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> a green monochrome MicroLED display measuring less than 0.5 mm (0.02 in) in diameter, with a resolution of 14,000 pixels per inch So, that's 280px at the wid

> a green monochrome MicroLED display measuring less than 0.5 mm (0.02 in) in diameter, with a resolution of 14,000 pixels per inch So, that's 280px at the widest part of the display, and a total of ~62k pixels, right? There's a decent amount you can do with that, but I think even the demo picture in the PR shot's probably ambitious for the amount of data this can display.

causality0 4y ago
The fact they didn't give an actual pixel count makes me think they mean 14,000 per square inch, which with these measurements means four pixels.
- jhardy54 4y ago
  Did I math wrong? 0.02 in * 0.02 in == 0.0004 sq in 0.0004 sq in * 14000 ppi == 5.6 pixels
  - bruckie 4y ago
    You assumed square. Round gives 4 pixels: π(.02/2)^2 * 14000 = 4.4
  - bornfreddy 4y ago
    It's probably 14000 ppi, not pixels per _square_ inch. So that's 0.02 in * 14000 in each direction (which is 280 x 280 pixels).
- zamadatix 4y ago
  "Much to my delight, I found I could interact with a compass to find my bearings, view images, and use an on-screen teleprompter to read a surprising but familiar quote." It would seem difficult to view images or read a teleprompter via 4 pixels. More likely they mean exactly what they wrote - 0.02 inches at 14,000 pixels per inch.
ghostly_s 4y ago
This has got to be a contender for most-uselessly framed statistic ever. I considered they might do some fancy interpolation w/ the saccades of the eye to simulate a larger display, mimicking our eyes own tricks, but it doesn't seem that's possible if the display is saccading along with the rest of the eyeball? I'm also confused about what useful 'stabilizing' their sensors can do when the display only covers a small point in the center of the pupil.
- ricardobeat 4y ago
  This video has some clips showing the viewing experience, as they are aiming for: https://youtu.be/d2fBBJRjccs https://youtu.be/d2fBBJRjccs

causality0 4y ago

The fact they didn't give an actual pixel count makes me think they mean 14,000 per square inch, which with these measurements means four pixels.

jhardy54 4y ago
Did I math wrong? 0.02 in * 0.02 in == 0.0004 sq in 0.0004 sq in * 14000 ppi == 5.6 pixels
- bruckie 4y ago
  You assumed square. Round gives 4 pixels: π(.02/2)^2 * 14000 = 4.4
- bornfreddy 4y ago
  It's probably 14000 ppi, not pixels per _square_ inch. So that's 0.02 in * 14000 in each direction (which is 280 x 280 pixels).
zamadatix 4y ago
"Much to my delight, I found I could interact with a compass to find my bearings, view images, and use an on-screen teleprompter to read a surprising but familiar quote." It would seem difficult to view images or read a teleprompter via 4 pixels. More likely they mean exactly what they wrote - 0.02 inches at 14,000 pixels per inch.

ghostly_s 4y ago

This has got to be a contender for most-uselessly framed statistic ever. I considered they might do some fancy interpolation w/ the saccades of the eye to simulate a larger display, mimicking our eyes own tricks, but it doesn't seem that's possible if the display is saccading along with the rest of the eyeball? I'm also confused about what useful 'stabilizing' their sensors can do when the display only covers a small point in the center of the pupil.

ricardobeat 4y ago
This video has some clips showing the viewing experience, as they are aiming for: https://youtu.be/d2fBBJRjccs https://youtu.be/d2fBBJRjccs