Resolution Vs. Enlargement

[Steven M. Anthony]

The discussions of format and crop factor got me thinking a lot about print size and the impact enlargement has on camera shake.

One issue that came up was that camera shake gets magnified not by the APS-C sensor, but because you have to enlarge the image more to get a print the comparable size as that from a full-frame sensor.

This logic certainly holds true for film. To get an 11 X 14 from a frame of APS film, you have to enlarge it more than you do for a frame of 35mm film.

But digital offers another dimension to play with: resolution.

I can print a photo from a Canon 1Ds (if someone gave me either the camera or an image file from one!) at a resolution of 245ppi and get an 11” X 14” print. I can get the same sized print from my 10D if I print it at 185ppi; and I can get the same sized print from my Sony 707 if I print it at 175ppi.

While the quality of the prints from the 3 cameras will be of different quality (technically), I have printed and sold several 11 X 14 prints from images shot with my Sony 707—at normal viewing distance, they look great. Even up close, it’s hard to distinguish the quality of the 11X14 vs an 8X10 of the same image. The point being, even at 175ppi, a print can look great--and clearly acceptable for purchase.

So my question: Does using a lower resolution (without re-sampling) have the effect of enlarging the image without magnifying effects like camera shake?

[Bodog]

Steve, not sure what you mean by magnify, but it would seem to me that whatever detail is in the image is not going to change at different pixel/inch ratios of the image. If the edges are blurred, then the blur will be there at whatever resolution you set the image. As the printed size increases with the lowered resolution, then the blur will increase along with the image size. I can see the Perceived sharpness increasing with with the higher resolutions (and smaller print size), but the blur would still be there. On the other hand, if the larger image was viewed from a greater distance, then the perceived sharpness should be the same as the smaller image viewed at a shorter distance. So, in the end I guess you could say that lowering resolution could magnify the effects of camera shake. But I don't believe it is from lowered the resolution itself, but from the increased the print size. It's all in how you view the printed image. This is from my own experience not a scientific analysis. :rolleyes:

[Steven M. Anthony]

Steve, not sure what you mean by magnify, but it would seem to me that whatever detail is in the image is not going to change at different pixel/inch ratios of the image.

I guess I'm not sure what I mean. In the discussion of 1/fl vs. 1/fl*crop factor (rule of thumb to avoid camera shake) the argument was made that one needed to use 1/fl*crop factor because of the extra magnification required to get the APS-C image to be as big as a full-frame image.

But the more I think about it, the less I''m convinced that there IS any magnification invloved.

With APS (or any) film, enlarging an image is an optical process--involving bending light through lenses, and projecting an image onto paper.

But I can't get my head around how this works with digital--if there is an optical analogy or not. And whether such an analogy even answers the "is there a magnification effect" question!

If the edges are blurred, then the blur will be there at whatever resolution you set the image. As the printed size increases with the lowered resolution, then the blur will increase along with the image size. I can see the Perceived sharpness increasing with with the higher resolutions (and smaller print size), but the blur would still be there.

That's where I've netted out, too. And a, say, 1Ds has the same issues. Granted, it can maintain a higher resolution for a given print size relative to a 10D, but from my own prints, 11 X 14's from my Sony 707 (5MP) look as good a c-prints at a resolution of about 175 (I think i used to do them at 180---so they were a tad under 11 X 14).

On the other hand, if the larger image was viewed from a greater distance, then the perceived sharpness should be the same as the smaller image viewed at a shorter distance. So, in the end I guess you could say that lowering resolution could magnify the effects of camera shake.

Okay, now you've lost me. I don't see how the first sentence leads to the second.

[Bodog]

Sorry, just coming back to the original question. Assuming someone looking at the larger might think that the blur was magnified when compared to the same image printed at a smaller size simply because it would not be as noticable in the smaller print.

[Steven M. Anthony]

Sorry, just coming back to the original question. Assuming someone looking at the larger might think that the blur was magnified when compared to the same image printed at a smaller size simply because it would not be as noticable in the smaller print.

Oh, okay. So, for a given camera, the camera shake visible in a small print will be less than in a big print, because the blurry area get bigger along with everything else. That makes sense. But that will be true for prints out of any camera, if I understand your clarification.

But what was put forth in the other thread was that when comparing equal-sized prints from, say, a full-frame camera (let's just say a 1Ds) with an 80mm to one from a 10D with a 50mm (which has the FoV of an 80mm on a 1Ds), the 10D print will show more camera shake because you'd have to magnify the 10D image more--because it was recorded on an APS-C-sized sensor (and further assuming equal shutter speeds across cameras).

That seemed logical when I first read it. But now it doesn't seem right--given I achieve my print size by manipulating resolution. There are, of course, many benefits to having a larger sensor in camera, but I don't think it impacts camera shake the way some people think. If both the 1Ds image and the 10D image contain the same elements in their respective frames, and both are printed at 11 X 14, the camera shake recorded by each will look the same size in each print.

Of course, as always, I'm open to additional info in support of either side of the discussion.

[Belmondo]

If both the 1Ds image and the 10D image contain the same elements in their respective frames, and both are printed at 11 X 14, the camera shake recorded by each will look the same size in each print.

That is correct.

The 1Ds image would be higher resolution, but the camera shake would be equal.

[rdenney]

That is correct.

The 1Ds image would be higher resolution, but the camera shake would be equal.

Yup. One should note that the camera shake and other faults may be masked by the pixelation of an image that is enlarged so that the pixels become apparent.

But if the pixels are small enough not to be visible as pixels on the print, then we can treat all analog faults as behaving as if it was a continuous medium.

It's interesting that that we strive for smaller and smaller dots on our printers, without realizing that sometimes those smaller dots make smooth what we might rather have hard edges. I have a Special Edition print taken by Ansel Adams that I bought at the Ansel Adams Gallery in Yosemite many years ago. Bob Ross made the print. The print is about the same size as the reproduction of that image in the full-size Yosemite and the Range of Light. Here's the point: The reproduction in the book is sharper. I have always looked as those images from six inches thinking--WOW! How did Adams get such sharp images with his coke-bottle-bottom lenses of the day? Turns out, even at 8x10 format, he didn't. The 11x14ish print from that 8x10 negative reveals a hint of fuzziness that the reproduction's screening process completely masks. By posterizing those very fine gradations, the screening process sharpened the image.

Enlargement magnifies all faults, and at some point, it also reveals the pixels themselves (just as over-enlarged prints from negatives reveal more grain than image). Also, one has less information being spread over more print, and this causes a loss of subtle gradation that becomes quite apparent when one compares it to prints made from larger formats.

Steven says his 10D can make saleable prints at 13x19. I wish mine could. They look good by themselves, untill I stand them next to prints enlarged from medium-format film that is scanned at relatively low resolution. I'm enlarging the pixels by the same amount from those two sources, but the larger format still looks better. Why? Each pixel has more information and less noise, with the result that all sorts of subtle effects emerge. And when I compare the medium-format prints to silver prints I made from 4x5 film when I was doing lots of large-format work, I see that the medium format images suffer by comparison.

Rick "who can see the difference in smaller prints, too" Denney

[Steven M. Anthony]

Yup. One should note that the camera shake and other faults may be masked by the pixelation of an image that is enlarged so that the pixels become apparent. But if the pixels are small enough not to be visible as pixels on the print, then we can treat all analog faults as behaving as if it was a continuous medium.


Right. That's where the benefit of the larger sensor comes in. YOu can make a larger print with more resolution that you can from a smaller sensor.


Enlargement magnifies all faults, and at some point, it also reveals the pixels themselves (just as over-enlarged prints from negatives reveal more grain than image). Also, one has less information being spread over more print, and this causes a loss of subtle gradation that becomes quite apparent when one compares it to prints made from larger formats.
Does this come through as banding?

Steven says his 10D can make saleable prints at 13x19. I wish mine could. They look good by themselves, untill I stand them next to prints enlarged from medium-format film that is scanned at relatively low resolution.
I wish! I actually said I make 11 X 14s (or just a tad smaller). And if you get really close, sure you can tell the difference in quality from an 8X10 (at higher resolution)--but non-photogs never get that close when looking at art!-at least in my experience.

[slin100]

One of the important points in your hypothesis, which I agree with the others is correct, is that the shutter speed was kept constant. If you had, instead, followed the old 1/fl rule, then camera shake would not be identical.

[Steven M. Anthony]

One of the important points in your hypothesis, which I agree with the others is correct, is that the shutter speed was kept constant. If you had, instead, followed the old 1/fl rule, then camera shake would not be identical.

Maybe I was too hasty in typing my shutter-speed assumption. I had, in fact, meant to say that each used the same shutter speed formula--1/fl. So the 10D w/50mm used a 1/50, and the 1Ds w/80mm used a 1/80.

If the rule holds up at all--and as Rick pointed out in another thread, it might not hold up under any conditions..--there should be no difference in camera shake: because the 50mm on a 10D is a 50mm (even though the 10D's sensor only records the FoV of an 80mm (on a 35mm-format camera)) and will contribute to camera shake accordingly. The 80mm on the 1Ds is an 80mm, and will contribute to camera shake accordingly. The 80mm should induce more shake than the 50mm and therefore needs to use a quicker shutter speed.

Thus, while print size might effect perceptions of absolute levels of camera shake, it seems that sensor size has nothing to do with it. So, if 1/fl works at all, it need not be adjusted for sensor size, because sensor size seems to have no impact on camera shake.

Another component of camera shake I just thought of has to do with the length and weight of the lens attached to the camera. A long, heavy telephoto will apply more angular momentum (help me here, Rick)--maybe torque?--to the camera/lens system, making it more difficult to hold steady. Adding the thrifty-fifty to a 10D is barely noticable!

[slin100]

If you agree that camera shake is magnified more with APS film than 35mm film, then it's true with digital. Resolution has nothing to do with this phenomenon. As Rick said, ...if the pixels are small enough not to be visible as pixels on the print, then we can treat all analog faults as behaving as if it was a continuous medium.
IOW, think of digital as you would film.

[slin100]

Steven,

Lest you think that I'm fixated on strictly following the 1/fl rule of thumb, I'm not. I simply believe that faster shutter speeds are needed when handholding a camera with a smaller sensor. I also agree with your points about camera/lens weight and bulk being major contributors.

[rdenney]

Does this come through as banding?


No. It comes through as large, noisy (or, if film, grainy) areas that from normal view distance appear to be all one color. If the gradient is sharp, you'll still have the same values, but the gradient will be less distinct because it will be clouded by noise/grain.

Larger formats produce smoother images where very subtle gradations become visible. It's most noticeable in large, flat areas. The larger formats will show all sorts of subtle variations that appear as one noisy color with smaller formats.

Someone on this forum posted a link to an excellent article on the subject of photon-counting error and photosite size. I still have it up on my computer at home, but can't recall the link here. It quite clearly demonstrated that smaller sensors have reached their potential, having run up against the limit imposed by photo-counting statistics.

Ah, here it is:

http://clarkvision.com/imagedetail/does.pixel.size.matter/

It should be noted that noise in the shadows is much better with digital sensors than with film. That's one reason I don't "shoot to the right". I expose such that my exposure gives me the most to work with, and there's better tonal separation a little away from the highest values, according to the characteristic curves shown on that website.

Rick "who thinks 10D images enlarged beyond 8x10 start to look flat, JUST LIKE 35mm images enlarged beyond 8x10" Denney

[rdenney]

Another component of camera shake I just thought of has to do with the length and weight of the lens attached to the camera. A long, heavy telephoto will apply more angular momentum (help me here, Rick)--maybe torque?--to the camera/lens system, making it more difficult to hold steady. Adding the thrifty-fifty to a 10D is barely noticable!

Actually, I find that a long, heavy telephoto works a lot better than a long, light telephoto. But with all telephotos, I brace if I'm going to use a long shutter speed.

One reason I kept the 75-300 only for a day is that there is no way to brace it. Just the part of the lens you want to lean up against a tree or grasp with your hand is the part that turns during focusing. In one day, I got so annoyed trying to use the lens that it didn't matter if the image quality was good or not. I replaced it with a 70-200/4L, mostly because it was the sort of lens I could hold, braced if needed. A monopod under the camera body doesn't help if you can't hold the front of the lens with your hand--it will wiggle back and forth.

I have a Pentacon Prakticar 500mm/5.6 lens designed for medium format. It's a huge monster. It uses 118mm filters, and it's at least 20 inches long. And it weighs a ton. I find it quite stable on a monopod; much moreso than any of my long lenses for small cameras.

Camera shake is a function of vibration, and vibration is a function of mass and damping. A heavier item requires more energy to set it to vibrating. We can damp vibration by holding the camera with our gooey (i.e., well damped) hands at the far ends of the lens. Then, all we need is a shutter speed fast enough to handle the slower movements of our bodies, rather than the fast vibration that victimizes we older folks. If I have to go slower, the camera will get braced up against something. The sandbag that you hang from the tripod or lay over the lens provides both mass and damping. A beanbag is a wonderful tool to help when bracing up against something.

IS helps, too, of course.

That thrifty-fifty is awfully light. I have found that I didn't like my camera-store trial of it. I'm using an old Super-Takumar on an adaptor for when I want a really fast 50, but if I'm going to shoot at a smaller aperture than 2.8, I'll use my 2.5 Compact Macro, which is at least twice as heavy as the 1.8. The bigger advantage of the thrifty-fifty compared with most cheaper Canon zoom lenses is that it is much faster, allowing a couple of stops worth of faster shutter speeds. Then, all you have to worry about enlarge is lack of depth of field and a range of uncorrected lens faults, heh, heh.

Rick "all photography is the art of compromise" Denney

[Steven M. Anthony]

If you agree that camera shake is magnified more with APS film than 35mm film, then it's true with digital.

Well, the more I think it through, the more I question whether it actually happens with film!

Consider:

You take a photo from spot X with an APC camera w/50mm. Let's say--for ease of illustration, the content of that photo is 2 light poles with a sofa between them--the light poles line up with the sides of the frame, the sofa is shown lengthwise. The sofa takes up half the width of the frame.

Now you take a photo of the same scene, from spot X, with a35mm-format AE-1/80mm.

With each shot, our camera "shakes." Because the 80mm lens magnifies more than a 50mm lens (because a lens is a lens no matter what sits behind it), camera shake is magnified more with the 80mm than it is with the 50mm. That's why you use a quicker shutter speed with the 80mm--so the film is exposed to camera shake for a shorter period of time. So we shoot the 50mm at 1/50, and the 80mm at 1/80. This help equate the 2 in terms of the amount of camera shake each records. Note how it has nothing to do with film size.

The negatives will contain the same scene. The negative from the AE-1 will be larger than the APS negative.

Even after compensating for lens fl with shutter speed (50mm @1/50th vs. 80mm @ 1/80th), if we made contact prints of each negative, the APS/50mm contact print would look sharper than the AE-1/80mm as less shake would be visible in the APS print because it's a smaller print.

We enlarge both to be 11 X 14.

While we are enlarging the APS negative more than the 35mm film to reach 11X14, the APS neg started with less visible camera shake. So the net/net, after enlargment, is the 2 look the same--with the same amount of camera shake visible.

Does this scenario hold?

[Steven M. Anthony]

No. It comes through as large, noisy (or, if film, grainy) areas that from normal view distance appear to be all one color.

Okay, I'm familiar with that.

[slin100]

Well, the more I think it through, the more I question whether it actually happens with film!

Consider:

You take a photo from spot X with an APC camera w/50mm. Let's say--for ease of illustration, the content of that photo is 2 light poles with a sofa between them--the light poles line up with the sides of the frame, the sofa is shown lengthwise. The sofa takes up half the width of the frame.

Now you take a photo of the same scene, from spot X, with a35mm-format AE-1/80mm.

With each shot, our camera "shakes." Because the 80mm lens magnifies more than a 50mm lens (because a lens is a lens no matter what sits behind it), camera shake is magnified more with the 80mm than it is with the 50mm. That's why you use a quicker shutter speed with the 80mm--so the film is exposed to camera shake for a shorter period of time. So we shoot the 50mm at 1/50, and the 80mm at 1/80. This help equate the 2 in terms of the amount of camera shake each records. Note how it has nothing to do with film size.

The negatives will contain the same scene. The negative from the AE-1 will be larger than the APS negative.

Even after compensating for lens fl with shutter speed (50mm @1/50th vs. 80mm @ 1/80th), if we made contact prints of each negative, the APS/50mm contact print would look sharper than the AE-1/80mm as less shake would be visible in the APS print because it's a smaller print.

We enlarge both to be 11 X 14.

While we are enlarging the APS negative more than the 35mm film to reach 11X14, the APS neg started with less visible camera shake. So the net/net, after enlargment, is the 2 look the same--with the same amount of camera shake visible.

Does this scenario hold?
This is a very interesting thought experiment!

I think the scenario doesn't hold. While the image on the 10D is smaller due to shorter focal length, the longer shutter speed means that the recorded shake is the physically the same size as that recorded on the 35mm frame.

If this is not clear, consider this. I'm going to switch to a tripod-mounted camera with a moving subject, without loss of generality, because I think it makes the problem easier to understand. Suppose the sofa were sliding along its length, laterally across the camera's field of view. The amount of movement recorded on the medium by the camera is going to be directly proportional to the speed at which the sofa is moving, the focal length of the lens and the shutter speed, and inversely proportional to the distance of the sofa from the camera. Reference for this assertion is from the book, Exposure: The Fundamentals of Camera Technique, by W.F Berg.

Let's suppose that the recorded blur from the moving sofa is 1mm long on the 35mm frame. Now, take the 10D. Because it has a 50mm, the blur will be magnified 1.6x less. However, the shutter speed is 1.6x longer. The end result is that the recorded blue is also 1mm long on the APS-C frame.

Now, if you print both recorded images at the same size, the recorded blur from the APS-C will be 1.6x longer than that of the 35mm frame. Therefore, you cannot slow down the shutter speed as you decrease focal length, because both shutter speed and focal length contribute equal weight to blur.

[Steven M. Anthony]

This is a very interesting thought experiment!

I think the scenario doesn't hold. While the image on the 10D is smaller due to shorter focal length, the longer shutter speed means that the recorded shake is the physically the same size as that recorded on the 35mm frame.
If this is not clear, consider this. I'm going to switch to a tripod-mounted camera with a moving subject, without loss of generality, because I think it makes the problem easier to understand. Suppose the sofa were sliding along its length, laterally across the camera's field of view. The amount of movement recorded on the medium by the camera is going to be directly proportional to the speed at which the sofa is moving, the focal length of the lens and the shutter speed, and inversely proportional to the distance of the sofa from the camera. Reference for this assertion is from the book, Exposure: The Fundamentals of Camera Technique, by W.F Berg.

Let's suppose that the recorded blur from the moving sofa is 1mm long on the 35mm frame. Now, take the 10D. Because it has a 50mm, the blur will be magnified 1.6x less. However, the shutter speed is 1.6x longer. The end result is that the recorded blue is also 1mm long on the APS-C frame.

But you're forgetting that the 35mm format film is 1.6 times bigger than the APS frame (in each direction) in addition to factors mentioned above.

Without this additional enlargement factor applied to the 35mm format, all the elements (not just the blur) would be the same size in the APS as they are in the 35mm negative. But if all the elements were the same size in both negatives, both would have to be 35mm format to fit! This is why the blur looks bigger on the contact print of the 35mm film relative to the APS--because it is, in fact, bigger.

Now, if you print both recorded images at the same size, the recorded blur from the APS-C will be 1.6x longer than that of the 35mm frame. Therefore, you cannot slow down the shutter speed as you decrease focal length, because both shutter speed and focal length contribute equal weight to blur.

But when you factor in your "missing" correction for frame size of the 35mm film before enlarging the images, the additional 1.6 times enlargement needed for the APS to reach a given enlargement for the 35mm film brings the size of the blur (and all the elements in the image) from the APS frame up to the same size as we have on the enlarged 35mm frame. So the 2 enlargements look identical--APS utilizing 1/fl and 35mm utilizing 1/fl!

Transferring this to the digital realm, there is no need to use the crop factor in calculating the 1/fl rule of thumb! (At least as far as I can tell...)

Rick, what's your take on this?

[slin100]

But you're forgetting that the 35mm format film is 1.6 times bigger than the APS frame (in each direction) in addition to factors mentioned above.

I didn't forget.

Without this additional enlargement factor applied to the 35mm format, all the elements (not just the blur) would be the same size in the APS as they are in the 35mm negative. But if all the elements were the same size in both negatives, both would have to be 35mm format to fit! This is why the blur looks bigger on the contact print of the 35mm film relative to the APS--because it is, in fact, bigger.

There is no missing enlargement factor. The blur is not, in fact, bigger on the 35mm frame; they are same size. Unless Rick chimes in on this, I may have to draw a picture to prove my point.
Transferring this to the digital realm, there is no need to use the crop factor in calculating the 1/fl rule of thumb! (At least as far as I can tell...)
Until I draw my picture, I'll have to go back and restate the digital P&S as a counterexample that this is not true.

[Steven M. Anthony]

I didn't forget.

There is no missing enlargement factor. The blur is not, in fact, bigger on the 35mm frame; they are same size. Unless Rick chimes in on this, I may have to draw a picture to prove my point.

Until I draw my picture, I'll have to go back and restate the digital P&S as a counterexample that this is not true.

If the blur is the same size on the APS neg as it is on the 35mm neg, what keeps the image of the sofa on the APS neg from being the same size as it is on the 35mm neg? And if all the elements in the photo are the same size on the APS neg as they are on the 35mm neg (and you need to consider the distance between physical elements as elements, too), how do they all fit? If all the elements on the APS neg are the same size as they are on the 35mm neg, the APS neg would also have to be the same size as the 35mm neg!! And we know that can't happen, right?

[slin100]

If the blur is the same size on the APS neg as it is on the 35mm neg, what keeps the image of the sofa on the APS neg from being the same size as it is on the 35mm neg? And if all the elements in the photo are the same size on the APS neg as they are on the 35mm neg (and you need to consider the distance between physical elements as elements, too), how do they all fit? If all the elements on the APS neg are the same size as they are on the 35mm neg, the APS neg would also have to be the same size as the 35mm neg!! And we know that can't happen, right?
The sofa on the APS neg is not the same size as the one on the 35mm neg, but the size of the blur is! I'm drawing my diagram...

[slin100]

Ok, here's the drawing. Assume object A is travelling toward point B at constant velocity. It reaches point B at time x and point C at time 1.6x. The blur, segment DE, is recorded at the focal plane by the object moving from A to B with a 80mm lens. Similarly, the blur is the same length DE segment recorded of the object moving from A to C with a 50mm lens.

This says nothing about the size of the object, only the size of the blur recorded. As you correctly stated, the object is 1.6x times smaller in the APS frame with the 50mm lens than the 35mm frame with the 80mm lens.


http://img58.echo.cx/img58/808/blur29tv.jpg (http://www.imageshack.us)

[Steven M. Anthony]

Ok, here's the drawing...

What I keep hitting up against is that when comparing the full frame 35mm format/80mm @1/80th scenario vs the APS-sized film/50mm @1/50th scenario, there seem to be 3 differences:

Shutter speed (1/80th vs. 1/50th--with 1/50th being 1.6 X longer than 1/80th)

Magnification of optics (80mm = 1.6 X 50mm)

Frame size (FF = 1.6 X APS)

Your graph only seems to account for 2 of the 3--shutter speed and magnification of optics.

Based on the 2 you include in your figure, the math works out to show the blur to be equal-size in both scenarios: Mag of 80mm relative to 50mm (1.6) X shorter shutter duration due to 1/fl (1/1.6) = 1. So, prior to projection onto the film plane, the blur is the same size.

But now add in the fact that the full frame is 1.6 times larger than the APS frame. So the image that is optically projected onto the 35mm frame is 1.6 times the size of the image projected on frame. So, now, the blur is 1.6 times larger on the 35mm neg as compared to the APS neg.

Another way to think about it is that with magnification and shutter speed accounted for (but prior to projection), the blur is the same for both (as you show in your figure). But the 35mm frame is a bigger enlargement of that image vs. the APS frame (1.6 times bigger). If we could keep the focus of the image plane of the APS-sized image inside the APS camera, and enlarge it to 35mm format frame size (by moving the back of the camera out away from the lens), the blur would grow to the size of the blur on the 35mm frame. This is precisely what happens in an enlarger: We take an infocus image plane (the neg) and enlarge it.

Note: If we used a lens that projected its full FoV onto the APS frame, and contained the same content as the 35mm frame shot with an 80mm (which would be the APS format equivalent of a 35mm-format 80mm lens), and used 1/80th for the 35mm shot and 1/50th for the APS shot, your math would work out (because then, there is no magnification difference)--as magnification is related to how much of the scene is included in the entire FoV, not just the part of the entire FoV that has a sensor behind it. In MY scenario, though (which is the 10D vs full-35mm frame scenario), the APS frame is capturing only a portion of an un-magnified frame--it's capturing only a portion of the lenses entire FoV, so it is not magnifying anything.

[slin100]

It's a bit ironic to hear this (that there is some missing magnification factor) from you, because you are the one who has been most fervent about pointing out that an APS-C camera is just a crop of a 35mm frame!

In any case, the size of the frame is immaterial. If it would help you understand better, just forget about the APS frame for now and assume that the experiment is being done solely on a 35mm frame. Now we're strictly in a 35mm domain and simply comparing the effects of blur between a 50mm and 80mm lens. Do you see how it doesn't change the result?

The objects are, indeed, smaller using the 50mm lens, but I have proven beyond a shadow of a doubt that the size of the blur, on the sensor, is the same. And they are the same because the longer shutter speed has allowed the reduced-size image to travel for a longer period of time across the frame.

[slin100]

Based on the 2 you include in your figure, the math works out to show the blur to be equal-size in both scenarios: Mag of 80mm relative to 50mm (1.6) X shorter shutter duration due to 1/fl (1/1.6) = 1. So, prior to projection onto the film plane, the blur is the same size.

No, the blur is the same size after projection onto the film plane. How could it be before? That would be before the optics.

But now add in the fact that the full frame is 1.6 times larger than the APS frame. So the image that is optically projected onto the 35mm frame is 1.6 times the size of the image projected on frame. So, now, the blur is 1.6 times larger on the 35mm neg as compared to the APS neg.

I think what you're getting hung up on is the relationship between the object size and the size of the blur. They are unrelated. To see this, use a point source, like a penlight.

[Steven M. Anthony]

No, the blur is the same size after projection onto the film plane. How could it be before? That would be before the optics.

I think what you're getting hung up on is the relationship between the object size and the size of the blur. They are unrelated. To see this, use a point source, like a penlight.

Okay. After this explanation of yours and drawing it out myself (pictures help me!), I see that the magnification "happens" at the projection. So there is no additional 1.6 factor. Thanks for your patience!

But if the recorded blur is the same size on the neg for 1/50 w/50mm, 1/80 w/80mm, and, presumably, 1/100 w/100mm and 1/500 for 500mm, etc..., what use is the rule?

[slin100]

Okay. After this explanation of yours and drawing it out myself (pictures help me!), I see that the magnification "happens" at the projection. So there is no additional 1.6 factor. Thanks for your patience!

I'm so giddy you agreed with me! You are most welcome, Steven! :D

But if the recorded blur is the same size on the neg for 1/50 w/50mm, 1/80 w/80mm, and, presumably, 1/100 w/100mm and 1/500 for 500mm, etc..., what use is the rule?
This is the reason the rule has be adjusted to take into account the format size. If you use 1/80 w/ 50mm on a 10D, then the blur on the neg is 1.6x smaller than the 35mm. Consequently, when you enlarge both APS and 35mm frames to the same print size, the blur ends up being the same on each print! Does it, now, make sense that the rule of thumb should be 1/effective focal length instead of 1/true focal length?

[Steven M. Anthony]

Yes, I see how across sensor formats the correction factor equates the size of the blur upon printing to a target size. But what use is the rule in general if, for a given format, the blur is the same size no matter the shutter speed?

[slin100]

The blur is not the same size regardless of the shutter speed. Shutter speed proportionally affects blur size.

[Jon]

Yes, I see how across sensor formats the correction factor equates the size of the blur upon printing to a target size. But what use is the rule in general if, for a given format, the blur is the same size no matter the shutter speed?

The blur isn't the same size no matter the shutter speed. The amount the camera moves in a given unit of time is (relatively) constant for a given photographer under normal conditions. But as you use progressively faster shutter speeds, the amount of movement that's captured changes. So, there's less blur on the original image taken at 1/125 than on the original taken at 1/60 with the same lens. It's the same as photographing a moving object in the distance, really. At higher shutter speeds, it gets progressively harder to tell from the cues in the picture if the subject was moving.

[Steven M. Anthony]

The blur is not the same size regardless of the shutter speed. Shutter speed proportionally affects blur size.

But if it's true for 80mm and 50mm, why not for other fls too?

[Steven M. Anthony]

The blur isn't the same size no matter the shutter speed. The amount the camera moves in a given unit of time is (relatively) constant for a given photographer under normal conditions. But as you use progressively faster shutter speeds, the amount of movement that's captured changes. So, there's less blur on the original image taken at 1/125 than on the original taken at 1/60 with the same lens. It's the same as photographing a moving object in the distance, really. At higher shutter speeds, it gets progressively harder to tell from the cues in the picture if the subject was moving.

Yes, I get that with the same lens, blur gets smaller as shutter speed gets quicker. But i'm talking across lenses.

If 1/50th with a 50mm creates the same size blur (on the neg/sensor) as an 80mm will at 1/80th, then the size of the blur (of the same scene) with a 1000mm at 1/1000th will also be the same as the 50mm at 1/50th. But when you blow up the 50mm shot and crop for the content of the 1000mm, the enlarged 50mm shot will look way more blury.

So it seems, as I think Rick said, print enlargement is a huge factor in determining what shutter speed to use.

[Jon]

But when you blow up the 50mm shot and crop for the content of the 1000mm, the enlarged 50mm shot will look way more blury.
When you overenlarge an image, you're introducing all the other aberrations the lens(and film/sensor) are prone to. Put a 1000 mm lens on an Eos 1V with Velvia, then replace it with a 50 mm f/1.8, and without moving the camera. Both at f/16. When you enlarge the 50 mm so that your subject (from the shot with the 1000 mm) is the same size and you'll see massive grain. You'd also see diffraction, chromatic aberration, and any number of other distortions that are reduced to an "acceptable" level under"normal" enlargements.

[Steven M. Anthony]

When you overenlarge an image, you're introducing all the other aberrations the lens(and film/sensor) are prone to. Put a 1000 mm lens on an Eos 1V with Velvia, then replace it with a 50 mm f/1.8, and without moving the camera. Both at f/16. When you enlarge the 50 mm so that your subject (from the shot with the 1000 mm) is the same size and you'll see massive grain. You'd also see diffraction, chromatic aberration, and any number of other distortions that are reduced to an "acceptable" level under"normal" enlargements.

Okay--maybe the 1000mm was a bad comparison. But if the blur due to camera shake is the same size on the negsensor, regardless of the fl of the lens if 1/fl is used to set shutter speed, it doesn't seem that helpful...

[slin100]

The 1/fl rule assumes some standard of enlargement. If you exceed that standard, as would happen when you enlarge an image captured with a 50mm by 20x to match a 1000mm lens, then the rule breaks down. If you intend to exceed the standard, you must compensate by using a higher shutter speed that is proportional to the amount by which the standard was exceeded.

It's really no different than DOF. DOF calculations are also predicated on some standard print size. For example, I understand that Canon marks the DOF scales on EF lenses assuming an 8x10 print size. If you are going to print 20x30, then the DOF is going to appear narrower assuming the viewing distance is kept constant, so it would behoove you to compensate by closing down the aperture.

[rdenney]

I'm going to go through some arithmetic to test my assumption. After all, I could be wrong, heh, heh.

Let's say that I have a palsy and my hand shakes the camera through an arc of 0.1 degrees in 10 milliseconds (1/100 of a second) of time. What might that produce on the image plane?

Let's assume, for now, a 43mm normal lens on a 35mm camera.

With a shutter speed of 1/100 of a second (yes, I know this isn't available), the movement will be closely approximated (at these tiny angles) by 43*sin(0.1). That is 0.075 mm on the image plane. (The actual number is 2*43*sin(0.05)/sin(89.95), which also equals 0.075mm--I'm just proving that my approximation is close enough at these tiny angles.)

If the focal length was 86 (twice normal), the movement would be 86*sin(0.1). That is 0.15mm on the image plane. So, doubling the focal length, as expected, doubles the movement at the image plane. This is true no matter what the format is (and this has been Steve's point all along, to which people have argued on grounds of relevancy rather than fact).

Now, let's compare it with a 10D that is using a 28mm normal lens. The movement would be 28*sin(0.1) = 0.049mm. If the focal length was 56 (twice normal), the movement at the image plane would be 56*sin(0.1) = 0.098mm.

Again, these distances are not controlled by format. If we remove format from the above descriptions, we find that the movement is related to the focal length only, assuming camera movement of the same amount.

Now, let's apply some shutter speed control.

At 28mm, we will use a shutter speed of 1/30. At 43, we'll use 1/45. At 56, we'll use 1/60. And at 86, we'll use 1/90. This is our 1/focal length rule.

The movement, in degrees of arc, at 1/30 is 100/30=3.3 times what it was a 1/100, which is 0.333 degrees. At 1/45, the movement is 0.222 degrees. At 1/60, it's 0.167 degrees, and at 1/90 it's 0.111 degrees.

The movement for the 28 at 1/30 will be 28*sin(.333)=0.163mm. With the 43 at 1/45, it's 43*sin(.222)=.167mm. With the 56 at 1/60, it's 56*sin(.167)=.163mm. Starting to see a trend? With the 86 at 1/90, the movement is 86*sin(.111)=.167mm. (Small differences are due to our rounding to the nearest shutter speed.)

Thus, camera movement is the same at the image plane, in absolute distance, if we follow the 1/fl rule of thumb. If we rotate the camera as a result of shake through an arc of 0.1 degrees per 10 milliseconds, the movement of the image plane will be about 0.165mm at all focal lengths, if we follow the 1/focal length rule. The movement of that camera shake does not, at this point, depend on format, but rather only on focal length and shutter speed.

Now, let's say we enlarge the image to a 13x19 print.

13 inches is 330mm. 330/15 is 22, and 330/13.75, calculating enlargement on the vertical dimension. Thus, that 0.165mm of movement will be enlarged 22 times if we start with a 15x23 image, and 13.75 times if we start with a 24x36 image.

So, 0.165mm times 22 is 3.6mm. And 0.165mm times 13.75 is 2.3mm.

Summing up so far:

28mm at 1/30 on 10D = 3.6mm shake smear on 13x19 print.
43mm at 1/45 on Elan = 2.3mm shake smear on 13x19 print.
56mm at 1/60 on 10D = 3.6mm shake smear on 13x19 print.
86mm at 1/90 on Elan = 2.3mm shake smear on 13x19 print.

Now, let's redo the 10D numbers, using 1/equivalent focal length.

28*1.6 = 43. So, if we shoot the 28 lens at 1/45, we get 28*sin(.222) = 0.108mm. And, if we shoot the 56 at 1/90, we get 56*sin(.111) = 0.108mm.

Now, let's enlarge again. 0.108mm times 22 is 2.4mm on the print. (I guarantee that the difference here between the 2.3 above and the 2.4 here is rounding error, but I ain't going back through it to prove it).

Thus, on a given size print, one must convert one's focal length to equivalent 35mm focal length, because the 1/fl rule only works (if it works at all) for the 35mm format.

Just for fun, I want to test my rule for a 127mm wide angle lens on the Speed Graphic shooting 4x5 film. The 127 is .78 time normal in the 4x5 format. Thus, it's equivalent 35mm format lens is .78*43 = 34mm. So, let's say I apply my rule above and use a shutter speed of 1/30 with the 127 on the 4x5 Speed Graphic. And my movement is the same as above, 0.1 degree per 10 milliseconds. At 1/30, I get 0.333 degrees of arc movement, same as above. 127*sin(0.333)=0.74mm of movement at the image plane.

But I only have to enlarge the 4x5 negative 3.25 times to get the 13x19ish print. 0.74*3.25 is 2.4mm, which is the same as I got for 35mm and for the 10D when converted to equivalent focal lengths. So, my rule works for this extreme case. 1/equivalent focal length provides the same smear on the final print, assuming all formats are compared at the same print size.

If all formats are compared at the same enlargement multiplier rather than at the same print size, then 1/focal length works for all formats. If we can enlarge the 35mm frame to 13x19, then we can only enlarge the 10D image to 8x12, and the 4x5 image can be blown up to a whopping 54x68, if we use 1/fl as our standard, and if camera shake is out only criterion of image quality. Most have argued in this discussion that it's more in line with general use to compare at the same print size, especially when merely comparing between APS and 35mm formats. But that is a convention that we can agree to (or not).

Of course, my example numbers are vastly larger than would be tolerable, but it makes the numbers big enough so that you don't have to count zeros.

Rick "hoping this clears it up once and for all" Denney

[Steven M. Anthony]

Thanks Rick.

My takeaway from this is to forget the dang rule and go with what my experience tells me to use!