In this particular case the "out of context" post was valid to use as an example to back your claim.
That, and the rest of my post, was 100% correct.
They ignore nothing. You simply don't understand what they are saying (a familiar story). Here -- I'll spell it out to you. For the figures above, the assumption is that we are taking a pic of the same scene with the same exposure, so that the area of the scene in the focal plane is the same for both systems.
That's wrong, right from the start. The aperture is most certainly not the diameter of the image circle.
No. A 4:3 rectangle inscribed in a circle with the same diameter as a 3:2 rectangle will have 4% more area. This does not mean "4% greater efficiency" unless the final photo is displayed at 4:3 or more square. If the photo is displayed 3:2 or more wide, then the 3:2 rectangle is "more efficient". This has been explained to you multiple times, but you still lack the cognitive capacity to understand it.
...simply incorrect. My post was completely correct, as written. Whether you understand that or not, however, is another matter entirely.
You really have a bee in your bonnet about this.
Which unfortunately for you displayed your utter ignorance.
Twice the area means √2 times better S/N ratio. Nothing apparent about that, Boggis.
Yes.
No, silly - f/4 is two stops more than f/2; quadrupling.
Sorry, didn't notice your post until just now. OK, let's consider two sensors of the sames size that are equally efficient (same QE, same read noise / area, same saturation / area), where Sensor A has four times the number of pixels as Sensor B.
Pretty much!I guess we've hashed this to death.
No, it was 96% correct.
4:3 covers a greater area than 3:2 for the same diagonal. It is the sensor diagonal, and thus the required image circle, that relates to relative aperture.
No, it isn't. Sorry for any confusion. I was trying to make a jump to the important consideration and worded that poorly. (An explanation of how the actual image circle -- which may not be circular due to internal baffles or lens design -- projected at a given f-number is calibrated to yield the correct exposure for the sensor -- circumscribed by the "image circle" -- seemed needlessly complicated, and I assumed that you'd make the same logical leap.)
It covers 4% more area of the image circle. The aperture setting adjusts the amount of light projected onto that circle.
That's a crop .
I seem to recall having this exact same argument with you years ago, where you finally agreed that there is a 4% difference but claimed that it isn't significant.
FF has 3.8 the area of FT. This fact gives FF a √3.8 ≈ 2.0 time, or 1 stop, noise advantage over FT.
if a coronal mass ejection hits earth while i'm taking pics will my pixels have more photon noise? if not will there be any effect on my pics versus normal conditions?Sorry, didn't notice your post until just now. OK, let's consider two sensors of the sames size that are equally efficient (same QE, same read noise / area, same saturation / area), where Sensor A has four times the number of pixels as Sensor B.
Without loss of generality, assume QE = 1 and the read noise for a pixel of Sensor A is 4 electrons. Let's also assume that 9 electrons fall on each pixel of Sensor A, and thus 9 x 4 = 36 electrons fall on each pixel of Sensor B (just to make the computations more simple).
The photon noise for one pixel on Sensor A is sqrt 9 = 3 electrons. This makes the total noise sqrt (3² + 4²) = 5 electrons / pixel. The total noise for four pixels on Sensor A is sqrt (5² + 5² + 5² + 5²) = 10 electrons, giving us an NSR of 10 / 36 = 28%.
Now let's consider one pixel of Sensor B. The photon noise will be sqrt 36 = 6 electrons, and the read noise is 4 x 4 = 16 electrons. This gives us a total noise of sqrt (6² + 16²) = sqrt 292 = 17 electrons, resulting in an NSR of 17 / 36 = 47%.
So, more pixels on an equally efficient sensor result in less noise / area. It is interesting to note that if the read noise were to scale with the linear dimension of the pixel, rather than the area, then the noise would be the same.
Pretty much!I guess we've hashed this to death.
Yes, if the read noise scales with the linear pixel dimensions, then the 'normalized' noise (like in DxO's 'print' figures) will be the same, and that's what I mean then talking about "same per area read noise" and "equally efficient" sensors. In your example, I'd say that sensor A is more efficient than sensor B.
Sorry, didn't notice your post until just now. OK, let's consider two sensors of the sames size that are equally efficient (same QE, same read noise / area, same saturation / area), where Sensor A has four times the number of pixels as Sensor B.
Without loss of generality, assume QE = 1 and the read noise for a pixel of Sensor A is 4 electrons. Let's also assume that 9 electrons fall on each pixel of Sensor A, and thus 9 x 4 = 36 electrons fall on each pixel of Sensor B (just to make the computations more simple).
The photon noise for one pixel on Sensor A is sqrt 9 = 3 electrons. This makes the total noise sqrt (3² + 4²) = 5 electrons / pixel. The total noise for four pixels on Sensor A is sqrt (5² + 5² + 5² + 5²) = 10 electrons, giving us an NSR of 10 / 36 = 28%.
Now let's consider one pixel of Sensor B. The photon noise will be sqrt 36 = 6 electrons, and the read noise is 4 x 4 = 16 electrons. This gives us a total noise of sqrt (6² + 16²) = sqrt 292 = 17 electrons, resulting in an NSR of 17 / 36 = 47%.
So, more pixels on an equally efficient sensor result in less noise / area. It is interesting to note that if the read noise were to scale with the linear dimension of the pixel, rather than the area, then the noise would be the same.
Pretty much!I guess we've hashed this to death.
Here's the answer to your question, given very soon after this thread was started:
All my calculations are correct, boggis -- here they are again:
...I really love that song -- one of my favorites, actually.All together now ...
Woh!!... Dynamic Range, woa!
What is it good for?
Absolutely nothing
Say it again y'all...
Great Bustard. Our very own IQ Starr.
which predictably answers nothing
and only raises the point that you when asked the simple question, how does an IMATEST RAW of 11.5 stops DR get redefined as 10.2 stops in DxO. Ive asked this several times in several ways, and you ran and hid under your flat rock in case you were pressed to answer.
Give me your PayPal ID -- I'll send you the 25 cents to cover the cost of your call.
quite typically, no defence
You're here, aren't you? And, aside from the "entertainment value" of your posts, what else is there to expect?
nothing to do with your likewise fraudulent claim
