Reading mode: Light Dark

We Should Applaud DPR

Go to last post
Giklab said:
Giklab wrote:
Giklab said:
Lee Jay wrote:
Giklab said:
AlphaTikal wrote:

And why should the buyer bother with this additional confusion (assuming the target is not interested in fullframe cameras)?
Click to expand...
I've actually had people tell me than an FZ20 is better in low light than what I was shooting with (Canon 5D, 24-105/4) because the Panasonic's lens is f/2.8 and mine is only f/4. When I tell them the Panasonic's equivalent f-stop is f/17.5, they "get it" (why my camera is so much better in low-light than theirs). In other words, it was the "36-432mm f/2.8 lens" (which is really a 6-72mm f/2.8 lens) that was confusing them.
Click to expand...
So you actually explained something incorrectly.
Click to expand...
No, I did not.
Giklab said:
(you even contradict yourself in your post). Your 5D wasn't better in low light because their lens let in f/17.5's worth of light compared to your f/4 (per sensor area),
Click to expand...
Yes, it was.
Giklab said:
but because the 5D's sensor is so much larger the high ISO performance is much better.
Click to expand...
That's an equivalent explanation, but a more confusing one.
Giklab said:
Your camera was better than theirs, not lens aperture.
Click to expand...
Incorrect. My camera is better exactly because of lens aperture. All formats perform the same with the same field of view and the same aperture (not f-stop).
Giklab said:
Even a 6-120mm 2.8 doesn't magically get to be f17.5 just because it's used on a tiny sensor
Click to expand...
It doesn't magically get to be 36-432mm either. The tiny sensor makes it "equivalent" to a 36-432/17.5.
Giklab said:
And this brings up another point some people don't seem to understand properly. It's true that a compact superzoom lens with a f/2.8 max aperture lets in less light than a FF f/4 lens, but it lets in more light per sensor area, which is what actually counts in practice.
Click to expand...
No, total light capture is actually what counts in practice. What you just said is one of the hardest to kill myths in all of photography.
Giklab said:
You can say all day long that a superzoom lets in f/17.5's worth of light compared to your FF lens' f/4, but if you need 1/200 at f/4 at ISO 200 to get a proper exposure on your FF 5D, I'll need 1/200 at f/4 at ISO 200 to get an identical shot from my superzoom.
Click to expand...
And you'll have way, way more noise, and it's so because of the smaller aperture, despite the faster f-stop.
 
Lee Jay said:
Lee Jay wrote:
Lee Jay said:
Trafford wrote:
Lee Jay said:
Lee Jay wrote:

The vertical axis should be inverted.
--
Lee Jay
(see profile for equipment)
Click to expand...
why? Please explain.
Click to expand...
It's a performance envelope (the envelope inside of which you can shoot) and higher should be better.
Click to expand...
I completely disagree. It's not "performance" it is a simple value; therefore, the higher the number should increase vertically.
 
RedFox88 said:
RedFox88 wrote:
RedFox88 said:
Lee Jay wrote:
RedFox88 said:
Trafford wrote:
RedFox88 said:
Lee Jay wrote:

The vertical axis should be inverted.
--
Lee Jay
(see profile for equipment)
Click to expand...
why? Please explain.
Click to expand...
It's a performance envelope (the envelope inside of which you can shoot) and higher should be better.
Click to expand...
I completely disagree. It's not "performance" it is a simple value; therefore, the higher the number should increase vertically.
Click to expand...
In DPReview's chart, down is better.
 
Lee Jay said:
Lee Jay said:
(you even contradict yourself in your post). Your 5D wasn't better in low light because their lens let in f/17.5's worth of light compared to your f/4 (per sensor area),
Click to expand...
Yes, it was.
Click to expand...
Let's start with this. Imagine you have a Sony a7, a FF mirrorless, so you could do this comparison IRL. You have a 28-70 f/4 FE mount lens (native) and a 5-13mm f/4 lens for a 1/2.3 sensor (5.5 times crop). The 1/2.3 lens should be a rough equivalent to a 28-70 f/22 by your logic.

Now imagine you mount them both on the a7, and then take a picture with both at, say 1/50, f/4, ISO 100 at 28mm eq so you get a properly exposed image with the native lens. Now, if your theory stands, you should get a completely dark frame with the other lens, since you are effectively shooting at f/22 with the 1/2.3 lens. But I think you'd get a dark frame because of lack of coverage, and a properly exposed circle at the centre of the image, because the lens is actually f/4 there.
 
Giklab said:
Giklab wrote:
Giklab said:
Giklab said:
(you even contradict yourself in your post). Your 5D wasn't better in low light because their lens let in f/17.5's worth of light compared to your f/4 (per sensor area),
Click to expand...
Yes, it was.
Click to expand...
Let's start with this. Imagine you have a Sony a7, a FF mirrorless, so you could do this comparison IRL. You have a 28-70 f/4 FE mount lens (native) and a 5-13mm f/4 lens for a 1/2.3 sensor (5.5 times crop). The 1/2.3 lens should be a rough equivalent to a 28-70 f/22 by your logic.
Click to expand...
Right. And it is.
Giklab said:
Now imagine you mount them both on the a7, and then take a picture with both at, say 1/50, f/4, ISO 100 at 28mm eq so you get a properly exposed image with the native lens. Now, if your theory stands, you should get a completely dark frame with the other lens, since you are effectively shooting at f/22 with the 1/2.3 lens. But I think you'd get a dark frame because of lack of coverage, and a properly exposed circle at the centre of the image, because the lens is actually f/4 there.
Click to expand...
Uh, no, you're entirely off the rails.

I'll help.

28mm and f/4 on full frame is an aperture diameter of 28mm/4 = 7mm. On a 1/2.3" sensor, the equivalent would be a 5mm focal length. At f/4, that's an aperture diameter of 5mm/4 = 1.25mm.

They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.

To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
 
Lee Jay said:
Lee Jay wrote:
Lee Jay said:
They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.
Click to expand...
To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
Click to expand...
While you ponder all the implications of 1.25mm holes, DOF with various formats (for professionals, full-frame is usually medium-format for superior DOF control), numbers of photons which are captured, and shot noise in images, remember that for the purpose of exposure, f/4 is simply f/4.
 
Member said:
Darrell Spreen wrote:
Member said:
Lee Jay wrote:
Member said:
They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.
Click to expand...
To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
Click to expand...
While you ponder all the implications of 1.25mm holes, DOF with various formats (for professionals, full-frame is usually medium-format for superior DOF control), numbers of photons which are captured, and shot noise in images, remember that for the purpose of exposure, f/4 is simply f/4.
Click to expand...
Yes, and exposure is a largely-irrelevant thing in the digital age, as it dictates basically nothing in this context.
 
Member said:
Darrell Spreen wrote:
Member said:
Lee Jay wrote:
Member said:
They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.
Click to expand...
To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
Click to expand...
While you ponder all the implications of 1.25mm holes, DOF with various formats (for professionals, full-frame is usually medium-format for superior DOF control), numbers of photons which are captured, and shot noise in images, remember that for the purpose of exposure, f/4 is simply f/4.
Click to expand...
...as a metric for comparing different formats? That is, why would we compare f/4 on one format to f/4 on another format unless f/4 was as wide as the lens opened up and we were interested in getting as low noise as possible or as narrow a DOF as possible?

For example, f/4 on FF puts 4x as much light on the sensor as f/4 on mFT, as well as having half the DOF and half the diffraction softening (not that diffraction really matters at such a wide aperture, though). On the other hand, f/4 on FF puts the same amount of light on the sensor as f/2 on mFT, has the same DOF, and the same diffraction softening.

So, it seems to me that comparing f/4 on FF to f/4 on mFT would an odd thing to do unless f/4 were as wide as the lens opened up on each system and we wanted as shallow a DOF as we could get and/or as low noise as we could get.
 
Giklab said:
Giklab wrote:
Giklab said:
Giklab said:
(you even contradict yourself in your post). Your 5D wasn't better in low light because their lens let in f/17.5's worth of light compared to your f/4 (per sensor area),
Click to expand...
Yes, it was.
Click to expand...
Let's start with this. Imagine you have a Sony a7, a FF mirrorless, so you could do this comparison IRL. You have a 28-70 f/4 FE mount lens (native) and a 5-13mm f/4 lens for a 1/2.3 sensor (5.5 times crop). The 1/2.3 lens should be a rough equivalent to a 28-70 f/22 by your logic.

Now imagine you mount them both on the a7, and then take a picture with both at, say 1/50, f/4, ISO 100 at 28mm eq so you get a properly exposed image with the native lens. Now, if your theory stands, you should get a completely dark frame with the other lens, since you are effectively shooting at f/22 with the 1/2.3 lens. But I think you'd get a dark frame because of lack of coverage, and a properly exposed circle at the centre of the image, because the lens is actually f/4 there.
Click to expand...
The small sensor has the same nominal aperture value, the same shutter speed and the same ISO. The exposure will be the same, but noise level will be different. Clearly the small sensor has much less light to work with, because the actual diameter of the aperture is much narrower. That's where the noise comes from, the signal (light) needs more amplification on a smaller sensor to brighten the image.

What we are concerned with as photographers are DOF, shutter speed and noise (grain). We all agree the DOF will be f22 equivalent. The shutter speed is what it says it is, no conversion required. The only question mark is ISO, but why should a photographer care about an artificial construct like ISO? We should care about noise, because that's what impacts IQ. And noise from these two cameras will be radically different.

The Pentax Q10 uses a high quality 1/2.3" BSI sensor. If we compare DXOMark SNR scores for the Q10 vs. RX1, we see ISO 183 for the Q vs. ISO 2534 for the RX1. So the actual conversion factor for noise with these two cameras would be 2534/183 = 13.8X, even worse than the the 5.5X one would expect. At base ISO, noise is not a concern. It is a huge concern as light level drops. I like aperture conversion, because it at least acknowledges that f4.0 on a cropped body is not truly equivalent to f4.0 on a FF body, despite what ISO says.

The aperture number is a real calculation, the ratio of focal length to effective aperture diameter. it is absolutely tied to sensor size, because sensor size changes focal length. Aperture should be acknowledged with a conversion factor, just as focal length is.

5-13mm f/4 lens for a 1/2.3 sensor = 28-70 f/22, and this is easily demonstrated by putting a Q lens on a FF camera. The actual aperture diameter is tiny and will impede the light reaching the FF sensor.
 
Great Bustard said:
Great Bustard wrote:
Great Bustard said:
Darrell Spreen wrote:
Great Bustard said:
Lee Jay wrote:
Great Bustard said:
They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.
Click to expand...
To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
Click to expand...
While you ponder all the implications of 1.25mm holes, DOF with various formats (for professionals, full-frame is usually medium-format for superior DOF control), numbers of photons which are captured, and shot noise in images, remember that for the purpose of exposure, f/4 is simply f/4.
Click to expand...
...as a metric for comparing different formats? That is, why would we compare f/4 on one format to f/4 on another format unless f/4 was as wide as the lens opened up and we were interested in getting as low noise as possible or as narrow a DOF as possible?
Click to expand...
I think most confused people do not realise that F/4 is only a value for light intensity, not light quantity. And light intensity really has very little value for all practical purposes. It is light quantity that matters from IQ point of view.
 
ultimitsu said:
ultimitsu wrote:
ultimitsu said:
Great Bustard wrote:
ultimitsu said:
Darrell Spreen wrote:
ultimitsu said:
Lee Jay wrote:
ultimitsu said:
They're both looking at the same field of view, but one is looking through a 7mm hole, the other through a 1.25mm hole. (7/1.25)^2 = 31.36x more light for the full-frame sensor, even though the f-stop is the same.
Click to expand...
To put it another way, how small an f-stop would the full-frame lens at 28mm have to select to get a 1.25mm aperture? 28/1.25 = 22.4.

So, FF, 28mm, f/22.4 = 1/2.3", 5mm, f/4. Both have the same field of view and the same aperture diameter. Aperture diameter controls both depth-of-field and the number of photons captured which controls shot noise in the final overall image.
Click to expand...
While you ponder all the implications of 1.25mm holes, DOF with various formats (for professionals, full-frame is usually medium-format for superior DOF control), numbers of photons which are captured, and shot noise in images, remember that for the purpose of exposure, f/4 is simply f/4.
Click to expand...
...as a metric for comparing different formats? That is, why would we compare f/4 on one format to f/4 on another format unless f/4 was as wide as the lens opened up and we were interested in getting as low noise as possible or as narrow a DOF as possible?
Click to expand...
I think most confused people do not realise that F/4 is only a value for light intensity, not light quantity.
Click to expand...
I think most people do not understand the difference. Some of those profess not to care, but the ironic thing is that the things the do care about hinge on the significance of this difference.
ultimitsu said:
And light intensity really has very little value for all practical purposes. It is light quantity that matters from IQ point of view.
Click to expand...
Indeed.
 
Great Bustard said:
Great Bustard wrote:

So, it seems to me that comparing f/4 on FF to f/4 on mFT would an odd thing to do unless f/4 were as wide as the lens opened up on each system and we wanted as shallow a DOF as we could get and/or as low noise as we could get.
Click to expand...

Your explanations always seem so clear and logical to me, but how many years have you been trying to educate four-thirds shooters about equivalence, and they still can't see? Whenever I say that f2.8 on m4/3 is equivalent to f3.7 on APS-C, it's a battle every time.

The origin of the problem is that aperture is stated in f-numbers instead of millimeters diameter.
 
The smaller the sensor, the more difficult it is to shoot at high ISO. If I'm shooting with, say, an Iphone, I'm not using it for the bokeh but just to grab shots. With the limited ISO wiggle room, knowing that I have a f2 lens attached in relation to the sensor is very relevant information.

The equivalency in DOF with a 35mm, not so important in the grand scheme of things and needlessly confuses matters.
 
bluevellet said:
The smaller the sensor, the more difficult it is to shoot at high ISO. If I'm shooting with, say, an Iphone, I'm not using it for the bokeh but just to grab shots. With the limited ISO wiggle room, knowing that I have a f2 lens attached in relation to the sensor is very relevant information.

The equivalency in DOF with a 35mm, not so important in the grand scheme of things and needlessly confuses matters.
Click to expand...
 
Lee Jay said:
Lee Jay wrote:
Lee Jay said:
The smaller the sensor, the more difficult it is to shoot at high ISO. If I'm shooting with, say, an Iphone, I'm not using it for the bokeh but just to grab shots. With the limited ISO wiggle room, knowing that I have a f2 lens attached in relation to the sensor is very relevant information.

The equivalency in DOF with a 35mm, not so important in the grand scheme of things and needlessly confuses matters.
Click to expand...
Click to expand...
 
audiobomber said:
audiobomber wrote:

The origin of the problem is that aperture is stated in f-numbers instead of millimeters diameter.
Click to expand...
Because the relative f-number is focal number and is about the lens' focal. Millimeters would not help much, as the problem begins with different sensor sizes. The problem is, light capture needs two parts, the lens (focal size) AND the sensor (size).

If I mount a lens for fullframe on a fullframe with f/2.8, I get f/2.8. If I mount it on APS-C, the f-stop f/2.8 is still f/2.8 for measurement, but the sensor only uses the middle and the light around gets lost. So it is effectively only f/4.0 on the sensor, but the lens still is set to f/2.8.

This is really not complicated. But once the head is in chaos, this looks complicated.

--
http://www.flickr.com/photos/blackhole_eater/
(All photos are creative common licensed. Check them out.)
 
Last edited:
AlphaTikal said:
AlphaTikal wrote:
AlphaTikal said:
audiobomber wrote:

The origin of the problem is that aperture is stated in f-numbers instead of millimeters diameter.
Click to expand...
Because the relative f-number is focal number and is about the lens' focal. Millimeters would not help much, as the problem begins with different sensor sizes. The problem is, light capture needs two parts, the lens (focal size) AND the sensor (size).
Click to expand...
No, it doesn't. Aperture diameter (mm) is all that's needed. When that is constant, light capture is constant regardless of sensor size. This is why we use aperture when doing astrophotography.
 
audiobomber said:
audiobomber wrote:
audiobomber said:
Great Bustard wrote:

So, it seems to me that comparing f/4 on FF to f/4 on mFT would an odd thing to do unless f/4 were as wide as the lens opened up on each system and we wanted as shallow a DOF as we could get and/or as low noise as we could get.
Click to expand...
Your explanations always seem so clear and logical to me, but how many years have you been trying to educate four-thirds shooters about equivalence, and they still can't see? Whenever I say that f2.8 on m4/3 is equivalent to f3.7 on APS-C, it's a battle every time.
Click to expand...
For what it's worth, more and more are coming to understand and not take objective facts as some sort of personal insult.
audiobomber said:
The origin of the problem is that aperture is stated in f-numbers instead of millimeters diameter.
Click to expand...
That's not the issue, since people are able to covert with ease the diagonal angle of views between systems via the crop factor. It is simply their refusal to understand that the crop factor applies to the f-ratio as well, in terms of both DOF and the total amount of light falling on the sensor for a given shutter speed.
 
You must log in or register to reply here.

Keyboard shortcuts

f
Forum
m
My threads
Mobile site
About
Editorial content
Cameras & Lenses
Community
All content, design, and layout are Copyright © 1998–2025 Digital Photography Review All Rights Reserved.
Reproduction in whole or part in any form or medium without specific written permission is prohibited.
When you use DPReview links to buy products, the site may earn a commission.
©GPS Media - Guides, Products, Services.
Back
Top