11 ms·

It's difficult to understand the math on the storage density. Four colors out of a possible 32 colors is about 15 bits of information, not 40,000 bits of infor

It's difficult to understand the math on the storage density. Four colors out of a possible 32 colors is about 15 bits of information, not 40,000 bits of information.[1] If it's 15 bits per pixel and 115M pixels, then the capacity is 1.7Gb, not 4.6Tb. Maybe I've misunderstood the coding. Corrections are welcome. [1] Crude overestimate, assume 5 bits per color for 20 bits per pixel. More accurate is log2(32 choose 4), which you can type into Google to get 15 bits.

jazzyjackson 2y ago
Here's the math in the cited paper... I feel like they're making an error of 35960 as 36kilobits instead of, yeah, 15bits ? https://ieeexplore.ieee.org/document/9438269 https://ieeexplore.ieee.org/document/9438269 If, in our previous example, the 4 wavelengths were to be selected from a palette of 32 different wavelengths, a single worfel location could store ~36 Kilobits of data. Thus, a 1 cm2 media with 10μ2 data locations (8μ2 worfels with 2μ spacing on all sides) = 1,000,000 worfels/cm2. For example, (32!/((32-4)! • 4!)) = 35,960 distinct states. (An analogous use of formula (1) is drawing a hand of 5 playing cards from a 52-card deck yields 2,598,960 distinct hands.) Applying the 35,960-state permutation table for k=4 (i.e., superimposing 4 wavelengths per worfel), and drawing from a palette, N, of 32 different wavelengths, yields 35,960,000,000 bits (≈35.9 gigabits) per cm2; or 35.9 x (6.42 cm2 per square inch) ≈ 230.4 gigabits/in2. And so for an example of a 4″x5″ media (20 in2), 20 x 230.4 ≈ 4.6 terabits per 4x5 inch media.
- oh_my_goodness 2y ago
  Thanks. I believe choosing from 35,960 possible states only takes 15 bits, not 35,960 bits. But it's late on a Friday.
  - jazzyjackson 2y ago
    Yeah, there might be one more combinatorial explosion, so they can choose one of 16 combinations with each of those 35960 combinations... breaks one's brain.
    - oh_my_goodness 2y ago
      I hope so. That would be cool.
- deleted 2y ago
  [deleted]
simne 2y ago
Looks like typo or "noisy phone". In current fiber-optics solutions using multi-wavelength, so on fiber input mixed 8 or more laser waves of different lengths (to be strict not wl but bands) and on output mix divided with optical filter and all wavelengths processed separately. But you don't have to use exact wavelength in this, you could use any wl within band (yes, 4 colors in each of 40000 bands looking possible), if you have light source with changeable wl and detector with enough precision, and number of wl's could be much larger than in for example LCD or OLED. Unfortunately, authors of article for some reason don't write about this important nuance, but it mean, in reality to make this technology, need precision light source(s) with adjustable wl's (classic RGB is combination of just 3 wl's, and modern amoleds usually 4 wl's) and precision compact spectrometer (for example, what I seen myself gives 1024 lines for visual spectrum, so just 10 bits). All these are not impossible, but will not easy to achieve.
- simne 2y ago
  Sorry, I made correction comment: https://news.ycombinator.com/item?id=41974605 https://news.ycombinator.com/item?id=41974605

jazzyjackson 2y ago

Here's the math in the cited paper... I feel like they're making an error of 35960 as 36kilobits instead of, yeah, 15bits ? https://ieeexplore.ieee.org/document/9438269 https://ieeexplore.ieee.org/document/9438269 If, in our previous example, the 4 wavelengths were to be selected from a palette of 32 different wavelengths, a single worfel location could store ~36 Kilobits of data. Thus, a 1 cm2 media with 10μ2 data locations (8μ2 worfels with 2μ spacing on all sides) = 1,000,000 worfels/cm2. For example, (32!/((32-4)! • 4!)) = 35,960 distinct states. (An analogous use of formula (1) is drawing a hand of 5 playing cards from a 52-card deck yields 2,598,960 distinct hands.) Applying the 35,960-state permutation table for k=4 (i.e., superimposing 4 wavelengths per worfel), and drawing from a palette, N, of 32 different wavelengths, yields 35,960,000,000 bits (≈35.9 gigabits) per cm2; or 35.9 x (6.42 cm2 per square inch) ≈ 230.4 gigabits/in2. And so for an example of a 4″x5″ media (20 in2), 20 x 230.4 ≈ 4.6 terabits per 4x5 inch media.

oh_my_goodness 2y ago
Thanks. I believe choosing from 35,960 possible states only takes 15 bits, not 35,960 bits. But it's late on a Friday.
- jazzyjackson 2y ago
  Yeah, there might be one more combinatorial explosion, so they can choose one of 16 combinations with each of those 35960 combinations... breaks one's brain.
  - oh_my_goodness 2y ago
    I hope so. That would be cool.
deleted 2y ago
[deleted]

simne 2y ago

Looks like typo or "noisy phone". In current fiber-optics solutions using multi-wavelength, so on fiber input mixed 8 or more laser waves of different lengths (to be strict not wl but bands) and on output mix divided with optical filter and all wavelengths processed separately. But you don't have to use exact wavelength in this, you could use any wl within band (yes, 4 colors in each of 40000 bands looking possible), if you have light source with changeable wl and detector with enough precision, and number of wl's could be much larger than in for example LCD or OLED. Unfortunately, authors of article for some reason don't write about this important nuance, but it mean, in reality to make this technology, need precision light source(s) with adjustable wl's (classic RGB is combination of just 3 wl's, and modern amoleds usually 4 wl's) and precision compact spectrometer (for example, what I seen myself gives 1024 lines for visual spectrum, so just 10 bits). All these are not impossible, but will not easy to achieve.

simne 2y ago
Sorry, I made correction comment: https://news.ycombinator.com/item?id=41974605 https://news.ycombinator.com/item?id=41974605