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Saturday, May 14, 2016

What resolution is needed for 360 video?

I got my basic 360 video viewer working and was not pleased with the resolution.   I've realized that people are really serious that they need very high res.   I was skeptical of these claims because I am not that impressed with 4K TVs relative to 2K TVs unless they are huge.   So what minimum res do we need?    Let's say I have the following 1080p TV (we'll call that 2K to conform to the 4K terminology-- 2K horizontal pixels):

Image from https://wallpaperscraft.com
If we wanted to tile the wall horizontally with that TV we would need 3-4 of them.   For a 360 surround we would need 12-20.   Let's call it 10 because we are after approximate minimum res.  So that's 20K pixels.   To get up to "good" surround video 20K pixels horizontally.   4K is much more like NTSC.   As we know, in some circumstances that is good enough.

Facebook engineers have a nice talk on some of the engineering issues these large numbers imply. 

Edit: Robert Menzel pointed out on Twitter that the same logic is why 8K does suffice for current HMDs.


5 comments:

Jared M Johnson said...

Don't know if my other post got lost, but I look at a lot of 360x180 panoramas everyday. Anything below 4000 pixels across the equator is visually crappy viewed at normal field of views on a computer monitor. I would say 4k in equirectangular format is the lower bound. I think the upper bound is 50 pixels per degree iirc, for most people, which is almost five times that...

Peter Shirley said...

50*360 = 18K so that does sound right. Are there any non-research sources for that? Playback sounds challenging at that bandwidth! What about vertical?

Jared M Johnson said...

For vertical, if you're just storing in equirectangular, then you'll have 9k pixels. Which will probably compress well with jpg or video. However, as far as giving each pixel as much information as any other pixel, I think you would only need ~6k pixels, with the azimuth mapping as x, being direct from 0 to 2*pi, and taking the sine of the elevation angle to get y, and an image with aspect PI:1, https://en.wikipedia.org/wiki/Lambert_cylindrical_equal-area_projection.

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