The Online Photographer

A full frame sensor has 3.8 times the area as a Micro 4/3 sensor, so with the same pixel density a full frame sensor would have 20MP*3.8 = 76MP.

Source: http://photoseek.com/2013/compare-digital-camera-sensor-sizes-full-frame-35mm-aps-c-micro-four-thirds-1-inch-type/

The quick answer is 80 MP. With a 2X crop the Micro 4/3 sensor has 1/4 the area of full frame. Unless there are subtleties I'm not aware of.

71MP, assuming MFT is 18x13.5mm and FF is 36x24mm.

I am slightly confused about sensor size vs imaging area: MFT seems to either be 18x13.5 or 17.3x13 and I am not sure what FF really is (although I think 35mm film really is 36x24 depending on the camera obviously.

If you assume the smaller area it is about 77MP.

The general formula is

C = ((H*W)/(h*w))*c

where C is pixel count on the larger format, H, W are the dimensions of the larger format, c is the pixel count of the smaller format & h, w are the dimensions of the smaller format. This works for any two formats of course.

80MP

80

A 20MP 4/3 sensor is 3872 pixels high spread over 13mm, or ~298 px/mm. A full frame sensor is 24mm high and at the same pixel pitch would be 7152 pixels high. At a 4/3 aspect ratio that would be 7152x9536 or ~68MP. At the normal FF aspect ratio of 1.5 it would be 7152x10728 or ~77MP.

As a follow-on comment I've been recently using a Sony RX100V with a 20MP 1" sensor (8.8mm x 13.2mm). If we scale that to full frame with the same pixel pitch we would be ~148MP. The image quality at base ISO on the RX100V is amazing, especially when you consider the pixel pitch.

That would let them build one heck of a versatile camera. Extra reach in a crop mode (though likely to be lens limited). Extra clean if pixel averaged to 36MP. Or high res for large detailed images. I will be surprised if we don't see such a camera in the next decade.

Oops, should have googled first. It is half the diagonal of full frame, so Pythagorem theorem gives me 43.27mm for the diagonal of full frame. Half of that is 21.635 for M43. The surface area of full frame is 864 square millimeters (24mmx36mm). Now , I will cheat because figuring out the width of the M43 sensor from the diagonal and the aspect ratio is beyond what I'm willing to do with math. I looked up the width as 17.3mm. So with the Pythagorean Theorem, I calculated the height to be the root of 168, which is approximately 12.96. This makes the surface area of M43 about 224 square millimeters. Dividing the surface area of full frame by the surface area of that of M43 gives me a factor of approximately 3.85.

So 20mpx3.85. I'll round that up to an 80mp sensor for full frame with the same pixel density as a 20mp M43 sensor.

That was a lot of work, and this is with leaving out calculating the width of the sensor from the diagonal and the aspect ratio, which is mathematically feasible but probably beyond what I remember personally from high school.

I've always done it this way:
m4/3 sensor area is 225 sqmm
FF is about 860 (24x36 = 864 but many are 23.9mm high)
860 / 225 = 3.82 times the area of m 4/3
If pixels were the same size....
20 x 3.82 = 76,4
So 76.4 MP would fit in the larger area.

But you probably wouldn't do that because there is so much benefit from larger pixels, you would use fewer larger ones.

I don't think the calculations using ratios of area are correct since FF and m43 do not have the same aspect ratios, so you'd be assuming non-square pixels. The 68MP calculation using pixel density is correct, but who makes a 32x24 mm^2 sensor anyway?

(But such a sensor exists, macro 4/3 might be a cool name for it!)

Well, the calculation is easy enough - it just scales by area (so, as others have observed, 76MP for full-frame, and roughly 130MP for the new Fuji and Hasselblad mini-medium format cameras).

However, try getting a sharp 76MP hand-held image without the amazing stabilisation provided by current micro 4/3 bodies. I routinely shoot both an E-M1.2 and Leica M 262 side by side, and usually the Olympus files are sharper.

This is partly because of the extremely precise and reliable AF, and partly because of the image stabiliser. But the files are so sharp that I wonder if Olympus is also applying some kind of sharpening algorithm to the RAW files.

Why do you ask? Do you want a full-frame sensor with tiny tiny sensors crammed onto it in the same manner they are in a micro-4/3rds camera?

Is it an African or a European sensor? Laden or unladen?

;-)

Rather than performing the number crunching comparison excercise, this is how I would make my decision. Going back to film days would you choose medium format film and camera/lens system if in fact you could get a long telephoto in medium format or would you choose 35mm film/camera/lens for bird, wildlife, etc. Today in digital I would most definitely choose M4/3 for any long telephoto work and remember back in the film days a 300mm lens was considered the minimum for bird or wildlife photography.

If one is shooting at base ISO and the resolution of the smaller sensor is sufficient, then the smaller sensor wins since the one can save on weight, size on money.

If, OTOH, one needs maximum resolution the larger sensor wins. Also, if one shoots in low light and the equivalent apertures of the lenses are the same, the large sensor wins.

These naturally describe best case scenarios where one uses the best bodies and lenses. Failing that the comparison changes. Size and weight should not be dismissed though, it's hard to move around with heavy packs and tripod shooting is not as agile as hand held.

If you are photographing small things like birds in flight, pixel density is a big help . Thats why both Canon & Nikon put versions of their best AF systems in the 7D mk II and D500. They are very popular cameras for that purpose.
If you already have the lenses, FF &APS c make a very strong combination.
If you don't already have the lenses, Cameras like the OLY take it one step further. I'm sure they are great for BIF.

You know, it was when people all over the world started going mental about "aperture equivalence" that I decided never to shoot digital again. If you start a "pixel equivalence" craze, well...

Some bird photographers compare setups by calculating PPD - pixels per duck. In other words, if I photograph the same small distant object and crop as required, how many pixels am I left with. Seems like a good, real world way of comparing the effective reach.

Same pixel density on the sensor?
Or same pixel density on the subject? (which seems more relevant to many people)

If you care about the subject then you need to be comparing focal length in this as well.

If your lens is a 400mm f/6.3 with a 20 MP micro 4/3 camera then you’re already against the diffraction limit wide open. Here is a useful calculator (not mine, just found it): https://www.pointsinfocus.com/tools/diffraction-limited-effective-resolutions . It seems trustworthy as the numbers stack up rather well with diffraction-limited empirical lens resolutions of the best lenses and sensors at photozone.de (see e.g. photozone lens test numbers for the 90mm Sony macro on the A7RII at f/11). Using this calculator, the “effective” maximum resolution of an f/6.3 m4/3 lens (green color, 550 nm wavelength) is about 13 MP wide open while the competing 800mm f/5.6 on full frame would be limited to about 51 MP. I don’t have a horse in this race--the only telephoto lens I own is a Jupiter-9 (85mm) used a few times a year on full frame. As the old saying goes, 4 MP should be enough resolution for everyone, so this diffraction-limited loss might not matter to most birders or sport photographers.

Correction. The "smaller difference" I was thinking of is the comparison between m43 and APS-C sensor sizes.

>>Eliot Porter, the great nature photographer, shot mostly with an 8x10 but did his bird photography with 35mm.

You're overthinking this, Mike. The most likely reason was that long telephoto lenses were available for 35mm cameras and not available for large format sheet film cameras. Can you imagine how huge a 400mm equivalent lens for 8x10 format would have to be?

How can there be many different answers? Oh you mean all but one are wrong!

You have a GX8 so let's assume that it is the 20 MP m43 sensor that you reference. It has 5184 pixels across its 17.3mm sensor width, so the pixel size is 3.3372 microns. You can fit 10,787 of those pixels along a 36mm long sensor, and 7192 along it's 24mm height, for a total of 77,580,104 pixels.

It's simple, a GX8 has 5184 by 3888 pixels on a 17.3 by 13 mm sensor. Taking the wide dimension, 5184 pixels / 17.3 mm = (about) 300 pixels per millimeter (ppm) pixel density. A full frame sensor is 36 by 24 mm. 36 mm x 300 ppm = 10800 pixels, 24 mm x 300 ppm = 7200 pixels. So full frame at the same pixel density as a GX8 would be 10800 by 7200 pixels, 77.76 MP.

A short digression on a famous related mathematical question:

The wording of the posed question suggests that areal rather than linear density is what is under consideration.

Assuming unit density, the accuracy of approximating, by the area of a region, the number of points having integer coordinates and lying within the region is highly dependent on the region's shape, and in the case of a disk leads to a fascinating and famous mathematical question:

Estimate, as a function of R, the difference between the area of the disk of radius R centered at the origin, and the number of lattice points with integer coordinates (positive, negative, or zero) that lie within it.

This problem interested Gauss, who showed that the difference is dominated by a multiple of R, which for large R is considerably smaller than the area, which increases quadratically with R. Much later, the error exponent was improved to R to the power 2/3, and still later, by R to a power slightly smaller than 2/3. Later the English mathematician Hardy (cf. the movie "The Man who knew Infinity") did work, with possible input from Ramanujan, suggesting, but not proving, that any exponent greater than 1/2 would work. Hardy also showed, however, that 1/2 itself does not work, so a valid exponent cannot be less than or equal to 1/2. The problem of unravelling the mystery of the exponent for the error estimate is known as the Gauss circle problem, and remains, along with the Riemann Hypothesis, one of the most famous unsettled questions in mathematics. The conjectured result is that any exponent greater than 1/2 is valid.