a6k Loc: Detroit & Sanibel

The usual version of equivalent focal length is based purely on the focal length number that is linearly related to the angle of view with the normal being the dimensions of a "full frame" sensor. To make an example that is well known, my "crop factor 1.5" Sony a6500 has an angle of view that is 1/1.5 as great as the same lens would show on an a7 or a9. Thus the equivalent focal length is increased by 50% in this case and my 500 mm lens has an apparent or equivalent length of 750 mm. I have no argument with any of this.

However, when it comes to comparing long telephoto capabilities and the frequent necessity of cropping, the number of pixels becomes a relevant factor. Example: a hypothetical full frame camera with a 400 mm lens. One version has a sensor that is best for low light - 20 max and the other version has a sensor for high resolution - 40 mpx. They are both 2:3 aspect ratio so the 40 mpx sensor has twice as many pixels linearly. Ignoring the limits of lens performance, a print of a given size and using 300 dpi, can produce an image of the subject twice as large linearly on the 40 mpx camera. The same logic can be used to compare two cameras with different size sensors.

My wife uses a Nikon Coolpix 900 which is said to have an equivalent focal length at maximum zoom of 2000 mm. I use a Sony a6500 with a 500 mm reflex lens so the equivalent focal length is 750 mm. You would expect from those numbers that the Nikon can produce an image that is (2000/750= 2.67) time as large, linearly.

But if I take into account that the Sony has 6000 pixels across and the Nikon has 4608 then the proposed alternate version of equivalent lengths is 2228 for Sony and 4654 for Nikon, a difference of 2.09 times, linearly. The correction of about 28% is in favor of the camera with more pixels. To be clear, lens sharpness and other measures of image quality are ignored here just as they are in any other crop factor discussion of this kind. The numbers I have used for the new version are NOT mm's but simply the result of a calculation that allows direct comparison.

So this means that the final result, either a print or a displayed image, has potential that is not simply the result of comparing sensor dimensions. This is one way to incorporate the pixel count, too. In this example, one calculation leads to an 8" print and the other leads to a 10" print.

For most "work" I doubt this has any utility but for birds or other small/distant targets where the longest lens wins, this approach to the potential for the final image has, I think, relevance. Your comments are invited.

Pablo8 Loc: Nottingham UK.

Too late...……….The wheel has already been invented!

selmslie Loc: Fernandina Beach, FL, USA

You might want to take a closer look at your numbers.

A 40 MP images has twice as many megapixels as 20 MP but that's an area measurement. The linear resolution is only 1.414 times as great.

A 40 MP sensor has twice as many pixels as a 10 MP image if you compare them linearly (in each dimension) and 4x as many by area.

a6k Loc: Detroit & Sanibel

Scotty, you are absolutely correct on my error in the example. But I think you will find the rest of the numbers are OK. I have a rather busy little spreadsheet where I did the actual calculations.
Thanks.

OddJobber Loc: Portland, OR

Move along, folks. Nothing new here.

DaveO Loc: Northeast CT

Old habits are hard to break.

a6k Loc: Detroit & Sanibel

Two already saying the wheel was already invented or nothing new here. Neither provided anything to back that up. I'd be interested in any previous write-ups on this.

BebuLamar

So tell us what is new in your post except the little mistake that someone already corrected?

a6k Loc: Detroit & Sanibel

What is new, at least in my own understanding is simply that when comparing the reach of two cameras with the intent of getting the best image at a given size or the largest image at a given resolution, a straight comparison of "equivalent" focal length may provide a less accurate answer than if we correct for the number of pixels in the sensor (linear in my calculations). The difference in the example of my two cameras is around 27+ percent or the difference between an 8" image and a 10" image. I consider that a non-trivial difference.

Isn't that what I said in the original OP?

BebuLamar

And that is not new.

a6k Loc: Detroit & Sanibel

You are now #3 to say it is not new but please refer me to where it was said or explained before.

BebuLamar

People have been talking about using APS-C camera for more reach when using the same lens on the APS-C and the FF. It's true when both have the same number of megapixels like the D7200 vs the D750. When compared to the D800/D810 it became less so. When the D500 and the D850 came out the reach advantage of the D500 compared to the D850 became almost none. People have been talking about that here UHH and other forum.
There is nothing new to what you said.

a6k Loc: Detroit & Sanibel

If it's nothing new, then where is the math?

BebuLamar

You call you elementary math something new?

Rongnongno Loc: FL

WHAT?????? SOMEONE INVENTED THE WHEEL????

I lost my life trying to create one then???