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11-11-2014, 08:33 AM   #211
bxf
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QuoteOriginally posted by normhead Quote
OK, here's how I understand it....it's not that the crop sensor of a D810 is better than a K-5, it's that if you shoot the image with a D810 and the same FoV with a K-5 (or D810 in crop mode) ... remember, it's the same field of view, there is twice as much light on the D810's sensor to make the same image...

So say you have noise in your K-5 image... the image has to be twice the size.. so the noise will be twice as apparent.

...
Yes, I understand this last point. In fact, I made a similar mention in Opening post.

But I don't see that this relates to Total Light.

Apart from the above, something suddenly occurred to me: the entire concept of Total Light (to me, at least) is based on the question of why more light on a larger sensor is better than less light on a smaller sensor. Someone has presented an analogy of more water on a larger field improving the crop quality of the entire field, as compared to a proportionally smaller quantity of water on a smaller field.

So now I'm thinking: are we correct in assuming that light acts like e.g. water? In other words, is it possible that this is not a proportional thing? We know that twice as much water, on twice as much land, results in the same quantity of water per unit area. Do we in fact know that the same applies with light?

---------- Post added 11-11-14 at 15:37 ----------

QuoteOriginally posted by Rondec Quote
I prefer to think of it as a magnification issue. The problem with the K5 is not that it is noisier, per se, it is that you see its pixels quicker when you enlarge the photo/put it on a bigger monitor.

If you aren't magnifying your image much, then the differences will not be visible.
This is the easiest concept to understand, but as I said to Norm above, I don't see where the concept of Total Light is related to this. So if this magnification is in fact the "bigger" answer, then it essentially trivializes the Total Light argument. Perhaps rightly so, I don't know.

---------- Post added 11-11-14 at 15:43 ----------

QuoteOriginally posted by jsherman999 Quote
...

It doesn't improve the quality of a given unit of area, it improves the entire image, because more photons were captured for the same exposure. The resulting image was created with more light.

Keep in mind that when someone says 'larger sensor gets more total light than smaller,' the same FOV and distance to subject - same "framing" - is being assumed in that comparison. It's assumed that the photographer is shooting the same subject from the same distance at the same exposure.

Ex: Take a portrait at 50mm 1/100s f/2.8 (FF) and one at 35mm 1/100s f/2.8 (aps-c) from the same distance, and the FF image has been created using a larger physical aperture (17.8mm) vs. the aps-c image (12.5mm.) The FF image was created with more total light as a result of that, the volume of photons that was collected in the frame over the same period of time was higher, and (if the sensors are similar-gen) that image will be less noisy.

...
But the way I see it, it always comes back to the question of why 1.5x photons per 1.5Y.area is better than x photons per Y.area. Unless what I question above in the comparison to water is valid.

11-11-2014, 09:00 AM   #212
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QuoteOriginally posted by bxf Quote
Apart from the above, something suddenly occurred to me: the entire concept of Total Light (to me, at least) is based on the question of why more light on a larger sensor is better than less light on a smaller sensor. Someone has presented an analogy of more water on a larger field improving the crop quality of the entire field, as compared to a proportionally smaller quantity of water on a smaller field.

So now I'm thinking: are we correct in assuming that light acts like e.g. water? In other words, is it possible that this is not a proportional thing? We know that twice as much water, on twice as much land, results in the same quantity of water per unit area. Do we in fact know that the same applies with light?
Yes, that analogy holds water People often use it.

You can think of a photon of light as a raindrop, four buckets on one square meter on a 50 square meter field and on a 100 square meter field - those four buckets catch the same amount of raindrops.

But the 100 square-meter field catches many more total raindrops.

In this link, the first graph ("screen") is the 4-buckets in one square meter comparison - looks the same. The second graph ("print") shows you the rain collected from the whole field.

Most people can get that. When a lot of people 'lose' that analogy is when they try to imagine how it applies to the image, because the image isn't bigger/wider than the smaller-format image, it's the same FOV. The link you have to make in your mind is that the same-FOV image is being projected onto that sensor with a larger physical aperture, and then when displayed, it's constructed with more light as the main ingredient. Think of being delivered two 50 gallon buckets, one containing all water collected from the 50-sq meter field, the other containing all water collected from the 100 sq meter field. Both are 50 gallon buckets (same FOV,) one is just more full. Those 50-gallon buckets are your resulting images.

A simple different example that a lot of people seem to 'get' - but 'lose' when it comes to their cameras - you know how space telescopes like the hubble have huge mirrors, and huge physical apertures? An Iris you could lay down in, longer than your body? It's to collect more total light. The more total light it collects, the deeper and more clearly it can see into space. (longer exposure time also collects more total light.) I think the primary mirror uses an iris 2.4 meters (not MM!), in diameter. That's a lot of total light.

Last edited by jsherman999; 11-11-2014 at 09:40 AM.
11-11-2014, 09:14 AM   #213
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QuoteOriginally posted by bxf Quote
This is the easiest concept to understand, but as I said to Norm above, I don't see where the concept of Total Light is related to this. So if this magnification is in fact the "bigger" answer, then it essentially trivializes the Total Light argument. Perhaps rightly so, I don't know.
The lower magnification is just a bi-product of using larger sensor to capture more light. But there are many other ways of capturing more light (larger aperture, longer exposure time, adding artificial light...).

You can use a smaller sensor and capturing more light to get better IQ than using a larger sensor capturing less light.
A user of a APS-C camera might use tripod capturing a landscape at night using base ISO, when a user of a FF camera shoots free hand and has to use much higher ISO (to get faster shutter speed to avoid camera shake).
11-11-2014, 09:18 AM   #214
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QuoteOriginally posted by bxf Quote
("magnification...")
This is the easiest concept to understand, but as I said to Norm above, I don't see where the concept of Total Light is related to this. So if this magnification is in fact the "bigger" answer, then it essentially trivializes the Total Light argument. Perhaps rightly so, I don't know.[COLOR="Silver"]
As I said before, magnification is the mechanism that shows you the noise more clearly, it's not the reason the noise is there in the first place. If you were to take an aps-c and FF shot with the same FOV and same total light (for example, 50mm f/2.8 on FF and 35mm f/1.8 on aps-c = same FOV, same total light,) then the magnification of that aps-c image that happens on display or print wouldn't look any noisier.

Magnification is the mechanism, Total Light is the reason.

Magnification is your gearshift, Total light is the gears.

.

---------- Post added 11-11-14 at 10:22 AM ----------

QuoteOriginally posted by Fogel70 Quote
You can use a smaller sensor and capturing more light to get better IQ than using a larger sensor capturing less light.
A user of a APS-C camera might use tripod capturing a landscape at night using base ISO, when a user of a FF camera shoots free hand and has to use much higher ISO (to get faster shutter speed to avoid camera shake).
Exactly. It doesn't really matter how you collect the total light - longer exposure times can work too.

Where larger formats help is when you're shutter-speed constrained and can't depend on longer exposure times (which is about 80% of my shooting.) A larger physical aperture sending light to a larger sensor allows you to get more total light for the same exposure time.


Last edited by jsherman999; 11-11-2014 at 09:31 AM.
11-11-2014, 10:03 AM - 2 Likes   #215
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QuoteOriginally posted by bxf Quote
Apart from the above, something suddenly occurred to me: the entire concept of Total Light (to me, at least) is based on the question of why more light on a larger sensor is better than less light on a smaller sensor.
(First, my apology for jumping in so late in this discussion - feel free to ignore).

Try this thought experiment. Take a properly exposed picture using a camera (camera A) with an ASP-C sensor. Now take a picture of the same subject using a hypothetical camera (camera B) which has a sensor identical to the first camera, except it is half its size. Use exactly the same exposure parameters on both. The images are the same except B has captured half the light as A.

If I understand what you are saying, camera A sensor noise per unit area is the same as camera B sensor noise per unit area, and therefore noise in a final print from both must be the same as well. Am I understanding your point? We could say A noise = B noise.

To continue the experiment. use a camera C, which is identical to camera B except its sensor is half the size. If A noise = B noise, then it follows that B noise = C noise. And that implies A noise = C noise.

Let's do a few more iterations, in our thought experiment. all the way to camera Z. Each iteration produces an image identical in noise per unit area, differing only in that total light captured is half as much. So A noise = B noise, and B noise = C noise ... Y noise = Z noise. Which gives us the result A noise = Z noise. Image A is identical to image Z except that image Z has captured 1/2^(n-1) = 1/33,554,432th total light as image A.

Now let's make an 8X10 glossy using a couple of the captured images and compare them. They should look the same, because noise per unit area in each camera is the same, right? The only difference is total light captured. So the question is, would you rather make your 8X10 glossy from image A or image Z, and why?

Hope this helps.
11-12-2014, 08:53 AM   #216
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QuoteOriginally posted by cfraz Quote
(First, my apology for jumping in so late in this discussion - feel free to ignore).
On the contrary what you wrote is essential reading, good contribution. Very smart/simple way to present it.

---------- Post added 11-12-14 at 10:01 AM ----------

If someone wants another way to visualize this, I've mentioned "The_Suede"'s contribution to this thread: link.

I'll paste some of what he wrote below:

"If you want a really simplified version...

This image is a part of the simplified object field. It's an infinitely large grid/wall of perfect point lights spaced at an even distance from each other. They all have the same intensity. We're aiming a 2x crop camera with a 50mm lens straight at it, and get this:


2x crop sensor - 50mm lens - f/4.0 aperture


Each individual point included in the image will send light to/through the optical system. How much light from each point that will actually pass through the optical system is determined by the aperture area, or more accurately the front pupil area. Nothing else - at least not in the simplified version.

Let's just say/assume that the aperture used was f/4.0, and that the intensity of the point lights in front of the system each contribute 1 (one) photon to the image with the front pupil area you get at 50mm F4.0. Then you have a total of 150 photons on the image plane (10x15 points are included, 1 photon from each). That was the "total light gathered".

Now - with the same camera - change the lens to a 25mm (half focal length), while keeping aperture at f/4.0. What happens? The field of view widens, and you get four times as many light points included in the image plane.


2x crop sensor - 25mm lens - f/4.0 aperture


-But since the front pupil area is now four times smaller (half focal length, same aperture > four times smaller front pupil area) - you still get the same total amount of light passed through!

Four times as many (150x4 = 600) points of light that each contribute four times less (1/4 = 0.25) photons to the image = still a total of (600x0.25) = 150 photons on the sensor / image plane! Same as with the 50mm F4.0 lens.

That's why the "f/stop" system works - it keeps image plane exposure (light per area on the image plane) constant if you keep the f/# constant - no matter what focal length you choose.
Longer focal length > larger front pupil area, but fewer points of light included.
Shorter focal length > smaller front pupil area, but more points of light included.

But what happens if we increase the included angle of view by using a 2x larger image plane/sensor (four times larger area) with the 50mm lens in stead? Well, you still include four times as many light points, but the front pupil area doesn't change. So each point still gives the same amount of light on the image plane.


1x sensor - 50mm lens - f/4.0 aperture


So now you have 600 (20x30) points of light included again - just as with the 2x crop sensor and a 25mm lens - but in this case each point still contributes 1 photon to the sensor/image plane - resulting in a total photon count of 600x1.0 = 600 in stead of 600x0.25 = 150.

...................

Then there are some losses, like cos4, mechanical vignetting, optical reflection/absorption and so on - but that's rather beside the point from a purely "simple system" point of view.

The two basic parts are really:

f/# determines image plane exposure
exposure is "amount of light energy per area unit"
-So you get the light energy amount that your system "gathered" by:

(scene average luminance) / (f-stop)^2 * (sensor area) * (exposure time)

And from that you can get that the system "light gathering ability" when keeping scene average luminance and exposure time constant is:

Light gathering ability = (sensor area) / (f-stop)^2

Since (sensor area) is the square function of crop ratio this can be further simplified by doing a square root on both, and then you get:

Light gathering ability = (crop ratio) / (f-stop)

In the end that means that to collect as many photons per second from a certain scene, you need to change f/# with the same factor as you change the crop ratio.



.

Last edited by jsherman999; 11-12-2014 at 09:02 AM.
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