[wim]

32? 36??

8k is 7680 horizontal and 4320 vertical.

7680 / 4 = 1920.
1920 * 3 = 5760.
7680 * 5760 = 44.2Mp.

?
8K width X "6K" width? Why?

8K width X 8K height is still 7680 (4x 1920) x 4320 (4x 1080). That's 33.2 Mpix, or if you define it old school. 31.6 Mpix (generally rounded to 32 Mpix).

Kind regards, Wim

Uhmm, no. You certainly do not get it.

8K width = 7680 pixels.

So, what is a 4/3rds sensor? 4x3 aspect ratio.
7680 x 4320 is 16 x 9, not 4x3. MFT is called MFT (MicroFourThirds, not MSN (MicroSixteenNineths))...
Now calculate how many MP the MFT sensor needs to have to do 8K video.

Indeed, exactly as I showed you: 44.2mp.

I actually did get that, but the question is whether a video-cam at 8K really needs 44 (or 42.3 old style) Mpix, or more specifically a 4:3 layout. It can still be a 4/3 sensor in size (diameter), and have an 8K layout. I should probably have been clearer on that.

So far, I've only seen it announced as a 32 Mpix camera, when a number is given, so I guess we'll have to wait and see.

However, density-wise it would indeed be a 42.3 or 44 Mpix sensor, so a bit over twice as dense as current implementations from Panasonic and Olympus.

Kind regards, Wim