is the same as a 80mm Lens... At the same aperture (say f=5.6) will the DOF still be the one for 50mm or 80mm? (We know a 50mm lens has more DOF than a 80mm Lens)
For DOF comparison (10D/300D/D30/D60 vs. 35mm camera) you can take a 1.26x factor (=square root of 1.6), for example:
A 100 mm lens on a 10D/300D/D30/D60 has the same angle of view as a 160mm lens on a 35mm camera, and the same DOF as a 126mm lens on a 35mm camera.

--Jens--

[James01]

Don’t quote me on this but this is how I see it:

I think it also has something to do with the distance of the CMOS to the rear of the lens and the size of the CMOS. On the 300D the lens has a special mount (EFS) where the lens sits closer to the chip and this compensates for the 1.6X. When you use an EF lens you get the same 1.6X as everyone else.

[Andy_T]

Hmm.

I'd propose a practical test to clear that issue....

- Take two shots of your object with Fuji Velvia and a 1.8/50 with your EOS 1V (or other film camera), using the same settings for both

- Develop both slides

- take a pair of scissors and crop one of the slides to the size of the 10D's sensor

- compare the DOF and check for changes :)

Regards,
Andy

[Jesper]

It is all explained in detail in the article "Understanding the DSLR Magnification Factor" from Luminous Landscape: http://www.luminous-landscape.com/tutorials/understanding-series/dslr-mag.shtml

But really, it's very simple.

The sensor of most digital cameras is smaller than the size of a 35mm film frame (which is 24 x 36 mm). When you use the same lens at the same focal length on a digital camera and on a film camera, they will record the same image, but in the digital camera the edges of the image are cut off - that's why it's called a crop factor. The depth of field will however be the same.

Photo.net has an explanation of what happens to DOF with digital cameras: http://www.photo.net/learn/optics/dofdigital/

The 10D has a sensor size of 15 x 22.5 mm. When you enlarge this to a 10 x 15 cm (= approx. 4 x 6 inch) image, you're enlarging the image 6.7 times. When you print a 24 x 36 mm film frame at 10 x 15 cm, you are only enlarging it 4.2 times. So there's your magical magnification factor! To print a digital and a film frame at the same size, you have to enlarge the digital image more than the film image.

[ilya]

Jesper wrote:
It is all explained in detail in the article "Understanding the DSLR Magnification Factor" from Luminous Landscape: http://www.luminous-landscape.com/tutorials/understanding-series/dslr-mag.shtml

But really, it's very simple.

The sensor of most digital cameras is smaller than the size of a 35mm film frame (which is 24 x 36 mm). When you use the same lens at the same focal length on a digital camera and on a film camera, they will record the same image, but in the digital camera the edges of the image are cut off - that's why it's called a crop factor. The depth of field will however be the same.

Photo.net has an explanation of what happens to DOF with digital cameras: http://www.photo.net/learn/optics/dofdigital/

The 10D has a sensor size of 15 x 22.5 mm. When you enlarge this to a 10 x 15 cm (= approx. 4 x 6 inch) image, you're enlarging the image 6.7 times. When you print a 24 x 36 mm film frame at 10 x 15 cm, you are only enlarging it 4.2 times. So there's your magical magnification factor! To print a digital and a film frame at the same size, you have to enlarge the digital image more than the film image.


best explanation i've seen to date. bravo

[CoolToolGuy]

Bravo to Jesper.

I will add one more perspective. Its a format. If you print a 35mm image to 11x14 you get one level of graininess, and if you print a medium format (6x6, 6x4.5, etc) at the same size you get a lesser amount of graininess. So the 10D format is a different format from 35mm just like 6x4.5 versus 6x6.

Have Fun

[Daytripper]

The depth of field changes because not only has the magnification increased, but the ratio of the aperture (in mm) to the focal length has changed... for the purposes of DOF calculation this is the same at stopping the lens down 0.6 fstops.

EXA1a wrote:
For DOF comparison (10D/300D/D30/D60 vs. 35mm camera) you can take a 1.26x factor (=square root of 1.6), for example:
A 100 mm lens on a 10D/300D/D30/D60 has the same angle of view as a 160mm lens on a 35mm camera, and the same DOF as a 126mm lens on a 35mm camera.

--Jens--

Yep, and the easiest way to incorporate this in DOF calculations is to apply the factor of 1.6 to the resolution input - Instead of using 28.6 lines/mm (CoC 0.035) use 45 lines/mm (CoC 0.023) with the uncorrected focal length and the calculations work fine.

Pete.

[hmhm]

Those who try to argue that depth of field between a full-frame and 1.6x sensor are "the same" don't really have a strong understanding of depth of field, and are incorrectly applying their intuition. Cutting the edges off your velvia slide _does_ change depth of field, as it requires greater enlargement to make a print of a given size. It also requires greater subject distance of shorter focal length to achieve the same framing, each of which also impact depth of field.

The "whatever" factor is just an issue of "format", which we've always had in photography. My 10D sensor is smaller than my 35mm film cameras which is smaller than my 6x7 medium format which is smaller than the 4x5 and 8x10 which I don't have. Medium and large format photographers never got caught up in the "crop" terminology, they just needed to understand that format impacts the angle of view that a lens of a given focal length achieves, and that it impacts depth of field.

The comparison of a smaller sensor to a large sensor with "digital zoom" is somewhat appropriate and somewhat inappropriate. If you imagine two sensors of different sizes but with identical pixel densities, then the smaller sensor is exactly the same as a digital zoom of the larger sensor. But no two sensors seem to ever have the same density, so this comparison is seldom very helpful, and since most people don't really understand the impact of digital zoom either (e.g. it changes depth of field, too), it's probably not a helpful analogy.
-harry

[Jesper]

hmhm wrote:
Those who try to argue that depth of field between a full-frame and 1.6x sensor are "the same" don't really have a strong understanding of depth of field, and are incorrectly applying their intuition. Cutting the edges off your velvia slide _does_ change depth of field, as it requires greater enlargement to make a print of a given size. It also requires greater subject distance of shorter focal length to achieve the same framing, each of which also impact depth of field.

You are right.

If you take a film frame and cut off the edges so that it's as large as the sensor of the digital camera, that cropped image will look the same (with the same depth of field) as the same image recorded with the digital camera (lens on the same aperture and focal length).

It does *not* look the same as zooming in 1.6x and then making the photo with the film camera. That is why the "magnification factor" is *not* the same as making the focal length of your lens 1.6 times as long!


The "whatever" factor is just an issue of "format", which we've always had in photography. My 10D sensor is smaller than my 35mm film cameras which is smaller ...

I agree that it's a question of format. But because the 10D etc. use the same lenses as the 35mm film format, people want to compare the format of digital SLRs like the 10D with the traditional film format.

[CyberDyneSystems]

Hah,. after my involvemnet in two historically long posts on this subject a few months back,.. I have nothing to add to this except this.

Well Said Jesper!!!! :D

[Daytripper]

Jesper wrote:
If you take a film frame and cut off the edges so that it's as large as the sensor of the digital camera, that cropped image will look the same (with the same depth of field) as the same image recorded with the digital camera (lens on the same aperture and focal length).

It does *not* look the same as zooming in 1.6x and then making the photo with the film camera. That is why the "magnification factor" is *not* the same as making the focal length of your lens 1.6 times as long!


DOF cannot be discussed without making reference to the magnification of the image being viewed.

1.) If you crop a 35mm transparency to the same dimensions as a 300D sensor, then yes, DOF of the transparency remains the same, but;

2) The cropped transparency must be enlarged by a factor of 6.7 to make a 6x4" print, whereas a full sized transparency must only be enlarged by a factor of 4.2

It is this difference in the enlargement that causes the difference in the DOF of the final image - As Jens said, the equivalent DOF of a lens on the 300D will be equivalent to a lens 1.26 x the focal length of the same lens on a full frame sensor.

Pete

[Jesper]

CyberDyneSystems wrote:
Hah,. after my involvemnet in two historically long posts on this subject a few months back,.. I have nothing to add to this except this.

Well Said Jesper!!!! :D

Thanks! :D This thread contains a lot of info that would be good for the Canon EOS Digital FAQ or Best of the Forum about which there was a post a few days ago...

[EXA1a]

Daytripper wrote:

DOF cannot be discussed without making reference to the magnification of the image being viewed.

1.) If you crop a 35mm transparency to the same dimensions as a 300D sensor, then yes, DOF of the transparency remains the same, but;


You are right in what you say further down but above is the important thinking error of many people:
When you crop a slide the DOF does indeed change!
Again: although the information on the piece of celluloid hasn't changed at all, the DOF very well changes!
For DOF calculations you assume a certain viewing distance which is usually the diagonal measure of a picture, no matter, how large it is (from approx. 4 inches up). You have to enlarge a slide and...read on further down!

2) The cropped transparency must be enlarged by a factor of 6.7 to make a 6x4" print, whereas a full sized transparency must only be enlarged by a factor of 4.2

It is this difference in the enlargement that causes the difference in the DOF of the final image - As Jens said, the equivalent DOF of a lens on the 300D will be equivalent to a lens 1.26 x the focal length of the same lens on a full frame sensor.

Pete

[Daytripper]

Daytripper wrote:

DOF cannot be discussed without making reference to the magnification of the image being viewed.

1.) If you crop a 35mm transparency to the same dimensions as a 300D sensor, then yes, DOF of the transparency remains the same, but;


EXA1a wrote:

You are right in what you say further down but above is the important thinking error of many people:
When you crop a slide the DOF does indeed change!


True enough - my original statement more correctly should have said "When you crop a transparency the Circle of Confusion does not change....

Pete

[rdenney]

vvizard wrote:
Could we please get this "whatever" demystified now? If it's already been done, please post the topicid for me to read.

This "Whatever" is called different things each time I've seen it, and the one I first saw write it "whatever" is the best I've found so far =) (don't remember who though). So how does this work? It's a smaller sensor. Therefore, it's a crop of a fullframe. No magic focal-length multiplier involved. Right? But, if you take the exact same picture from a 1Ds and a 10D, and print it in a given size (A4), the image from the 10D is then "stretched" to fit the media it's printed on. And when you stretch a crop, what do you get? Yeah, you get magnification. And what does Canon call this in a 10D? They call it "focal-length-multiplier". What do they call it in P&S? Digital-zoom!

Is the 1.6x-whatnot just a digital-zoom with a fancy name? Could it be they figured their sensor was so good that it could handle 1.6x magnification without quality-loss, and therefore they shrunk it 1.6x and called it focal-length-multiplier?

Am I totally off now? Or is this a decent way to put it? If it is, I'll be gratfull, cause this thing have chewed on my mind from the day I understood I was going to get a 10D or 300D. If it isn't, somebody please explain.

The 1.6 is an equivalence factor. It gives you a notion of how the smaller format compares to the bigger format in terms of focal-length application.

People should think of the 10D/300D as a 15x23-format camera that allows you to use lenses originally intended for 24x36-format cameras.

For all cameras from Minox spy cameras to 20x24 studio view cameras, the normal lens is that which provides the normal field of view--that is--the field of view that magnifies the scene similarly to your eyes. This is taken to be a focal length equal to the diagonal size of the frame. For a 24x36 camera (i.e. full 35mm frame), this "normal" lens is 43mm. For a 4x5 view camera, this normal lens is 160mm. For a 15x23-sensored Canon 10D, this normal lens is 27mm.

43 divided by 27 is 1.6. Therefore, if you know what a lens does on a full-frame 35mm camera, divide it by 1.6 to get an equivalent lens on a 10D. Thus, it is an equivalence factor--nothing more, nothing less.

For wide angle or telephoto, just think of them in terms of how much longer or shorter they are than the normal focal length. Thus, a 300mm lens on a 10D is about 11 times the normal focal length. Thus, it will behave exactly like any lens on any camera that is 11 times the normal focal length for that camera. A 20mm lens is around three-quarters the normal focal length on a 10D, and thus it will behave exactly as any lens that is three-quarters the focal length of the normal lens for any camera.

Thus, to get the same image on a 10D with that 11-times-normal 300mm lens, you'd need (43 times 11 =) 473mm on a full-frame 35mm camera. These relationships are linear, which means that, conveniently, the 1.6 equivalence applies here, too. 300 times 1.6 = 480 (the small difference is just rounding error). To get the same frame on my 6x9 medium-format view camera, I'd need a lens 11 times the normal focal length of 100mm, which is 1100mm. That's why you don't see much extreme telephoto work done in the larger formats.

Those of us who shoot in many different formats get used to this in a hurry. It's no big deal. We think of a lens as, say, a short telephoto for portraits. A short telephoto for medium format might be 150 or 180mm. For a 35mm full-frame camera, it might be 80 or 100mm. For an even smaller format camera like one that uses a 15x23 sensor, that portrait-length telephoto would be a 50 or a 60. For a Minox 8mm camera, it might be 20mm (I'm guessing on that one).

It has nothing whatever to do with the back-focus distance, which is the distance between the lens mount and the sensor surface. That is a result of the camera design. The focal length is still the focal length--a 50mm lens still has to be 50mm away (at its center of focus) from the sensor, no matter how big that sensor or film is.

When we call it a magnification factor, we think we are magnifying something. We aren't. When we call it a crop factor, we think we are starting with one image and doing something to make it different (such as cropping). We aren't. All we are doing is projecting the image from a lens of a given focal length on a sensor or film frame of a given size, and seeing what we see. The only purpose of that number is to give people trained to know what the focal lengths do in 35mm a quick way to determine what those same lenses will do with a smaller sensor.

When it comes to making prints, forget all that. The only number that matters in digital is the number of pixels you have (unless pixel density exceeds lens quality, which it nearly does with the 10D). 3000x2000 pixels will make a print 11x16 inches at 180 pixels/inch. It doesn't matter how much space those 3000 pixels took up in the camera. Thus, a 10D can make a print of 11"x16" at 180 pixels/inch, while a 1Ds can make a print of 15"x23" at 180 pixels/inch. 180 pixels/inch (or whatever number suits you) becomes your minimum quality standard for resolution, just as an 11x enlargement from a negative (or 8x or whatever) was a minimum quality standard for prints from film. Bigger film allows bigger prints at the same quality standard. But if you think of one being a "digital" zoom of the other, you will surely drive yourself batty.

Rick "already batty as it is" Denney

[2new]

Excellent information - Thank you all!!!