I think the notion of FF = heavier lens may not be true

Started Nov 8, 2013 | Discussions thread
Great Bustard Forum Pro • Posts: 42,723
Re: Still "No comprendo", eh?

EinsteinsGhost wrote:

Great Bustard wrote:

EinsteinsGhost wrote:

Good. Let me continue to help you walk through this... and we will even assume the metering logic to be identical (based on a light meter) and the scene of course:

You have 35mm f/4 lens on APS-C and 50mm f/4 lens on FF. Now, what do you expect the shutter speeds to be?

Let me direct you to this post upthread:


which answers all that, and is followed by a miserable and utter comprehension fail on your part.

No, it resumes the Great Bustard run around (in circles). The question is simple enough, and shouldn't take much more than a sentence. Heck, I will make it even simpler for you to answer with a yes, or a no:

35mm f/4 on APS-C for a scene versus 50mm f/4 lens on FF for the scene, do you expect shutter speed to be about twice as fast for the FF?

Since you are both incapable and unwilling, I shall quote from the link for you:


The answer I gave before was:


This section will answer the following four questions:

  • For a given scene, what is the difference in exposure, if any, between f/2.8 1/200 ISO 400 and f/5.6 1/200 ISO 1600?
  • What role does the ISO setting play?
  • What role does the sensor size play?
  • What does any of this have to do with the visual properties of the photo?

As mentioned in the introduction of this essay, the concept of Equivalence is controversial because it replaces the paradigm of exposure, and its agent, f-ratio, with a new paradigm of total light, and its agent, aperture. The first step in explaining this paradigm shift is to define exposure, brightness, and total light.

Now let me flesh it out by quoting more from the link:

Mathematically, we can express these four quantities rather simply:

  • Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)
  • Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)
  • Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)
  • Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance, regardless of the focal length or the sensor size. However, the brightness for the two photos will be the same since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

If you are still confused, please let us know.


Now, are you honestly telling us that you lack the cognitive capacity to understand that the above says that 35mm f/4 on APS-C and 50mm f/4 on FF will meter for the same shutter speed assuming the camera's meters meter in the same manner?

Are you honestly telling us that you cannot understand that the exposures, while the same, will result in over twice as much light falling on the FF sensor (as well as a more narrow DOF on FF), and that this is why FF has less noise than APS-C for a given exposure and sensor efficiency?

Are you honestly going to dig your hole that much deeper with such self-harming posts? Wow. Well, it takes all kinds, I suppose.

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